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Search: id:A065876
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| A065876 |
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a(n) is the smallest m > n such that n^2+1 divides m^2+1. |
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+0 5
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| 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
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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a(n) exists because ((n^2-n+1)^2+1)= 0 mod(n^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. Adams-Watters (FrankTAW(AT)Netscape.net), 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. Adams-Watters (FrankTAW(AT)Netscape.net), Oct 17 2006
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LINKS
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Franklin T. Adams-Watters, Table of n, a(n) for n = 0..1000
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MATHEMATICA
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Do[k = 1; While[m = (k^2 + 1)/(n^2 + 1); m < 2 || !IntegerQ[m], k++ ]; Print[k], {n, 1, 40 } ]
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PROGRAM
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(PARI) { for (n=0, 1000, a=n+1; while ((a^2 + 1)%(n^2 + 1) != 0, a++); write("b065876.txt", n, " ", a) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Nov 03 2009]
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CROSSREFS
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See also A088165, A005574, A002731.
Sequence in context: A116880 A051123 A096188 this_sequence A095008 A134346 A049772
Adjacent sequences: A065873 A065874 A065875 this_sequence A065877 A065878 A065879
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KEYWORD
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nonn,new
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Dec 07 2001
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EXTENSIONS
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More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Dec 11 2001
Further terms from Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Oct 17 2006
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