A098931 a(0) = 1, a(n) = 1 + 2*3 + 4*5 + 6*7 + ... + (2n)*(2n+1) for n > 0.
1, 7, 27, 69, 141, 251, 407, 617, 889, 1231, 1651, 2157, 2757, 3459, 4271, 5201, 6257, 7447, 8779, 10261, 11901, 13707, 15687, 17849, 20201, 22751, 25507, 28477, 31669, 35091, 38751, 42657, 46817, 51239, 55931, 60901, 66157, 71707, 77559, 83721
Offset: 0
Examples
a(0) = 1; a(1) = 1 + 2*3 = 7; a(2) = 1 + 2*3 + 4*5 = 27, etc.
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Cf. A068377.
Programs
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Magma
[1+3*n^2+n*(5+4*n^2)/3: n in [0..40]]; // Vincenzo Librandi, Jul 28 2015
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Maple
seq((4/3)*n^3+3*n^2+(5/3)*n+1, n=0..100); # Robert Israel, Jul 28 2015
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Mathematica
Table[1 + 3 n^2 + n (5 + 4 n^2)/3, {n, 0, 40}] (* Robert G. Wilson v, Oct 23 2004 *) LinearRecurrence[{4, -6, 4, -1}, {1, 7, 27, 69}, 40] (* Vincenzo Librandi, Jul 28 2015 *) -
PARI
a(n)=n*(4*n^2+9*n+5)/3+1 \\ Charles R Greathouse IV, Jul 28 2015
Formula
a(n) = 1 + 3*n^2 + n*(5 + 4*n^2)/3.
G.f.: (1 + 3*x + 5*x^2 - x^3)/(1-x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Vincenzo Librandi, Jul 28 2015
From Robert Israel, Jul 28 2015: (Start)
E.g.f.: (1+6*x+7*x^2+(4/3)*x^3)*exp(x).
a(n) = 1 + Sum(A068377(i),i=1..n+1). (End)
Extensions
Edited and extended by Robert G. Wilson v, Oct 23 2004
Comments