

A242753


Number of ordered ways to write n = k + m with 0 < k <= m such that the inverse of k mod prime(k) among 1, ..., prime(k)  1 is prime and the inverse of m mod prime(m) among 1, ..., prime(m)  1 is also prime.


6



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

1,6


COMMENTS

Conjecture: a(n) > 0 for all n > 3.
This implies that there are infinitely many positive integers k such that k*q == 1 (mod prime(k)) for some prime q < prime(k).


LINKS

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


EXAMPLE

a(11) = 1 since 11 = 4 + 7, 4*2 == 1 (mod prime(4)=7) with 2 prime, and 7*5 == 1 (mod Prime(7)=17) with 5 prime.
a(36) = 1 since 36 = 18 + 18, and 18*17 == 1 (mod 61) with 17 prime.
a(46) = 1 since 46 = 6 + 40, 6*11 == 1 (mod prime(6)= 13) with 11 prime, and 40*13 == 1 (mod prime(40)=173) with 13 prime.


MATHEMATICA

p[n_]:=PrimeQ[PowerMod[n, 1, Prime[n]]]
Do[m=0; Do[If[p[k]&&p[nk], m=m+1], {k, 1, n/2}]; Print[n, " ", m]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000040, A000720, A242425, A242748, A242750, A242752, A242754, A242755.
Sequence in context: A099910 A213325 A043555 * A232443 A118821 A118824
Adjacent sequences: A242750 A242751 A242752 * A242754 A242755 A242756


KEYWORD

nonn


AUTHOR

ZhiWei Sun, May 22 2014


STATUS

approved



