

A237016


a(n) = {0 < k < n: phi(k)*sigma(nk) is a square}, where phi(.) is Euler's totient function and sigma(j) is the sum of all positive divisors of j.


4



0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 3, 1, 4, 2, 2, 1, 0, 1, 2, 2, 2, 2, 6, 4, 2, 2, 4, 2, 2, 4, 1, 6, 5, 6, 3, 3, 8, 3, 2, 4, 6, 1, 2, 4, 3, 3, 3, 5, 6, 5, 5, 3, 2, 5, 4, 4, 3, 6, 5, 7, 10, 7, 4, 2, 1, 4, 6, 7, 9, 6, 12, 3, 3, 4, 12, 6, 6, 5, 6, 4, 5, 8, 6, 5, 10, 7, 7, 2, 5, 8, 4, 2, 4, 3, 8, 4, 4, 11, 6, 6
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OFFSET

1,11


COMMENTS

Conjecture: (i) a(n) > 0 except for n = 1, 7, 17.
(ii) If n > 5, then phi(k)*sigma(nk) + 1 is a square for some 0 < k < n.
(iii) If n > 309, then there is a positive integer k < n/2 such that sigma(k)*sigma(nk) is a square.
See also A236998 for a similar conjecture.


LINKS



EXAMPLE

a(9) = 1 since phi(8)*sigma(1) = 4*1 = 2^2.
a(16) = 1 since phi(6)*sigma(10) = 2*18 = 6^2.
a(31) = 1 since phi(24)*sigma(7) = 8*8 = 8^2.
a(65) = 1 since phi(19)*sigma(46) = 18*72 = 36^2.


MATHEMATICA

sigma[n_]:=DivisorSigma[1, n]
SQ[n_]:=IntegerQ[Sqrt[n]]
p[n_, k_]:=SQ[EulerPhi[k]*sigma[nk]]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



