A179441 Number of solutions to a+b+c < d+e with each of a,b,c,d,e in {1..n+1}.

1, 21, 121, 432, 1182, 2723, 5558, 10368, 18039, 29689, 46695, 70720, 103740, 148071, 206396, 281792, 377757, 498237, 647653, 830928, 1053514, 1321419, 1641234, 2020160, 2466035, 2987361, 3593331, 4293856, 5099592, 6021967, 7073208, 8266368, 9615353, 11134949, 12840849

Offset

1,2

References

Mathematics and Computer Education 1988 - 89 #261 Unsolved.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Formula

a(n) = (1/120)*(27*n^5 + 80*n^4 + 65*n^3 - 20*n^2 - 32*n).

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n > 6.

G.f.: x*(1 + 15*x + 10*x^2 + x^3)/(1 - x)^6.

Example

a(1) = 1 since from {1,2} there is only one solution: {1,1,1} for a,b,c and {2,2} for d,e.
a(2) = 21 since from {1,2,3} there are 21 ways to select a,b,c,d,e such that a+b+c < d+e.

Mathematica

k=10;
Table[p=Expand[Sum[x^k, {k, 1, n}]^2 Sum[1/x^k, {k, 1, n}]^3];
Twowins=Drop[CoefficientList[p, x], 1]//Total, {n, 2, k}]

Prog

(PARI) a(n)=(27*n^5 + 80*n^4 + 65*n^3 - 20*n^2 - 32*n)/120 \\ Andrew Howroyd, Apr 15 2021

Crossrefs

Cf. A197083.

Sequence in context: A365205 A361699 A200888 * A164785 A179956 A117388

Adjacent sequences: A179438 A179439 A179440 * A179442 A179443 A179444

Keyword

nonn,easy

Author

Bobby Milazzo, Jul 14 2010

Extensions

Name edited and terms a(24) and beyond from Andrew Howroyd, Apr 15 2021

Status

approved