A342826 -id:A342826 - cp's OEIS frontend

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, 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

Carole Dubois, Mar 30 2021

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

Cf. A009994, A342945, A342826, A178354.

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

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

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

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.

Original entry on oeis.org

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

Carole Dubois, Mar 30 2021

			3265 is in the sequence because 3^1 + 2^2 + 6^3 + 5^4 = 3! + 2! + 6! + 5! = 848.

Cf. A002275, A342944, A342826, A178354.

Select[Range@40000,Total[(a=IntegerDigits@#)^Range@Length@a]==Total[a!]&] (* Giorgos Kalogeropoulos, Mar 30 2021 *)

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

cp's OEIS Frontend

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.

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