A132109 a(n) = (2^(n+1) + n^2 + n)/2.
1, 3, 7, 14, 26, 47, 85, 156, 292, 557, 1079, 2114, 4174, 8283, 16489, 32888, 65672, 131225, 262315, 524478, 1048786, 2097383, 4194557, 8388884, 16777516, 33554757, 67109215, 134218106, 268435862, 536871347, 1073742289, 2147484144
Offset: 0
Examples
a(3) = 14 = sum of row 3 terms of triangle A132108: (4 + 5 + 4 + 1). a(3) = 14 = (1, 3, 3, 1) dot (1, 2, 2, 1) = (1 + 6 + 6 + 1).
Links
- Index entries for linear recurrences with constant coefficients, signature (5,-9,7,-2).
Programs
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Mathematica
Table[2^n + n^2 - n, {n, 20}]/2 (* Vladimir Joseph Stephan Orlovsky, May 07 2010 *) LinearRecurrence[{5, -9, 7, -2}, {1, 3, 7, 14}, 20] (* Eric W. Weisstein, Aug 09 2017 *) CoefficientList[Series[(1 - 2 x + x^2 - x^3)/((-1 + x)^3 (-1 + 2 x)), {x, 0, 20}], x] (* Eric W. Weisstein, Aug 09 2017 *) -
PARI
a(n) = n*(n+1)/2 + 2^n; \\ Altug Alkan, Aug 10 2017
Formula
a(n) = (2^(n+1) + n^2 + n)/2. - Eric W. Weisstein, Aug 09 2017
a(n) = A290699(n+1)/2. - Eric W. Weisstein, Aug 09 2017
G.f.: ( 1-2*x+x^2-x^3 ) / ( (2*x-1)*(x-1)^3 ). - R. J. Mathar, May 23 2016
a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4). - Eric W. Weisstein, Aug 09 2017
Extensions
More terms from Vladimir Joseph Stephan Orlovsky, May 07 2010
Name changed to the simple formula by Eric W. Weisstein, Aug 09 2017
Comments