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.

A049988 Number of nondecreasing arithmetic progressions of positive integers with sum n.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 7, 5, 7, 9, 9, 7, 14, 8, 11, 16, 13, 10, 20, 11, 17, 21, 16, 13, 27, 17, 18, 26, 22, 16, 35, 17, 23, 31, 23, 25, 41, 20, 25, 36, 33, 22, 46, 23, 31, 48, 30, 25, 52, 29, 38, 47, 36, 28, 57, 37, 41, 52, 37, 31, 71, 32, 39, 62, 44, 43, 69, 35, 45, 62, 57, 37, 79, 38
Offset: 0

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Comments

From Gus Wiseman, May 03 2019: (Start)
a(n) is the number of integer partitions of n with equal differences. The Heinz numbers of these partitions are given by A325328. For example, the a(1) = 1 through a(9) = 9 partitions are:
1 2 3 4 5 6 7 8 9
11 21 22 32 33 43 44 54
111 31 41 42 52 53 63
1111 11111 51 61 62 72
222 1111111 71 81
321 2222 333
111111 11111111 432
531
111111111
(End)
From Petros Hadjicostas, Sep 29 2019: (Start)
We show how Leroy Quet's g.f. Sum_{n >= 0} a(n)*x^n = 1/(1-x) + Sum_{k >= 2} x^k/(1-x^(k*(k-1)/2))/(1-x^k) in the Formula section below can be derived from Graeme McRae's g.f. for A049982 (see one of the links below).
Let b(n) = A049982(n) for n >= 1. Then Graeme McRae proved that Sum_{n >= 1} b(n)*x^n = Sum_{k >= 2} x^t(k)/(x^t(k) - x^t(k-1) - x^k + 1) = Sum_{k >= 2} x^t(k)/((1 - x^k) * (1 - x^t(k-1))), where t(k) = A000217(k) = k*(k+1)/2.
Since a(n) - b(n) = A000005(n) for n >= 1, to finish the proof, we only need to show that K(x) := 1 + Sum_{n >= 1} a(n)*x^n - Sum_{n >= 1} b(n)*x^n is the g.f. of A000005 (= number of divisors). But it is easy to show that K(x) = 1 + Sum_{k >= 1} x^k/(1 - x^k) = 1 + Sum_{n >= 1} A000005(n)*x^n (Lambert series for the number of divisors function). (End)

Crossrefs

Programs

  • Mathematica
    a[n_]:=If[n==0,1,Block[{i,c=Floor[(n-1)/2]+DivisorSigma[0,n]},Do[i=1;While[i*kGus Wiseman, May 07 2019 *)
    Table[Length[Select[IntegerPartitions[n],SameQ@@Differences[#]&]],{n,0,30}] (* Gus Wiseman, May 03 2019 *)
  • PARI
    seq(n)={Vec(1/(1-x) + sum(k=2, n, x^k/(1 - x^(k*(k-1)/2))/(1-x^k) + O(x*x^n)))} \\ Andrew Howroyd, Sep 28 2019

Formula

G.f.: 1/(1-x) + Sum_{k>=2} x^k/(1-x^(k*(k-1)/2))/(1-x^k). - Leroy Quet, Apr 08 2010. [Edited by Gus Wiseman, May 03 2019]
a(n) = A049982(n) + A000005(n) = A049980(n) + A000005(n) - 1 for n >= 1. - Petros Hadjicostas, Sep 28 2019

Extensions

Edited by Max Alekseyev, May 03 2010
a(0) = 1 prepended by Gus Wiseman, May 03 2019