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Second pentagonal numbers with odd index: a(n) = (2*n-1)*(3*n-1).
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%I #42 Feb 18 2022 05:16:27

%S 1,2,15,40,77,126,187,260,345,442,551,672,805,950,1107,1276,1457,1650,

%T 1855,2072,2301,2542,2795,3060,3337,3626,3927,4240,4565,4902,5251,

%U 5612,5985,6370,6767,7176,7597,8030,8475,8932,9401,9882,10375,10880,11397,11926

%N Second pentagonal numbers with odd index: a(n) = (2*n-1)*(3*n-1).

%C Sequence found by reading the segment (1, 2) together with the line (one of the diagonal axes) from 2, in the direction 2, 15, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - _Omar E. Pol_, Sep 08 2011

%H G. C. Greubel, <a href="/A033568/b033568.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: (1-x+12*x^2)/(1-x)^3.

%F a(n) = a(n-1) + 12*n - 11 (with a(0)=1). - _Vincenzo Librandi_, Nov 17 2010

%F a(n) = 6*n^2 - 5*n + 1 = A051866(n) + 1. - _Omar E. Pol_, Jul 18 2012

%F E.g.f.: (1 + x + 6*x^2)*exp(x). - _G. C. Greubel_, Oct 12 2019

%F From _Amiram Eldar_, Feb 18 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 1 + Pi/(2*sqrt(3)) + 2*log(2) - 3*log(3)/2.

%F Sum_{n>=0} (-1)^n/a(n) = 1 + (1/sqrt(3) - 1/2)*Pi - log(2). (End)

%p seq((2*n-1)*(3*n-1), n=0..50); # _G. C. Greubel_, Oct 12 2019

%t Table[(2*n-1)*(3*n-1),{n,0,50}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 28 2010 *)

%t LinearRecurrence[{3,-3,1},{1,2,15},50] (* _Ray Chandler_, Dec 08 2011 *)

%o (PARI) a(n)=(2*n-1)*(3*n-1) \\ _Charles R Greathouse IV_, Sep 24 2015

%o (Magma) [(2*n-1)*(3*n-1): n in [0..50]]; // _G. C. Greubel_, Oct 12 2019

%o (Sage) [(2*n-1)*(3*n-1) for n in range(50)] # _G. C. Greubel_, Oct 12 2019

%o (GAP) List([0..50], n-> (2*n-1)*(3*n-1)); # _G. C. Greubel_, Oct 12 2019

%Y Cf. A001318, A005449, A033570, A049452, A049453, A051866.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Ray Chandler_, Dec 08 2011