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A357502 a(n) = ((1 + sqrt(n))^n - (1 - sqrt(n))^n)/(2*sqrt(n)).

%I A357502 #26 Oct 14 2022 16:29:47
%S A357502 1,2,6,20,80,342,1624,8136,43776,246410,1463264,9033180,58200064,
%T A357502 387905182,2679200640,19068105488,139929124864,1054773505170,
%U A357502 8167509816832,64795371984420,526534098026496,4374163243287398,37135913476691968,321727849480560600
%N A357502 a(n) = ((1 + sqrt(n))^n - (1 - sqrt(n))^n)/(2*sqrt(n)).
%F A357502 From _Alois P. Heinz_, Oct 01 2022: (Start)
%F A357502 a(n) = [x^n] x/(1-2*x-(n-1)*x^2).
%F A357502 a(n) = Sum_{j=0..floor(n/2)} n^j * binomial(n,2*j+1).
%F A357502 a(n) = A099173(n,n). (End)
%p A357502 b:= proc(n, k) option remember;
%p A357502      `if`(n<2, n, 2*b(n-1, k)+(k-1)*b(n-2, k))
%p A357502     end:
%p A357502 a:= n-> b(n$2):
%p A357502 seq(a(n), n=1..26);  # _Alois P. Heinz_, Oct 01 2022
%t A357502 Simplify[Table[((1 + Sqrt[n])^n - (1 - Sqrt[n])^n)/(2*Sqrt[n]),{n,26}]] (* _Stefano Spezia_, Oct 03 2022 *)
%o A357502 (PARI) a(n) = polcoeff(lift( Mod('x,'x^2-2*'x+1-n)^n ),1); \\ _Kevin Ryde_, Oct 01 2022
%o A357502 (Python)
%o A357502 from sympy import simplify, sqrt
%o A357502 def A357502(n): return simplify(((1+sqrt(n))**n-(1-sqrt(n))**n)/sqrt(n))>>1 # _Chai Wah Wu_, Oct 14 2022
%Y A357502 Main diagonal of A099173.
%K A357502 nonn,easy
%O A357502 1,2
%A A357502 _Alexander R. Povolotsky_, Oct 01 2022