A029007 Expansion of 1/((1-x)(1-x^2)(1-x^4)(1-x^5)).

1, 1, 2, 2, 4, 5, 7, 8, 11, 13, 17, 19, 24, 27, 33, 37, 44, 49, 57, 63, 73, 80, 91, 99, 112, 122, 136, 147, 163, 176, 194, 208, 228, 244, 266, 284, 308, 328, 354, 376, 405, 429, 460, 486, 520, 549, 585, 616, 655, 689

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Partitions of n into parts 1, 2, 4 and 5. - David Neil McGrath, Sep 14 2014

Examples

			There are 7 partitions of 6 from 1,2,4 and 5. These are (51)(42)(411)(222)(2211)(21111)(111111). - _David Neil McGrath_, Sep 14 2014
G.f. = 1 + x + 2*x^2 + 2*x^3 + 4*x^4 + 5*x^5 + 7*x^6 + 8*x^7 + 11*x^8 + ...
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Xiangyu Ding and Lisa Hui Sun, Truncated theta series from the Bailey lattice, arXiv:2403.11608 [math.CO], 2024. See p. 18.
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,0,-2,0,1,-1,1,1,-1).

Programs

Maple

f:= gfun[rectoproc]({a(0)=1, a(1)=1, a(2)=2, a(3)=2, a(4)=4, a(5)=5, a(6)=7, a(7)=8, a(8)=11, a(9)=13, a(10)=17, a(11)=19, a(n)=a(n-1)+a(n-2)-a(n-3)+a(n-4)- 2*a(n-6)+ a(n-8)-a(n-9)+a(n-10)+a(n-11)-a(n-12)},a(n)):
seq(f(n),n=0..100); # Robert Israel, Sep 14 2014

Mathematica

CoefficientList[Series[1/((1-x)(1-x^2)(1-x^4)(1-x^5)),{x,0,50}],x] (* or *) LinearRecurrence[{1,1,-1,1,0,-2,0,1,-1,1,1,-1},{1,1,2,2,4,5,7,8,11,13,17,19},50] (* Harvey P. Dale, Dec 06 2013 *)

PARI

a(n)=my(v=apply(u->for(i=1,#u,if(u[i]==3,return(0)));1,partitions(n,5)));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Sep 15 2014

Formula

a(0)=1, a(1)=1, a(2)=2, a(3)=2, a(4)=4, a(5)=5, a(6)=7, a(7)=8, a(8)=11, a(9)=13, a(10)=17, a(11)=19, a(n)=a(n-1)+a(n-2)-a(n-3)+a(n-4)- 2*a(n-6)+ a(n-8)-a(n-9)+a(n-10)+a(n-11)-a(n-12). - Harvey P. Dale, Dec 06 2013

a(n) = floor((2*n^3+36*n^2+193*n+525)/480+(n+1)*(-1)^n/32). - Tani Akinari, Sep 30 2014

Euler transform of length 5 sequence [ 1, 1, 0, 1, 1]. - Michael Somos, Sep 30 2014

a(n) = -a(-12-n) for all n in Z. - Michael Somos, Sep 30 2014

0 = a(n) - a(n+1) - a(n+5) + a(n+6) for all odd n in Z. - Michael Somos, Sep 30 2014

0 = a(n) - a(n+1) - a(n+5) + a(n+6) - floor((n+10)/4) for all even n in Z. - Michael Somos, Sep 30 2014