A279508 a(n) = smallest number k such that floor(phi(k)/tau(k)) = n.
2, 1, 5, 7, 27, 11, 13, 58, 17, 19, 55, 23, 65, 106, 29, 31, 85, 142, 37, 158, 41, 43, 115, 47, 119, 125, 53, 133, 145, 59, 61, 254, 262, 67, 274, 71, 73, 298, 1180, 79, 187, 83, 203, 346, 89, 209, 235, 382, 97, 394, 101, 103, 169, 107, 109, 253, 113, 458, 295
Offset: 0
Keywords
Examples
For n = 2; a(2) = 5 because 5 is the smallest number with floor(phi(5) / tau(5)) = floor(4/2) = 2.
Programs
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Magma
[Min([n: n in[1..100000] | Floor(EulerPhi(n)/NumberOfDivisors(n)) eq k]): k in [0..60]]
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Mathematica
Table[k = 1; While[Floor[EulerPhi[k]/DivisorSigma[0, k]] != n, k++]; k, {n, 0, 58}] (* Michael De Vlieger, Dec 14 2016 *)
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PARI
a(n) = my(k=1); while(floor((eulerphi(k)/numdiv(k)))!=n, k++); k \\ Felix Fröhlich, Dec 14 2016
Formula
a((p-1)/2) = p for p = prime > 3.
Comments