A303935 Size of orbit of n under repeated application of sum of factorial of digits of n.

2, 1, 1, 16, 8, 10, 15, 32, 36, 35, 2, 2, 17, 33, 13, 10, 15, 32, 36, 35, 17, 17, 9, 37, 7, 12, 6, 8, 33, 31, 33, 33, 37, 18, 34, 31, 48, 39, 24, 8, 13, 13, 7, 34, 30, 54, 42, 39, 29, 52, 10, 10, 12, 31, 54, 10, 24, 21, 41, 24, 15, 15, 6, 48, 42, 24, 12, 42

Offset

0,1

Comments

Numbers n for which a(n)=1 are called factorions (A014080).

Apart from factorions, only 3 cycles exist:

169 -> 363601 -> 1454 -> 169, so a(169) = a(363601) = a(1454) = 3.

871 -> 45361 -> 871, so a(871) = a(45361) = 2.

872 -> 45362 -> 872, so a(872) = a(45362) = 2.

All other n produce a chain reaching either a factorion or a cycle.

Links

Philippe Guglielmetti, Table of n, a(n) for n = 0..1000

S. S. Gupta, Sum of the factorials of the digits of integers, The Mathematical Gazette, 88-512 (2004), 258-261.

Project Euler, Problem 74: Digit factorial chains

Example

For n = 4, 4!=24, 2!+4!=26, 2!+6!=722, 7!+2!+2!=5044, 5!+0!+4!+4!=169, 1!+6!+9!=363601, 3!+6!+3!+6!+0!+1!=1454, then 1!+4!+5!+4!=169 which already belongs to the chain, so a(4) = length of [4, 24, 26, 722, 5044, 169, 363601, 1454] = 8.

Mathematica

Array[Length@ NestWhileList[Total@ Factorial@ IntegerDigits@ # &, #, UnsameQ, All, 100, -1] &, 68, 0] (* Michael De Vlieger, May 10 2018 *)

Prog

(Python)
for n in count(0):
    l=[]
    i=n
    while i not in l:
        l.append(i)
        i=sum(map(factorial, map(int, str(i))))
    print(n, len(l))
(PARI) f(n) = if (!n, n=1); my(d=digits(n)); sum(k=1, #d~, d[k]!);
a(n) = {my(v = [n], vs = Set(v)); for (k=1, oo, new = f(n); if (vecsearch(vs, new), return (#vs)); v = concat(v, new); vs = Set(v); n = new; ); } \\ Michel Marcus, May 18 2018

Crossrefs

Cf. A061602, A014080 (contains n for which a(n) = 1).

Sequence in context: A078089 A095836 A296524 * A156697 A173504 A322621

Adjacent sequences: A303932 A303933 A303934 * A303936 A303937 A303938

Keyword

nonn,base

Author

Philippe Guglielmetti, May 03 2018

Status

approved