A283034 - OEIS

%I #20 Jul 24 2021 01:11:35

%S 1,4913,19683,52521875,24794911296,68719476736,271818611107,

%T 1174711139837

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

%C 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).

%e 1 = 1^1,

%e 4913 = (4+9+1+3)^3,

%e 19683 = (1+9+6+8+3)^3,

%e 52521875 = (5+2+5+2+1+8+7+5)^5.

%t 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 *)

%o (VBA)

%o Sub calcul()

%o    Sheets("Result").Select

%o    Range("A1").Select

%o    For i = 1 To 10000000

%o       Sum = 0

%o       For k = 1 To Len(i)

%o          Sum = Sum + Mid(i, k, 1)

%o       Next

%o       If Sum ^Mid(i, len(i), 1)= i Then

%o          ActiveCell.Value = i

%o          ActiveCell.Offset(1, 0).Select

%o       End If

%o    Next

%o End Sub

%Y Cf. A007953, A010879, A023106.

%K nonn,base,fini,full

%O 1,2

%A _Shmelev Aleksei_, Feb 27 2017

%E a(5)-a(8) from _Giovanni Resta_, Feb 27 2017