%I #19 May 22 2021 04:34:59

%S 1,11,12,111,121,133,202,1020,1111,1211,1331,1403,2021,2030,2120,2220,

%T 2305,2413,3012,3102,3115,3202,3215,3322,3335,4033,4123,4223,4434,

%U 10165,10201,10210,10300,10533,11065,11111,11200,12065,12111,12200,13050,13265,13311

%N 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.

%e 2413 is in this sequence because 2^0 + 4^1 + 1^2 + 3^3 = 2! + 4! + 1! + 3! = 33.

%t Select[Range@20000,Total[(a=IntegerDigits@#)^Range[0,Length@a-1]]==Total[a!]&] (* _Giorgos Kalogeropoulos_, Mar 30 2021 *)

%o (Python)

%o from math import factorial

%o def digfac(s): return sum(factorial(int(d)) for d in s)

%o def digpow(s): return sum(int(d)**i for i, d in enumerate(s))

%o def aupto(limit):

%o alst = []

%o for k in range(1, limit+1):

%o s = str(k)

%o if digpow(s) == digfac(s): alst.append(k)

%o return alst

%o print(aupto(14000)) # _Michael S. Branicky_, Mar 30 2021

%o (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

%Y Cf. A009994, A342945, A342826, A178354.

%K nonn,base

%O 1,2

%A _Carole Dubois_, Mar 30 2021