A238869 Number of partitions of n where the difference between consecutive parts is at most 9.

1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 76, 99, 131, 170, 221, 283, 364, 461, 586, 737, 926, 1154, 1439, 1779, 2199, 2703, 3317, 4051, 4942, 6001, 7278, 8796, 10610, 12760, 15323, 18344, 21928, 26148, 31127, 36971, 43848, 51890, 61321, 72327, 85183, 100149, 117588, 137827, 161343, 188583, 220139, 256607, 298761, 347360

Offset: 0

Author

Joerg Arndt, Mar 08 2014

Keywords

• nonn

Comments

Also the number of partitions of n such that all parts, with the possible exception of the largest are repeated at most nine times (by taking conjugates).

In general, for d > 0, "number of partitions with max diff d" is asymptotic to d^(1/4) * exp(Pi*sqrt(2*d*n/(3*(d+1)))) / (2^(5/4) * 3^(1/4) * (d+1)^(3/4) * n^(3/4)). - Vaclav Kotesovec, Jan 26 2022

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)

Crossrefs

Sequences "number of partitions with max diff d": A000005 (d=0, for n>=1), A034296 (d=1), A224956 (d=2), A238863 (d=3), A238864 (d=4), A238865 (d=5), A238866 (d=6), A238867 (d=7), A238868 (d=8), this sequence, A000041 (d --> infinity).

Programs

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(9, 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

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[9, 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[#]]]<10&)],{n,0,60}] (* _Harvey P. Dale, Nov 24 2024 *)

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^(10*i))/(1-q^i) ) ) )

Formula

G.f.: 1 + sum(k>=1, q^k/(1-q^k) * prod(i=1..k-1, (1-q^(10*i))/(1-q^i) ) ).

a(n) = Sum_{k=0..9} A238353(n,k). - Alois P. Heinz, Mar 09 2014

a(n) ~ 3^(1/4) * exp(Pi*sqrt(3*n/5)) / (4 * 5^(3/4) * n^(3/4)). - Vaclav Kotesovec, Jan 26 2022