%I #61 Jan 23 2025 03:46:31

%S 0,4,144,4900,166464,5654884,192099600,6525731524,221682772224,

%T 7530688524100,255821727047184,8690408031080164,295218051329678400,

%U 10028723337177985444,340681375412721826704,11573138040695364122500,393146012008229658338304,13355391270239113019379844

%N Squares k such that 2*k+1 is also a square.

%C With the exception of 0, a subsequence of A075114. - _R. J. Mathar_, Dec 15 2008

%C Consequently, A014105(k) is a square if and only if k = a(n). - _Bruno Berselli_, Oct 14 2011

%C From _M. F. Hasler_, Jan 17 2012: (Start)

%C Bisection of A079291. The squares 2*k+1 are given in A055792.

%C A204576 is this sequence written in binary. (End)

%C a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - _George F. Johnson_, Nov 02 2012

%H D. W. Wilson, <a href="/A084703/a084703.txt">Table of n, a(n-1) for n = 1..100</a> (offset=1)

%H Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Omur/omur6.html">A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters</a>, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.

%H Thomas Koshy, <a href="https://www.fq.math.ca/Papers/60-1/koshy12122020.pdf">Products Involving Reciprocals of Gibonacci Polynomials</a>, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.

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

%F a(n) = 4*A001110(n) = A001542(n)^2.

%F a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - _Charlie Marion_, Jul 01 2003

%F a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - _Charlie Marion_, Jul 16 2003

%F G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - _R. J. Mathar_, Dec 15 2008

%F a(n) = A079291(2n). - _M. F. Hasler_, Jan 16 2012

%F From _George F. Johnson_, Nov 02 2012: (Start)

%F a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.

%F a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).

%F a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).

%F a(n-1)*a(n+1) = (a(n) - 4)^2.

%F 2*a(n) + 1 = (A001541(n))^2.

%F a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.

%F a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.

%F a(n)*a(n+1) = (4*A029549(n))^2.

%F a(n+1) - a(n) = 4*A046176(n).

%F a(n) + a(n+1) = 4*(6*A029549(n) + 1).

%F a(n) = (2*A001333(n)*A000129(n))^2.

%F Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)

%F Empirical: a(n) = A089928(4*n-2), for n > 0. - _Alex Ratushnyak_, Apr 12 2013

%F a(n) = 4*A001109(n)^2. - _G. C. Greubel_, Aug 18 2022

%F Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - _Amiram Eldar_, Jan 23 2025

%t b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)

%t a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]

%t 4*ChebyshevU[Range[-1,30], 3]^2 (* _G. C. Greubel_, Aug 18 2022 *)

%o (Magma) [4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // _G. C. Greubel_, Aug 18 2022

%o (SageMath) [4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # _G. C. Greubel_, Aug 18 2022

%Y Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.

%Y Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.

%Y Cf. A089928, A204576.

%Y Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

%K nonn,easy

%O 0,2

%A _Amarnath Murthy_, Jun 08 2003

%E Edited and extended by _Robert G. Wilson v_, Jun 15 2003