A168659 Number of partitions of n such that the number of parts is divisible by the greatest part. Also number of partitions of n such that the greatest part is divisible by the number of parts.
1, 1, 2, 2, 3, 3, 6, 6, 8, 9, 14, 16, 22, 25, 33, 39, 51, 60, 79, 92, 116, 137, 174, 204, 254, 300, 368, 435, 530, 625, 760, 896, 1076, 1267, 1518, 1780, 2121, 2484, 2946, 3444, 4070, 4749, 5594, 6514, 7637, 8879, 10384, 12043, 14040, 16255
Offset: 1
Keywords
Examples
a(5)=3 because in the partitions [1,1,1,1,1], [1,1,1,2], [1,1,3] the number of parts is divisible by the greatest part; not true for the partitions [1,2,2],[2,3], [1,4], and [5]. - _Emeric Deutsch_, Dec 04 2009 From _Gus Wiseman_, Feb 08 2021: (Start) The a(1) = 1 through a(10) = 9 partitions of the first type: 1 11 21 22 311 321 322 332 333 4222 111 1111 2111 2211 331 2222 4221 4321 11111 111111 2221 4211 4311 4411 4111 221111 51111 52111 211111 311111 222111 222211 1111111 11111111 321111 322111 21111111 331111 111111111 22111111 1111111111 The a(1) = 1 through a(11) = 14 partitions of the second type (A=10, B=11): 1 2 3 4 5 6 7 8 9 A B 21 22 41 42 43 44 63 64 65 311 321 61 62 81 82 83 322 332 333 622 A1 331 611 621 631 632 4111 4211 4221 4222 641 4311 4321 911 51111 4411 4322 52111 4331 4421 8111 52211 53111 611111 (End)
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..5000 (terms 1..301 from Vladeta Jovovic corrected by N. J. A. Sloane, Oct 05 2010, terms 302..1000 from Seiichi Manyama)
Crossrefs
Note: A-numbers of Heinz-number sequences are in parentheses below.
Row sums of A350879.
Programs
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Maple
a := proc (n) local pn, ct, j: with(combinat): pn := partition(n): ct := 0: for j to numbpart(n) do if `mod`(nops(pn[j]), max(seq(pn[j][i], i = 1 .. nops(pn[j])))) = 0 then ct := ct+1 else end if end do: ct end proc: seq(a(n), n = 1 .. 50); # Emeric Deutsch, Dec 04 2009
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Mathematica
Table[Length[Select[IntegerPartitions[n],Divisible[Length[#],Max[#]]&]],{n,30}] (* Gus Wiseman, Feb 08 2021 *) nmax = 100; s = 0; Do[s += Normal[Series[Sum[x^((m+1)*k - 1) * Product[(1 - x^(m*k + j - 1))/(1 - x^j), {j, 1, k-1}], {k, 1, (1 + nmax)/(1 + m) + 1}], {x, 0, nmax}]], {m, 1, nmax}]; Rest[CoefficientList[s, x]] (* Vaclav Kotesovec, Oct 18 2024 *)
Formula
G.f.: Sum_{i>=1} Sum_{j>=1} x^((i+1)*j-1) * Product_{k=1..j-1} (1-x^(i*j+k-1))/(1-x^k). - Seiichi Manyama, Jan 24 2022
a(n) ~ c * exp(Pi*sqrt(2*n/3)) / n^(3/2), where c = 0.04628003... - Vaclav Kotesovec, Nov 16 2024
Extensions
Extended by Emeric Deutsch, Dec 04 2009
Comments