A238865 Number of partitions of n where the difference between consecutive parts is at most 5.
1, 1, 2, 3, 5, 7, 11, 15, 21, 28, 38, 50, 67, 87, 114, 146, 188, 238, 302, 379, 476, 593, 737, 911, 1124, 1379, 1688, 2058, 2504, 3034, 3669, 4422, 5319, 6378, 7634, 9114, 10859, 12908, 15316, 18134, 21434, 25283, 29775, 35001, 41080, 48133, 56312, 65778, 76727, 89366, 103947, 120739, 140065, 162271, 187769, 217006, 250504
Offset: 0
Keywords
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)
Crossrefs
Programs
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Maple
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(b(n-i*j, i-1), j=0..min(5, n/i)))) end: g:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(b(n-i*j, i-1), j=1..n/i))) end: a:= n-> add(g(n, k), k=0..n): seq(a(n), n=0..60); # Alois P. Heinz, Mar 09 2014
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Mathematica
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<1, 0, Sum[b[n - i*j, i-1], {j, 0, Min[5, n/i]}]]]; g[n_, i_] := g[n, i] = If[n == 0, 1, If[i<1, 0, Sum[b[n - i*j, i-1], {j, 1, n/i}]]]; a[n_] := Sum[g[n, k], {k, 0, n}]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Feb 18 2015, after Alois P. Heinz *) Table[Count[IntegerPartitions[n],?(Max[Abs[Differences[#]]]<6&)],{n,0,60}] (* _Harvey P. Dale, Feb 04 2017 *)
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PARI
N=66; q = 'q + O('q^N); Vec( 1 + sum(k=1, N, q^k/(1-q^k) * prod(i=1,k-1, (1-q^(6*i))/(1-q^i) ) ) )
Formula
G.f.: 1 + sum(k>=1, q^k/(1-q^k) * prod(i=1..k-1, (1-q^(6*i))/(1-q^i) ) ).
a(n) = Sum_{k=0..5} A238353(n,k). - Alois P. Heinz, Mar 09 2014
a(n) ~ 5^(1/4) * exp(Pi*sqrt(5*n/9)) / (12 * n^(3/4)). - Vaclav Kotesovec, Jan 26 2022
Comments