A178213 Smith numbers of order 3.
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
Offset: 1
Examples
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.
Links
- Ely Golden and Donovan Johnson, Table of n, a(n) for n = 1..10000 (terms 1 to 1000 by Donovan Johnson)
- Ely Golden, Smith number sequence generator optimized for order 3
- Eric W. Weisstein, Smith Number
Programs
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Mathematica
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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