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a(n) = (4n+1)^5.
1

%I #23 Apr 21 2023 05:46:21

%S 1,3125,59049,371293,1419857,4084101,9765625,20511149,39135393,

%T 69343957,115856201,184528125,282475249,418195493,601692057,844596301,

%U 1160290625,1564031349,2073071593,2706784157,3486784401,4437053125,5584059449,6956883693,8587340257,10510100501

%N a(n) = (4n+1)^5.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F 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). - _Wesley Ivan Hurt_, Apr 12 2023

%F From _Amiram Eldar_, Apr 21 2023: (Start)

%F a(n) = A016813(n)^5.

%F Sum_{n>=0} 1/a(n) = 5*Pi^5/3072 + 31*zeta(5)/64. (End)

%t Table[(4n+1)^5,{n,0,100}] (* _Mohammad K. Azarian_, Jun 18 2016 *)

%t LinearRecurrence[{6,-15,20,-15,6,-1},{1,3125,59049,371293,1419857,4084101},30] (* _Harvey P. Dale_, Oct 02 2022 *)

%Y Cf. A013663, A016813.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_