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A283034
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Numbers k such that k = (sum of digits of k)^(last digit of k).
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0
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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