|
%I #47 Jul 08 2020 08:06:46
%S 2,5,10,17,26,37,65,82,101,122,145,170,197,226,257,290,362,401,442,
%T 485,530,577,626,677,730,785,842,901,962,1090,1157,1226,1297,1370,
%U 1522,1601,1765,1937,2026,2117,2210,2305,2402,2501,2602,2705,2810,2917,3026
%N Squarefree numbers of form k^2 + 1.
%C Heath-Brown (following Estermann) shows that, for any e > 0, there are k sqrt(x) + O(x^{7/24 + e}) members of this sequence up to x, for k = Product(1 - 2/p^2) = 0.8948412245... (A335963) where the product is over primes p = 1 mod 4. - _Charles R Greathouse IV_, Nov 19 2012, corrected by _Amiram Eldar_, Jul 08 2020
%C Integers k for which the period of the continued fraction of sqrt(k) is 1. - _Michel Marcus_, Apr 12 2019
%H Vincenzo Librandi, <a href="/A069987/b069987.txt">Table of n, a(n) for n = 1..1000</a>
%H T. Estermann, <a href="http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002275163">Einige Sätze über quadratfreie Zahlen</a>, Math. Ann. 105 (1931), pp. 653-662.
%H D. R. Heath-Brown, <a href="https://arxiv.org/abs/1010.6217">Square-free values of n^2 + 1</a>, arXiv:1010.6217 [math.NT], 2010-2012.
%H D. R. Heath-Brown, <a href="https://doi.org/10.4064/aa155-1-1">Square-free values of n^2 + 1</a>, Acta Arithmetica 155 (2012), pp. 1-13.
%F a(n) = A049533(n)^2 + 1.
%p select(numtheory:-issqrfree, [seq(n^2+1, n=1..100)]); # _Robert Israel_, Feb 09 2016
%t Select[ Range[10^4], IntegerQ[ Sqrt[ # - 1]] && Union[ Transpose[ FactorInteger[ # ]] [[2]]] [[ -1]] == 1 &]
%t Select[Range[60]^2+1,SquareFreeQ] (* _Harvey P. Dale_, Mar 21 2013 *)
%o (PARI) for(n=1,100,if(issquarefree(n^2+1),print1(n^2+1,",")))
%Y Cf. A059591, A002496, A124809, A005117, A002522, A335963.
%K nonn
%O 1,1
%A Sharon Sela (sharonsela(AT)hotmail.com), May 01 2002
%E Edited and extended by _Robert G. Wilson v_, _Benoit Cloitre_ and _Vladeta Jovovic_, May 04 2002
|
