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a(n) = (n+1)*(2*n+1)*(4*n+1).
5

%I #33 Aug 01 2022 10:54:46

%S 1,30,135,364,765,1386,2275,3480,5049,7030,9471,12420,15925,20034,

%T 24795,30256,36465,43470,51319,60060,69741,80410,92115,104904,118825,

%U 133926,150255,167860,186789,207090,228811,252000,276705,302974,330855,360396,391645

%N a(n) = (n+1)*(2*n+1)*(4*n+1).

%C Apart from offset, same as A100147.

%D R. Tijdeman, Some applications of Diophantine approximation, pp. 261-284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.

%H Reinhard Zumkeller, <a href="/A079588/b079588.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F Sum_{n>=0} 1/a(n) = Pi/3 (cf. Tijdeman).

%F G.f.: (1+26*x+21*x^2)/(1-x)^4. - _L. Edson Jeffery_, Mar 25 2013

%F Sum_{n>=0} a(n)/2^n = 308; Sum_{n>=0} (-1)^n*a(n)/2^n = -4/3. - _L. Edson Jeffery_, Mar 25 2013

%F a(n) = 8*n^3 + 14*n^2 + 7*n + 1. - _Reinhard Zumkeller_, Jun 08 2015

%F Sum_{n>=0} (-1)^n/a(n) = log(2)/3 - Pi/2 + sqrt(2)*Pi/3 + 2*sqrt(2)*arcsin(1)/3. - _Amiram Eldar_, Jan 13 2021

%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Wesley Ivan Hurt_, Jun 23 2021

%t Table[(n + 1)*(2*n + 1)*(4*n + 1), {n, 0, 40}] (* _Amiram Eldar_, Jan 13 2021 *)

%t LinearRecurrence[{4,-6,4,-1},{1,30,135,364},40] (* _Harvey P. Dale_, Aug 01 2022 *)

%o (Haskell)

%o a079588 n = product $ map ((+ 1) . (* n)) [1, 2, 4]

%o -- _Reinhard Zumkeller_, Jun 08 2015

%Y Cf. A100147.

%Y Cf. A258721 (first differences), A011199.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Jan 26 2003