A335961 Alternating factorions: Numbers m such that m = S_af(m) = af(d_1)+af(d_2)+...+af(d_k) where d_1 d_2 ... d_n is the decimal expansion of m and af(m) = m!-(m-1)!+(m-2)!+...1! (alternating factorial) with af(0) = 0 (base 10).

0, 1, 620, 621, 643

Offset

1,3

Comments

Largest k such that S_af(k) > k is 1599999. That's why there are only five numbers such that S_af(m) = m. Proved by computer calculations.

If m has eight or more digits then S_af(m) < m. Proved directly.

Links

Table of n, a(n) for n=1..5.

Example

For m = 620, S_af(620) = af(6)+af(2)+af(0) = 619+1+0 = 620.

Mathematica

af[0] = 0; af[n_] := af[n] = n! - af[n - 1]; Select[Range[1000], Total[af /@ IntegerDigits[#]] == # &] (* Amiram Eldar, Jul 02 2020 *)

Prog

(C++)
#include <iostream>
using namespace std;
int S_af(int n) { const int af[]={0, 1, 1, 5, 19, 101, 619, 4421, 35899, 326981}; int s=0; while (n) {s+=af[n%10]; n/=10; } return s; }
int main() {int n=0; while(n<=1599999){if(n==S_af(n)){cout<<n<<endl; }n++; }}

Crossrefs

Cf. A005165 (alternating factorial), A014080 (factorions).

Sequence in context: A200812 A133207 A182351 * A202239 A010031 A252527

Adjacent sequences: A335958 A335959 A335960 * A335962 A335963 A335964

Keyword

nonn,base,fini,full

Author

Andrzej Kukla, Jul 01 2020

Status

approved