

A233867


a(n) = {0 < m < 2*n: m is a square with 2*n  1  phi(m) prime}, where phi(.) is Euler's totient function (A000010).


8



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

1,5


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) For any odd number 2*n  1 > 4, there is a positive integer k < 2*n such that 2*n  1  phi(k) and 2*n  1 + phi(k) are both prime.
By Goldbach's conjecture, 2*n > 2 could be written as p + q with p and q both prime, and hence 2*n  1 = p + (q  1) = p + phi(q).
By induction, phi(k^2) (k = 1,2,3,...) are pairwise distinct.


LINKS



EXAMPLE

a(29) = 1 since 2*29  1 = 37 + phi(5^2) with 37 prime.
a(39) = 1 since 2*39  1 = 71 + phi(3^2) with 71 prime.
a(66) = 1 since 2*66  1 = 89 + phi(7^2) with 89 prime.
a(128) = 1 since 2*128  1 = 223 + phi(8^2) with 223 prime.
a(182) = 1 since 2*182  1 = 331 + phi(8^2) with 331 prime.
a(413) = 1 since 2*413  1 = 823 + phi(2^2) with 823 prime.
a(171) = 3 since 2*171  1 = 233 + phi(18^2) = 257 + phi(14^2) = 293 + phi(12^2) with 233, 257, 293 all prime.


MATHEMATICA

a[n_]:=Sum[If[PrimeQ[2n1EulerPhi[k^2]], 1, 0], {k, 1, Sqrt[2n1]}]
Table[a[n], {n, 1, 100}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



