

A065876


a(n) is the smallest m > n such that n^2 + 1 divides m^2 + 1.


7



1, 3, 3, 7, 13, 21, 31, 43, 18, 73, 91, 111, 17, 47, 183, 211, 241, 133, 57, 343, 381, 47, 172, 83, 553, 601, 651, 173, 342, 813, 242, 265, 132, 403, 411, 1191, 1261, 237, 327, 1483, 1561, 1641, 748, 857, 850, 1981, 684, 463, 413, 2353, 255, 2551, 593, 1177, 2863, 123, 3081, 307, 1288, 3423
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OFFSET

0,2


COMMENTS

a(n) exists because n^2 + 1 divides (n^2  n + 1)^2 + 1. The set of n such a(n) = n^2  n + 1 is S = (2, 3, 4, 5, 6, 7, 9, 11, 14, 15, ...).
a(n) = n^2  n + 1 whenever n^2 + 1 is prime or twice a prime. Up to n=1000, the only other n for which a(n) = n^2  n + 1 are 7, 41 and 239. Is it a coincidence that these are NSW primes (A088165)?  Franklin T. AdamsWatters, Oct 17 2006
It appears that the density of even numbers in this sequence approaches a limit near 1/4. It appears that the density of even values for indices where a(n) != n^2  n + 1 is approaching a number near 1/4 and based on the previous comment the density of n for which a(n) = n^2  n + 1 is almost certainly 0.  Franklin T. AdamsWatters, Oct 17 2006


LINKS



MATHEMATICA

Do[k = 1; While[m = (k^2 + 1)/(n^2 + 1); m < 2  !IntegerQ[m], k++ ]; Print[k], {n, 1, 40 } ]


PROG

(PARI) { for (n=0, 1000, a=n+1; while ((a^2 + 1)%(n^2 + 1) != 0, a++); write("b065876.txt", n, " ", a) ) } \\ Harry J. Smith, Nov 03 2009


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



