cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Robin Chapman

Robin Chapman's wiki page.

Robin Chapman has authored 2 sequences.

A238741 Number of strokes needed to draw all Young diagrams with n cells.

Original entry on oeis.org

0, 1, 2, 5, 12, 21, 40, 63, 105, 159, 245, 355, 526, 739, 1052, 1452, 2011, 2718, 3691, 4909, 6546
Offset: 0

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Author

Robin Chapman, Mar 04 2014

Keywords

Comments

a(n) is the number of strokes needed to draw the Young diagrams (considered as an array of boxes) of all the partitions of n. (The generating function is omitted for the present, as finding it is a current Monthly problem).

Links

  • Richard P. Stanley, Problem 11762, The American Mathematical Monthly, 121 (2014), 266.

A087910 Exponent of the greatest power of 2 dividing the numerator of 2^1/1 + 2^2/2 + 2^3/3 + ... + 2^n/n.

Original entry on oeis.org

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
Offset: 1

Views

Author

Robin Chapman, Oct 17 2003

Keywords

Comments

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

Examples

			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.

References

  • A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see p. 278.

Links

Crossrefs

Cf. A108866, A007814, A119387, A000523.

Programs

  • Maple
    S:= 0:
for n from 1 to 100 do
  S:= S + 2^n/n;
  a[n]:= padic:-ordp(numer(S),2);
od:
seq(a[n],n=1..100); # Robert Israel, Jun 09 2015
  • Mathematica
    s[n_] := -2^(n + 1) LerchPhi[2, 1, n + 1] - I Pi;
a[n_] := IntegerExponent[Numerator[Simplify[s[n]]], 2];
Array[a, 62] (* Peter Luschny, Feb 22 2020 *)
  • PARI
    a(n) = valuation(sum(k=1,n,2^k/k), 2) \\ Jianing Song, Feb 22 2020

Formula

a(n) = A007814(A108866(n)). - Michel Marcus, Feb 22 2020
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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