A209874 Least m > 0 such that the prime p=A002313(n+1) divides m^2+1.

1, 2, 8, 4, 12, 6, 32, 30, 50, 46, 34, 22, 10, 76, 98, 100, 44, 28, 80, 162, 112, 14, 122, 144, 64, 16, 82, 60, 228, 138, 288, 114, 148, 136, 42, 104, 274, 334, 20, 266, 392, 254, 382, 348, 48, 208, 286, 52, 118, 86, 24, 516, 476, 578, 194, 154, 504, 106, 58, 26, 566, 96, 380, 670, 722, 62, 456, 582, 318, 526, 246, 520, 650, 726, 494, 324

Offset

0,2

Comments

This yields the prime factors of numbers of the form N^2+1, cf. formula in A089120: For n=0,1,2,... check whether N = +/- a(n) [mod 2*A002313(n+1)], if so, then A002313(n+1) is a prime factor of N^2+1.

Obviously, p then divides (2kp +/- a(n))^2+1 for all k >=0 ; in particular it will be the least prime factor of such numbers if there is no earlier match.

Alternatively one could deal separately with the case of odd N, for which p=2 divides N^2+1, and even N, for which only Pythagorean primes A002144(n)=A002313(n+1) can be prime factors of N^2+1.

Links

Table of n, a(n) for n=0..75.

Formula

For n>0, A209874(n) = 2*sqrt(-1/4 mod A002144(n)), where sqrt(a mod p) stands for the positive x < p/2 such that x^2=a in Z/pZ.

A209874(n) = A209877(n)*2 for n>0.

Prog

(PARI) A209874(n)=if( n, 2*lift(sqrt(Mod(-1, A002144[n])/4)), 1)
(PARI) /* for illustrative purpose: a(n) is the smaller of the 2 possible remainders mod 2*p of numbers N such that N^2+1 has p as smallest prime factor */ forprime( p=1, 199, p>2 & p%4 != 1 & next; my(c=[]); for(i=1, 9e9, factor(i^2+1)[1, 1]==p |next; c=vecsort(concat(c, i%(2*p)), , 8); #c==1 || print1(", "c[1]) || break))

Crossrefs

Cf. A002496, A014442, A085722, A144255, A089120, A193432.

Cf. A002314, A177979.

Sequence in context: A323986 A334447 A192034 * A175181 A110003 A035302

Adjacent sequences: A209871 A209872 A209873 * A209875 A209876 A209877

Keyword

nonn

Author

M. F. Hasler, Mar 11 2012

Status

approved