A016781 a(n) = (3*n+1)^5.
1, 1024, 16807, 100000, 371293, 1048576, 2476099, 5153632, 9765625, 17210368, 28629151, 45435424, 69343957, 102400000, 147008443, 205962976, 282475249, 380204032, 503284375, 656356768, 844596301, 1073741824, 1350125107, 1680700000, 2073071593, 2535525376
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
Programs
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Magma
[(3*n+1)^5: n in [0..30]]; // Vincenzo Librandi, Sep 21 2011
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Mathematica
(3Range[0,20]+1)^5 (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{1,1024,16807,100000,371293,1048576},30] (* Harvey P. Dale, May 13 2012 *)
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Maxima
A016781(n):=(3*n+1)^5$ makelist(A016781(n),n,0,20); /* Martin Ettl, Nov 12 2012 */
Formula
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, May 13 2012
From Wolfdieter Lang, Apr 02 2017: (Start)
O.g.f.: (1+1018*x+10678*x^2+14498*x^3+2933*x^4+32*x^5)/(1-x)^6.
E.g.f: exp(x)*(1+1023*x+7380*x^2+8775*x^3+2835*x^4+243*x^5). (End)
Sum_{n>=0} 1/a(n) = 2*Pi^5/(3^6*sqrt(3)) + 121*zeta(5)/3^5. - Amiram Eldar, Mar 29 2022
Comments