A283034 Numbers k such that k = (sum of digits of k)^(last digit of k).

1, 4913, 19683, 52521875, 24794911296, 68719476736, 271818611107, 1174711139837

Offset: 1

Author

Shmelev Aleksei, Feb 27 2017

Keywords

• nonn
• base
• fini
• full

Comments

The check must be done up to 10^22 (then for 23 digits in a number max result can be (23*10)^9 = 4, 6*10^20 < 10^22).

Examples

			1 = 1^1,
4913 = (4+9+1+3)^3,
19683 = (1+9+6+8+3)^3,
52521875 = (5+2+5+2+1+8+7+5)^5.
		

Crossrefs

Cf. A007953, A010879, A023106.

Programs

Mathematica

Union[Reap[nd=1; Sow[1]; While[Ceiling[(10^(nd-1))^(1/9)] <= 9 nd, Do[ Do[v = s^e; If[Mod[v, 10] == e && Plus @@ IntegerDigits@ v == s, Sow[v]], {s, Ceiling[ (10^(nd-1))^(1/e)], Min[ Floor[10^(nd/e)], 9 nd]}], {e, 2, 9}]; nd++]][[2, 1]]] (* all terms, Giovanni Resta, Feb 27 2017 *)

VBA

Sub calcul()
   Sheets("Result").Select
   Range("A1").Select
   For i = 1 To 10000000
      Sum = 0
      For k = 1 To Len(i)
         Sum = Sum + Mid(i, k, 1)
      Next
      If Sum ^Mid(i, len(i), 1)= i Then
         ActiveCell.Value = i
         ActiveCell.Offset(1, 0).Select
      End If
   Next
End Sub

Extensions

a(5)-a(8) from Giovanni Resta, Feb 27 2017