A026812 Number of partitions of n in which the greatest part is 6.

0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 14, 20, 26, 35, 44, 58, 71, 90, 110, 136, 163, 199, 235, 282, 331, 391, 454, 532, 612, 709, 811, 931, 1057, 1206, 1360, 1540, 1729, 1945, 2172, 2432, 2702, 3009, 3331, 3692, 4070, 4494, 4935, 5427, 5942, 6510, 7104, 7760

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Also number of partitions of n into 6 parts. - Washington Bomfim, Jan 15 2021

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 1..1000 from Vincenzo Librandi)
• G. E. Andrews, Partitions: At the Interface of q-Series and Modular Forms, The Ramanujan Journal 7, 385-400 (2003), Eq.(3.10).
• Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-1,0,-2,0,1,1,1,1,0,-2,0,-1,0,0,1,1,-1).

Crossrefs

Essentially same as A001402.

Cf. A026810, A026811, A026813, A026814, A026815, A026816.

Cf. A001401, A001402.

Programs

GAP

List([0..70],n->NrPartitions(n,6)); # Muniru A Asiru, May 17 2018

Mathematica

Table[ Length[ Select[ Partitions[n], First[ # ] == 6 & ]], {n, 1, 60} ]
CoefficientList[Series[x^6/((1 - x) (1 - x^2) (1 - x^3) (1 - x^4) (1 - x^5) (1 - x^6)), {x, 0, 60}], x] (* Vincenzo Librandi, Oct 18 2013 *)
Drop[LinearRecurrence[{1,1,0,0,-1,0,-2,0,1,1,1,1,0,-2,0,-1,0,0,1,1,-1}, Append[Table[0,{20}],1],115],14] (* Robert A. Russell, May 17 2018 *)

PARI

my(x='x+O('x^99)); concat(vector(6), Vec(x^6/prod(k=1, 6, 1-x^k))) \\ Altug Alkan, May 17 2018

PARI

a = vector(60,n,n--; round((n+11)*((6*n^4+249*n^3+2071*n^2 -4931*n+40621) /518400 +n\2*(n+10)/192+((n+1)\3+n\3*2)/54))); a = concat([0,0,0,0,0,0], a) \\ Washington Bomfim, Jan 16 2021

Formula

G.f.: x^6 / ((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)*(1-x^6)). - Colin Barker, Dec 20 2012

a(n) = A008284(n,6). - Robert A. Russell, May 13 2018

a(n) = Sum_{m=1..floor(n/6)} Sum_{l=m..floor((n-m)/5)} Sum_{k=l..floor((n-l-m)/4)} Sum_{j=k..floor((n-k-l-m)/3)} Sum_{i=j..floor((n-j-k-l-m)/2)} 1. - Wesley Ivan Hurt, Jun 29 2019

a(n) = A001402(n) - A001401(n). a(n) = A001402(n-6). - Washington Bomfim, Jan 15 2021

a(n) = round((1/86400)*n^5 + (1/3840)*n^4 + (19/12960)*n^3 - (n mod 2)*(1/384)*n^2 + (1/17280)*b(n mod 6)*n), where b(0)=96, b(1)=b(5)=-629, b(2)=b(4)=-224, and b(3)=-309. - Washington Bomfim and Jon E. Schoenfield, Jan 16 2021

Extensions

More terms from Robert G. Wilson v, Jan 11 2002

a(0)=0 prepended by Seiichi Manyama, Jun 08 2017