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A087910
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Exponent of the greatest power of 2 dividing the numerator of 2^1/1 + 2^2/2 + 2^3/3 + ... + 2^n/n.
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2
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1, 2, 2, 5, 8, 5, 5, 13, 9, 10, 10, 12, 12, 12, 12, 22, 17, 18, 18, 21, 22, 21, 21, 27, 25, 26, 26, 27, 27, 27, 27, 40, 33, 34, 34, 37, 39, 37, 37, 48, 41, 42, 42, 44, 44, 44, 44, 54, 49, 50, 50, 53, 54, 53, 53, 58, 57, 59, 62, 58, 58, 58
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OFFSET
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1,2
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COMMENTS
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Problem 9 of the 2002 Sydney University Mathematical Society Problems competition asked for a proof that a(n) tends to infinity with n. While this is immediate from the theory of the 2-adic logarithm, elementary proofs are available.
a(n) tends to infinity with n implies that log(-1) = 0 in the 2-adic field, by setting x = 2 in -log(1-x) = Sum_{k>=1} x^k/k. - Jianing Song, Aug 05 2019
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REFERENCES
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A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see p. 278.
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LINKS
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FORMULA
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Sum_{k=1..n} 2^k/k = (2^n/n)*Sum_{k=0..n-1} 1/binomial(n-1,k), so a(n) >= n - v(n,2) - max_{k=0..n-1} v(binomial(n-1,k),2) = n - A007814(n) - A119387(n) = n - floor(log_2(n)), where v(n,2) is the 2-adic valuation of n. It seems that the equality holds if and only if n = 2^m - 1 for some m. - Jianing Song, Feb 22 2020
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EXAMPLE
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a(5) = 8 as 2^1/1 + 2^2/2 + 2^3/3 + 2^4/4 + 2^5/5 = 256/15 whose numerator is divisible by 2^8 but not by 2^9.
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MAPLE
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S:= 0:
for n from 1 to 100 do
S:= S + 2^n/n;
a[n]:= padic:-ordp(numer(S), 2);
od:
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MATHEMATICA
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s[n_] := -2^(n + 1) LerchPhi[2, 1, n + 1] - I Pi;
a[n_] := IntegerExponent[Numerator[Simplify[s[n]]], 2];
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PROG
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(PARI) a(n) = valuation(sum(k=1, n, 2^k/k), 2) \\ Jianing Song, Feb 22 2020
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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