A052335 Number of partitions of n into at most 1 copy of 1, 2 copies of 2, 3 copies of 3, ... .
1, 1, 1, 2, 3, 4, 5, 7, 10, 13, 17, 22, 28, 36, 46, 58, 73, 91, 114, 141, 173, 213, 261, 318, 387, 469, 567, 683, 821, 984, 1176, 1403, 1671, 1984, 2351, 2781, 3284, 3869, 4550, 5343, 6264, 7330, 8565, 9993, 11642, 13543, 15733, 18252, 21148, 24471, 28282, 32646, 37640, 43348, 49867, 57302, 65776, 75426, 86405, 98882
Offset: 0
Keywords
Examples
a(5)=4 because we have [5], [4,1], [3,2] and [2,2,1] ([3,1,1], [2,1,1,1] and [1,1,1,1,1] do not qualify).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 129 terms from Reinhard Zumkeller)
- Mircea Merca and Emil Simion, n-Color Partitions into Distinct Parts as Sums over Partitions, Symmetry (2023) Vol. 15, Iss. 11.
Programs
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Maple
g:=product((1-x^(j*(j+1)))/(1-x^j),j=1..53): gser:=series(g,x=0,55): seq(coeff(gser,x,n),n=0..49); # Emeric Deutsch, Mar 04 2006 # second Maple program: with(numtheory): a:= proc(n) option remember; `if`(n=0, 1, add(add( `if`(issqr(4*d+1), 0, d), d=divisors(j))*a(n-j), j=1..n)/n) end: seq(a(n), n=0..80); # Alois P. Heinz, Apr 01 2014
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Mathematica
CoefficientList[Series[Product[Sum[x^(i j ), {i, 0, j}], {j, 1, 49}], {x, 0, 49}], x] (* Second program: *) a[n_] := a[n] = If[n == 0, 1, Sum[Sum[If[IntegerQ @ Sqrt[4*d+1], 0, d], {d, Divisors[j]}]*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Jan 30 2017, after Alois P. Heinz *) -
PARI
N=66; q='q+O('q^N); Vec( prod(n=1,N, sum(k=0,n,q^(k*n)) ) ) \\ Joerg Arndt, Apr 01 2014
Formula
G.f.: Product_{i>=1} (1-x^(i*(i+1)))/(1-x^i).
G.f.: (1+x) * (1+x^2+x^4) * (1+x^3+x^6+x^9) * (1+x^4+x^8+x^12+x^16) * ... (g.f. above, expanded). - Joerg Arndt, Apr 01 2014
G.f.: Product_{n>=1} (1 - q^(n*(n+1))) / Product_{n>=1} (1 - q^n). - Joerg Arndt, Apr 01 2014
a(n) = p(n,1,1) with p(n,t,k) = if t<0 then 0 else if k<=n then p(n-k,t-1,k)+p(n,k+1,k+1) else 0^n. - Reinhard Zumkeller, Jan 20 2010
a(n) ~ exp(Pi*sqrt(2*n/3) - 3^(1/4) * Zeta(3/2) * n^(1/4) / 2^(3/4) - 3*Zeta(3/2)^2/(32*Pi)) / sqrt(2*n). - Vaclav Kotesovec, Jan 01 2017
Comments