A036469 Partial sums of A000009 (partitions into distinct parts).
1, 2, 3, 5, 7, 10, 14, 19, 25, 33, 43, 55, 70, 88, 110, 137, 169, 207, 253, 307, 371, 447, 536, 640, 762, 904, 1069, 1261, 1483, 1739, 2035, 2375, 2765, 3213, 3725, 4310, 4978, 5738, 6602, 7584, 8697, 9957, 11383, 12993, 14809, 16857, 19161, 21751, 24661
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000
- Kevin Beanland and Hung Viet Chu, On Schreier-type Sets, Partitions, and Compositions, arXiv:2311.01926 [math.CO], 2023.
- A. V. Chekhonadskikh, Some classical number sequences in control system design, Siberian Electronic Mathematical Reports, Volume 14, p. 620-628. See Theorem 2.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 774
Crossrefs
Cf. A166240. - Gary W. Adamson, Oct 09 2009
Column k=1 of A346520.
Programs
-
Maple
g:=1/(1-x)/product(1-x^(2*j-1),j=1..30): gser:=series(g,x=0,50): seq(coeff(gser,x,n),n=0..46); # Emeric Deutsch, Mar 29 2006 # second Maple program: b:= proc(n, i) b(n, i):= `if`(n=0, 1, `if`(i<1, 0, b(n, i-1)+`if`(i>n, 0, b(n-i, min(n-i, i-1))))) end: a:= proc(n) option remember; b(n, n) +`if`(n>0, a(n-1), 0) end: seq(a(n), n=0..60); # Alois P. Heinz, Nov 21 2012
-
Mathematica
CoefficientList[ Series[Product[(1 + t^i), {i, 1, Infinity}]/(1 - t), {t, 0, 46}], t] (* Geoffrey Critzer, May 16 2010 *) b[n_, i_] := If[n == 0, 1, If[i<1, 0, b[n, i-1]+If[i>n, 0, b[n-i, Min[n-i, i-1]]]]]; a[n_] := a[n] = b[n, n]+If[n>0, a[n-1], 0]; Table[a[n], {n, 0, 60}] // Flatten (* Jean-François Alcover, Mar 10 2014, after Alois P. Heinz *) Accumulate[Table[PartitionsQ[n], {n, 0, 50}]] (* Vaclav Kotesovec, Oct 26 2016 *)
Formula
G.f.: 1/[(1-x)*product(1-x^(2j-1), j=1..infinity)]. - Emeric Deutsch, Mar 29 2006
a(n) ~ 3^(1/4) * exp(Pi*sqrt(n/3)) / (2*Pi*n^(1/4)) * (1 + (18+13*Pi^2) / (48*Pi*sqrt(3*n)) + (2916 - 1404*Pi^2 + 121*Pi^4)/(13824*Pi^2*n)). - Vaclav Kotesovec, Feb 26 2015, updated Oct 26 2016
For n > 0, a(n) = A026906(n) + 1. - Vaclav Kotesovec, Oct 26 2016
Faster converging g.f.: A(x) = (1/(1 - x))*Sum_{n >= 0} x^(n*(2*n-1))/Product_{k = 1..2*n} (1 - x^k). - Peter Bala, Feb 02 2021
Comments