A056771 -id:A056771 - cp's OEIS frontend

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720

Offset: 0

N. J. A. Sloane

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002
For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003
This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]
n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003
For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003
a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004
Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005
Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005
Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005
One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006
Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006
Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006
If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007
If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006
If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007
a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007
For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011
Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009
a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009
Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010
If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
  for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
  for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1,  64 = 35*2 - 6, 373 = 204*2 - 35;
  for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1;  99 = 35*3 - 6; 577 = 204*3 - 35;
  for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;
  for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.
  See also A002315, A054488, A038761, A054489, A054490.
  - Charlie Marion, Dec 08 2010
See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012
a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012
a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012
From Richard R. Forberg, Aug 30 2013: (Start)
The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:
a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;
a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015
Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015
The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and  2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016
a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016
Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017
In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021
Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

			G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - _Michael B. Porter_, Jul 02 2016

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Index entries for sequences related to Chebyshev polynomials.
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Cf. A000217, A000290, A001108, A001542, A001653, A001850, A002315, A002965, A278310.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

a:=[0,1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018

a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
   (map (* 6) $ tail a001109_list) a001109_list
-- Reinhard Zumkeller, Dec 17 2011

[n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015

a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n],n=0..26); # Emeric Deutsch
with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008

Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {0,1}, 30]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
CoefficientList[Series[x/(1-6x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Mar 23 2011 *)
LinearRecurrence[{6, -1}, {0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
a[ n_]:= ChebyshevU[n-1, 3]; (* Michael Somos, Sep 02 2012 *)
Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)

{a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

{a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */

{a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */

is(n)=ispolygonal(n^2,3) \\ Charles R Greathouse IV, Nov 03 2016

[lucas_number1(n,6,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).
a(n) = sqrt(A001110(n)).
a(n) = A001542(n)/2.
a(n) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000
a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - Henry Bottomley, Apr 19 2000
a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002
a(n) = ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) / (4*sqrt(2)). - Gregory V. Richardson, Oct 13 2002. Corrected for offset 0, and rewritten. - Wolfdieter Lang, Feb 10 2015
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - Lekraj Beedassy, Apr 23 2003
a(-n) = -a(n). - Michael Somos, Apr 07 2003
For n >= 1, a(n) = Sum_{k=0..n-1} A001653(k). - Charlie Marion, Jul 01 2003
For n > 0, 4*a(2*n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003
For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - Charlie Marion, Jul 18 2003
a(2*n+1) = a(n+1)^2 - a(n)^2. - Charlie Marion, Jan 12 2004
a(k)*a(2*n+k) = a(n+k)^2 - a(n)^2; e.g., 204*7997214 = 40391^2 - 35^2. - Charlie Marion, Jan 15 2004
For j < n+1, a(k+j)*a(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(2*n)/2;
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004
A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
a(n) = A001653(n+1) - A038723(n); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004, modified by Davide Colazingari, Jun 24 2016
a(n+1) = Sum_{k=0..n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - Benoit Cloitre, Sep 28 2005
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = ( (1 + sqrt(2) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - Mohamed Bouhamida, Sep 20 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),3), see second formula. - Marcos Carreira, Dec 27 2006
The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P(n + 2) * (P(n + 2) + P(n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007
For k = 0..n, a(2*n-k) - a(k) = 2*a(n-k)*A001541(n). Also, a(2*n+1-k) - a(k) = A002315(n-k)*A001653(n). - Charlie Marion, Jul 18 2007
[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009
a(n+1)^2 - 6*a(n+1)*a(n) + a(n)^2 = 1. - Charlie Marion, Dec 14 2010
a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1) - a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m < n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) = (A002315(n) - A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011
16*a(n)^2 + 1 = A056771(n). - James R. Buddenhagen, Dec 09 2011
A010054(A000290(a(n))) = 1. - Reinhard Zumkeller, Dec 17 2011
In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - Charlie Marion, Jan 11 2012
a(n+1) = Sum_{k=0..n} A101950(n,k)*5^k. - Philippe Deléham, Feb 10 2012
PSUM transform of a(n+1) is A053142. PSUMSIGN transform of a(n+1) is A084158. BINOMIAL transform of a(n+1) is A164591. BINOMIAL transform of A086347 is a(n+1). BINOMIAL transform of A057087(n-1). - Michael Somos, May 11 2012
a(n+k) = A001541(k)*a(n) + sqrt(A132592(k)*a(n)^2 + a(k)^2). Generalizes formula dated Oct 09 2000. - Charlie Marion, Nov 27 2012
a(n) + a(n+2*k) = A003499(k)*a(n+k); a(n) + a(n+2*k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
G.f.: G(0)*x/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014
a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015
a(n-k)*a(n+k) + a(k)^2 = a(n)^2, a(n+k) + a(n-k) = A003499(k)*a(n), for n >= k >= 0. - Alexander Samokrutov, Sep 30 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
4*a(n)^2 - 1 = A278310(n) for n > 0. - Bruno Berselli, Nov 24 2016
From Klaus Purath, Jan 18 2020: (Start)
a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1.
a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021
(End)
a(n) = A002965(2*n)*A002965(2*n+1). - Jon E. Schoenfield, Jan 08 2022
a(n) = A002965(4*n)/2. - Gerry Martens, Jul 14 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023

Additional comments from Wolfdieter Lang, Feb 10 2000

Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015

A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657

Offset: 0

N. J. A. Sloane

Chebyshev polynomials of the first kind evaluated at 3.
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 8*y^2 = 1, the corresponding values of y are in A001109. For n > 0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003 [edited by Jon E. Schoenfield, May 04 2014]
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r = sqrt(8). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions greater than 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r = sqrt(2). - Benoit Cloitre, Feb 24 2004
This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark: (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i = sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005
a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 2006
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators=A001541 and denominators=A001542. - Clark Kimberling, Aug 26 2008
Also index of sequence A082532 for which A082532(n) = 1. - Carmine Suriano, Sep 07 2010
Numbers n such that sigma(n-1) and sigma(n+1) are both odd numbers. - Juri-Stepan Gerasimov, Mar 28 2011
Also, numbers such that floor(a(n)^2/2) is a square: base 2 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001075. - M. F. Hasler, Jan 15 2012
Numbers such that 2n^2 - 2 is a square. Also integer square roots of the expression 2*n^2 + 1, at values of n given by A001542. Also see A228405 regarding 2n^2 -+ 2^k generally for k >= 0. - Richard R. Forberg, Aug 20 2013
Values of x (or y) in the solutions to x^2 - 6xy + y^2 + 8 = 0. - Colin Barker, Feb 04 2014
Panda and Ray call the numbers in this sequence the Lucas-balancing numbers C_n (see references and links).
Partial sums of X or X+1 of Pythagorean triples (X,X+1,Z). - Peter M. Chema, Feb 03 2017
a(n)/A001542(n) is the closest rational approximation to sqrt(2) with a numerator not larger than a(n), and 2*A001542(n)/a(n) is the closest rational approximation to sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations to sqrt(2) with restricted numerator or denominator. a(n)/A001542(n) > sqrt(2) > 2*A001542(n)/a(n). - A.H.M. Smeets, May 28 2017
x = a(n), y = A001542(n) are solutions of the Diophantine equation x^2 - 2y^2 = 1 (Pell equation). x = 2*A001542(n), y = a(n) are solutions of the Diophantine equation x^2 - 2y^2 = -2. Both together give the set of fractional approximations for sqrt(2) obtained from limited fractions obtained from continued fraction representation to sqrt(2). - A.H.M. Smeets, Jun 22 2017
a(n) is the radius of the n-th circle among the sequence of circles generated as follows: Starting with a unit circle centered at the origin, every subsequent circle touches the previous circle as well as the two limbs of hyperbola x^2 - y^2 = 1, and lies in the region y > 0. - Kaushal Agrawal, Nov 10 2018
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022

			99^2 + 99^2 = 140^2 + 2. - _Carmine Suriano_, Jan 05 2015
G.f. = 1 + 3*x + 17*x^2 + 99*x^3 + 577*x^4 + 3363*x^5 + 19601*x^6 + 114243*x^7 + ...

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
J. W. L. Glaisher, On Eulerian numbers (formulas, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.
G. K. Panda, Some fascinating properties of balancing numbers, In Proc. of Eleventh Internat. Conference on Fibonacci Numbers and Their Applications, Cong. Numerantium 194 (2009), 185-189.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

T. D. Noe, Table of n, a(n) for n = 0..200
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Jean-Paul Allouche, Zeta-regularization of arithmetic sequences, EPJ Web of Conferences (2020) Vol. 244, 01008.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
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John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Morgan Fiebig, aBa Mbirika, and Jürgen Spilker, Period patterns, entry points, and orders in the Lucas sequences: theory and applications, arXiv:2408.14632 [math.NT], 2024. See p. 5.
Robert Frontczak, A Note on Hybrid Convolutions Involving Balancing and Lucas-Balancing Numbers, Applied Mathematical Sciences, Vol. 12, 2018, No. 25, 1201-1208.
Robert Frontczak, Sums of Balancing and Lucas-Balancing Numbers with Binomial Coefficients, International Journal of Mathematical Analysis (2018) Vol. 12, No. 12, 585-594.
Robert Frontczak, Powers of Balancing Polynomials and Some Consequences for Fibonacci Sums, International Journal of Mathematical Analysis (2019) Vol. 13, No. 3, 109-115.
Robert Frontczak, Identities for generalized balancing numbers, Notes on Number Theory and Discrete Mathematics (2019) Vol. 25, No. 2, 169-180.
Robert Frontczak and Taras Goy, Additional close links between balancing and Lucas-balancing polynomials, arXiv:2007.14048 [math.NT], 2020.
Robert Frontczak and Taras Goy, More Fibonacci-Bernoulli relations with and without balancing polynomials, arXiv:2007.14618 [math.NT], 2020.
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Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Bisection of A001333. A003499(n) = 2a(n).
Cf. A055997 = numbers n such that n(n-1)/2 is a square.
Row 1 of array A188645.
Cf. A046090, A001109, A053142, A084130, A001601, A056771, A077420, A005319, A082532, A001542.
Cf. A055792 (terms squared), A132592.

a001541 n = a001541_list !! (n-1)
a001541_list =
1 : 3 : zipWith (-) (map (* 6) $ tail a001541_list) a001541_list
-- Reinhard Zumkeller, Oct 06 2011
(Scheme, with memoization-macro definec)
(definec (A001541 n) (cond ((zero? n) 1) ((= 1 n) 3) (else (- (* 6 (A001541 (- n 1))) (A001541 (- n 2))))))
;; Antti Karttunen, Oct 04 2016

[n: n in [1..10000000] |IsSquare(8*(n^2-1))]; // Vincenzo Librandi, Nov 18 2010

a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
A001541:=-(-1+3*z)/(1-6*z+z**2); # Simon Plouffe in his 1992 dissertation

Table[Simplify[(1/2) (3 + 2 Sqrt[2])^n + (1/2) (3 - 2 Sqrt[2])^n], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
a[ n_] := If[n == 0, 1, With[{m = Abs @ n}, m Sum[4^i Binomial[m + i, 2 i]/(m + i), {i, 0, m}]]]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := ChebyshevT[ n, 3]; (* Michael Somos, Jul 11 2011 *)
LinearRecurrence[{6, -1}, {1, 3}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)

{a(n) = real((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

{a(n) = subst( poltchebi( abs(n)), x, 3)}; /* Michael Somos, Apr 07 2003 */

{a(n) = if( n<0, a(-n), polsym(1 - 6*x + x^2, n) [n+1] / 2)}; /* Michael Somos, Apr 07 2003 */

{a(n) = polchebyshev( n, 1, 3)}; /* Michael Somos, Jul 11 2011 */

a(n)=([1,2,2;2,1,2;2,2,3]^n)[3,3] \\ Vim Wenders, Mar 28 2007

G.f.: (1-3*x)/(1-6*x+x^2). - Barry E. Williams and Wolfdieter Lang, May 05 2000
E.g.f.: exp(3*x)*cosh(2*sqrt(2)*x). Binomial transform of A084128. - Paul Barry, May 16 2003
From N. J. A. Sloane, May 16 2003: (Start)
a(n) = sqrt(8*((A001109(n))^2) + 1).
a(n) = T(n, 3), with Chebyshev's T-polynomials A053120. (End)
a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.
a(n) = cosh(2*n*arcsinh(1)). - Herbert Kociemba, Apr 24 2008
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all elements x of the sequence, 2*x^2 - 2 is a square. Limit_{n -> infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002 [corrected by Peter Pein, Mar 09 2009]
a(n) = 3*A001109(n) - A001109(n-1), n >= 1. - Barry E. Williams and Wolfdieter Lang, May 05 2000
For n >= 1, a(n) = A001652(n) - A001652(n-1). - Charlie Marion, Jul 01 2003
From Paul Barry, Sep 18 2003: (Start)
a(n) = ((-1+sqrt(2))^n + (1+sqrt(2))^n + (1-sqrt(2))^n + (-1-sqrt(2))^n)/4 (with interpolated zeros).
E.g.f.: cosh(x)*cosh(sqrt(2)x) (with interpolated zeros). (End)
For n > 0, a(n)^2 + 1 = 2*A001653(n-1)*A001653(n). - Charlie Marion, Dec 21 2003
a(n)^2 + a(n+1)^2 = 2*(A001653(2*n+1) - A001652(2*n)). - Charlie Marion, Mar 17 2003
a(n) = Sum_{k >= 0} binomial(2*n, 2*k)*2^k = Sum_{k >= 0} A086645(n, k)*2^k. - Philippe Deléham, Feb 29 2004
a(n)*A002315(n+k) = A001652(2*n+k) + A001652(k) + 1; for k > 0, a(n+k)*A002315(n) = A001652(2*n+k) - A001652(k-1). - Charlie Marion, Mar 17 2003
For n > k, a(n)*A001653(k) = A011900(n+k) + A053141(n-k-1). For n <= k, a(n)*A001653(k) = A011900(n+k) + A053141(k-n). - Charlie Marion, Oct 18 2004
A053141(n+1) + A055997(n+1) = a(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e. - Creighton Dement, Nov 16 2004
a(n) = sqrt( A055997(2*n) ). - Alexander Adamchuk, Nov 24 2006
a(2n) = A056771(n). a(2*n+1) = 3*A077420(n). - Alexander Adamchuk, Feb 01 2007
a(n) = (A000129(n)^2)*4 + (-1)^n. - Vim Wenders, Mar 28 2007
2*a(k)*A001653(n)*A001653(n+k) = A001653(n)^2 + A001653(n+k)^2 + A001542(k)^2. - Charlie Marion, Oct 12 2007
a(n) = A001333(2*n). - Ctibor O. Zizka, Aug 13 2008
A028982(a(n)-1) + 2 = A028982(a(n)+1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n) = 2*A001108(n) + 1. - Paul Weisenhorn, Dec 17 2011
a(n) = sqrt(2*x^2 + 1) with x being A001542(n). - Zak Seidov, Jan 30 2013
a(2n) = 2*a(n)^2 - 1 = a(n)^2 + 2*A001542(n)^2. a(2*n+1) = 1 + 2*A002315(n)^2. - Steven J. Haker, Dec 04 2013
a(n) = 3*a(n-1) + 4*A001542(n-1); e.g., a(4) = 99 = 3*17 + 4*12. - Zak Seidov, Dec 19 2013
a(n) = cos(n * arccos(3)) = cosh(n * log(3 + 2*sqrt(2))). - Daniel Suteu, Jul 28 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
Inverse binomial transform of A084130.
Exponential convolution of A000079 and A084058.
Sum_{n>=0} (-1)^n*a(n)/n! = cosh(2*sqrt(2))/exp(3) = 0.4226407909842764637... (End)
a(2*n+1) = 2*a(n)*a(n+1) - 3. - Timothy L. Tiffin, Oct 12 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Jan 20 2017
a(2^n) = A001601(n+1). - A.H.M. Smeets, May 28 2017
a(A298210(n)) = A002350(2*n^2). - A.H.M. Smeets, Jan 25 2018
a(n) = S(n, 6) - 3*S(n-1, 6), for n >= 0, with S(n, 6) = A001109(n+1), (Chebyshev S of A049310). See the first comment and the formula a(n) = T(n, 3). - Wolfdieter Lang, Nov 22 2020
From Peter Bala, Dec 31 2021: (Start)
a(n) = [x^n] (3*x + sqrt(1 + 8*x^2))^n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) hold for all prime p and positive integers n and k.
O.g.f. A(x) = 1 + x*d/dx(log(B(x))), where B(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. of A001850. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - 2/a(n)) = 1/2.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 1/a(n)) = 1/4.
Sum_{n >= 1} 1/(a(n)^2 - 2) = 1/2 - 1/sqrt(8). (End)
From  Peter Bala, Jun 23 2025: (Start)
Product_{n >= 0} (1 + 1/a(2^n)) = sqrt(2).
Product_{n >= 0} (1 - 1/(2*a(2^n))) = (4/7)*sqrt(2). See A002812. (End)

A029547 Expansion of g.f. 1/(1 - 34*x + x^2).

Original entry on oeis.org

1, 34, 1155, 39236, 1332869, 45278310, 1538129671, 52251130504, 1775000307465, 60297759323306, 2048348816684939, 69583562007964620, 2363792759454112141, 80299370259431848174, 2727814796061228725775, 92665403695822344828176, 3147895910861898495432209

Offset: 0

N. J. A. Sloane

Chebyshev sequence U(n,17)=S(n,34) with Diophantine property.
b(n)^2 - 2*(12*a(n))^2 = 1 with the companion sequence b(n)=A056771(n+1). - Wolfdieter Lang, Dec 11 2002
More generally, for t(m) = m + sqrt(m^2-1) and u(n) = (t(m)^(n+1) - 1/t(m)^(n+1))/(t(m) - 1/t(m)), we can verify that ((u(n+1) - u(n-1))/2)^2 - (m^2-1)*u(n)^2 = 1. - Bruno Berselli, Nov 21 2011
a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,33}. - Milan Janjic, Jan 26 2015

Indranil Ghosh, Table of n, a(n) for n = 0..651 (terms 0..200 from Vincenzo Librandi)
Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas numbers, INTEGERS 6 (2006) #A15.
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

A091761 is an essentially identical sequence.
Cf. A200441, A200724.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).

m:=17;; a:=[1,2*m];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 22 2019

I:=[1,34]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

with (combinat):seq(fibonacci(4*n+4,2)/12, n=0..15); # Zerinvary Lajos, Apr 21 2008

Table[GegenbauerC[n, 1, 17], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{34,-1},{1,34},20] (* Vincenzo Librandi, Nov 22 2011 *)
ChebyshevU[Range[21] -1, 17] (* G. C. Greubel, Dec 22 2019 *)

A029547(n, x=[0,1],A=[17,72*4;1,17]) = vector(n,i,(x*=A)[1]) \\ M. F. Hasler, May 26 2007

vector( 21, n, polchebyshev(n-1, 2, 17) ) \\ G. C. Greubel, Dec 22 2019

[lucas_number1(n,34,1) for n in range(1, 16)] # Zerinvary Lajos, Nov 07 2009

[chebyshev_U(n,17) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = 34*a(n-1) - a(n-2), a(-1)=0, a(0)=1.
a(n) = S(n, 34) with S(n, x):= U(n, x/2) Chebyshev's polynomials of the 2nd kind. See A049310. - Wolfdieter Lang, Dec 11 2002
a(n) = (ap^(n+1) - am^(n+1))/(ap - am) with ap = 17+12*sqrt(2) and am = 17-12*sqrt(2).
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*binomial(n-k, k)*34^(n-2*k).
a(n) = sqrt((A056771(n+1)^2 -1)/2)/12.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) with a(-1)=0, a(0)=1, a(1)=34. Also a(n) = (sqrt(2)/48)*((17+12*sqrt(2))^n-(17-12*sqrt(2))^n) = (sqrt(2)/48)*((3+2*sqrt(2))^(2n+2)-(3-2*sqrt(2))^(2n+2)) = (sqrt(2)/48)*((1+sqrt(2))^(4n+4)-(1-sqrt(2))^(4n+4)). - Antonio Alberto Olivares, Mar 19 2008
a(n) = Sum_{k=0..n} A101950(n,k)*33^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 0} (1 + 1/a(n)) = 1/4*(4 + 3*sqrt(2)).
Product {n >= 1} (1 - 1/a(n)) = 2/17*(4 + 3*sqrt(2)). (End)
E.g.f.: exp(17*x)*(24*cosh(12*sqrt(2)*x) + 17*sqrt(2)*sinh(12*sqrt(2)*x))/24. - Stefano Spezia, Apr 16 2023

A132592 X-values of solutions to the equation X(X + 1) - 8Y^2 = 0.

Original entry on oeis.org

0, 8, 288, 9800, 332928, 11309768, 384199200, 13051463048, 443365544448, 15061377048200, 511643454094368, 17380816062160328, 590436102659356800, 20057446674355970888, 681362750825443653408, 23146276081390728245000, 786292024016459316676608, 26710782540478226038759688

Offset: 0

Mohamed Bouhamida, Nov 14 2007

Equivalently, numbers k such that both k/2 and k+1 are squares. - Karl-Heinz Hofmann, Sep 20 2022

Equivalently, numbers k such that the k-dimensional volume and total (k-1)-dimensional volume are equal, with side length being a positive integer, for all regular polyhedra constructible in k dimensions. - Matt Moir, Jul 09 2024

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Intersection between A132411 and A001105.
Cf. A007654.
Cf. A001541, A058331, A001079, A037270, A055792, A071253, A108741, A132592, A146311, A146312, A146313, A173115, A173116, A173121.

I:=[0,8,288]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+ Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 24 2018

Table[Round[N[Sinh[2 n ArcCosh[Sqrt[2]]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 8, 288}, 30] (* Vincenzo Librandi, Dec 24 2018 *)

A132592 = [0, 8]
for n in range(2, 18): A132592.append(34 * A132592[-1] - A132592[-2] + 16)
print(A132592) # Karl-Heinz Hofmann, Sep 20 2022

a(0)=0, a(1)=8 and a(n) = 34*a(n-1) - a(n-2) + 16.
a(n) = (A056771(n) - 1)/2. - Max Alekseyev, Nov 13 2009
a(n) = sinh(2*n*arccosh(sqrt(2))^2) (n=0,1,2,3,...). - Artur Jasinski, Feb 10 2010
G.f.: -8*x*(x+1)/((x-1)*(x^2-34*x+1)). - Colin Barker, Oct 24 2012
a(n) = A055792(n+1)-1 = A001541(n)^2 - 1. - Antti Karttunen, Oct 03 2016

More terms from Max Alekseyev, Nov 13 2009

A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

Original entry on oeis.org

1, 33, 1121, 38081, 1293633, 43945441, 1492851361, 50713000833, 1722749176961, 58522759015841, 1988051057361633, 67535213191279681, 2294209197446147521, 77935577499977736033, 2647515425801796877601

Offset: 0

Wolfdieter Lang, Nov 29 2002

(3*a(n))^2 - 2*(2*b(n))^2 = 1 with companion sequence b(n)= A046176(n+1), n>=0 (special solutions of Pell equation).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas Numbers, INTEGERS 6 (2006) #A15
Tanya Khovanova, Recursive Sequences
S. Vidhyalakshmi, V. Krithika, and K. Agalya, On The Negative Pell Equation y^2 = 72x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER) Volume 4, Issue 2, February (2016).
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A056771 (even part).
Row 34 of array A094954.
Row 3 of array A188646.
Cf. similar sequences listed in A238379.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329. This is the case k=3.

I:=[1,33]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

LinearRecurrence[{34,-1},{1,33},20] (* Vincenzo Librandi, Nov 22 2011 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041027 *)
a[18, 20] (* Gerry Martens, Jun 07 2015 *)

makelist(expand(((1+sqrt(2))^(4*n+2)+(1-sqrt(2))^(4*n+2))/6),n,0,14);  /* _Bruno Berselli, Nov 22 2011 */

Vec((1-x)/(1-34*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Nov 22 2011

a(n) = 34*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = T(2*n+1, 3)/3 = S(n, 34) - S(n-1, 34), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 34)= A029547(n), T(n, 3)=A001541(n).
G.f.: (1-x)/(1-34*x+x^2).
a(n) = sqrt(8*A046176(n+1)^2 + 1)/3.
a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(2*n)-a(n-1), where k = (sqrt(2)+1)^4 = 17+12*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
a(n) = a(-n-1) = A029547(n)-A029547(n-1) = ((1+sqrt(2))^(4n+2)+(1-sqrt(2))^(4n+2))/6. - Bruno Berselli, Nov 22 2011

A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 97, 17, 1, 1, 1351, 577, 31, 1, 1, 18817, 19601, 1921, 49, 1, 1, 262087, 665857, 119071, 4801, 71, 1, 1, 3650401, 22619537, 7380481, 470449, 10081, 97, 1, 1, 50843527, 768398401, 457470751, 46099201, 1431431, 18817, 127, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given the function f(x,y) = (sqrt(x^2+y) + x)^2 and constant k=f(x,y), then for all integers x >= 1 and y=[+-]1, k may be irrational, but (k^n + k^(-n))/2 always produces integer sequences; y=-1 results shown here; y=1 results are A188645.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{2*k}(x), evaluated at x=n. - Seiichi Manyama, Dec 30 2018

			Row 2 gives {( (2+sqrt(3))^(2*n) + (2-sqrt(3))^(2*n) )/2}.
Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,   7,     97,      1351,        18817, ...
   3 | 1,  17,    577,     19601,       665857, ...
   4 | 1,  31,   1921,    119071,      7380481, ...
   5 | 1,  49,   4801,    470449,     46099201, ...
   6 | 1,  71,  10081,   1431431,    203253121, ...
   7 | 1,  97,  18817,   3650401,    708158977, ...
   8 | 1, 127,  32257,   8193151,   2081028097, ...
   9 | 1, 161,  51841,  16692641,   5374978561, ...
  10 | 1, 199,  79201,  31521799,  12545596801, ...
  11 | 1, 241, 116161,  55989361,  26986755841, ...
  12 | 1, 287, 164737,  94558751,  54276558337, ...
  13 | 1, 337, 227137, 153090001, 103182433537, ...
  14 | 1, 391, 305761, 239104711, 186979578241, ...
  15 | 1, 449, 403201, 362074049, 325142092801, ...
  ...

Row 2 is A011943, row 3 is A056771, row 8 is A175633, (row 2)*2 is A067902, (row 9)*2 is A089775.
Column 0-5 give A000012, A056220, A144130, A243132, A243134, A243136.
(column 1)*2 is A060626.
Cf. A188645 (f(x, y) as above with y=1).
Diagonals give A173129, A322899.
Cf. A188646, A322836.

max = 9; y = -1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188646(n,k-1) + A188646(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited by Seiichi Manyama, Dec 30 2018

More terms from Seiichi Manyama, Jan 01 2019

A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

Original entry on oeis.org

1, 1, 0, 1, 1, -1, 1, 2, 1, 0, 1, 3, 7, 1, 1, 1, 4, 17, 26, 1, 0, 1, 5, 31, 99, 97, 1, -1, 1, 6, 49, 244, 577, 362, 1, 0, 1, 7, 71, 485, 1921, 3363, 1351, 1, 1, 1, 8, 97, 846, 4801, 15124, 19601, 5042, 1, 0, 1, 9, 127, 1351, 10081, 47525, 119071, 114243, 18817, 1, -1

Offset: 0

Seiichi Manyama, Dec 28 2018

			Square array begins:
   1, 1,    1,     1,      1,      1,       1, ...
   0, 1,    2,     3,      4,      5,       6, ...
  -1, 1,    7,    17,     31,     49,      71, ...
   0, 1,   26,    99,    244,    485,     846, ...
   1, 1,   97,   577,   1921,   4801,   10081, ...
   0, 1,  362,  3363,  15124,  47525,  120126, ...
  -1, 1, 1351, 19601, 119071, 470449, 1431431, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Mirror of A101124.
Columns 0-20 give A056594, A000012, A001075, A001541, A001091, A001079, A023038, A011943(n+1), A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203, A322888, A056771, A322889, A078986, A322890.
Rows 0-10 give A000012, A001477, A056220, A144129, A144130, A243131, A243132, A243133, A243134, A243135, A243136.
Main diagonal gives A115066.
Cf. A323182 (Chebyshev polynomial of the second kind).

Table[ChebyshevT[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 28 2018 *)

T(n,k) = polchebyshev(n,1,k);
matrix(7, 7, n, k, T(n-1,k-1)) \\ Michel Marcus, Dec 28 2018

T(n, k) = round(cos(n*acos(k)));\\ Seiichi Manyama, Mar 05 2021

T(n, k) = if(n==0, 1, n*sum(j=0, n, (2*k-2)^j*binomial(n+j, 2*j)/(n+j))); \\ Seiichi Manyama, Mar 05 2021

A(0,k) = 1, A(1,k) = k and A(n,k) = 2 * k * A(n-1,k) - A(n-2,k) for n > 1.
A(n,k) = cos(n*arccos(k)). - Seiichi Manyama, Mar 05 2021
A(n,k) = n * Sum_{j=0..n} (2*k-2)^j * binomial(n+j,2*j)/(n+j) for n > 0. - Seiichi Manyama, Mar 05 2021

A091761 a(n) = Pell(4n) / Pell(4).

Original entry on oeis.org

0, 1, 34, 1155, 39236, 1332869, 45278310, 1538129671, 52251130504, 1775000307465, 60297759323306, 2048348816684939, 69583562007964620, 2363792759454112141, 80299370259431848174, 2727814796061228725775, 92665403695822344828176, 3147895910861898495432209

Offset: 0

Paul Barry, Feb 06 2004

A000129(k*n)/A000129(k) = ((sqrt(2)-1)^k(-1)^k-(sqrt(2)+1)^k)((sqrt(2)-1)^(k*n)(-1)^(k*n)-(sqrt(2)+1)^(k*n))/((sqrt(2)-1)^(2k)+(sqrt(2)+1)^(2k)-2(-1)^k).
All squares of the form (3m-1)^3 + (3m)^3 + (3m+1)^3 (cf. A116108) are given for m = 24 b, where b is a square of this sequence. From Ribenboim & McDaniel, it follows there are no squares > 1 in this sequence. - M. F. Hasler, Jun 05 2007
A divisibility sequence, cf. R. K. Guy's post to the SeqFan list. - M. F. Hasler, Feb 05 2013
a(n) gives all nonnegative solutions of the Pell equation b(n)^2 - 32*(3*a(n))^2 = +1, together with b(n) = A056771(n). - Wolfdieter Lang, Mar 09 2019

M. F. Hasler, Table of n, a(n) for n = 0..99
R. K. Guy, A new sequence, post to the SeqFan list, Feb 05 2013.
Tanya Khovanova, Recursive Sequences
Paulo Ribenboim and Wayne L. McDaniel, The Square Terms in Lucas Sequences, Journal of Number Theory 58, 104-123 (1996).
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (34,-1).

A029547 is an essentially identical sequence, cf. formula.

Cf. A000129, A001109, A002965, A029547, A041085, A056771, A116108.

I:=[0,1]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // G. C. Greubel, Mar 11 2019

with (combinat):seq(fibonacci(4*n,2)/12, n=0..17); # Zerinvary Lajos, Apr 21 2008

LinearRecurrence[{34,-1}, {0,1}, 20] (* G. C. Greubel, Mar 11 2019 *)

A091761(n, x=[ -1,17],A=[17,72*4;1,17]) = vector(n,i,(x*=A)[1]) \\ M. F. Hasler, May 26 2007

A091761(n)=([34,1;-1,0]^(n-1))[1,1] \\ M. F. Hasler, Jun 05 2007

[lucas_number1(n,34,1) for n in range(0, 16)]# Zerinvary Lajos, Nov 07 2009

G.f.: x/(1-34*x+x^2).
a(n) = A000129(4n)/A000129(4).
a(n) = ((1+sqrt(2))^(4n) - (1-sqrt(2))^(4n))*sqrt(2)/48.
From M. F. Hasler, Jun 05 2007: (Start)
a(n) = n (mod 2^m) for any m >= 0.
a(n) = sinh(4*n*log(sqrt(2)+1))/(12*sqrt(2)).
a(n) = A[1,1], first element of the 2 X 2 matrix A = (34,1;-1,0)^(n-1). (End)
a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=1. - Philippe Deléham, Nov 03 2008
A029547(n) = a(n+1). - M. F. Hasler, Feb 05 2013
a(n) = sqrt((A056771(n)^2 - 1)/(32*9)), n >= 0. See the Pell remark above. - Wolfdieter Lang, Mar 11 2019
E.g.f.: exp(17*x)*sinh(12*sqrt(2)*x)/(12*sqrt(2)). - Stefano Spezia, Apr 16 2023
a(n) = A002965(8*n)/12. - Gerry Martens, Jul 14 2023

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1

Offset: 0

Seiichi Manyama, Dec 26 2018

			Square array begins:
   1,  1,   1,    1,      1,       1,         1, ...
   1,  3,  17,   99,    577,    3363,     19601, ...
   1,  5,  49,  485,   4801,   47525,    470449, ...
   1,  7,  97, 1351,  18817,  262087,   3650401, ...
   1,  9, 161, 2889,  51841,  930249,  16692641, ...
   1, 11, 241, 5291, 116161, 2550251,  55989361, ...
   1, 13, 337, 8749, 227137, 5896813, 153090001, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-3 give A000012, A005408, A069129(n+1), A322830.
Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986.
Main diagonal gives A173174.
A(n-1,n) gives A173148(n).
Cf. A322699, A322747.

A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *)

a(n) = 2 * A322699(n) + 1.
A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k).
A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k).
A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1.
A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = 2*n+1.

A202299 y-values in the solution to x^2 - 18*y^2 = 1.

Original entry on oeis.org

0, 4, 136, 4620, 156944, 5331476, 181113240, 6152518684, 209004522016, 7100001229860, 241191037293224, 8193395266739756, 278334248031858480, 9455171037816448564, 321197481037727392696, 10911259184244914903100

Offset: 1

Vincenzo Librandi, Dec 18 2011

The corresponding values of x of this Pell equation are in A056771.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A001542, A056771.

I:=[0, 4]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]];

LinearRecurrence[{34,-1},{0,4},30]
With[{c=6*Sqrt[2]},Table[((17-2c)^n-(17+2c)^n)/-c,{n,0,20}]]//Simplify (* Harvey P. Dale, Dec 16 2024 *)

makelist(expand(((3+2*sqrt(2))^(2*n-2)-(3-2*sqrt(2))^(2*n-2))/(6*sqrt(2))), n, 1, 16); /* _Bruno Berselli, Dec 19 2011 */

a(n) = 34*a(n-1)-a(n-2) with a(1)=0, a(2)=4.
G.f.: 4*x^2/(1-34*x+x^2).
a(n) = (1/3)*A001542(2n-2). - Bruno Berselli, Dec 19 2011

cp's OEIS Frontend

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

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A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

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A029547 Expansion of g.f. 1/(1 - 34*x + x^2).

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A132592 X-values of solutions to the equation X*(X + 1) - 8*Y^2 = 0.

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A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

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A132592 X-values of solutions to the equation X(X + 1) - 8Y^2 = 0.

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.