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.

A215513 spt(n) - p(n): total number of smallest parts in all partitions of n minus the number of partitions of n.

Original entry on oeis.org

0, 1, 2, 5, 7, 15, 20, 35, 50, 77, 105, 161, 214, 305, 413, 570, 751, 1022, 1330, 1772, 2295, 2996, 3837, 4970, 6305, 8050, 10155, 12844, 16065, 20169, 25055, 31197, 38549, 47650, 58540, 71960, 87916, 107424, 130655, 158830, 192260, 232642, 280406
Offset: 1

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Author

Omar E. Pol, Jan 13 2013

Keywords

Comments

Also total number of smallest parts that are not on the right border in all partitions of n.

Examples

			For n = 6 the partitions of 6 with the smallest parts that are not in the right border in brackets are
-----------------------------------------
.      Partitions of 6            Value
-----------------------------------------
.                       6           0
.                  [3]+ 3           1
.                   4 + 2           0
.              [2]+[2]+ 2           2
.                   5 + 1           0
.               3 + 2 + 1           0
.               4 +[1]+ 1           1
.           2 + 2 +[1]+ 1           1
.           3 +[1]+[1]+ 1           2
.       2 +[1]+[1]+[1]+ 1           3
.  [1]+[1]+[1]+[1]+[1]+ 1           5
--------------------------------------
.                           Total: 15
On the other hand the total number of smallest parts in all partitions of 6 is 26 and the number of partitions of 6 is 11, so a(6) = 26 - 11 = 15.
		

Crossrefs

Programs

  • Mathematica
    b[n_, i_] := b[n, i] = If[n==0 || i==1, n, {q, r} = QuotientRemainder[n, i]; If[r == 0, q, 0] + Sum[b[n - i*j, i - 1], {j, 0, n/i}]];
    a[n_] := b[n, n] - PartitionsP[n];
    Array[a, 50] (* Jean-François Alcover, Jun 05 2021, using Alois P. Heinz's code for A092269 *)

Formula

a(n) = A092269(n) - A000041(n).
a(n) = A000070(n-2) + A220479(n), n >= 2.
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2*Pi*sqrt(2*n)) * (1 - 25*Pi/(24*sqrt(6*n)) + (25/48 + 49*Pi^2/6912)/n). - Vaclav Kotesovec, Jul 31 2017