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A276040
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Least k such that n^n divides phi(k^k) (k > 0).
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1
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1, 4, 6, 8, 10, 12, 14, 12, 18, 20, 22, 24, 26, 28, 30, 24, 34, 36, 38, 40, 42, 44, 46, 36, 50, 52, 45, 56, 58, 60, 62, 48, 66, 68, 70, 72, 74, 76, 78, 60, 82, 84, 86, 88, 90, 92, 94, 72, 98, 100, 102, 104, 106, 90, 110, 84, 114, 116, 118, 120
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OFFSET
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1,2
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COMMENTS
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Indices of odd terms in this sequence are 1, 27, 81, 135, 189, 297, 343, 351, 405, 459, 513, 621, 625, 675, 783, ...
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LINKS
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EXAMPLE
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a(3) = 6 because 3^3 divides phi(6^6) = 15552.
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MATHEMATICA
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Table[k = 1; While[! Divisible[EulerPhi[k^k], n^n], k++]; k, {n, 60}] (* Michael De Vlieger, Aug 21 2016 *)
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PROG
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(PARI) a(n) = {my(k = 1); while(eulerphi(k^k) % n^n, k++); k; }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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