A011932 a(n) = floor( n*(n-1)*(n-2)*(n-3)/22 ).

0, 0, 0, 0, 1, 5, 16, 38, 76, 137, 229, 360, 540, 780, 1092, 1489, 1985, 2596, 3338, 4228, 5285, 6529, 7980, 9660, 11592, 13800, 16309, 19145, 22336, 25910, 29896, 34325, 39229, 44640, 50592, 57120, 64260, 72049, 80525, 89728, 99698, 110476, 122105, 134629, 148092, 162540, 178020, 194580, 212269, 231137

Offset

0,6

Links

G. C. Greubel, Table of n, a(n) for n = 0..2500

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1,0,0,0,0,0,0,1,-4,6,-4,1).

Formula

From Chai Wah Wu, Aug 02 2020: (Start)

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + a(n-11) - 4*a(n-12) + 6*a(n-13) - 4*a(n-14) + a(n-15) for n > 14.

G.f.: x^4*(1 + x + 2*x^2 + x^4 + 2*x^5 + x^6 + 2*x^8 + x^9 + x^10)/((1-x)^4*(1-x^11)). (End)

Mathematica

Table[Floor[(n(n-1)(n-2)(n-3))/22], {n, 0, 60}] (* Harvey P. Dale, Nov 25 2017 *)

Prog

(PARI) a(n)=n*(n-1)*(n-2)*(n-3)\22 \\ Charles R Greathouse IV, Oct 18 2022
(Magma) [Floor(12*Binomial(n, 4)/11): n in [0..80]]; // G. C. Greubel, Nov 03 2024
(SageMath) [12*binomial(n, 4)//11 for n in range(81)] # G. C. Greubel, Nov 03 2024

Crossrefs

Cf. A011915.

Sequence in context: A218259 A211807 A174723 * A357289 A131283 A082199

Adjacent sequences: A011929 A011930 A011931 * A011933 A011934 A011935

Keyword

nonn,easy

Author

N. J. A. Sloane

Extensions

More terms added by G. C. Greubel, Nov 03 2024

Status

approved