A242425 Number of squares k^2 < prime(n) with k^2*p == 1 (mod prime(n)) for some prime p < prime(n).

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

Offset

1,5

Comments

Conjecture: a(n) > 0 for all n > 3. In other words, for any prime p > 5, there exists a positive square k^2 < p such that the inverse of k^2 mod p among 1, ..., p-1 is prime.

We have verified this for all n = 4, ..., 10^7. See also A242441 for an extension of the conjecture.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(4) = 1 since 2^2*2 == 1 (mod prime(4)=7).
a(6) = 1 since 3^2*3 == 1 (mod prime(6)=13).
a(9) = 1 since 4^2*13 == 1 (mod prime(9)=23).
a(18) = 1 since 7^2*5 == 1 (mod prime(18)=61).
a(30) = 1 since 8^2*83 == 1 (mod prime(30)=113).
a(46) = 1 since 10^2*2 == 1 (mod prime(46)=199).
a(52) = 1 since 15^2*17 == 1 (mod prime(52)=239).
a(97) = 1 since 18^2*11 == 1 (mod prime(97)=509).

Mathematica

r[k_, n_]:=PowerMod[k^2, -1, Prime[n]]
Do[m=0; Do[If[PrimeQ[r[k, n]], m=m+1], {k, 1, Sqrt[Prime[n]-1]}]; Print[n, " ", m]; Continue, {n, 1, 80}]

Crossrefs

Cf. A000040, A000290, A242441, A242444.

Sequence in context: A118382 A007723 A067437 * A263104 A282518 A230241

Adjacent sequences: A242422 A242423 A242424 * A242426 A242427 A242428

Keyword

nonn

Author

Zhi-Wei Sun, May 13 2014

Status

approved