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A342944
Numbers m such that d(1)^0 + d(2)^1 + ... + d(k)^(k-1) = d(1)! + d(2)! + ... + d(k)!, where d(i), i=1..k, are the digits of m.
1
1, 11, 12, 111, 121, 133, 202, 1020, 1111, 1211, 1331, 1403, 2021, 2030, 2120, 2220, 2305, 2413, 3012, 3102, 3115, 3202, 3215, 3322, 3335, 4033, 4123, 4223, 4434, 10165, 10201, 10210, 10300, 10533, 11065, 11111, 11200, 12065, 12111, 12200, 13050, 13265, 13311
OFFSET
1,2
EXAMPLE
2413 is in this sequence because 2^0 + 4^1 + 1^2 + 3^3 = 2! + 4! + 1! + 3! = 33.
MATHEMATICA
Select[Range@20000, Total[(a=IntegerDigits@#)^Range[0, Length@a-1]]==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))
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(14000)) # Michael S. Branicky, Mar 30 2021
(PARI) is(n) = my(d = digits(n)); sum(i = 1, #d, d[i]!) == sum(i = 1, #d, d[i]^(i-1)) \\ David A. Corneth, Mar 30 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Carole Dubois, Mar 30 2021
STATUS
approved