A204467 Number of 3-element subsets that can be chosen from {1,2,...,6*n+3} having element sum 9*n+6.
1, 8, 25, 50, 85, 128, 181, 242, 313, 392, 481, 578, 685, 800, 925, 1058, 1201, 1352, 1513, 1682, 1861, 2048, 2245, 2450, 2665, 2888, 3121, 3362, 3613, 3872, 4141, 4418, 4705, 5000, 5305, 5618, 5941, 6272, 6613, 6962, 7321, 7688, 8065, 8450, 8845, 9248, 9661
Offset: 0
Examples
a(1) = 8 because there are 8 3-element subsets that can be chosen from {1,2,...,9} having element sum 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (2, 0, -2, 1).
Programs
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Maple
a:= n-> 1 +floor((3+9/2*n)*n): seq(a(n), n=0..50);
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Mathematica
Table[(6n(3n+2)+(-1)^n+3)/4,{n,0,50}] (* or *) LinearRecurrence[{2,0,-2,1},{1,8,25,50},50] (* Harvey P. Dale, May 25 2015 *)
Formula
a(n) = 1+floor((3+9/2*n)*n).
G.f.: -(2*x+1)*(x^2+4*x+1)/((x+1)*(x-1)^3).
a(n) = (6*n*(3*n+2)+(-1)^n+3)/4. - Bruno Berselli, Jan 17 2012
a(0)=1, a(1)=8, a(2)=25, a(3)=50, a(n)=2*a(n-1)-2*a(n-3)+a(n-4). - Harvey P. Dale, May 25 2015
Comments