A121822 Number of closed walks of length 2*n on the 5-cube.
1, 5, 65, 1205, 26465, 628805, 15424865, 382964405, 9550195265, 238539648005, 5961554097665, 149021418519605, 3725378557692065, 93133051794619205, 2328313585536338465, 58207725254446186805, 1455192101905494196865
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..700
- G. R. Franssens, On a number pyramid related to the binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
- L. Reyzin, Mathoverflow, Number of closed walks on an n-cube
- Index entries for linear recurrences with constant coefficients, signature (35,-259,225).
Programs
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GAP
List([0..20], n-> (25^n +5*9^n +10)/16); # G. C. Greubel, Jun 07 2019
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Magma
[(25^n +5*9^n +10)/16: n in [0..20]]; // G. C. Greubel, Jun 07 2019
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Mathematica
Table[(25^n +5*9^n +10)/16, {n,0,20}] (* G. C. Greubel, Jun 07 2019 *) -
PARI
a(n)=(25^n+5*9^n+10)>>4 \\ Charles R Greathouse IV, Jan 17 2012
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Sage
[(25^n +5*9^n +10)/16 for n in (0..20)] # G. C. Greubel, Jun 07 2019
Formula
a(n) = (25^n + 5*9^n + 10)/16.
G.f.: (1 - 30*x + 149*x^2)/(1 - 35*x + 259*x^2 - 225*x^3).
From Peter Bala, Nov 13 2006: (Start)
E.g.f.: cosh^5(x).
O.g.f.: 1/(1-5*1x/(1-4*2x/(1-3*3x/(1-2*4x/(1-1*5x))))) (continued fraction). (End)
(-1)^n*a(n) = Sum_{k=0..n} A086872(n,k)*(-6)^(n-k). - Philippe Deléham, Aug 17 2007
a(n) = (1/2^5)*Sum_{j = 0..5} binomial(5,j)*(5 - 2*j)^(2*n). See Reyzin link. - Peter Bala, Jun 03 2019
Extensions
Corrected by T. D. Noe, Nov 07 2006