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Numbers n m such that the digits of n m are also digits of nm^3.
Presumably if a digit d appears k times in n, m, then it should appear at least k times in nm^3. - N. J. A. Sloane, Nov 24 2018
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Cf. A029776.
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Harvey P. Dale, <a href="/A064931/b064931.txt">Table of n, a(n) for n = 1..1000</a>
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1, 4, 5, 6, 9, 10, 11, 12, 21, 24, 25, 29, 32, 33, 34, 39, 40, 49, 50, 51, 54, 56, 59, 60, 61, 64, 65, 67, 71, 72, 73, 75, 76, 90, 97, 99, 100, 101, 102, 106, 109, 110, 114, 119, 120, 124, 125, 129, 137, 153, 176, 201, 202, 210, 212, 224, 228, 231, 233, 236, 238, 245, 249, 279, 345, 346, 347, 368, 389, 467, 689, 1235, 1236, 1238, 1249, 1256, 1258, 1345, 1368, 1459, 1467, 1567, 1679, 1689, 2349, 2359, 2457, 2458, 2589, 3459, 3467
Do[a = IntegerDigits[n^3]; b = IntegerDigits[n]; If[Intersection[a, b] == b, Print[n]], {n, 1, 10^4}]
Select[Range[400], Min[DigitCount[#^3]-DigitCount[#]]>-1&] (* Harvey P. Dale, Nov 24 2018 *)
Corrected and Mathematica program replaced by Harvey P. Dale, Nov 24 2018
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