

A303935


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


1



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
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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, 88512 (2004), 258261.
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



