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A202279
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Numbers k such that the sum of digits^3 of k equals Sum_{d|k, 1<d<k} d.
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4
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142, 160, 1375, 6127, 12643, 51703, 86833, 103039, 104647, 112093, 137317, 218269, 261883, 266923, 449881, 505891, 617569, 907873
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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The sequence is finite because the restricted sum of divisors of n, for n composite, is at least sqrt(n), while the sum of the cubes of the digits of n is at most 9^3*log_10(n+1). - Giovanni Resta, Oct 05 2018
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LINKS
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FORMULA
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EXAMPLE
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160 is in the sequence because 1^3 + 6^3 + 0^3 = 217, and the sum of the divisors 1< d<160 is 2 + 4 + 5 + 8 + 10 + 16 + 20 + 32 + 40 + 80 = 217.
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MAPLE
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add(d^3, d=convert(n, base, 10)) ;
end proc:
if n > 1 then
numtheory[sigma](n)-1-n ;
else
0;
end if;
end proc:
isA202279 := proc(n)
end proc:
for n from 1 do
if isA202279(n) then
printf("%d, \n", n);
end if;
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MATHEMATICA
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Q[n_]:=Module[{a=Total[Rest[Most[Divisors[n]]]]}, a == Total[IntegerDigits[n]^3]]; Select[Range[2, 5*10^7], Q]
Select[Range[1000000], DivisorSigma[1, #]-#-1==Total[IntegerDigits[#]^3]&] (* Harvey P. Dale, Jul 19 2014 *)
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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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STATUS
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approved
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