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 A289585 Quotients as they appear as k increases when tau(k) divides phi(k). 5
 1, 1, 2, 3, 1, 2, 1, 5, 6, 2, 8, 1, 9, 3, 11, 1, 3, 2, 14, 1, 15, 5, 4, 6, 18, 6, 2, 20, 21, 4, 23, 14, 8, 4, 26, 10, 3, 9, 7, 29, 30, 6, 12, 33, 11, 3, 35, 2, 36, 9, 6, 15, 3, 39, 10, 41, 2, 16, 14, 5, 44, 2, 18, 15, 18, 48, 7, 10, 50, 4, 51, 6, 6, 13, 53, 3, 54, 5, 18, 56, 22, 12, 24, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Numbers k such that tau(k) divides phi(k) are in A020491. Only for seven integers which are in A020488, we have a(n) = 1. The integers such that a(n) = 2, 3, 4 are respectively in A062516, A063469, A063470. When p is an odd prime then phi(p) = p-1, tau(p) = 2, so phi(p)/tau(p) = (p-1)/2 and A005097 is an infinite subsequence. For k = A058891(m+1), that is 2^A000225(m), with m>=2, the corresponding quotient phi(k)/tau(k) is the integer A076688(m). - Bernard Schott, Aug 15 2020 LINKS Robert G. Wilson v, Table of n, a(n) for n = 1..1000 FORMULA a(n) = A000010(A020491(n)) / A000005(A020491(n)). - David A. Corneth, Jul 09 2017 EXAMPLE a(10) = 2 because A020491(10) = 15 and phi(15)/tau(15) = 8/4 = 2. MAPLE for n from 1 to 50 do q:=phi(n)/tau(n); if q=floor(q) then print(n, q, phi(n), tau(n)) else fi; od: MATHEMATICA f[n_] := Block[{d = EulerPhi[n]/DivisorSigma[0, n]}, If[ IntegerQ@d, d, Nothing]]; Array[f, 120] (* Robert G. Wilson v, Jul 09 2017 ) PROG (PARI) lista(nn) = {for (n=1, nn, q = eulerphi(n)/numdiv(n); if (denominator(q)==1, print1(q, ", ")); ); } \\ Michel Marcus, Jul 10 2017 CROSSREFS Cf. A000010, A000005, A020491, A015733, A020488, A058891, A062516, A063469, A063470, A175667, A112955, A112954. Cf. A000225, A058891, A076688. Cf. A005097 (a subsequence). Cf. A290634 (complement). Sequence in context: A260451 A328749 A036848 * A128864 A106798 A214640 Adjacent sequences:  A289582 A289583 A289584 * A289586 A289587 A289588 KEYWORD nonn AUTHOR Bernard Schott, Jul 08 2017 STATUS approved

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Last modified June 24 05:58 EDT 2021. Contains 345416 sequences. (Running on oeis4.)