History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford’s Law: Tables of Logarithms, TaxCheats, and The Leading Digit
Phenomenon
Michelle Manes ([email protected])
USC Women in Math24 April, 2008
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
History
(1881) Simon Newcomb publishes “Note on thefrequency of use of the different digits in naturalnumbers.” The world ignores it.
(1938) Frank Benford (unaware of Newcomb’s work,presumably) publishes “The law of anomalousnumbers.”
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
History
(1881) Simon Newcomb publishes “Note on thefrequency of use of the different digits in naturalnumbers.” The world ignores it.
(1938) Frank Benford (unaware of Newcomb’s work,presumably) publishes “The law of anomalousnumbers.”
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Statement of Benford’s Law
Newcomb noticed that the early pages of the book oftables of logarithms were much dirtier than the laterpages, so were presumably referenced more often.
He stated the rule this way:
Prob(first significant digit = d) = log10
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Statement of Benford’s Law
Newcomb noticed that the early pages of the book oftables of logarithms were much dirtier than the laterpages, so were presumably referenced more often.
He stated the rule this way:
Prob(first significant digit = d) = log10
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford Base b
DefinitionA sequence of positive numbers {xn} is Benford(base b) if
Prob(first significant digit = d) = logb
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford’s Law
Base 10 Predictionsdigit probability it occurs as a leading digit
1 30.1%2 17.6%3 12.5%4 9.7%5 7.9%6 6.7%7 5.8%8 5.1%9 4.6%
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford’s Data
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
More Data
Benford’s Law compared with: numbers from the frontpages of newspapers, U.S. county populations, and theDow Jones Industrial Average.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake five years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and five more years) to get back to first digit 2.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake five years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and five more years) to get back to first digit 2.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Dow Illustrates Benford’s Law
Suppose the Dow Jones average is about $1,000. If theaverage goes up at a rate of about 20% a year, it wouldtake five years to get from 1 to 2 as a first digit.
If we start with a first digit 5, it only requires a 20%increase to get from $5,000 to $6,000, and that isachieved in one year.
When the Dow reaches $9,000, it takes only an 11%increase and just seven months to reach the $10,000mark. This again has first digit 1, so it will take anotherdoubling (and five more years) to get back to first digit 2.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford’s Law and Tax Fraud (Nigrini, 1992)
Most people can’t fake data convincingly.
Many states (including California) and the IRS now usefraud-detection software based on Benford’s Law.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford’s Law and Tax Fraud (Nigrini, 1992)
Most people can’t fake data convincingly.
Many states (including California) and the IRS now usefraud-detection software based on Benford’s Law.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Benford’s Law and Tax Fraud (Nigrini, 1992)
Most people can’t fake data convincingly.
Many states (including California) and the IRS now usefraud-detection software based on Benford’s Law.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
True Life Tale
Manager from Arizona State Treasurer wasembezzling funds.
Most amounts were below $100,000 (criticalthreshold for checks that would require morescrutiny).Over 90% of the checks had a first digit 7, 8, or 9.(Trying to get close to the threshold without goingover — artificially changes the data and so breaks fitwith Benford’s law.)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
True Life Tale
Manager from Arizona State Treasurer wasembezzling funds.Most amounts were below $100,000 (criticalthreshold for checks that would require morescrutiny).
Over 90% of the checks had a first digit 7, 8, or 9.(Trying to get close to the threshold without goingover — artificially changes the data and so breaks fitwith Benford’s law.)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
True Life Tale
Manager from Arizona State Treasurer wasembezzling funds.Most amounts were below $100,000 (criticalthreshold for checks that would require morescrutiny).Over 90% of the checks had a first digit 7, 8, or 9.(Trying to get close to the threshold without goingover — artificially changes the data and so breaks fitwith Benford’s law.)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
True Life Tale
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Problems with “Proofs” of Benford’s Law
Discrete density and summability methods.
Continuous density and summability methods. (Sameproblem.)
Scale invariance.
If there is a reasonable first-digit law, it should bescale-invariant. That is, it shouldn’t matter if themeasurements are in feet or meters, pounds orkilograms, etc.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Problems with “Proofs” of Benford’s Law
Discrete density and summability methods.Fd = {x ∈ N | first digit of x is d}. No natural density.
That is,
limn→∞
Fd ∩ {1,2, . . . ,n}n
does not exist.
Continuous density and summability methods. (Sameproblem.)
Scale invariance.
If there is a reasonable first-digit law, it should bescale-invariant. That is, it shouldn’t matter if themeasurements are in feet or meters, pounds orkilograms, etc.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Problems with “Proofs” of Benford’s Law
Discrete density and summability methods.
Continuous density and summability methods. (Sameproblem.)
Scale invariance.
If there is a reasonable first-digit law, it should bescale-invariant. That is, it shouldn’t matter if themeasurements are in feet or meters, pounds orkilograms, etc.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Problems with “Proofs” of Benford’s Law
Discrete density and summability methods.
Continuous density and summability methods. (Sameproblem.)
Scale invariance.
If there is a reasonable first-digit law, it should bescale-invariant. That is, it shouldn’t matter if themeasurements are in feet or meters, pounds orkilograms, etc.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Problems with “Proofs” of Benford’s Law
Discrete density and summability methods.
Continuous density and summability methods. (Sameproblem.)
Scale invariance.
If there is a reasonable first-digit law, it should bescale-invariant. That is, it shouldn’t matter if themeasurements are in feet or meters, pounds orkilograms, etc.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionFor each integer b > 1, define the mantissa function
Mb : R+ → [1,b)
x 7→ r
where r is the unique number in [1,b) such that x = rbn
for some n ∈ Z.
ExamplesM10(9) = 9 = M100(9).M2(9) = 9/8 = 1.001 (base 2).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionFor each integer b > 1, define the mantissa function
Mb : R+ → [1,b)
x 7→ r
where r is the unique number in [1,b) such that x = rbn
for some n ∈ Z.
ExamplesM10(9) = 9 = M100(9).M2(9) = 9/8 = 1.001 (base 2).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionFor E ⊂ [1,b), let
〈E〉b = M−1b (E) =
⋃n∈Z
bnE ⊂ R+.
DefinitionMb = {〈E〉b | E ⊂ B(1,b)}is the σ-algebra on R+
generated by Mb.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionFor E ⊂ [1,b), let
〈E〉b = M−1b (E) =
⋃n∈Z
bnE ⊂ R+.
DefinitionMb = {〈E〉b | E ⊂ B(1,b)}is the σ-algebra on R+
generated by Mb.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionLet Pb be the probability measure on (R+,Mb) defined by
Pb(〈[1, γ)〉b) = logb γ.
This probability measure:Agrees with Benford’s law.Is the unique scale-invariant probability measure on(R+,Mb).
Proof comes down to uniqueness of Haar measure.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionLet Pb be the probability measure on (R+,Mb) defined by
Pb(〈[1, γ)〉b) = logb γ.
This probability measure:
Agrees with Benford’s law.Is the unique scale-invariant probability measure on(R+,Mb).
Proof comes down to uniqueness of Haar measure.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionLet Pb be the probability measure on (R+,Mb) defined by
Pb(〈[1, γ)〉b) = logb γ.
This probability measure:Agrees with Benford’s law.
Is the unique scale-invariant probability measure on(R+,Mb).
Proof comes down to uniqueness of Haar measure.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionLet Pb be the probability measure on (R+,Mb) defined by
Pb(〈[1, γ)〉b) = logb γ.
This probability measure:Agrees with Benford’s law.Is the unique scale-invariant probability measure on(R+,Mb).
Proof comes down to uniqueness of Haar measure.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Hill’s Formulation (1988)
DefinitionLet Pb be the probability measure on (R+,Mb) defined by
Pb(〈[1, γ)〉b) = logb γ.
This probability measure:Agrees with Benford’s law.Is the unique scale-invariant probability measure on(R+,Mb).
Proof comes down to uniqueness of Haar measure.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benforddistributed base-10:
1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benforddistributed base-10:
1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benforddistributed base-10:
1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
What types of sequences are Benford?
Real-world data can be a good fit or not, depending onthe type of data. Data that is a good fit is “suitablyrandom” — comes in many different scales, and is a largeand randomly distributed data set, with no artificial orexternal limitations on the range of the numbers.
Some numerical sequences are clearly not Benforddistributed base-10:
1,2,3,4,5,6,7, . . . (uniform distribution)
1,10,100,1000, . . . (first digit is always 1)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Some numerical sequences seem to be a good fit
Powers of Two38
History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Some numerical sequences seem to be a good fit
Fibonacci Numbers39
History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
Proof:x = Mb(x) · bk for some k ∈ Z.First digit of x in base b is d iff d ≤ Mb(x) < d + 1.logb d ≤ y < logb(d + 1), wherey = logb(Mb(x)) = logb x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
logb(d +1)−logb(d) = logb
(d + 1
d
)= logb
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
Proof:x = Mb(x) · bk for some k ∈ Z.
First digit of x in base b is d iff d ≤ Mb(x) < d + 1.logb d ≤ y < logb(d + 1), wherey = logb(Mb(x)) = logb x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
logb(d +1)−logb(d) = logb
(d + 1
d
)= logb
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
Proof:x = Mb(x) · bk for some k ∈ Z.First digit of x in base b is d iff d ≤ Mb(x) < d + 1.
logb d ≤ y < logb(d + 1), wherey = logb(Mb(x)) = logb x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
logb(d +1)−logb(d) = logb
(d + 1
d
)= logb
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
Proof:x = Mb(x) · bk for some k ∈ Z.First digit of x in base b is d iff d ≤ Mb(x) < d + 1.logb d ≤ y < logb(d + 1), wherey = logb(Mb(x)) = logb x mod 1.
If the distribution is uniform (mod 1), then theprobability y is in this range is
logb(d +1)−logb(d) = logb
(d + 1
d
)= logb
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
Proof:x = Mb(x) · bk for some k ∈ Z.First digit of x in base b is d iff d ≤ Mb(x) < d + 1.logb d ≤ y < logb(d + 1), wherey = logb(Mb(x)) = logb x mod 1.If the distribution is uniform (mod 1), then theprobability y is in this range is
logb(d +1)−logb(d) = logb
(d + 1
d
)= logb
(1 +
1d
).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
0 1
1 102
log 2 ! log 10
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base b if {yi} is equidistributedmod 1, where yi = logb xi .
Kronecker-Weyl TheoremIf β 6∈ Q then nβ mod 1 is equidistributed.(Thus if logb α 6∈ Q, then αn is Benford.)
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Powers of 2
TheoremThe sequence {2n} for n ≥ 0 is Benford base b for any bthat is not a rational power of 2.
Proof:Consider the sequence of logarithms {n(logb 2)}.By the Kronecker-Weyl Theorem, this is uniform(mod 1) as long as logb 2 6∈ Q.If b is not a rational power of 2, then the sequence oflogarithms is uniformly distributed (mod 1), so theoriginal sequence is Benford base b.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Powers of 2
TheoremThe sequence {2n} for n ≥ 0 is Benford base b for any bthat is not a rational power of 2.
Proof:Consider the sequence of logarithms {n(logb 2)}.By the Kronecker-Weyl Theorem, this is uniform(mod 1) as long as logb 2 6∈ Q.If b is not a rational power of 2, then the sequence oflogarithms is uniformly distributed (mod 1), so theoriginal sequence is Benford base b.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers Benford base bfor almost every b.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].∣∣∣(1−
√5
2
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence will be Benford base-b for any b wherelogb
(1+√
52
)6∈ Q.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers Benford base bfor almost every b.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].
∣∣∣(1−√
52
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence will be Benford base-b for any b wherelogb
(1+√
52
)6∈ Q.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers Benford base bfor almost every b.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].∣∣∣(1−
√5
2
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence will be Benford base-b for any b wherelogb
(1+√
52
)6∈ Q.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Fibonacci Numbers
TheoremThe sequence {Fn} of Fibonacci numbers Benford base bfor almost every b.
Heuristic Argument:Closed form for Fibonacci numbers:
Fn =1√5
[(1 +√
52
)n
−
(1−√
52
)n].∣∣∣(1−
√5
2
)∣∣∣ < 1, so the leading digits are completely
determined by 1√5
(1+√
52
)n.
This sequence will be Benford base-b for any b wherelogb
(1+√
52
)6∈ Q.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
Consider the sequence {an} given by some initialconditions a0,a1, . . . ,ak−1 and then a recurrence relation
an+k = c1an+k−1 + c2an+k−2 + · · ·+ ckan,
with c1, c2, . . . , ck fixed real numbers.
Find the eigenvalues of the recurrence relation and orderthem so that |λ1| ≥ |λ2| ≥ · · · ≥ |λk |.
There exist number u1,u2, . . . ,uk (which depend on theinitial conditions) so that an = u1λ
n1 + u2λ
n2 + · · ·+ ukλ
nk .
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
Consider the sequence {an} given by some initialconditions a0,a1, . . . ,ak−1 and then a recurrence relation
an+k = c1an+k−1 + c2an+k−2 + · · ·+ ckan,
with c1, c2, . . . , ck fixed real numbers.
Find the eigenvalues of the recurrence relation and orderthem so that |λ1| ≥ |λ2| ≥ · · · ≥ |λk |.
There exist number u1,u2, . . . ,uk (which depend on theinitial conditions) so that an = u1λ
n1 + u2λ
n2 + · · ·+ ukλ
nk .
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
Consider the sequence {an} given by some initialconditions a0,a1, . . . ,ak−1 and then a recurrence relation
an+k = c1an+k−1 + c2an+k−2 + · · ·+ ckan,
with c1, c2, . . . , ck fixed real numbers.
Find the eigenvalues of the recurrence relation and orderthem so that |λ1| ≥ |λ2| ≥ · · · ≥ |λk |.
There exist number u1,u2, . . . ,uk (which depend on theinitial conditions) so that an = u1λ
n1 + u2λ
n2 + · · ·+ ukλ
nk .
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflogb |λ1| 6∈ Qand the initial conditions are such that u1 6= 0,then the sequence {an} is Benford base b.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to rewritethis as an = u1λ
n1 (1 +O(βn)).
Then yn = logb(an) = n logb λ1 + logb u1 +O(βn).Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflogb |λ1| 6∈ Qand the initial conditions are such that u1 6= 0,then the sequence {an} is Benford base b.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.
Some clever algebra using our assumptions to rewritethis as an = u1λ
n1 (1 +O(βn)).
Then yn = logb(an) = n logb λ1 + logb u1 +O(βn).Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflogb |λ1| 6∈ Qand the initial conditions are such that u1 6= 0,then the sequence {an} is Benford base b.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to rewritethis as an = u1λ
n1 (1 +O(βn)).
Then yn = logb(an) = n logb λ1 + logb u1 +O(βn).Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflogb |λ1| 6∈ Qand the initial conditions are such that u1 6= 0,then the sequence {an} is Benford base b.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to rewritethis as an = u1λ
n1 (1 +O(βn)).
Then yn = logb(an) = n logb λ1 + logb u1 +O(βn).
Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Linear Recurrence Sequences
TheoremWith a linear recurrence sequence as described, iflogb |λ1| 6∈ Qand the initial conditions are such that u1 6= 0,then the sequence {an} is Benford base b.
Sketch of Proof:
Rewrite the closed form as an = u1λn1
(1 +O
(kuλn
2λn
1
))where u = maxi |ui |+ 1.Some clever algebra using our assumptions to rewritethis as an = u1λ
n1 (1 +O(βn)).
Then yn = logb(an) = n logb λ1 + logb u1 +O(βn).Show in the limit the error term affects a vanishinglysmall portion of the distribution.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Elliptic Divisibility Sequences
DefinitionAn integral divisibility sequence is a sequence of integers{un} satisfying
un | um whenever n | m.
An elliptic divisibility sequence is an integral divisibilitysequence which satisfies the following recurrence relationfor all m ≥ n ≥ 1:
um+num−nu21
= um+1um−1u2n − un+1un−1u2
m.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Boring Elliptic Divisibility Sequences
The sequences of integers, where un = n.
The sequence 0,1,−1,0,1,−1, . . ..
The sequence1,3,8,21,55,144,377,987,2584,6765, . . . (this isevery-other Fibonacci number).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Boring Elliptic Divisibility Sequences
The sequences of integers, where un = n.
The sequence 0,1,−1,0,1,−1, . . ..
The sequence1,3,8,21,55,144,377,987,2584,6765, . . . (this isevery-other Fibonacci number).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Boring Elliptic Divisibility Sequences
The sequences of integers, where un = n.
The sequence 0,1,−1,0,1,−1, . . ..
The sequence1,3,8,21,55,144,377,987,2584,6765, . . . (this isevery-other Fibonacci number).
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Not-So-Boring Elliptic Divisibility Sequences
The sequences which begins0,1,1,−1,1,2,−1,−3,−5,7,−4,−28,29,59,129,−314,−65,1529,−3689,−8209,−16264,833313,113689,−620297,2382785,7869898,7001471,−126742987,−398035821,168705471, . . .(This is sequence A006769 in the On-LineEncyclopedia of Integer Sequences.)
The sequence which begins1,1,−3,11,38,249,−2357,8767,496036,−3769372,−299154043,−12064147359, . . ..
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Not-So-Boring Elliptic Divisibility Sequences
The sequences which begins0,1,1,−1,1,2,−1,−3,−5,7,−4,−28,29,59,129,−314,−65,1529,−3689,−8209,−16264,833313,113689,−620297,2382785,7869898,7001471,−126742987,−398035821,168705471, . . .(This is sequence A006769 in the On-LineEncyclopedia of Integer Sequences.)
The sequence which begins1,1,−3,11,38,249,−2357,8767,496036,−3769372,−299154043,−12064147359, . . ..
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Why We Like Elliptic Divisibility Sequences
Special case of Somos sequences, which areinteresting and an active area of research.
Connection to elliptic curves, also an active area ofresearch.
Elliptic curve
E over Qwith rational point
P on E
↔ EDS: denominators
of the sequenceof points {P,2P,3P, . . .}
Applications to elliptic curve cryptography.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Why We Like Elliptic Divisibility Sequences
Special case of Somos sequences, which areinteresting and an active area of research.
Connection to elliptic curves, also an active area ofresearch.
Elliptic curve
E over Qwith rational point
P on E
↔ EDS: denominators
of the sequenceof points {P,2P,3P, . . .}
Applications to elliptic curve cryptography.
69
History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Why We Like Elliptic Divisibility Sequences
Special case of Somos sequences, which areinteresting and an active area of research.
Connection to elliptic curves, also an active area ofresearch.
Elliptic curve
E over Qwith rational point
P on E
↔ EDS: denominators
of the sequenceof points {P,2P,3P, . . .}
Applications to elliptic curve cryptography.70
History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Elliptic Divisibility Sequences are Benford?
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Elliptic Divisibility Sequences are Benford?
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Elliptic Divisibility Sequences are Benford?
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
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History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
75
History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
76
History Applications Formalism Benford and Integer Sequences Benford and Recurrence Relations
Heuristic Argument
It’s well-known that elliptic divisibility sequencessatisfy a growth condition like un ≈ cn2 where theconstant c depends on the arithmetic height of thepoint P and on the curve E .
Weyl’s theorem tells us that {nkα} is uniformdistributed (mod 1) iff α 6∈ Q.
So we should at least be able to conclude that a givenEDS is Benford base b for almost every b.
But: The argument with the big-O error terms isdelicate, and not enough is known in the case of EDS.
77
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