A025767 Expansion of 1/((1-x)*(1-x^3)*(1-x^4)).
1, 1, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 17, 18, 20, 22, 24, 26, 28, 30, 33, 35, 37, 40, 43, 45, 48, 51, 54, 57, 60, 63, 67, 70, 73, 77, 81, 84, 88, 92, 96, 100, 104, 108, 113, 117, 121, 126, 131, 135, 140, 145, 150, 155, 160, 165, 171, 176, 181, 187, 193, 198
Offset: 0
Examples
The a(4)=3 partitions of 4 into parts 1, 3, and 4 are (4), (3,1), and (1,1,1,1). - _David Neil McGrath_, Aug 30 2014 From _John M. Campbell_, Jan 29 2016: (Start) Letting n=12, there are a(n-4)=a(8)=6 partitions mu of n=12 of length 3 such that mu_1-mu_2 is even and mu_2-mu_3 is odd or vice versa: (10,1,1) |- n (8,3,1) |- n (7,3,2) |- n (6,5,1) |- n (6,3,3) |- n (5,5,2) |- n (End)
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,-1,0,-1,1).
Programs
-
Magma
[Floor(n^2/24 + n/3 + 1): n in [0..70]]; // Vincenzo Librandi, Aug 31 2014
-
Maple
A056594 := proc(n) op(1+(n mod 4),[1,0,-1,0]) ; end proc: A061347 := proc(n) op(1+(n mod 3),[-2,1,1]) ; end proc: A025767 := proc(n) n^2/24+n/3+83/144+(-1)^n/16 +A061347(n+1)/9 +A056594(n)/4 ; end proc: # R. J. Mathar, Mar 31 2011
-
Mathematica
Table[Floor[n^2/24 + n/3 + 1], {n, 0, 60}] (* Vincenzo Librandi, Aug 31 2014 *)
-
PARI
a(n)=if(n<0,0,(n^2+8*n)\24+1)
-
PARI
{a(n) = round( ((n + 4)^2 - 1) / 24 )}; /* Michael Somos, Nov 09 2007 */
-
PARI
Vec(1/((1-x)*(1-x^3)*(1-x^4)) + O(x^80)) \\ Michel Marcus, Jan 29 2016
Formula
G.f.: 1/((1-x)*(1-x^3)*(1-x^4)).
a(n) = floor(n^2/24+n/3+1).
a(n) = Sum_{k=0..floor(n/4)} floor((n-4*k+3)/3). - Paul Barry, Jan 20 2006
Euler transform of length 4 sequence [1, 0, 1, 1]. - Michael Somos, Nov 09 2007
a(n) = a(-8 - n) for all n in Z. - Michael Somos, Nov 09 2007
a(n) = n^2/24 + n/3 + 83/144 + (-1)^n/16 + A061347(n+1)/9 + A056594(n)/4. - R. J. Mathar, Mar 31 2011
a(n) = a(n-1)+a(n-3)-a(n-5)-a(n-7)+a(n-8). - David Neil McGrath, Aug 30 2014
a(n) = Sum_{k=1..floor((n+4)/4)} Sum_{j=k..floor((n+4-k)/3)} Sum_{i=j..floor((n+4-j-k)/2)} [j = i], where [ ] is the Iverson bracket. - Wesley Ivan Hurt, Jan 17 2021
a(n)-a(n-1) = A008679(n). - R. J. Mathar, Jun 23 2021
a(n)-a(n-4) = A008620(n). - R. J. Mathar, Jun 23 2021
Comments