A052481 a(n) = 2^n*(binomial(n,2) + 1).
1, 2, 8, 32, 112, 352, 1024, 2816, 7424, 18944, 47104, 114688, 274432, 647168, 1507328, 3473408, 7929856, 17956864, 40370176, 90177536, 200278016, 442499072, 973078528, 2130706432, 4647288832, 10099884032, 21877489664, 47244640256, 101737037824, 218506461184
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Jonathan F. Mason and Richard H. Hudson, A Generalization of Euler's Formula and its Connection to Fibonacci Numbers, in: Frederic T. Howard (ed.), Applications of Fibonacci Numbers, Volume 9: Proceedings of The Tenth International Research Conference on Fibonacci Numbers and Their Applications, Springer, Dordrecht, 2004, pp. 177-185, alternative link.
- Index entries for linear recurrences with constant coefficients, signature (6,-12,8).
Programs
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GAP
List([0..30], n-> 2^(n-1)*(n^2-n+2)); # G. C. Greubel, May 16 2019
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Magma
[2^n*(Binomial(n,2)+1): n in [0..30]]; // Vincenzo Librandi, Dec 22 2016
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Mathematica
Table[2^n (Binomial[n, 2]+1), {n,0,30}] (* Vincenzo Librandi, Dec 22 2016 *) LinearRecurrence[{6,-12,8},{1,2,8},30] (* Harvey P. Dale, May 16 2019 *)
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PARI
{a(n) = 2^(n-1)*(n^2-n+2)}; \\ G. C. Greubel, May 16 2019
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Sage
[2^(n-1)*(n^2-n+2) for n in (0..30)] # G. C. Greubel, May 16 2019
Formula
For the sequence 1, 1, 1, 2, 8, 32, ... we have a(n) = 2^n*(n^2-5n+8)/8. - Paul Barry, Jun 26 2003
From R. J. Mathar, Jan 04 2011: (Start)
a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3).
G.f.: (1-4*x+8*x^2)/(1-2*x)^3. (End)
E.g.f.: (1 + 2*x^2)*exp(2*x). - G. C. Greubel, May 16 2019
Comments