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a(n) = floor(Sum_{k=n^4..(n+1)^4} k^(1/4)).
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%I #28 Nov 07 2024 17:42:03

%S 1,26,171,628,1685,3726,7231,12776,21033,32770,48851,70236,97981,

%T 133238,177255,231376,297041,375786,469243,579140,707301,855646,

%U 1026191,1221048,1442425,1692626,1974051,2289196,2640653,3031110,3463351,3940256,4464801,5040058,5669195

%N a(n) = floor(Sum_{k=n^4..(n+1)^4} k^(1/4)).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = 4*n^4+8*n^3+8*n^2+5*n+1.

%F From _Stefano Spezia_, Jul 09 2024: (Start)

%F G.f.: (1 + 21*x + 51*x^2 + 23*x^3)/(1 - x)^5.

%F E.g.f.: exp(x)*(1 + 25*x + 60*x^2 + 32*x^3 + 4*x^4). (End)

%t LinearRecurrence[{5,-10,10,-5,1},{1,26,171,628,1685},40] (* _Harvey P. Dale_, Nov 07 2024 *)

%Y Cf. A248698, A248575, A374384.

%K nonn,easy

%O 0,2

%A _Amrit Awasthi_, Jul 09 2024