(42S) Subfactorial

[Albert Chan]

(09-06-2021 06:50 PM)John Keith Wrote:  The second formula can also be simplified to round(n!/e). More information and formulas at A000166.

Another perspective is with n! numerator, n!/!n is best approximation for e

With (n-1)! numerator, (n-1)!/!(n-1) is also best approximation for e

Difference of two fraction should be as tiny as possible, but not zero.

numer(n!/!n - (n-1)!/!(n-1)) = n! * !(n-1) - (n-1)! * !n = (n-1)! * (n * !(n-1) - !n)

Minimizing numerator (but not zero), last term should be ±1 (depends on parity of n)

!n = n * !(n-1) ± 1

With !1 = round(1!/e) = 0, !2 = round(2!/e) = 1, !2 = 2 * !1 + 1

!n = n * !(n-1) + (-1)^n