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A178213
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Smith numbers of order 3.
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7
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6606, 8540, 13086, 16866, 21080, 26637, 27468, 33387, 34790, 35364, 35377, 40908, 44652, 48154, 48860, 52798, 54814, 55055, 57726, 57894, 66438, 67297, 67356, 67594, 69549, 72465, 72598, 73026, 74371, 74785, 77485, 78745, 81546, 83175, 85927, 90174, 91208
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OFFSET
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1,1
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COMMENTS
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Composite numbers n not in A176670 such that the sum of the cubes of the digits of n equals the sum of the cubes of the digits of the prime factors of n (with multiplicity). A176670 lists composite numbers having the same digits as their prime factors (with multiplicity), excluding zero digits.
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LINKS
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EXAMPLE
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6606 = 2*3*3*367 is composite; sum of cubes of the digits is 6^3+6^3+0^3+6^3 = 648. Sum of cubes of the digits of the prime factors 2, 3, 3, 367 (with multiplicity) is 2^3+3^3+3^3+3^3+6^3+7^3 = 648. The sums are equal, so 6606 is in the sequence.
21080 = 2*2*2*5*17*31 is composite; sum of cubes of the digits is 2^3+1^3+0^3+8^3+0^3 = 521. Sum of cubes of the digits of the prime factors 2, 2, 2, 5, 17, 31 (with multiplicity) is 2^3+2^3+2^3+5^3+1^3+7^3+3^3+1^3 = 521. The sums are equal, so 21080 is in the sequence.
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MATHEMATICA
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fQ[n_] := Block[{id = Sort@ IntegerDigits@ n, fid = Sort@ Flatten[ IntegerDigits@ Table[ #[[1]], {#[[2]]}] & /@ FactorInteger@ n]}, While[ id[[1]] == 0, id = Drop[id, 1]]; While[ fid[[1]] == 0, fid = Drop[fid, 1]]; !PrimeQ@ n && id != fid && Plus @@ (id^3) == Plus @@ (fid^3)]; k = 2; lst = {}; While[k < 22002, If[fQ@ k, AppendTo[ lst, k]; Print@ k]; k++]; lst
With[{k = 3}, Select[Range[10^5], Function[n, And[Total@ Map[#^k &, IntegerDigits@ n] == Total@ Map[#^k &, Flatten@ IntegerDigits[#]], Not[Sort@ DeleteCases[#, 0] &@ IntegerDigits@ n == Sort@ DeleteCases[#, 0] &@#]] &@ Flatten@ Map[IntegerDigits@ ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]]]] (* Michael De Vlieger, Dec 10 2016 *)
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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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STATUS
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approved
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