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Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.
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%I #51 Aug 17 2024 18:20:29

%S 1,49,1681,57121,1940449,65918161,2239277041,76069501249,

%T 2584123765441,87784138523761,2982076586042449,101302819786919521,

%U 3441313796169221281,116903366249966604049,3971273138702695316401,134906383349641674153601,4582845760749114225906049,155681849482120242006652081

%N Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.

%C Also indices of triangular numbers (A000217) which are also centered octagonal numbers (A016754). - _Colin Barker_, Jan 16 2015

%C a(n)-th triangular number is a square; subsequence of A001108. - _Jaroslav Krizek_, Aug 05 2016

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

%D P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 288.

%D P. F. Teilhet, Note #2094, L'Intermédiaire des Mathématiciens, 10 (1903), pp. 235-238.

%H M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, <a href="http://www.jstor.org/stable/2968551">Problem 47</a>, Amer. Math. Monthly, 4 (1897), 25-28.

%H D. H. Lehmer, <a href="http://www.jstor.org/stable/1968647">Lacunary recurrence formulas for the numbers of Bernoulli and Euler</a>, Annals Math., 36 (1935), 637-649.

%H <a href="/index/Be#Bernoulli">Index entries for sequences related to Bernoulli numbers.</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).

%F a(n) = 34*a(n-1) - a(n-2) + 16 = A002315(n)^2 = 2*A001653(n)^2 - 1 = 2*A008844(n) - 1 = floor(A046176(n)*sqrt(2)) = 6*A055792(n+1) - a(n-1) + 4 = (6*A055792(n+2) + a(n-1) - 20)/35. - _Henry Bottomley_, Nov 13 2001

%F a(n) = A001108(2n+1). - _Ira M. Gessel_, Nov 05 2014

%F a(n) = Sum_{k=1..2*n+1} 2^(k-1)*binomial(4*n+2, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 03 2003

%F O.g.f.: -(1+14*x+x^2)/((-1+x)*(1-34*x+x^2)). - _R. J. Mathar_, Nov 23 2007

%F a(n) = -(cosh((2*n - 1)*arctanh(sqrt(2))))^2 = -1 - (sinh((2*n - 1)*arctanh(sqrt(2))))^2. - _Artur Jasinski_, Oct 30 2008

%F a(n) = Sum_{k=0..4n+1} A000129(k), see Santana and Diaz-Barrero link at A002315. - _Ivan N. Ianakiev_, Jul 15 2022

%t LinearRecurrence[{35,-35,1},{1,49,1681},17] (* _Stefano Spezia_, Aug 17 2024 *)

%Y Cf. A000217, A002315, A016754, A146313, A253826.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E a(14)-a(17) from _Stefano Spezia_, Aug 17 2024