

A236256


a(n) = {0 < k < n: m = phi(k) + phi(nk)/4 is an integer with C(2*m, m)  prime(m) prime}, where C(2*m, m) = (2*m)!/(m!)^2.


9



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

1,10


COMMENTS

Conjecture: a(n) > 0 for all n > 410.
This implies that there are infinitely many positive integers m with C(2*m, m)  prime(m) prime. We have verified the conjecture for n up to 51000.
See A236248 for a list of known numbers m with C(2*m, m)  prime(m) prime.
See also A236249 for those primes of the form C(2*m, m)  prime(m).


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000


EXAMPLE

a(12) = 1 since phi(2) + phi(10)/4 = 1 + 1 = 2 with C(2*2, 2)  prime(2) = 6  3 = 3 prime.
a(33) = 1 since phi(1) + phi(32)/4 = 1 + 4 = 5 with C(2*5, 5)  prime(5) = 252  11 = 241 prime.
a(697) = 1 since phi(452) + phi(697452)/4 = 224 + 42 = 266 with C(2*266, 266)  prime(266) = C(532, 266)  1699 prime.


MATHEMATICA

p[n_]:=IntegerQ[n]&&PrimeQ[Binomial[2n, n]Prime[n]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[nk]/4
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A000984, A236241, A236242, A236245, A236248, A236249.
Sequence in context: A037034 A229897 A139462 * A317086 A131376 A025840
Adjacent sequences: A236253 A236254 A236255 * A236257 A236258 A236259


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 21 2014


STATUS

approved



