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A342945
Numbers m such that d(1)^1 + d(2)^2 + ... + d(p)^k = d(1)! + d(2)! + ... + d(k)!, where d(i), i=1..k, are the digits of m.
1
1, 2, 11, 21, 33, 111, 211, 331, 403, 1065, 1111, 1200, 2065, 2111, 2200, 3050, 3265, 3311, 4031, 4122, 4130, 4543, 5143, 10651, 11111, 11650, 12001, 12010, 12100, 13000, 15330, 20651, 21111, 21650, 22001, 22010, 22100, 23000, 25330, 30200, 30501, 30510, 31500
OFFSET
1,2
EXAMPLE
3265 is in the sequence because 3^1 + 2^2 + 6^3 + 5^4 = 3! + 2! + 6! + 5! = 848.
MATHEMATICA
Select[Range@40000, Total[(a=IntegerDigits@#)^Range@Length@a]==Total[a!]&] (* Giorgos Kalogeropoulos, Mar 30 2021 *)
PROG
(Python)
from math import factorial
def digfac(s): return sum(factorial(int(d)) for d in s)
def digpow(s): return sum(int(d)**i for i, d in enumerate(s, start=1))
def aupto(limit):
alst = []
for k in range(1, limit+1):
s = str(k)
if digpow(s) == digfac(s): alst.append(k)
return alst
print(aupto(32000)) # Michael S. Branicky, Mar 30 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Carole Dubois, Mar 30 2021
STATUS
approved