A095311 47-gonal numbers.
1, 47, 138, 274, 455, 681, 952, 1268, 1629, 2035, 2486, 2982, 3523, 4109, 4740, 5416, 6137, 6903, 7714, 8570, 9471, 10417, 11408, 12444, 13525, 14651, 15822, 17038, 18299, 19605, 20956, 22352, 23793, 25279, 26810, 28386, 30007, 31673, 33384
Offset: 1
Examples
a(6) = 681 = 3*a(5) - 3*a(4) + a(3) = 3*455 - 3*274 + 138. a(37) = 30007 since M^37 * [1 0 0] = [1 37 30007].
References
- Albert H. Beiler, "Recreations in the Theory of Numbers", Dover, 1966, pp. 185-194.
Links
Programs
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Magma
I:=[1,47,138]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jul 25 2015
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Mathematica
a[n_] := (MatrixPower[{{1, 0, 0}, {1, 1, 0}, {1, 45, 1}}, n].{{1}, {0}, {0}})[[3, 1]]; Table[ a[n], {n, 40}] (* Robert G. Wilson v, Jun 05 2004 *) LinearRecurrence[{3, -3, 1}, {1, 47, 138}, 40] (* Vincenzo Librandi, Jul 25 2015 *) PolygonalNumber[47,Range[40]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 01 2016 *)
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PARI
a(n)=([0,1,0; 0,0,1; 1,-3,3]^(n-1)*[1;47;138])[1,1] \\ Charles R Greathouse IV, Jun 17 2017
Formula
a(n+3) = 3*a(n+2) - 3*a(n+1) - a(n); a(1) = 1, a(2) = 47, a(3) = 138.
Let M = the 3 X 3 matrix [1 0 0 / 1 1 0 / 1 45 1]. Then M^n * [1 0 0] = [1 n a(n)].
From Colin Barker, Jul 27 2013: (Start)
a(n) = (n*(45*n-43))/2.
G.f.: -x*(44*x+1) / (x-1)^3. (End)
E.g.f.: exp(x)*(x + 45*x^2/2). - Nikolaos Pantelidis, Feb 10 2023
Extensions
Edited by N. J. A. Sloane and Robert G. Wilson v, Jun 05 2004