

A061602


Sum of factorials of the digits of n.


35



1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 2, 2, 3, 7, 25, 121, 721, 5041, 40321, 362881, 3, 3, 4, 8, 26, 122, 722, 5042, 40322, 362882, 7, 7, 8, 12, 30, 126, 726, 5046, 40326, 362886, 25, 25, 26, 30, 48, 144, 744, 5064, 40344, 362904, 121, 121, 122, 126
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OFFSET

0,3


COMMENTS

Numbers n such that a(n) = n are known as factorions. It is known that there are exactly four of these [in base 10]: 1, 2, 145, 40585.  Amarnath Murthy
The sum of factorials of the digits is the same for 0, 1, 2 in any base.  Alonso del Arte, Oct 21 2012


LINKS

Harry J. Smith and Indranil Ghosh, Table of n, a(n) for n = 0..10000 (first 1001 terms from Harry J. Smith)
Project Euler Problem 74: Determine the number of factorial chains that contain exactly sixty nonrepeating terms. [From Dremov Dmitry (dremovd(AT)gmail.com), May 21 2009]
Eric Weisstein's World of Mathematics, Factorion.


EXAMPLE

a(24) = (2!) + (4!) = 2 + 24 = 26.
a(153) = 127 because 1! + 5! + 3! = 1 + 120 + 6 = 127.


MAPLE

A061602 := proc(n)
add(factorial(d), d=convert(n, base, 10)) ;
end proc: # R. J. Mathar, Dec 18 2011


MATHEMATICA

a[n_] := Total[IntegerDigits[n]! ]; Table[a[n], {n, 1, 53}] (* Saif Hakim (saif7463(AT)gmail.com), Apr 23 2006 *)


PROG

(PARI) { for (n=0, 1000, a=0; x=n; until (x==0, a+=(x  10*(x\10))!; x=x\10); write("b061602.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 25 2009
(MAGMA) a061602:=func< n  n eq 0 select 1 else &+[ Factorial(d): d in Intseq(n) ] >; [ a061602(n): n in [0..60] ]; // Klaus Brockhaus, Nov 23 2010
(Python)
import math
def A061602(n):
s=0
for i in str(n):
s+=math.factorial(int(i))
return s # Indranil Ghosh, Jan 11 2017


CROSSREFS

Cf. A061603, A108911, A193163.
Sequence in context: A066459 A269221 A071937 * A182287 A248778 A033647
Adjacent sequences: A061599 A061600 A061601 * A061603 A061604 A061605


KEYWORD

nonn,base,easy


AUTHOR

Amarnath Murthy, May 19 2001


EXTENSIONS

Corrected and extended by Vladeta Jovovic, May 19 2001
Link and amended comment by Mark Hudson (mrmarkhudson(AT)hotmail.com), Nov 12 2004


STATUS

approved



