A362113 Truncate Stirling's asymptotic series for 1! after n terms and round to the nearest integer.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 12, 13, -131, -144, 1878, 2047, -31243, -34023, 603493, 656720, -13392786, -14565501, 338472513, 367934625, -9665776360, -10502979551, 309738982467, 336455915833, -11068897604205, -12020303454921, 438669580592210

Offset: 0

Author

N. J. A. Sloane, Apr 15 2023

Keywords

• sign

Comments

Stirling's series for N! is an asymptotic expansion. It does not converge to N! as more terms are included in the sum.

Links

• G. Marsaglia and J. C. W. Marsaglia, A new derivation of Stirling's approximation to n!, Amer. Math. Monthly, 97 (1990), 827-829. MR1080390 (92b:41049)

Crossrefs

Cf. A001163/A001164, A362114-A362116.

Programs

Maple

h := proc(k) option remember; local j; `if`(k=0, 1, (h(k-1)/k-add((h(k-j)*h(j))/(j+1), j=1..k-1))/(1+1/(k+1))) end:
StirlingAsympt := proc(n) option remember; h(2*n)*2^n*pochhammer(1/2, n) end:
c := n -> StirlingAsympt(n); # # Peter Luschny, Feb 08 2011 (This is A001163(n)/A001164(n)).
S:=proc(k,N) local i; global c; sqrt(2*Pi)*N^(N+1/2)*exp(-N)*add(c(i)/N^i,i=0..k); end;
Digits:=200;
T:=proc(N,M) local k; [seq(round(evalf(S(k,N))),k=0..M)]; end;
T(1,40);

Formula

In general, we take Stirling's asymptotic series for N! (N >= 1, with N = 1 for the present sequence) and truncate it after n terms. This has the value

sqrt(2*Pi)*N^(N+1/2)*exp(-N)*(Sum_{j = 0..n} c(j)/N^j),

where c(j) = A001163(j)/A001164(j).

We then round this to the nearest integer to get a(n).