A026811 Number of partitions of n in which the greatest part is 5.

0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 84, 101, 119, 141, 164, 192, 221, 255, 291, 333, 377, 427, 480, 540, 603, 674, 748, 831, 918, 1014, 1115, 1226, 1342, 1469, 1602, 1747, 1898, 2062, 2233, 2418, 2611, 2818, 3034, 3266, 3507, 3765

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Essentially same as A001401: five zeros followed by A001401.

Also number of partitions of n into exactly 5 parts.

References

• D. E. Knuth, The Art of Computer Programming, vol. 4, fascicle 3, Generating All Combinations and Partitions, Section 7.2.1.4., p. 56, exercise 31.

Links

• Washington Bomfim, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-1,-1,-1,1,1,1,0,0,-1,-1,1).

Crossrefs

Cf. A026810, A026812, A026813, A026814, A026815, A026816, A002622 (partial sums), A008667 (first differences).

Programs

GAP

List([0..70],n->NrPartitions(n,5)); # Muniru A Asiru, May 17 2018

Mathematica

Table[Count[IntegerPartitions[n], {5, _}], {n, 0, 55}] (* corrected by Harvey P. Dale, Oct 24 2011 *)
Table[Length[IntegerPartitions[n, {5}]], {n, 0, 55}] (* Eric Rowland, Mar 02 2017 *)
CoefficientList[Series[x^5/Product[1 - x^k, {k, 1, 5}], {x, 0, 65}], x] (* Robert A. Russell, May 13 2018 *)
Drop[LinearRecurrence[{1,1,0,0,-1,-1,-1,1,1,1,0,0,-1,-1,1}, Append[Table[0,{14}],1],110],9] (* Robert A. Russell, May 17 2018 *)

PARI

a(n)=round((n^4+10*(n^3+n^2)-75*n-45*(-1)^n*n)/2880);
for(n=0,10000,print(n," ",a(n))); /* b-file format */
/* Washington Bomfim, Jul 03 2012 */

PARI

x='x+O('x^99); concat(vector(5), Vec(x^5/prod(k=1, 5, 1-x^k))) \\ Altug Alkan, May 17 2018

Formula

a(n) = round( ((n^4+10*(n^3+n^2)-75*n -45*n*(-1)^n)) / 2880 ). - Washington Bomfim, Jul 03 2012

G.f.: x^5/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)). - Joerg Arndt, Jul 04 2012

a(n) = A008284(n,5). - Robert A. Russell, May 13 2018

From Gregory L. Simay, Jul 28 2019: (Start)

a(2n) = a(2n-1) + a(n+1) + a(n) - a(n-3) - a(n-4);

a(2n+1) = a(2n) + a(n+3) - a(n-5). (End)

From R. J. Mathar, Jun 23 2021: (Start)

a(n) - a(n-5) = A001400(n-5).

a(n) - a(n-4) = A008669(n-5).

a(n) - a(n-3) = A029007(n-5).

a(n) - a(n-2) = A029032(n-5).

a(n) = +a(n-1) +a(n-2) -a(n-5) -a(n-6) -a(n-7) +a(n-8) +a(n-9) +a(n-10) -a(n-13) -a(n-14) +a(n-15). (End)

Extensions

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

a(0)=0 inserted by Joerg Arndt, Jul 04 2012