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 A248058 Least positive integer m such that m*n divides phi(m^2+n^2), where phi(.) is Euler's totient function. 4
 1, 1, 2, 1, 4, 1, 8, 1, 10, 1, 726, 2, 12, 1, 4, 1, 18, 3, 20, 1, 96, 23, 22, 1, 24, 1, 72, 2, 30, 8, 30, 1, 32, 35, 34, 1, 222, 40, 26, 1, 1312, 43, 42, 46, 360, 44, 48, 2, 588, 1, 50, 2, 5100, 1, 88, 1, 19152, 60, 8, 16 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Conjecture: (i) a(n) exists for any n > 0. (ii) For each n > 0, there is a positive integer m such that m*n divides sigma(m^2+n^2), where sigma(k) is the sum of all positive divisors of k. Note that a(n) = 1 if n^2 + 1 is prime. When n^2 + (n+1)^2 is prime, n*(n+1) divides phi(n^2 + (n+1)^2) = n^2 + (n+1)^2 - 1 and hence a(n) <= n + 1. If (n*q)^2 + 1 is prime for some q > 0, then for m = n^2*q the number phi(m^2+n^2) = phi(n^2)*phi((n*q)^2+1) = phi(n^2)*n^2 *q^2 is divisible by m*n = n^3*q. - Zhi-Wei Sun, Oct 03 2014 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..1242 EXAMPLE a(5) = 4 since 4*5 divides phi(4^2 + 5^2) = phi(41) = 40. a(919) = 37160684 since the product 919*37160684 = 34150668596 divides phi(919^2 + 37160684^2) = phi(1380916436192417) = 1379413805929632 = 40392*34150668596. MATHEMATICA Do[m=1; Label[aa]; If[Mod[EulerPhi[m^2+n^2], m*n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}] PROG (PARI) a(n)=m=1; while(eulerphi(m^2+n^2)%(m*n), m++); m vector(100, n, a(n)) \\ Derek Orr, Oct 01 2014 CROSSREFS Cf. A000010, A000203, A248035, A248036, A248054. Sequence in context: A097360 A325348 A307683 * A072345 A200583 A115120 Adjacent sequences:  A248055 A248056 A248057 * A248059 A248060 A248061 KEYWORD nonn AUTHOR Zhi-Wei Sun, Sep 30 2014 STATUS approved

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Last modified March 28 07:59 EDT 2020. Contains 333079 sequences. (Running on oeis4.)