A038725 -id:A038725 - cp's OEIS frontend

A117719 a(2n) = A001653(n) (Numbers n such that 2*n^2 - 1 is a square), a(2n+1) = A038725(n+1).

1, 2, 5, 11, 29, 64, 169, 373, 985, 2174, 5741, 12671, 33461, 73852, 195025, 430441, 1136689, 2508794, 6625109, 14622323, 38613965, 85225144, 225058681, 496728541, 1311738121, 2895146102, 7645370045, 16874148071, 44560482149

Offset: 0

Creighton Dement, Apr 13 2006

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A001653, A038725.

I:= [1,2,5,11]; [n le 4 select I[n] else 6*Self(n-2) -Self(n-4): n in [1..40]]; // G. C. Greubel, Jul 23 2023

LinearRecurrence[{0,6,0,-1}, {1,2,5,11}, 40] (* G. C. Greubel, Jul 23 2023 *)

A000129=BinaryRecurrenceSequence(2,1,0,1)
def A117719(n): return (3*A000129(n+1) +2*A000129(n) +(-1)^n*A000129(n-1))/4
[A117719(n) for n in range(41)] # G. C. Greubel, Jul 23 2023

G.f.: (1+2*x-x^2-x^3)/((1-2*x-x^2)*(1+2*x-x^2)).

a(n) = (1/4)*( 3*P(n+1) + 2*P(n) + (-1)^n*P(n-1) ), where P(n) = A000129(n). - G. C. Greubel, Jul 23 2023

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

Original entry on oeis.org

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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Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Triangular square number
Rick Young, Relevant quotation from biography of Ramanujan
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).

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

A156035 Decimal expansion of 3 + 2*sqrt(2).

Original entry on oeis.org

5, 8, 2, 8, 4, 2, 7, 1, 2, 4, 7, 4, 6, 1, 9, 0, 0, 9, 7, 6, 0, 3, 3, 7, 7, 4, 4, 8, 4, 1, 9, 3, 9, 6, 1, 5, 7, 1, 3, 9, 3, 4, 3, 7, 5, 0, 7, 5, 3, 8, 9, 6, 1, 4, 6, 3, 5, 3, 3, 5, 9, 4, 7, 5, 9, 8, 1, 4, 6, 4, 9, 5, 6, 9, 2, 4, 2, 1, 4, 0, 7, 7, 7, 0, 0, 7, 7, 5, 0, 6, 8, 6, 5, 5, 2, 8, 3, 1, 4, 5, 4, 7, 0, 0, 2

Offset: 1

Klaus Brockhaus, Feb 02 2009

Limit_{n -> oo} b(n+1)/b(n) = 3+2*sqrt(2) for b = A155464, A155465, A155466.
Limit_{n -> oo} b(n)/b(n-1) = 3+2*sqrt(2) for b = A001652, A001653, A002315, A156156, A156157, A156158. - Klaus Brockhaus, Sep 23 2009
From Richard R. Forberg, Aug 14 2013: (Start)
Ratios b(n+1)/b(n) for all sequences of the form b(n) = 6*b(n-1) - b(n-2), for any initial values of b(0) and b(1), converge to this ratio.
Ratios b(n+1)/b(n) for all sequences of the form b(n) = 5*b(n-1) + 5*b(n-2) + b(n-3), for all b(0), b(1) and b(2) also converge to 3 + 2*sqrt(2). For example see A084158 (Pell Triangles).
Ratios of alternating values, b(n+2)/b(n), for all sequences of the form b(n) = 2*b(n-1) + b(n-2), also converge to 3 + 2*sqrt(2). These include A000129 (Pell Numbers). Also see A014176. (End)
Let ABCD be a square inscribed in a circle. When P is the midpoint of the arc AB, then the ratio (PC*PD)/(PA*PB) is equal to 3+2*sqrt(2). See the Mathematical Reflections link. - Michel Marcus, Jan 10 2017
Limit of ratios of successive terms of A001652 when n-> infinity. - Harvey P. Dale, Jun 16 2017; improved by Bernard Schott, Feb 28 2022
A quadratic integer with minimal polynomial x^2 - 6x + 1. - Charles R Greathouse IV, Jul 11 2020
Ratio between radii of the large circumscribed circle R and the small internal circle r drawn on the Sangaku tablet at Isaniwa Jinjya shrine in Ehime Prefecture (pictures in links). - Bernard Schott, Feb 25 2022

			3 + 2*sqrt(2) = 5.828427124746190097603377448...

Diogo Queiros-Condé and Michel Feidt, Fractal and Trans-scale Nature of Entropy, Iste Press and Elsevier, 2018, page 45.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Mathematical Reflections, Solution to Problem J286, Issue 1, 2014, p. 5.
Bernard Schott, Sangaku at Isaniwa Jinya, The six circles.
Terakoya Suzu, Sangaku (mathematics tablet) II, Sangaku at Isaniwa Jinya shrine.
Wikipedia, Sangaku.
Bernard Ycart, Les Sangakus, Sangaku du Temple Isaniwa Jinya (in French).
Index entries for algebraic numbers, degree

Cf. A002193 (sqrt(2)), A090488, A010466, A014176.
Cf. A155464, A155465, A155466.
Cf. A104178 (decimal expansion of log_10(3+2*sqrt(2))).
Cf. A000129, A001109, A001541, A001542, A001652, A001653, A002315, A005319, A075870, A038723, A038725, A038761, A054488, A054489, A075848, A077413, A084158, A106328, A156156, A156157, A156158.
Cf. A242412 (sangaku).

SetDefaultRealField(RealField(100));  3 + 2*Sqrt(2); // G. C. Greubel, Aug 21 2018

RealDigits[N[3+2*Sqrt[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)

3+sqrt(8) \\ Charles R Greathouse IV, Jun 10 2011

Equals 1 + A090488 = 3 + A010466. - R. J. Mathar, Feb 19 2009
Equals exp(arccosh(3)), since arccosh(x) = log(x+sqrt(x^2-1)). - Stanislav Sykora, Nov 01 2013
Equals (1+sqrt(2))^2, that is, A014176^2. - Michel Marcus, May 08 2016
The periodic continued fraction is [5; [1, 4]]. - Stefano Spezia, Mar 17 2024

A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 23, 134, 781, 4552, 26531, 154634, 901273, 5253004, 30616751, 178447502, 1040068261, 6061962064, 35331704123, 205928262674, 1200237871921, 6995498968852, 40772755941191, 237641036678294, 1385073464128573, 8072799748093144, 47051725024430291, 274237550398488602

Offset: 0

Barry E. Williams, May 02 2000

This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - Wolfdieter Lang, Feb 26 2015

			n = 2: A054490(2)^2 - 2*(2*a(2))^2 =
       65^2 - 2*(2*23)^2 = -7,
      (4*a(2))^2 - 2*A054490(2)^2 =
      (4*23)^2 - 2*65^2 = 14. - _Wolfdieter Lang_, Feb 26 2015
a(2) = (A253811(1) + A101386(1))/2 = (19 + 27)/2 = 23. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A001541, A001653, A054490.

A038725(n) = a(-n).

a[0]:=1: a[1]:=4: 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

LinearRecurrence[{6,-1},{1,4},30] (* Harvey P. Dale, Aug 06 2020 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
a(n) = A001653(n+1) - A001109(n). - Antonio Alberto Olivares, Mar 29 2008
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008
G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*3^k. - Philippe Deléham, Mar 05 2014
a(n) = S(n,6) - 2*S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-1, x) = 0 evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell comment above. - Wolfdieter Lang, Feb 26 2015
a(0) = -(A038761(0) - A038762(0))/2, a(n) = (A253811(n-1) + A101386(n-1))/2, n >= 1. See the Mar 19 2015 comment on A054490. - Wolfdieter Lang, Mar 19 2015
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Apr 30 2020

More terms from James Sellers, May 03 2000

A054490 Expansion of (1+5x)/(1-6x+x^2).

Original entry on oeis.org

1, 11, 65, 379, 2209, 12875, 75041, 437371, 2549185, 14857739, 86597249, 504725755, 2941757281, 17145817931, 99933150305, 582453083899, 3394785353089, 19786259034635, 115322768854721, 672150354093691, 3917579355707425, 22833325780150859

Offset: 0

Barry E. Williams, May 04 2000

A Pellian-related second-order recursive sequence.
Third binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
Binomial transform of A164607. - R. J. Mathar, Oct 26 2011
Pisano period lengths: 1, 1, 4, 2, 6, 4, 3, 2, 12, 6, 12, 4, 14, 3, 12, 2, 8, 12, 20, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = x1 = a(n) of the first class of the (generalized) Pell equation x^2 - 2*y^2 = -7. For the corresponding y1 terms see 2*A038723(n). All positive solutions of the second class are given by (x2(n), y2(n)) = (A255236(n), A038725(n+1)), n >= 0. See (A254938(1), 2*A255232(1)) for the fundamental solution (1, 2) of the first class. See the Nagell reference, Theorem 111, p. 210, Theorem 110, p. 208, Theorem 108a, pp. 206-207.
This sequence also gives all positive solutions y = y1 of the first class of the Pell equation x^2 - 2*y^2 = 14. The corresponding solutions x1 are given in 4*A038723. This follows from the preceding comment. (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(0) = -(2*A038761(0) - A038762(0)), a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1.
This follows from the general trivial fact that if X^2 - D*Y^2 = N (X, Y positive integers, D > 1, not a square, and N a non-vanishing integer) then x:= D*Y +/- X and y:= Y +/- X (correlated signs) satisfy x^2 - D*y^2 = -(D-1)*N. with integers x and y. Here D = 2 and N = 7. (End)

			n = 2: sqrt(8*23^2-7) = 65.
2*19 + 27  = 65. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See y(n) in Lemma 7.9 p. 21.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000129, A001109, A038723, A038725, A054488, A054489, A101386, A253811, A255236.

a:=[1,11];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 20 2020

I:=[1,11]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

a[0]:=1: a[1]:=11: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006

CoefficientList[Series[(1+5x)/(1-6x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)
LinearRecurrence[{6, -1}, {1, 11}, 30] (* G. C. Greubel, Jul 26 2018 *)

my(x='x+O('x^30)); Vec((1+5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

[lucas_number1(2*n+1,2,-1) + 3*lucas_number1(2*n,2,-1) for n in (0..30)] # G. C. Greubel, Jan 20 2020

a(n) = 6*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=11.
a(n) = sqrt(8*A038723(n)^2 - 7).
a(n) = (11*((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) - ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = 11*S(n, 6) + 5*S(n-1, 6), n >= 0, with Chebyshev's polynomials S(n, x) (A049310) evaluated at x=6: S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comments above. - Wolfdieter Lang, Feb 26 2015
a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1. See the Mar 19 2015 comment above. - Wolfdieter Lang, Mar 19 2015
From G. C. Greubel, Jan 20 2020: (Start)
a(n) = Pell(2*n+1) + 3*Pell(2*n).
a(n) = ChebyshevU(n,3) + 5*ChebyshevU(n-1,3).
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 2*sqrt(2)*sinh(2*sqrt(2)*x) ). (End)

More terms from James Sellers, May 05 2000

More terms from Vincenzo Librandi, Mar 20 2015

A038762 a(n) = 6*a(n-1) - a(n-2) for n >= 2, with a(0)=3, a(1)=13.

Original entry on oeis.org

3, 13, 75, 437, 2547, 14845, 86523, 504293, 2939235, 17131117, 99847467, 581953685, 3391874643, 19769294173, 115223890395, 671574048197, 3914220398787, 22813748344525, 132968269668363, 774995869665653, 4517006948325555, 26327045820287677, 153445267973400507

Offset: 0

Barry E. Williams, May 03 2000

This gives part of the (increasingly sorted) positive solutions x to the Pell equation x^2 - 2*y^2 = +7. For the y solutions see A038761. The other part of solutions is found in A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			a(3)^2 - 2*A038761(3)^2 = 437^2 - 2*309^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
M. J. DeLeon, Pell's Equation and Pell Number Triples, Fib. Quart., 14(1976), pp. 456-460.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A038725, A038761, A077443, A101386, A253811.

I:=[3, 13]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

LinearRecurrence[{6,-1},{3,13},40] (* Vincenzo Librandi, Nov 16 2011 *)

x='x+O('x^30); Vec((3-5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

a(n) = sqrt(2*(A038761(n))^2+7).
a(n) = (13*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-3*((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = A077443(2n) = A038725(n)+A038725(n+1).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(n-1)+(1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Apr 20 2008
G.f.: (3-5*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008, corrected by R. J. Mathar, Nov 06 2011
a(n) = -5*A001109(n) +3*A001109(n+1). - R. J. Mathar, Nov 06 2011
a(n) = rational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only one part of the positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
E.g.f.: exp(3*x)*(3*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from James Sellers, May 04 2000

Unspecific Pell comment replaced by Wolfdieter Lang, Feb 05 2015

A077446 Numbers k such that 2*k^2 + 14 is a square.

Original entry on oeis.org

1, 5, 11, 31, 65, 181, 379, 1055, 2209, 6149, 12875, 35839, 75041, 208885, 437371, 1217471, 2549185, 7095941, 14857739, 41358175, 86597249, 241053109, 504725755, 1404960479, 2941757281, 8188709765, 17145817931, 47727298111

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "2*n^2 + 14 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = (7+n^2)/2. - Ctibor O. Zizka, Nov 09 2009
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = a(n+1), n >= 0, of the Pell equation x^2 - 2*y^2 = -7. For the corresponding y-solutions see y(n) = 2*A006452(n+2) = A077447(n+1)/2. This implies that X^2 - 2*Y^2 = 14 has the general solutions (X(n),Y(n)) = (2*y(n), x(n)). See the first comment above.
For the positive first class solutions see (A054490(n), 2*A038723(n)) and for the second class solutions (A255236(n), 2*A038725(n+1)). (End)
For n > 0, a(n) is the n-th almost Lucas-balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: (A077447(3))^2 - 2*a(3)^2 = 16^2 - 2*11^2  = 14;
a(3)^2 - 2*(2*A006452(3+1))^2 = 11^2 - 2*(2*4)^2 = -7. - _Wolfdieter Lang_, Feb 26 2015

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Michael De Vlieger, Table of n, a(n) for n = 1..2612
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
J. J. O'Connor and E. F. Robertson, Pell's Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001109, A006452, A038723, A038725, A049310, A054490, A077447, A155765, A156035, A156649, A255236.

LinearRecurrence[{0,6,0,-1},{1,5,11,31},50] (* Sture Sjöstedt, Oct 08 2012 *)

2*(a(n))^2 + 14 = (A077447(n))^2.
Lim. n-> Inf. a(n)/a(n-2) = 5.8284271247461... = 3 + 2*sqrt(2) = A156035 = RG (Great Ratio).
Lim. k-> Inf. a(2*k+1)/a(2*k) = 2.09383632135605... = (9 + 4*sqrt(2))/7 = A156649 = R1 (Ratio 1).
Lim. k -> Inf. a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2 (Ratio 2); RG = R1*R2.
a(2*k-1) = [ 2*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] - [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] + [(3+2*Sqrt(2))^(n-2) - (3-2*Sqrt(2))^(n-2)] ] / (4*Sqrt(2)) a(2*k) = [ 5*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] + [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] ] / (4*Sqrt(2)).
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1+x)*(x^2+4*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=5, a(3)=11, a(4)=31. - Sture Sjöstedt, Oct 08 2012
Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015
E.g.f.: 1 - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x). - Stefano Spezia, Nov 26 2022
a(n) = a(n-1) + 2*A217975(n-1) + A123335(n-2) - A123335(n-3) for n > 1 and with A123335(-1) = 1. - Vladimir Pletser, Aug 30 2025

A100525 Bisection of A048654.

Original entry on oeis.org

4, 22, 128, 746, 4348, 25342, 147704, 860882, 5017588, 29244646, 170450288, 993457082, 5790292204, 33748296142, 196699484648, 1146448611746, 6681992185828, 38945504503222, 226991034833504, 1323000704497802, 7711013192153308, 44943078448422046

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 24 2004

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A001109, A038761, A048654.

I:=[4,22,128]; [n le 3 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 13 2015

CoefficientList[Series[(4-2x)/(1-6x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Oct 13 2015 *)
LinearRecurrence[{6,-1},{4,22},30] (* Harvey P. Dale, Mar 25 2016 *)

Vec((4-2*x)/(1-6*x+x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

[2*(2*chebyshev_U(n,3) - chebyshev_U(n-1,3)) for n in (0..30)] # G. C. Greubel, Jun 28 2022

G.f.: 2*(2-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(0)=4, a(1)=22, a(n) = 6*a(n-1) - a(n-2) for n>1. - Philippe Deléham, Sep 19 2009
a(n) = 2*A038725(n+1). - R. J. Mathar, Sep 27 2014
a(n) = ( (5 + 4*sqrt(2))*(3 + 2*sqrt(2))^n - (5 - 4*sqrt(2))*(3 - 2*sqrt(2))^n )/(2*sqrt(2)). - Colin Barker, Oct 13 2015
From G. C. Greubel, Jun 28 2022: (Start)
a(n) = 2*( 2*ChebyshevU(n, 3) - ChenyshevU(n-1, 3) ).
E.g.f.: 2*exp(3*x)*( 2*cosh(2*sqrt(2)*x) + (5/(2*sqrt(2)))*sinh(2*sqrt(2)*x) ). (End)

A255236 All positive solutions x of the second class of the Pell equation x^2 - 2*y^2 = -7.

Original entry on oeis.org

5, 31, 181, 1055, 6149, 35839, 208885, 1217471, 7095941, 41358175, 241053109, 1404960479, 8188709765, 47727298111, 278175078901, 1621323175295, 9449763972869, 55077260661919, 321013799998645, 1871005539329951, 10905019435981061, 63559111076556415

Offset: 0

Wolfdieter Lang, Feb 26 2015

For the corresponding y = y2 terms see 2*A038725(n+1).
The Pell equation x^2 - 2*y^2 = 7 has two classes of solutions. See, e.g., the Nagell reference and comments under A254938 and A255233. Here the positive solutions based on the fundamental solution (5, 4) (the second largest positive solution) are considered.
The positive solutions of the first class are given in (A054490(n), 2*A038723(n)), n >= 0.
The combined solutions of both classes are given in (A077446, 4*A077447).
The solutions (x(n), y(n)) of x^2 - 2*y^2 = -7 translate to the solutions (X(n), Y(n)) = (2*y(n) , x(n)) of the Pell equation X^2 - 2*Y^2 = 14.

			n = 2: 181^2 - 2*(2*64)^2  = -7; (4*64)^2 - 2*181^2 = 14.
n = 2: 2*53 + 75 = 181. - _Wolfdieter Lang_, Mar 19 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A038725, A054490, A038723, A077446, 4*A077447, A254938, A255233.

I:=[5,31]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

CoefficientList[Series[(5 + x) / (1 - 6 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)

Vec((5 + x)/(1 - 6*x + x^2) + O(x^30)) \\ Michel Marcus, Mar 20 2015

a(n) = 5*S(n, 6) + S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310), with S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1).
G.f.: (5 + x)/(1 - 6*x + x^2).
a(n) = 6*a(n-1) - a(n-2), n >= 2, with a(-1) = -1 and a(0) = 5.
a(n) = 2*A038761(n) + A038762(n), n >= 0. See the Mar 19 comment on A054490. - Wolfdieter Lang, Mar 19 2015
a(n) = ((3-2*sqrt(2))^n*(-8+5*sqrt(2)) + (3+2*sqrt(2))^n*(8+5*sqrt(2))) / (2*sqrt(2)). - Colin Barker, Oct 13 2015

A207606 Triangle of coefficients of polynomials u(n,x) jointly generated with A207607; see the Formula section.

Original entry on oeis.org

1, 2, 3, 2, 4, 7, 2, 5, 16, 11, 2, 6, 30, 36, 15, 2, 7, 50, 91, 64, 19, 2, 8, 77, 196, 204, 100, 23, 2, 9, 112, 378, 540, 385, 144, 27, 2, 10, 156, 672, 1254, 1210, 650, 196, 31, 2, 11, 210, 1122, 2640, 3289, 2366, 1015, 256, 35, 2, 12, 275, 1782, 5148, 8008

Offset: 1

Clark Kimberling, Feb 19 2012

As triangle T(n,k) with 0 <= k <= n, it is (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 03 2012

			First five rows:
  1;
  2;
  3,  2;
  4,  7,  2;
  5, 16, 11,  2;
Triangle (2, -1/2, 1/2, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, ...), 0 <= k <= n, begins:
  1;
  2,   0;
  3,   2,   0;
  4,   7,   2,   0;
  5,  16,  11,   2,   0;
  6,  30,  36,  15,   2,   0;
  7,  50,  91,  64,  19,   2,   0;
  8,  77, 196, 204, 100,  23,   2,   0;

G. C. Greubel, Rows n = 1..101 of the triangle, flattened

Cf. A207607.

T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=0 then n+2
    elif k=n then 2
    else 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k)
      fi; end:
1, seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Mar 15 2020

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207606 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A207607 *)
(* Second program *)
T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, n+2, If[k==n, 2, 2*T[n-1, k] - T[n-2, k] + T[n-1, k-1] ]]]; Join[{1}, Table[T[n, k], {n, 0, 10}, {k, 0, n}]]//Flatten (* G. C. Greubel, Mar 15 2020 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else x*u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(u(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==1): return n+1
    elif (k==n): return 2
    else: return 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k)
[1]+[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Mar 15 2020

u(n,x) = u(n-1,x) + v(n-1,x), v(n,x) = x*u(n-1,x) + (x+1)v(n-1,x), where u(1,x)=1, v(1,x)=1.
As triangle T(n,k) with 0 <= k <= n: g.f.: (1-y*x)/(1-(2+y)*x+x^2). - Philippe Deléham, Mar 03 2012
As triangle T(n,k) with 0 <= k <= n: Sum_{k=0..n} T(n,k)*x^k = A132677(n), A000034(n)*A057077(n), A057079(n), A000027(n+1), A001519(n+1), A001075(n), A002310(n), A038725(n), A172968(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Mar 03 2012
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Mar 03 2012
T(n,k) = C(n+k-1,2*k+1) + 2*C(n+k-1,2*k), where C is binomial. - Yuchun Ji, May 23 2019
T(n,k) = T(n-1,k) + A207607(n-1,k). - Yuchun Ji, May 28 2019

cp's OEIS Frontend

A117719 a(2n) = A001653(n) (Numbers n such that 2*n^2 - 1 is a square), a(2n+1) = A038725(n+1).

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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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A156035 Decimal expansion of 3 + 2*sqrt(2).

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A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

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A054490 Expansion of (1+5*x)/(1-6*x+x^2).

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A054490 Expansion of (1+5x)/(1-6x+x^2).