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A274770
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Cube analog to Keith numbers.
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11
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1, 8, 17, 18, 26, 27, 44, 55, 63, 80, 105, 187, 326, 776, 1095, 2196, 6338, 13031, 13131, 25562, 27223, 70825, 140791, 553076, 632489, 1402680, 1404312, 3183253, 11311424, 50783292, 51231313, 182252596, 255246098, 522599548, 1180697763, 2025114819, 2137581414
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OFFSET
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1,2
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COMMENTS
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Like Keith numbers but starting from n^3 digits to reach n.
Consider the digits of the cube of a number n . Take their sum and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.
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LINKS
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EXAMPLE
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776^3 = 467288576 :
4 + 6 + 7 + 2 + 8 + 8 + 5 + 7 + 6 = 53;
6 + 7 + 2 + 8 + 8 + 5 + 7 + 6 + 53 = 102;
7 + 2 + 8 + 8 + 5 + 7 + 6 + 53 + 102 = 198;
2 + 8 + 8 + 5 + 7 + 6 + 53 + 102 + 198 = 389;
8 + 8 + 5 + 7 + 6 + 53 + 102 + 198 + 389 = 776.
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MAPLE
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with(numtheory): P:=proc(q, h) local a, b, k, n, t, v; v:=array(1..h);
for n from 1 to q do b:=n^3; a:=[];
for k from 1 to ilog10(b)+1 do a:=[(b mod 10), op(a)]; b:=trunc(b/10); od;
for k from 1 to nops(a) do v[k]:=a[k]; od; b:=ilog10(n^3)+1;
t:=nops(a)+1; v[t]:=add(v[k], k=1..b); while v[t]<n do t:=t+1; v[t]:=add(v[k], k=t-b..t-1);
od; if v[t]=n then print(n); fi; od; end: P(10^6, 10000);
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MATHEMATICA
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(* function keithQ[ ] is defined in A007629 *)
a274770[n_] := Join[{1, 8}, Select[Range[10, n], keithQ[#, 3]&]]
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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