A188647 -id:A188647 - cp's OEIS frontend

A001653 Numbers k such that 2*k^2 - 1 is a square.

1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409, 10181446324101389, 59341817924539925

Offset: 1

N. J. A. Sloane

Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1 (A069894).
(x,y) = (a(n), a(n+1)) are the solutions with x < y of x/(yz) + y/(xz) + z/(xy)=3 with z=2. - Floor van Lamoen, Nov 29 2001
Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - Lekraj Beedassy, Jun 05 2002
Numbers n such that 2*n^2 = ceiling(sqrt(2)*n*floor(sqrt(2)*n)). - Benoit Cloitre, May 10 2003
Also, number of domino tilings in S_5 X P_2n. - Ralf Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.
If x is in the sequence then so is x*(8*x^2-3). - James R. Buddenhagen, Jan 13 2005
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller, Jun 01 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 largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - Charlie Marion, Sep 14 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova, Jan 10 2007
The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators = A002315 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008
Apparently Ljunggren shows that 169 is the last square term.
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p < r then s-r = p+q+1. - Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = Y^2 with p < r then r = 3p+2q+1 and s = 4p+3q+2. - Mohamed Bouhamida, Sep 02 2009
Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500, ...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5, ..., a(k-1), 0, a(k)-a(k-1), ..., a(k)-1, a(k)-1, ..., a(k)-a(k-1), 0, a(k-1), ..., 5, 1. See Bouhamida's Sep 01 2009 comment. - Charlie Marion, May 02 2011
Apart from initial 1: subsequence of A198389, see also A198385. - Reinhard Zumkeller, Oct 25 2011
(a(n+1), 2*b(n+1)) and (a(n+2), 2*b(n+1)), n >= 0, with b(n):= A001109(n), give the (u(2*n), v(2*n)) and (u(2*n+1), v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1, v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u < v and (u,v) -> (u,2*u+v) if u > v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, Mar 06 2012
Area of the Fibonacci snowflake of order n. - José Luis Ramírez Ramírez, Dec 13 2012
Area of the 3-generalized Fibonacci snowflake of order n, n >= 3. - José Luis Ramírez Ramírez, Dec 13 2012
For the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section see a comment under A077445. - Wolfdieter Lang, Jan 18 2013
Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - Colin Barker, Feb 04 2014
Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - Ralf Stephan, Feb 20 2014
Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - Colin Barker, Mar 04 2014
The value of the hypotenuse in each triple of the Tree of primitive Pythagorean triples (cf. Wikipedia link) starting with root (3,4,5) and recursively selecting the central branch at each triple node of the tree. - Stuart E Anderson, Feb 05 2015
Positive integers z such that z^2 is a centered square number (A001844). - Colin Barker, Feb 12 2015
The aerated sequence (b(n)) n >= 1 = [1, 0, 5, 0, 29, 0, 169, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. - Peter Bala, Mar 25 2015
A002315(n-1)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. A002315(n-1)/a(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = y^2 + (y+1)^2. y-values are listed in A001652. Example: for x=29 and y=20, 28*29/2 + 29*30/2 = 20^2 + 21^2. - Bruno Berselli, Mar 19 2018
From Wolfdieter Lang, Jun 13 2018: (Start)
(a(n), a(n+1)), with a(0):= 1, give all proper positive solutions m1 = m1(n) and m2 = m2(n), with m1 < m2 and n >= 0, of the Markoff triple (m, m1, m2) (see A002559) for m = 2, i.e., m1^2 - 6*m1*m2 + m2^2 = -4. Hence the unique Markoff triple with largest value m = 2 is (1, 1, 2) (for general m from A002559 this is the famous uniqueness conjecture).
For X = m2 - m1 and Y = m2 this becomes the reduced indefinite quadratic form representation X^2 + 4*X*Y - 4*Y^2 = -4, with discriminant 32, and the only proper fundamental solution (X(0), Y(0)) = (0, 1). For all nonnegative proper (X(n), Y(n)) solutions see (A005319(n) = a(n+1) - a(n), a(n+1)), for n >= 0. (End)
Each Pell(2*k+1) = a(k+1) number with k >= 3 appears as largest number of an ordered Markoff (Markov) triple [x, y, m] with smallest value x = 2 as [2, Pell(2*k-1), Pell(2*k+1)]. This known result follows also from all positive proper solutions of the Pell equation q^2 - 2*m^2 = -1 which are q = q(k) = A002315(k)  and m = m(k) = Pell(2*k+1), for k >= 0. y = y(k) = m(k) - 2*q(k) = Pell(2*k-1), with Pell(-1) = 1. The k = 0 and 1 cases do not satisfy x=2 <= y(k) <= m(k). The numbers 1 and 5 appear also as largest Markoff triple members because they are also Fibonacci numbers, and for these triples x=1. - Wolfdieter Lang, Jul 11 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, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=a(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

			From _Muniru A Asiru_, Mar 19 2018: (Start)
For k=1, 2*1^2 - 1 = 2 - 1 = 1 = 1^2.
For k=5, 2*5^2 - 1 = 50 - 1 = 49 = 7^2.
For k=29, 2*29^2 - 1 = 1682 - 1 = 1681 = 41^2.
... (End)
G.f. = x + 5*x^2 + 29*x^3 + 169*x^4 + 985*x^5 + 5741*x^6 + ... - _Michael Somos_, Jun 26 2022

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P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.

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Index entries for two-way infinite sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Other two sides are A001652, A046090.
Cf. A001519, A001109, A005054, A122074, A056220, A056869 (subset of primes).
Cf. A000217, A000290, A002315, A002559, A005319, A069894.
Row 6 of array A094954.
Row 1 of array A188647.
Cf. similar sequences listed in A238379.

a:=[1,5];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018

a001653 n = a001653_list !! n
a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
-- Reinhard Zumkeller, May 07 2013

I:=[1,5]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 22 2014

a[0]:=1: a[1]:=5: 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
A001653:=-(-1+5*z)/(z**2-6*z+1); # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's

LinearRecurrence[{6,-1}, {1,5}, 40] (* Harvey P. Dale, Jul 12 2011 *)
a[ n_] := -(-1)^n ChebyshevU[2 n - 2, I]; (* Michael Somos, Jul 22 2018 *)
Numerator[{1} ~Join~
Table[FromContinuedFraction[Flatten[Table[{1, 4}, n]]], {n, 1, 40}]]; (* Greg Dresden, Sep 10 2019 *)

{a(n) = subst(poltchebi(n-1) + poltchebi(n), x, 3)/4}; /* Michael Somos, Nov 02 2002 */

a(n)=([5,2;2,1]^(n-1))[1,1] \\ Lambert Klasen (lambert.klasen(AT)gmx.de), corrected by Eric Chen, Jun 14 2018

{a(n) = -(-1)^n * polchebyshev(2*n-2, 2, I)}; /* Michael Somos, Jun 26 2022 */

G.f.: x*(1-x)/(1-6*x+x^2).
a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.
4*a(n) = A077445(n).
Can be extended backwards by a(-n+1) = a(n).
a(n) = sqrt((A002315(n)^2 + 1)/2). [Inserted by N. J. A. Sloane, May 08 2000]
a(n+1) = S(n, 6)-S(n-1, 6), n>=0, with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n+1) = T(2*n+1, sqrt(2))/sqrt(2), n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by Wolfdieter Lang, Mar 06 2012]
a(n) = A000129(2n+1). - Ira M. Gessel, Sep 27 2002
a(n) ~ (1/4)*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = (((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1)) - ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n)) / (4*sqrt(2)). Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 12 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 4) = a(n). - Benoit Cloitre, Nov 10 2002
For n and j >= 1, Sum_{k=0..j} a(k)*a(n) - Sum_{k=0..j-1} a(k)*a(n-1) = A001109(j+1)*a(n) - A001109(j)*a(n-1) = a(n+j); e.g., (1+5+29)*5 - (1+5)*1=169. - Charlie Marion, Jul 07 2003
From Charlie Marion, Jul 16 2003: (Start)
For n >= k >= 0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144.
For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0..2*n-1} a(k) = 4*A001109(2n); e.g., 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416.
Sum_{k=0..n} ((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144.
A001652(n) + A046090(n) - a(n) = A001542(n); e.g., 119 + 120 - 169 = 70.
(End)
Sum_{k=0...n} ((2k+1)*a(n-k)) = A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g., 1*169 + 3*29 + 5*5 + 7*1 = 288 = 17^2 - 1; 1*29 + 3*5 + 5*1 = 49 = 7^2. - Charlie Marion, Jul 18 2003
Sum_{k=0...n} a(k)*a(n) = Sum_{k=0..n} a(2k) and Sum_{k=0..n} a(k)*a(n+1) = Sum_{k=0..n} a(2k+1); e.g., (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741. - Charlie Marion, Sep 22 2003
For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n > 0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n)) = Sum_{k=0..n} c(2*k+1); e.g., 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1)) = Sum_{k=0..n} c(2*k); e.g., 119*120*169/(20+21+29) = 1+29+985+33461 = 34476. - Charlie Marion, Dec 01 2003
Also solutions x > 0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2). - Benoit Cloitre, Feb 15 2004
a(n)*a(n+3) = 24 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
For n >= k, a(n)*a(n+2*k+1) - a(n+k)*a(n+k+1) = a(k)^2-1; e.g., 29*195025-985*5741 = 840 = 29^2-1; 1*169-5*29 = 24 = 5^2-1; a(n)*a(n+2*k)-a(n+k)^2 = A001542(k)^2; e.g., 169*195025-5741^2 = 144 = 12^2; 1*29-5^2 = 4 = 2^2. - Charlie Marion Jun 02 2004
For all k, a(n) is a factor of a((2n+1)*k+n). a((2*n+1)*k+n) = a(n)*(Sum_{j=0..k-1} (-1)^j*(a((2*n+1)*(k-j)) + a((2*n+1)*(k-j)-1))+(-1)^k); e.g., 195025 = 5*(33461+5741-169-29+1); 7645370045 = 169*(6625109+1136689-1).- Charlie Marion, Jun 04 2004
a(n) = Sum_{k=0..n} binomial(n+k, 2*k)4^k. - Paul Barry, Aug 30 2004 [offset 0]
a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k+1)*2^k. - Paul Barry, Sep 30 2004 [offset 0]
For n < k, a(n)*A001541(k) = A011900(n+k)+A053141(k-n-1); e.g., 5*99 = 495 = 493+2. For n >= k, a(n)*A001541(k) = A011900(n+k)+A053141(n-k); e.g., 29*3 = 87 = 85+2. - Charlie Marion, Oct 18 2004
a(n) = (-1)^n*U(2*n, i*sqrt(4)/2) = (-1)^n*U(2*n, i), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0]
a(n) = Pell(2*n+1) = Pell(n)^2 + Pell(n+1)^2. - Paul Barry, Jul 18 2005 [offset 0]
a(n)*a(n+k) = A000129(k)^2 + A000129(2n+k+1)^2; e.g., 29*5741 = 12^2+169^2. - Charlie Marion, Aug 02 2005
Let a(n)*a(n+k) = x. Then 2*x^2-A001541(k)*x+A001109(k)^2 = A001109(2*n+k+1)^2; e.g., let x=29*985; then 2x^2-17x+6^2 = 40391^2; cf. A076218. - Charlie Marion, Aug 02 2005
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)+b^((2n+1)/2))/(2*sqrt(2)). a(n) = A001109(n+1)-A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
If k is in the sequence, then the next term is floor(k*(3+2*sqrt(2))). - Lekraj Beedassy, Jul 19 2005
a(n) = Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0]
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*Pell(n-j+1), where Pell = A000129. - Paul Barry, May 19 2006 [offset 0]
a(n) = round(sqrt(A002315(n)^2/2)). - Lekraj Beedassy, Jul 15 2006
a(n) = A079291(n) + A079291(n+1). - Lekraj Beedassy, Aug 14 2006
a(n+1) = 3*a(n) + sqrt(8*a(n)^2-4), a(1)=1. - Richard Choulet, Sep 18 2007
6*a(n)*a(n+1) = a(n)^2+a(n+1)^2+4; e.g., 6*5*29 = 29^2+5^2+4; 6*169*985 = 169^2+985^2+4. - Charlie Marion, Oct 07 2007
2*A001541(k)*a(n)*a(n+k) = a(n)^2+a(n+k)^2+A001542(k)^2; e.g., 2*3*5*29 = 5^2+29^2+2^2; 2*99*29*5741 = 2*99*29*5741=29^2+5741^2+70^2. - Charlie Marion, Oct 12 2007
[a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
From Charlie Marion, Apr 10 2009: (Start)
In general, for n >= k, a(n+k) = 2*A001541(k)*a(n)-a(n-k);
e.g., a(n+0) = 2*1*a(n)-a(n); a(n+1) = 6*a(n)-a(n-1); a(6+0) = 33461 = 2*33461-33461; a(5+1) = 33461 = 6*5741-985; a(4+2) = 33461 = 34*985-29; a(3+3) = 33461 = 198*169-1.
(End)
G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - Johannes W. Meijer, Aug 01 2010
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = a(n-1)*k-((k-1)/(k^n)). - Charles L. Hohn, Mar 06 2011
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(n) - a(n-1). - Charles L. Hohn, Apr 04 2011
Let T(n) be the n-th triangular number; then, for n > 0, T(a(n)) + A001109(n-1) = A046090(n)^2.  See also A046090. - Charlie Marion, Apr 25 2011
For k > 0, a(n+2*k-1) - a(n) = 4*A001109(n+k-1)*A002315(k-1); a(n+2*k) - a(n) = 4*A001109(k)*A002315(n+k-1). - Charlie Marion, Jan 06 2012
a(k+j+1) = (A001541(k)*A001541(j) + A002315(k)*A002315(j))/2. - Charlie Marion, Jun 25 2012
a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. - Bruno Berselli, Oct 23 2012
a(n) = A143608(n-1)*A143608(n) + 1 = A182190(n-1)+1. - Charlie Marion, Dec 11 2012
G.f.: G(0)*(1-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
a(n+1) = 4*A001652(n) + 3*a(n) + 2 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 = A001652(n-1)^2 + (A001652(n-1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
Sum_{n >= 2} 1/( a(n) - 1/a(n) ) = 1/4. - Peter Bala, Mar 25 2015
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^floor(k/2). - David Pasino, Jul 09 2016
E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + 2*cosh(2*sqrt(2)*x))*exp(3*x)/2. - Ilya Gutkovskiy, Jul 09 2016
a(n+2) = (a(n+1)^2 + 4)/a(n). - Vladimir M. Zarubin, Sep 06 2016
a(n) = 2*A053141(n)+1. - R. J. Mathar, Aug 16 2019
For n>1, a(n) is the numerator of the continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the denominators see A005319. - Greg Dresden, Sep 10 2019
a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/4). - Paul Weisenhorn, May 23 2020
a(n+1) = Sum_{k >= n} binomial(2*k,2*n)*(1/2)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
a(n+1) = 3*a(n) + A077444(n). - César Aguilera, Jul 13 2023

Additional comments from Wolfdieter Lang, Feb 10 2000
Better description from Harvey P. Dale, Jan 15 2002
Edited by N. J. A. Sloane, Nov 02 2002

A053755 a(n) = 4*n^2 + 1.

Original entry on oeis.org

1, 5, 17, 37, 65, 101, 145, 197, 257, 325, 401, 485, 577, 677, 785, 901, 1025, 1157, 1297, 1445, 1601, 1765, 1937, 2117, 2305, 2501, 2705, 2917, 3137, 3365, 3601, 3845, 4097, 4357, 4625, 4901, 5185, 5477, 5777, 6085, 6401, 6725, 7057

Offset: 0

Stuart M. Ellerstein (ellerstein(AT)aol.com), Apr 06 2000

Subsequence of A004613: all numbers in this sequence have all prime factors of the form 4k+1. E.g., 40001 = 13*17*181, 13 = 4*3 + 1, 17 = 4*4 + 1, 181 = 4*45 + 1. - Cino Hilliard, Aug 26 2006, corrected by Franklin T. Adams-Watters, Mar 22 2011
A000466(n), A008586(n) and a(n) are Pythagorean triples. - Zak Seidov, Jan 16 2007
Solutions x of the Mordell equation y^2 = x^3 - 3a^2 - 1 for a = 0, 1, 2, ... - Michel Lagneau, Feb 12 2010
Ulam's spiral (NW spoke). - Robert G. Wilson v, Oct 31 2011
For n >= 1, a(n) is numerator of radius r(n) of circle with sagitta = n and cord length = 1. The denominator is A008590(n). - Kival Ngaokrajang, Jun 13 2014
a(n)+6 is prime for n = 0..6 and for n = 15..20. - Altug Alkan, Sep 28 2015

Donald E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1997, Vol. 1, exercise 1.2.1 Nr. 11, p. 19.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
Kival Ngaokrajang, Illustration of initial terms.
Robert G. Wilson v, Cover of the March 1964 issue of Scientific American.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Column 2 of array A188647.
Cf. A016742, A256970 (smallest prime factors), A214345.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A000466, A001844, A004613, A008586, A008590, A017113, A046092, A087475, A156701, A176271.

List([0..45],n->4*n^2+1); # Muniru A Asiru, Nov 01 2018

a053755 = (+ 1) . (* 4) . (^ 2)  -- Reinhard Zumkeller, Apr 20 2015

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+5*x^2)/((1-x)^3))); /* or */ I:=[1,5]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+8: n in [1..50]]; // Vincenzo Librandi, Jun 26 2013

with (combinat):seq(fibonacci(3,2*n), n=0..42); # Zerinvary Lajos, Apr 21 2008

f[n_] := 4n^2 +1; Array[f, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 02 2008 *)
CoefficientList[Series[(1 + 2 x + 5 x^2) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 26 2013 *)
LinearRecurrence[{3,-3,1},{1,5,17},50] (* Harvey P. Dale, Dec 28 2021 *)

for(x=0,100,print1(4*x^2+1",")) \\ Cino Hilliard, Aug 26 2006

for n in range(0,50): print(4*n**2+1, end=', ') # Stefano Spezia, Nov 01 2018

a(n) = A000466(n) + 2. - Zak Seidov, Jan 16 2007
From R. J. Mathar, Apr 28 2008: (Start)
O.g.f.: (1 + 2*x + 5*x^2)/(1-x)^3.
a(n) = 3a(n-1) - 3a(n-2) + a(n-3). (End)
Equals binomial transform of [1, 4, 8, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
a(n) = A156701(n)/A087475(n). - Reinhard Zumkeller, Feb 13 2009
For n>0: a(n) = A176271(2*n,n+1); cf. A016754, A000466. - Reinhard Zumkeller, Apr 13 2010
a(n+1) = denominator of Sum_{k=0..n} (-1)^n*(2*n + 1)^3/((2*n + 1)^4 + 4), see Knuth reference. - Reinhard Zumkeller, Apr 11 2010
a(n) = 8*n + a(n-1) - 4. with a(0)=1. - Vincenzo Librandi, Aug 06 2010
a(n) = ((2*n - 1)^2 + (2*n + 1)^2)/2. - J. M. Bergot, May 31 2012
a(n) = 2*a(n-1) - a(n-2) + 8 with a(0)=1, a(1)=5. - Vincenzo Librandi, Jun 26 2013
a(n+1) = a(n) + A017113(n), a(0) = 1. - Altug Alkan, Sep 26 2015
a(n) = A001844(n) + A046092(n-1) = A001844(n-1) + A046092(n). - Bruce J. Nicholson, Aug 07 2017
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/2)*coth(Pi/2))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/2)*csch(Pi/2))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/2)*sinh(Pi/sqrt(2)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/2)*csch(Pi/2). (End)
E.g.f.: exp(x)*(1 + 2*x)^2. - Stefano Spezia, Jun 10 2021

Equation corrected, and examples that were based on a different offset removed, by R. J. Mathar, Mar 18 2010

A007805 a(n) = Fibonacci(6*n + 3)/2.

Original entry on oeis.org

1, 17, 305, 5473, 98209, 1762289, 31622993, 567451585, 10182505537, 182717648081, 3278735159921, 58834515230497, 1055742538989025, 18944531186571953, 339945818819306129, 6100080207560938369

Offset: 0

James A. Raymond, Clark Kimberling

Hypotenuse (z) of Pythagorean triples (x,y,z) with |2x-y|=1.
x(n) := 2*A049629(n) and y(n) := a(n), n >= 0, give all positive solutions of the Pell equation x^2 - 5*y^2 = -1. See the Gregory V. Richardson formula, where his x is the y here and A075796(n+1) = x(n). - Wolfdieter Lang, Jun 20 2013
Positive numbers n such that 5*n^2 - 1 is a square (A075796(n+1)^2). - Gregory V. Richardson, Oct 13 2002

G. C. Greubel, Table of n, a(n) for n = 0..795 (terms 0..100 from T. D. Noe)
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045.
Row 18 of array A094954.
Row 2 of array A188647.
Cf. similar sequences listed in A238379.

a007805 = (`div` 2) . a000045 . (* 3) . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 26 2013

I:=[1, 17]; [n le 2 select I[n] else 18*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

seq(combinat:-fibonacci(6*n+3)/2, n=0..30); # Robert Israel, Sep 10 2014

LinearRecurrence[{18, -1}, {1, 17}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
Table[Fibonacci[6n+3]/2, {n, 0, 20}] (* Harvey P. Dale, Dec 17 2011 *)
CoefficientList[Series[(1-x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

a(n)=fibonacci(6*n+3)/2 \\ Edward Jiang, Sep 09 2014

x='x+O('x^30); Vec((1-x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

G.f.: (1-x)/(1-18*x+x^2).
a(n) = 18*a(n-1) - a(n-2), n > 1, a(0)=1, a(1)=17.
a(n) = A134495(n)/2 = A001076(2n+1).
a(n+1) = 9*a(n) + 4*sqrt(5*a(n)^2-1). - Richard Choulet, Aug 30 2007, Dec 28 2007
a(n) = ((2+sqrt(5))^(2*n+1) - (2-sqrt(5))^(2*n+1))/(2*sqrt(5)). - Dean Hickerson, Dec 09 2002
a(n) ~ (1/10)*sqrt(5)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = q(n, 16). - Benoit Cloitre, Dec 06 2002
a(n) = 19*a(n-1)- 19*a(n-2) + a(n-3); f(x) = (sqrt(5)/10)*((2+sqrt(5))*(9+4*sqrt(5))^(x-1) - (2-sqrt(5))*(9-4*sqrt(5))^(x-1)). - Antonio Alberto Olivares, May 15 2008
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3). - Antonio Alberto Olivares, Jun 19 2008
a(n) = b(n+1) - b(n), n >= 0, with b(n) := F(6*n)/F(6) = A049660(n). First differences. See the o.g.f.s. - Wolfdieter Lang, 2012
a(n) = S(n,18) - S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/4^2. Compare with A001519 and A097843. - Peter Bala, Nov 29 2013
a(n) = 9*a(n-1) + 8*A049629(n-1), n >= 1, a(0) = 1. This is just the rewritten Chebyshev S(n, 18) recurrence. - Wolfdieter Lang, Aug 26 2014
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) - Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) + Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n)) n>=1 = [1, 0, 17, 0, 305, 0, 5473, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -20, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = sqrt(2 + (9-4*sqrt(5))^(1+2*n) + (9+4*sqrt(5))^(1+2*n))/(2*sqrt(5)). - Gerry Martens, Jun 04 2015

Better description and more terms from Michael Somos

A097315 Pell equation solutions (3b(n))^2 - 10a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

Original entry on oeis.org

1, 37, 1405, 53353, 2026009, 76934989, 2921503573, 110940200785, 4212806126257, 159975692596981, 6074863512559021, 230684837784645817, 8759948972303982025, 332647376109766671133, 12631840343198829521029, 479677285665445755127969, 18215105014943739865341793, 691694313282196669127860165

Offset: 0

Wolfdieter Lang, Aug 31 2004

Hypotenuses of primitive Pythagorean triples in A195616 and A195617. - Clark Kimberling, Sep 22 2011

			(x,y) = (3,1), (117,37), (4443,1405), ... give the positive integer solutions to x^2 - 10*y^2 = -1.
G.f. = 1 + 37*x + 1405*x^2 + 53353*x^3 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..631
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A078987 (partial sums), A097314, A195616, A195617, A049310, A053120, A005668.
Row 3 of array A188647.
Cf. A221874.
Cf. similar sequences listed in A238379.

a:=[1,37];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1, 37]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 01 2019

CoefficientList[Series[(1-x)/(1-38x+x^2), {x,0,20}], x] (* Michael De Vlieger, Feb 04 2017 *)
LinearRecurrence[{38,-1}, {1,37}, 21] (* G. C. Greubel, Aug 01 2019 *)

Vec((1-x)/(1-38*x+x^2) + O(x^20)) \\ Michel Marcus, Jun 04 2015

from itertools import islice
def A097315_gen(): # generator of terms
    x, y = 30, 10
    while True:
        yield y//10
        x, y = x*19+y*60, x*6+y*19
A097315_list = list(islice(A097315_gen(),20)) # Chai Wah Wu, Apr 24 2025

((1-x)/(1-38*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 38) - S(n-1, 38) = T(2*n+1, sqrt(10))/sqrt(10), with Chebyshev polynomials of the second and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 6*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-38*x+x^2).
a(n) = 38*a(n-1) - a(n-2) for n > 1. - Philippe Deléham, Nov 18 2008
a(n) = sqrt(2+(19-6*sqrt(10))^(1+2*n) + (19+6*sqrt(10))^(1+2*n))/(2*sqrt(10)). - Gerry Martens, Jun 04 2015
a(n) = A078987(n) - A078987(n-1). - R. J. Mathar, Dec 05 2015
a(n) = A005668(2*n+1). - Michael Somos, Feb 24 2023
E.g.f.: exp(19*x)*(10*cosh(6*sqrt(10)*x) + 3*sqrt(10)*sinh(6*sqrt(10)*x))/10. - Stefano Spezia, Apr 24 2025

Typo in recurrence formula corrected by Laurent Bonaventure (bonave(AT)free.fr), Oct 03 2010

More terms added by Indranil Ghosh, Feb 04 2017

A188645 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, 3, 1, 17, 9, 1, 99, 161, 19, 1, 577, 2889, 721, 33, 1, 3363, 51841, 27379, 2177, 51, 1, 19601, 930249, 1039681, 143649, 5201, 73, 1, 114243, 16692641, 39480499, 9478657, 530451, 10657, 99, 1, 665857, 299537289, 1499219281, 625447713, 54100801, 1555849, 19601, 129, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given 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 A188644.

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_{k}(x), evaluated at x=2*n^2+1. - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   3,     17,        99,          577, ...
   2 | 1,   9,    161,      2889,        51841, ...
   3 | 1,  19,    721,     27379,      1039681, ...
   4 | 1,  33,   2177,    143649,      9478657, ...
   5 | 1,  51,   5201,    530451,     54100801, ...
   6 | 1,  73,  10657,   1555849,    227143297, ...
   7 | 1,  99,  19601,   3880899,    768398401, ...
   8 | 1, 129,  33281,   8586369,   2215249921, ...
   9 | 1, 163,  53137,  17322499,   5647081537, ...
  10 | 1, 201,  80801,  32481801,  13057603201, ...
  11 | 1, 243, 118097,  57394899,  27893802817, ...
  12 | 1, 289, 167041,  96549409,  55805391361, ...
  13 | 1, 339, 229841, 155831859, 105653770561, ...
  14 | 1, 393, 308897, 242792649, 190834713217, ...
  15 | 1, 451, 406801, 366934051, 330974107201, ...
  ...

Row 1 is A001541, row 2 is A023039, row 3 is A078986, row 4 is A099370, row 5 is A099397, row 6 is A174747, row 8 is A176368, (row 1)*2 is A003499, (row 2)*2 is A087215.
Column 1 is A058331, (column 1)*2 is A005899.
A188644 (f(x, y) as above with y=-1).
Diagonal gives A173128.
Cf. A188647.

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) = (A188647(n,k-1) + A188647(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 and extended by Seiichi Manyama, Jan 01 2019

A188646 Array of a(n)=a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2-1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 181, 33, 1, 1, 2521, 1121, 61, 1, 1, 35113, 38081, 3781, 97, 1, 1, 489061, 1293633, 234361, 9505, 141, 1, 1, 6811741, 43945441, 14526601, 931393, 20021, 193, 1, 1, 94875313, 1492851361, 900414901, 91267009, 2842841, 37441, 253, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=-1 results shown here; y=1 results are A188647.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/n) * T_{2*k+1}(n), with the Chebyshev polynomials of the first kind (type T). - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,  13,    181,      2521,        35113, ...
   3 | 1,  33,   1121,     38081,      1293633, ...
   4 | 1,  61,   3781,    234361,     14526601, ...
   5 | 1,  97,   9505,    931393,     91267009, ...
   6 | 1, 141,  20021,   2842841,    403663401, ...
   7 | 1, 193,  37441,   7263361,   1409054593, ...
   8 | 1, 253,  64261,  16322041,   4145734153, ...
   9 | 1, 321, 103361,  33281921,  10716675201, ...
  10 | 1, 397, 158005,  62885593,  25028308009, ...
  11 | 1, 481, 231841, 111746881,  53861764801, ...
  12 | 1, 573, 328901, 188788601, 108364328073, ...
  13 | 1, 673, 453601, 305726401, 206059140673, ...
  14 | 1, 781, 610741, 477598681, 373481557801, ...
  15 | 1, 897, 805505, 723342593, 649560843009, ...
  ...

Row 1-8 give A000012, A001570, A077420, A302329, A302330, A302331, A302332, A253880.
Column 1 is A082109(n-1).
Cf. A188644, A188647 (f(x, y) as above with y=1).
Diagonal gives A322904.

A[n_, k_] := 1/n ChebyshevT[2k+1, n];
Table[A[n-k, k], {n, 1, 9}, {k, n-1, 0, -1}] // Flatten (* Jean-François Alcover, Jan 02 2019, after Seiichi Manyama *)

A(n,k) = 2 * A188644(n,k) - A(n,k-1).

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

Edited and extended by Seiichi Manyama, Jan 01 2019

A078988 Chebyshev sequence with Diophantine property.

Original entry on oeis.org

1, 65, 4289, 283009, 18674305, 1232221121, 81307919681, 5365090477825, 354014663616769, 23359602708228929, 1541379764079492545, 101707704826538279041, 6711167138787446924161, 442835323455144958715585, 29220420180900779828304449, 1928104896615996323709378049

Offset: 0

Wolfdieter Lang, Jan 10 2003

Bisection (even part) of A041025.
(4*A078989(n))^2 - 17*a(n)^2 = -1 (Pell -1 equation, see A077232-3).
Starting with a(1), hypotenuses of primitive Pythagorean triples in A195619 and A195620. - Clark Kimberling, Sep 22 2011

			(x,y) = (4,1), (268,65), (17684,4289), ... give the positive integer solutions to x^2 - 17*y^2 =-1.

Colin Barker, Table of n, a(n) for n = 0..549
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (66,-1).

Row 66 of array A094954.
Cf. A097316 for S(n, 66).
Row 4 of array A188647.

a:=[1,65];; for n in [3..20] do a[n]:=66*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1, 65]; [n le 2 select I[n] else 66*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019

CoefficientList[Series[(1-x)/(1-66x+x^2), {x,0,20}], x] (* Michael De Vlieger, Apr 15 2019 *)
LinearRecurrence[{66,-1}, {1,65}, 21] (* G. C. Greubel, Aug 01 2019 *)

Vec((1-x)/(1-66*x+x^2) + O(x^20)) \\ Colin Barker, Jun 15 2015

((1-x)/(1-66*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

G.f.: (1-x)/(1-66*x+x^2).
a(n) = T(2*n+1, sqrt(17))/sqrt(17) = ((-1)^n)*S(2*n, 8*i) = S(n, 66) - S(n-1, 66) with i^2=-1 and T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310.
a(n) = A041025(2*n).
a(n) = 66*a(n-1) - a(n-2) for n>1 ; a(0)=1, a(1)=65. - Philippe Deléham, Nov 18 2008

A097727 Pell equation solutions (5b(n))^2 - 26a(n)^2 = -1 with b(n)=A097726(n), n >= 0.

Original entry on oeis.org

1, 101, 10301, 1050601, 107151001, 10928351501, 1114584702101, 113676711262801, 11593909964103601, 1182465139627304501, 120599850332020955501, 12300002268726510156601, 1254479631559772015017801, 127944622416828019021659101, 13049097006884898168194210501

Offset: 0

Wolfdieter Lang, Aug 31 2004

Hypotenuses of primitive Pythagorean triples in A195622 and A195623. - Clark Kimberling, Sep 22 2011

			(x,y) = (5,1), (515,101), (52525,10301), ... give the positive integer solutions to x^2 - 26*y^2 =-1.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (102,-1).

Cf. A097725 for S(n, 102).

Row 5 of array A188647.

a:=[1,101];; for n in [3..20] do a[n]:=102*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1,101]; [n le 2 select I[n] else 102*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019

LinearRecurrence[{102,-1},{1,101},20] (* Harvey P. Dale, Apr 12 2014 *)
CoefficientList[Series[(1-x)/(1-102x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Apr 13 2014 *)

my(x='x+O('x^20)); Vec((1-x)/(1-102*x+x^2)) \\ G. C. Greubel, Aug 01 2019

((1-x)/(1-102*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 2*51) - S(n-1, 2*51) = T(2*n+1, sqrt(26))/sqrt(26), with Chebyshev polynomials of the 2nd and first kind. See A049310 for the triangle of S(n, x) = U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 10*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-102*x+x^2).
a(n) = 102*a(n-1) - a(n-2) for n > 1; a(0)=1, a(1)=101. - Philippe Deléham, Nov 18 2008

More terms from Harvey P. Dale, Apr 12 2014

A097730 Pell equation solutions (6b(n))^2 - 37a(n)^2 = -1 with b(n)=A097729(n), n >= 0.

Original entry on oeis.org

1, 145, 21169, 3090529, 451196065, 65871534961, 9616792908241, 1403985893068225, 204972323595052609, 29924555258984612689, 4368780095488158399985, 637811969386012141785121, 93116178750262284542227681, 13594324285568907531023456305

Offset: 0

Wolfdieter Lang, Aug 31 2004

			(x,y) = (6,1), (882,145), (128766,21169), ... give the positive integer solutions to x^2 - 37*y^2 = -1.

Indranil Ghosh, Table of n, a(n) for n = 0..461
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (146,-1).

Cf. A097729 for S(n, 146).

Row 6 of array A188647.

a:=[1,145];; for n in [3..20] do a[n]:=146*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1,145]; [n le 2 select I[n] else 146*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019

LinearRecurrence[{146, -1},{1, 145},12] (* Ray Chandler, Aug 12 2015 *)

my(x='x+O('x^20)); Vec((1-x)/(1-146*x+x^2)) \\ G. C. Greubel, Aug 01 2019

((1-x)/(1-146*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 2*73) - S(n-1, 2*73) = T(2*n+1, sqrt(37))/sqrt(37), with Chebyshev polynomials of the 2nd and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 12*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-146*x+x^2).
a(n) = 146*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=145. - Philippe Deléham, Nov 18 2008

A097733 Pell equation solutions (7b(n))^2 - 2(5a(n))^2 = -1 with b(n)=A097732(n), n >= 0. Note that D=50=25^2 is not squarefree.

Original entry on oeis.org

1, 197, 39005, 7722793, 1529074009, 302748930989, 59942759261813, 11868363584907985, 2349876047052519217, 465263588952813896981, 92119840736610099083021, 18239263202259846804541177

Offset: 0

Wolfdieter Lang, Aug 31 2004

			(x,y) = (7,1), (1393,197), (275807,39005), ... give the positive integer solutions to x^2 - 50*y^2 =-1.

Indranil Ghosh, Table of n, a(n) for n = 0..434
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (198,-1).

Cf. A097731 for S(n, 198).
Row 7 of array A188647.
Cf. A000129, A100047.

a:=[1,197];; for n in [3..20] do a[n]:=198*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1,197]; [n le 2 select I[n] else 198*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019

LinearRecurrence[{198, -1},{1, 197},20] (* Ray Chandler, Aug 11 2015 *)

my(x='x+O('x^20)); Vec((1-x)/(1-198*x+x^2)) \\ G. C. Greubel, Aug 01 2019

((1-x)/(1-198*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 2*99) - S(n-1, 2*99) = T(2*n+1, 5*sqrt(2))/(5*sqrt(2)), with Chebyshev polynomials of the 2nd and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 14*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-198*x+x^2).
a(n) = 198*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=197. - Philippe Deléham, Nov 18 2008
a(n) = k^n + k^(-n) - a(n-1) = A003499(3n) - a(n-1), where k = (sqrt(2)+1)^6 = 99 + 70*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Pell(6*n + 6 - 2*k) - Pell(6*n + 2*k) )/( Pell(6 - 2*k) - Pell(2*k) ), for k an arbitrary integer.
a(n) = ( Pell(6*n + 6 - 2*k - 1) + Pell(6*n + 2*k + 1) )/( Pell(6 - 2*k - 1) + Pell(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 197, 0, 39005, 0, 7722793, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -200, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/196. - Peter Bala, Mar 26 2015

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A001653 Numbers k such that 2*k^2 - 1 is a square.

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A053755 a(n) = 4*n^2 + 1.

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A007805 a(n) = Fibonacci(6*n + 3)/2.

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A097315 Pell equation solutions (3*b(n))^2 - 10*a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

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A188645 Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.

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A097315 Pell equation solutions (3b(n))^2 - 10a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

A097727 Pell equation solutions (5b(n))^2 - 26a(n)^2 = -1 with b(n)=A097726(n), n >= 0.

A097730 Pell equation solutions (6b(n))^2 - 37a(n)^2 = -1 with b(n)=A097729(n), n >= 0.

A097733 Pell equation solutions (7b(n))^2 - 2(5a(n))^2 = -1 with b(n)=A097732(n), n >= 0. Note that D=50=25^2 is not squarefree.