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a(n) = (4n+1)^11.
2

%I #19 Apr 21 2023 05:46:30

%S 1,48828125,31381059609,1792160394037,34271896307633,350277500542221,

%T 2384185791015625,12200509765705829,50542106513726817,

%U 177917621779460413,550329031716248441,1532278301220703125,3909821048582988049,9269035929372191597,20635899893042801193

%N a(n) = (4n+1)^11.

%H T. D. Noe, <a href="/A016823/b016823.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

%F From _Wesley Ivan Hurt_, Oct 10 2014 : (Start)

%F G.f.: (1 + 48828113*x + 30795122175*x^2 + 1418810334759*x^3 + 14826379326378*x^4 + 50417667664170*x^5 + 64020606756990*x^6 + 31088834650350*x^7 + 5356480404741*x^8 + 261595441397*x^9 + 1975200979*x^10 + 177147*x^11) / (x - 1)^12.

%F Recurrence: a(n) = 12*a(n-1)-66*a(n-2)+220*a(n-3)-495*a(n-4)+792*a(n-5)-924*a(n-6)+792*a(n-7)-495*a(n-8)+220*a(n-9)-66*a(n-10)+12*a(n-11)-a(n-12).

%F a(n) = A016813(n)^11 = A001020(A016813(n)). (End)

%F Sum_{n>=0} 1/a(n) = 50521*Pi^11/29727129600 + 2047*zeta(11)/4096. - _Amiram Eldar_, Apr 21 2023

%p A016823:=n->(4*n+1)^11: seq(A016823(n), n=0..20); # _Wesley Ivan Hurt_, Oct 10 2014

%t Table[(4 n + 1)^11, {n, 0, 20}] (* _Wesley Ivan Hurt_, Oct 10 2014 *)

%t CoefficientList[Series[(1 + 48828113 x + 30795122175 x^2 + 1418810334759 x^3 + 14826379326378 x^4 + 50417667664170 x^5 + 64020606756990 x^6 + 31088834650350 x^7 + 5356480404741 x^8 + 261595441397 x^9 + 1975200979 x^10 + 177147 x^11)/(x - 1)^12, {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Oct 10 2014 *)

%o (Magma) [(4*n+1)^11 : n in [0..20]]; // _Wesley Ivan Hurt_, Oct 10 2014

%Y Cf. A001020, A013669, A016813.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_