A056118 a(n) = (11*n+5)*(n+4)*(n+3)*(n+2)*(n+1)/120.
1, 16, 81, 266, 686, 1512, 2982, 5412, 9207, 14872, 23023, 34398, 49868, 70448, 97308, 131784, 175389, 229824, 296989, 378994, 478170, 597080, 738530, 905580, 1101555, 1330056, 1594971, 1900486, 2251096, 2651616, 3107192, 3623312
Offset: 0
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
Crossrefs
Cf. A055268.
Programs
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GAP
List([0..40], n-> (11*n+5)*Binomial(n+4, 4)/5 ); # G. C. Greubel, Jan 17 2020
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Magma
[(11*n+5)*Binomial(n+4, 4)/5: n in [0..40]]; // G. C. Greubel, Jan 17 2020
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Maple
seq( (11*n+5)*binomial(n+4, 4)/5, n=0..40); # G. C. Greubel, Jan 17 2020
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Mathematica
Table[((11n+5)Times@@(n+Range[4]))/120,{n,0,40}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1}, {1,16,81,266,686,1512}, 40] (* Harvey P. Dale, Oct 18 2013 *) Table[11*Binomial[n+5,5] -8*Binomial[n+4,4], {n,0,40}] (* G. C. Greubel, Jan 17 2020 *)
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PARI
vector(41, n, (11*n-6)*binomial(n+3,4)/5 ) \\ G. C. Greubel, Jan 17 2020
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Sage
[(11*n+5)*binomial(n+4, 4)/5 for n in (0..40)] # G. C. Greubel, Jan 17 2020
Formula
a(n) = (11*n+5)*binomial(n+4,4)/5.
G.f.: (1+10*x)/(1-x)^6.
a(0)=1, a(1)=16, a(2)=81, a(3)=266, a(4)=686, a(5)=1512; for n>5, 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). - Harvey P. Dale, Oct 18 2013
From G. C. Greubel, Jan 17 2020: (Start)
a(n) = 11*binomial(n+5,5) - 8*binomial(n+4,4).
E.g.f.: (360 +2760*x +3720*x^2 +1560*x^3 +235*x^4 +11*x^5)*exp(x)/120. (End)