A113224 -id:A113224 - cp's OEIS frontend

A113281 Self-convolution equals A113224.

1, 1, 3, 5, 11, 23, 51, 113, 255, 579, 1325, 3047, 7039, 16319, 37951, 88489, 206799, 484255, 1135969, 2668951, 6279509, 14793169, 34889553, 82372723, 194664283, 460436551, 1089943229, 2582033519, 6120967411, 14519686915, 34463104999

Offset: 0

Paul D. Hanna, Oct 22 2005

nonn

Convolution of bisections A113283 and A113284 yields A082639, which is a bisection of the self-convolution of this sequence. Terms of the logarithmic derivative A113282 are related to the self-convolution A113224 by: A113282(2*n) = A113224(2*n), A113282(4*n+1) = 3 + A113224(4*n+1), A113282(4*n+3) = 1 + A113224(4*n+3).

Cf. A113282, A113283, A113284.

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

a(n) ~ (1 + sqrt(2))^(n + 1/2) / sqrt(2*Pi*n). - Vaclav Kotesovec, Oct 18 2020

A002315 NSW numbers: a(n) = 6a(n-1) - a(n-2); also a(n)^2 - 2b(n)^2 = -1 with b(n) = A001653(n+1).

Original entry on oeis.org

1, 7, 41, 239, 1393, 8119, 47321, 275807, 1607521, 9369319, 54608393, 318281039, 1855077841, 10812186007, 63018038201, 367296043199, 2140758220993, 12477253282759, 72722761475561, 423859315570607, 2470433131948081, 14398739476117879, 83922003724759193

Offset: 0

N. J. A. Sloane

Named after the Newman-Shanks-Williams reference.
Also numbers k such that A125650(3*k^2) is an odd perfect square. Such numbers 3*k^2 form a bisection of A125651. - Alexander Adamchuk, Nov 30 2006
For positive n, a(n) corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs). - Lekraj Beedassy, Feb 06 2007
Also numbers m such that m^2 is a centered 16-gonal number; or a number of the form 8k(k+1)+1, where k = A053141(m) = {0, 2, 14, 84, 492, 2870, ...}. - Alexander Adamchuk, Apr 21 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=A001653. - Clark Kimberling, Aug 27 2008
The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators=A075870, denominators=A002315. - Clark Kimberling, Aug 27 2008
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->oo} a(n) = x*(k*x+1)^n, k = (a(1)-3), x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Numbers k such that (ceiling(sqrt(k*k/2)))^2 = (1+k*k)/2. - Ctibor O. Zizka, Nov 09 2009
A001109(n)/a(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4. - Gary Detlefs, Nov 25 2009
The values 2(a(n)^2+1) are all perfect squares, whose square root is given by A075870. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010
a(n) represents all positive integers K for which 2(K^2+1) is a perfect square. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(8)'s along the main diagonal, and i's along the superdiagonal and subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
Integers k such that A000217(k-2) + A000217(k-1) + A000217(k) + A000217(k+1) is a square (cf. A202391). - Max Alekseyev, Dec 19 2011
Integer square roots of floor(k^2/2 - 1) or A047838. - Richard R. Forberg, Aug 01 2013
Remark: x^2 - 2*y^2 = +2*k^2, with positive k, and X^2 - 2*Y^2 = +2 reduce to the present Pell equation a^2 - 2*b^2 = -1 with x = k*X = 2*k*b and y = k*Y = k*a. (After a proposed solution for k = 3 by Alexander Samokrutov.) - Wolfdieter Lang, Aug 21 2015
If p is an odd prime, a((p-1)/2) == 1 (mod p). - Altug Alkan, Mar 17 2016
a(n)^2 + 1 = 2*b(n)^2, with b(n) = A001653(n), is the necessary and sufficient condition for a(n) to be a number k for which the diagonal of a 1 X k rectangle is an integer multiple of the diagonal of a 1 X 1 square. If squares are laid out thus along one diagonal of a horizontal 1 X a(n) rectangle, from the lower left corner to the upper right, the number of squares is b(n), and there will always be a square whose top corner lies exactly within the top edge of the rectangle. Numbering the squares 1 to b(n) from left to right, the number of the one square that has a corner in the top edge of the rectangle is c(n) = (2*b(n) - a(n) + 1)/2, which is A055997(n). The horizontal component of the corner of the square in the edge of the rectangle is also an integer, namely d(n) = a(n) - b(n), which is A001542(n). - David Pasino, Jun 30 2016
(a(n)^2)-th triangular number is a square; a(n)^2 = A008843(n) is a subsequence of A001108. - Jaroslav Krizek, Aug 05 2016
a(n-1)/A001653(n) is the closest rational approximation of sqrt(2) with a numerator not larger than a(n-1). 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. a(n-1)/A001653(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Consider the quadrant of a circle with center (0,0) bounded by the positive x and y axes. Now consider, as the start of a series, the circle contained within this quadrant which kisses both axes and the outer bounding circle. Consider further a succession of circles, each kissing the x-axis, the outer bounding circle, and the previous circle in the series. See Holmes link. The center of the n-th circle in this series is ((A001653(n)*sqrt(2)-1)/a(n-1), (A001653(n)*sqrt(2)-1)/a(n-1)^2), the y-coordinate also being its radius. It follows that a(n-1) is the cotangent of the angle subtended at point (0,0) by the center of the n-th circle in the series with respect to the x-axis. - Graham Holmes, Aug 31 2019
There is a link between the two sequences present at the numerator and at the denominator of the fractions that give the coordinates of the center of the kissing circles. A001653 is the sequence of numbers k such that 2*k^2 - 1 is a square, and here, we have 2*A001653(n)^2 - 1 = a(n-1)^2. - Bernard Schott, Sep 02 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i+4*n+2) - G(i))/(2*G(i+2*n+1)) as long as G(i+2*n+1) != 0. - Klaus Purath, Mar 25 2021
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=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022
3*a(n-1) is the n-th almost Lucas-cobalancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022
In Moret-Blanc (1881) on page 259 some solution of m^2 - 2n^2 = -1 are listed. The values of m give this sequence, and the values of n give A001653. - Michael Somos, Oct 25 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 6xy + y^2 = 8 = A028884(1). In general, the following applies to all sequences (t) satisfying t(i) = 6t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 6xy + y^2 = A028884(t(1)-6). This includes and interprets the Feb 04 2014 comment on A001541 by Colin Barker as well as the Mar 17 2021 comment on A054489 by John O. Oladokun and the Sep 28 2008 formula on A038723 by Michael Somos. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028884(t(1)-6) always applies.
If (t) is a sequence satisfying t(k) = 7t(k-1) - 7t(k-2) + t(k-3) or t(k) = 6t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) - t(k))/(t(k+n+1) - t(k+n)) always applies, as long as t(k+n+1) - t(k+n) != 0 for integer k and n >= 0. (End)

			G.f. = 1 + 7*x + 41*x^2 + 239*x^3 + 1393*x^4 + 8119*x^5 + 17321*x^6 + ... - _Michael Somos_, Jun 26 2022

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)
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P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

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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).

Bisection of A001333. Cf. A001109, A001653. A065513(n)=a(n)-1.
First differences of A001108 and A055997. Bisection of A084068 and A088014. Cf. A077444.
Cf. A125650, A125651, A125652.
Row sums of unsigned triangle A127675.
Cf. A053141, A075870. Cf. A000045, A002878, A004146, A026003, A100047, A119915, A192425, A088165 (prime subsequence), A057084 (binomial transform), A108051 (inverse binomial transform).
See comments in A301383.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.
Cf. A000129, A001653, A003499, A358682, A002203.

a002315 n = a002315_list !! n
a002315_list = 1 : 7 : zipWith (-) (map (* 6) (tail a002315_list)) a002315_list
-- Reinhard Zumkeller, Jan 10 2012

I:=[1,7]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A002315 := proc(n)
    option remember;
    if n = 0 then
        1 ;
    elif n = 1 then
        7;
    else
        6*procname(n-1)-procname(n-2) ;
    end if;
end proc: # Zerinvary Lajos, Jul 26 2006, modified R. J. Mathar, Apr 30 2017
a:=n->abs(Im(simplify(ChebyshevT(2*n+1,I)))):seq(a(n),n=0..20); # Leonid Bedratyuk, Dec 17 2017
# third Maple program:
a:= n-> (<<0|1>, <-1|6>>^n. <<1, 7>>)[1, 1]:
seq(a(n), n=0..22);  # Alois P. Heinz, Aug 25 2024

a[0] = 1; a[1] = 7; a[n_] := a[n] = 6a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 20}] (* Robert G. Wilson v, Jun 09 2004 *)
Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {1,7},20]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
Table[ If[n>0, a=b; b=c; c=6b-a, b=-1; c=1], {n, 0, 20}] (* Jean-François Alcover, Oct 19 2012 *)
LinearRecurrence[{6, -1}, {1, 7}, 20] (* Bruno Berselli, Apr 03 2018 *)
a[ n_] := -I*(-1)^n*ChebyshevT[2*n + 1, I]; (* Michael Somos, Jun 26 2022 *)

{a(n) = subst(poltchebi(abs(n+1)) - poltchebi(abs(n)), x, 3)/2};

{a(n) = if(n<0, -a(-1-n), polsym(x^2-2*x-1, 2*n+1)[2*n+2]/2)};

{a(n) = my(w=3+quadgen(32)); imag((1+w)*w^n)};

for (i=1,10000,if(Mod(sigma(i^2+1,2),2)==1,print1(i,",")))

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

a(n) = (1/2)*((1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1)).
a(n) = A001109(n)+A001109(n+1).
a(n) = (1+sqrt(2))/2*(3+sqrt(8))^n+(1-sqrt(2))/2*(3-sqrt(8))^n. - Ralf Stephan, Feb 23 2003
a(n) = sqrt(2*(A001653(n+1))^2-1), n >= 0. [Pell equation a(n)^2 - 2*Pell(2*n+1)^2 = -1. - Wolfdieter Lang, Jul 11 2018]
G.f.: (1 + x)/(1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(n, 6)+S(n-1, 6) = S(2*n, sqrt(8)), S(n, x) = U(n, x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n, 6)= A001109(n+1).
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n+1). - 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 06 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -8) = a(n). - Benoit Cloitre, Nov 10 2002
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)-b^((2n+1)/2))/2. a(n) = A077444(n)/2. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
a(n) = Sum_{k=0..n} 2^k*binomial(2*n+1, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 08 2003
Same as: i such that sigma(i^2+1, 2) mod 2 = 1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
a(n) = L(n, -6)*(-1)^n, where L is defined as in A108299; see also A001653 for L(n, +6). - Reinhard Zumkeller, Jun 01 2005
a(n) = A001652(n)+A046090(n); e.g., 239=119+120. - Charlie Marion, Nov 20 2003
A001541(n)*a(n+k) = A001652(2n+k) + A001652(k)+1; e.g., 3*1393 = 4069 + 119 + 1; for k > 0, A001541(n+k)*a(n) = A001652(2n+k) - A001652(k-1); e.g., 99*7 = 696 - 3. - Charlie Marion, Mar 17 2003
a(n) = Jacobi_P(n,1/2,-1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
P_{2n}+P_{2n+1} where P_i are the Pell numbers (A000129). Also the square root of the partial sums of Pell numbers: P_{2n}+P_{2n+1} = sqrt(Sum_{i=0..4n+1} P_i) (Santana and Diaz-Barrero, 2006). - David Eppstein, Jan 28 2007
a(n) = 2*A001652(n) + 1 = 2*A046729(n) + (-1)^n. - Lekraj Beedassy, Feb 06 2007
a(n) = sqrt(A001108(2*n+1)). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
a(n) = sqrt(8*A053141(n)*(A053141(n) + 1) + 1). - Alexander Adamchuk, Apr 21 2007
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8), a(1)=1. - Richard Choulet, Sep 18 2007
a(n) = A001333(2*n+1). - Ctibor O. Zizka, Aug 13 2008
a(n) = third binomial transform of 1, 4, 8, 32, 64, 256, 512, ... . - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n) = (-1)^(n-1)*(1/sqrt(-1))*cos((2*n - 1)*arcsin(sqrt(2)). - Artur Jasinski, Feb 17 2010 *WRONG*
a(n+k) = A001541(k)*a(n) + 4*A001109(k)*A001653(n); e.g., 8119 = 17*239 + 4*6*169. - Charlie Marion, Feb 04 2011
In general, a(n+k) = A001541(k)*a(n)) + sqrt(A001108(2k)*(a(n)^2+1)). See Sep 18 2007 entry above. - Charlie Marion, Dec 07 2011
a(n) = floor((1+sqrt(2))^(2n+1))/2. - Thomas Ordowski, Jun 12 2012
(a(2n-1) + a(2n) + 8)/(8*a(n)) = A001653(n). - Ignacio Larrosa Cañestro, Jan 02 2015
(a(2n) + a(2n-1))/a(n) = 2*sqrt(2)*( (1 + sqrt(2))^(4*n) - (1 - sqrt(2))^(4*n))/((1 + sqrt(2))^(2*n+1) + (1 - sqrt(2))^(2*n+1)). [This was my solution to problem 5325, School Science and Mathematics 114 (No. 8, Dec 2014).] - Henry Ricardo, Feb 05 2015
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 7, 0, 41, 0, 239, 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 = -4, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047.
b(n) = 1/2*((-1)^n - 1)*Pell(n) + 1/2*(1 + (-1)^(n+1))*Pell(n+1). The o.g.f. is x*(1 + x^2)/(1 - 6*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*A026003(n-1)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*A026003(n-1)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*Pell(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*Pell(n)*(-x)^n.
Exp( Sum_{n >= 1} 8*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 8*A119915(n)*x^n.
Exp( Sum_{n >= 1} (-8)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 8*A119915(n)*(-x)^n. Cf. A002878, A004146, A113224, and A192425. (End)
E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + cosh(2*sqrt(2)*x))*exp(3*x). - Ilya Gutkovskiy, Jun 30 2016
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^ceiling(k/2). - David Pasino, Jul 09 2016
a(n) = A001541(n) + 2*A001542(n). - A.H.M. Smeets, May 28 2017
a(n+1) = 3*a(n) + 4*b(n), b(n+1) = 2*a(n) + 3*b(n), with b(n)=A001653(n). - Zak Seidov, Jul 13 2017
a(n) = |Im(T(2n-1,i))|, i=sqrt(-1), T(n,x) is the Chebyshev polynomial of the first kind, Im is the imaginary part of a complex number, || is the absolute value. - Leonid Bedratyuk, Dec 17 2017
a(n) = sinh((2*n + 1)*arcsinh(1)). - Bruno Berselli, Apr 03 2018
a(n) = 5*a(n-1) + A003499(n-1), a(0) = 1. - Ivan N. Ianakiev, Aug 09 2019
From Klaus Purath, Mar 25 2021: (Start)
a(n) = A046090(2*n)/A001541(n).
a(n+1)*a(n+2) = a(n)*a(n+3) + 48.
a(n)^2 + a(n+1)^2 = 6*a(n)*a(n+1) + 8.
a(n+1)^2 = a(n)*a(n+2) + 8.
a(n+1) = a(n) + 2*A001541(n+1).
a(n) = 2*A046090(n) - 1. (End)
3*a(n-1) = sqrt(8*b(n)^2 + 8*b(n) - 7), where b(n) = A358682(n). - Stefano Spezia, Nov 26 2022
a(n) = -(-1)^n - 2 + Sum_{i=0..n} A002203(i)^2. - Adam Mohamed, Aug 22 2024
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 3), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
For arbitrary x, a(n+x)^2 - 6*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 8 with a(n) := (1/2)*((1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1)) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(2) * A001542(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/8 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A081554(n) + 1/A081554(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 2*(1 - 1/A055997(k+2))). (End)

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

Original entry on oeis.org

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) =  x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
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).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
Alex Fink, Richard K. Guy and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 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 (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 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 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A097783 Chebyshev polynomials S(n,11) + S(n-1,11) with Diophantine property.

Original entry on oeis.org

1, 12, 131, 1429, 15588, 170039, 1854841, 20233212, 220710491, 2407582189, 26262693588, 286482047279, 3125039826481, 34088956044012, 371853476657651, 4056299287190149, 44247438682433988, 482665526219583719, 5265073349732986921, 57433141320843272412

Offset: 0

Wolfdieter Lang, Aug 31 2004

All positive integer solutions of Pell equation (3*a(n))^2 - 13*b(n)^2 = -4 together with b(n)=A078922(n+1), n>=0.

			All positive solutions to the Pell equation x^2 - 13*y^2 = -4 are (3=3*1,1), (36=3*12,10), (393=3*131,109), (4287=3*1429,1189 ), ...

Colin Barker, Table of n, a(n) for n = 0..963
Andersen, K., Carbone, L. and Penta, D., Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
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 (11,-1).

Cf. S(n, 11) = A004190(n).

Cf. A000045, A002315, A004146, A006190, A006497, A100047, A113224, A192425.

I:=[1,12]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1 + x) / (1 - 11 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-11*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,11,1)-lucas_number2(n-1,11,1))/9 for n in range(1, 19)] # Zerinvary Lajos, Nov 10 2009

a(n) = S(n, 11) + S(n-1, 11) = S(2*n, sqrt(13)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x) = 0 = U(-1, x).
a(n) = (-2/3)*i*((-1)^n)*T(2*n+1, 3*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-11*x+x^2).
a(n) = L(n,-11)*(-1)^n, where L is defined as in A108299; see also A078922 for L(n,+11). - Reinhard Zumkeller, Jun 01 2005
a(n) = 11*a(n-1) - a(n-2) with a(0)=1 and a(1)=12. - Philippe Deléham, Nov 17 2008
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 12, 0, 131, 0, 1429, 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 = -9, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = 1/2*( (-1)^n - 1 )*F(n,3) + 1/3*( 1 + (-1)^(n+1) )*F(n+1,3), where F(n,x) is the n-th Fibonacci polynomial. The o.g.f. is x*(1 + x^2)/(1 - 11*x^2 + x^4).
Exp( Sum_{n >= 1} 6*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*x^n.
Exp( Sum_{n >= 1} (-6)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = A006497(2n+1)/3. - Adam Mohamed, Aug 22 2024

A114689 Expansion of (1 +4x -x^2)/((1-x^2)(1-2*x-x^2)); a Pellian-related sequence.

Original entry on oeis.org

1, 6, 13, 36, 85, 210, 505, 1224, 2953, 7134, 17221, 41580, 100381, 242346, 585073, 1412496, 3410065, 8232630, 19875325, 47983284, 115841893, 279667074, 675176041, 1630019160, 3935214361, 9500447886, 22936110133, 55372668156, 133681446445, 322735561050

Offset: 0

Creighton Dement, Feb 18 2006

Elements of odd index give match to A075848: 2*n^2 + 9 is a square. Generating floretion: - 1.5'i + 'j + 'k - .5i' + j' + k' + .5'ii' - .5'jj' - .5'kk' - 'ij' + 'ik' - 'ji' + .5'jk' + 2'ki' - .5'kj' + .5e

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A000129, A005409, A100828, A111954, A113224.

Cf. A114647, A114688, A114695, A114696, A114697.

I:=[1,6,13,36]; [n le 4 select I[n] else 2*Self(n-1) +2*Self(n-2) -2*Self(n-3) -Self(n-4): n in [1..31]]; // G. C. Greubel, May 24 2021

Table[3*Fibonacci[n+1, 2] -1-(-1)^n, {n, 0, 30}] (* G. C. Greubel, May 24 2021 *)

Vec((-1-4*x+x^2)/((1-x)*(x+1)*(x^2+2*x-1)) + O(x^30)) \\ Colin Barker, May 26 2016

[3*lucas_number1(n+1,2,-1) -(1+(-1)^n) for n in (0..30)] # G. C. Greubel, May 24 2021

G.f.: (1 +4*x -x^2)/((1-x)*(1+x)*(1-2*x-x^2)).
From Colin Barker, May 26 2016: (Start)
a(n) = (-1 - (-1)^n) + 3*((1+sqrt(2))^(1+n) - (1-sqrt(2))^(1+n))/(2*sqrt(2)).
a(n) = 2*a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) for n>3.
(End)
a(n) = 3*A000129(n+1) - (1 + (-1)^n). - G. C. Greubel, May 24 2021

A114697 Expansion of (1+x+x^2)/((1-x^2)(1-2x-x^2)); a Pellian-related sequence.

Original entry on oeis.org

1, 3, 9, 22, 55, 133, 323, 780, 1885, 4551, 10989, 26530, 64051, 154633, 373319, 901272, 2175865, 5253003, 12681873, 30616750, 73915375, 178447501, 430810379, 1040068260, 2510946901, 6061962063, 14634871029, 35331704122, 85298279275, 205928262673

Offset: 0

Creighton Dement, Feb 18 2006

Generating floretion: (- .5'j + .5'k - .5j' + .5k' + 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki')*('i + 'j + i').

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A000129, A002203, A005409, A100828, A111954, A113224.

Cf. A114647, A114688, A114689, A114695, A116699.

Table[(3*LucasL[n, 2] +10*Fibonacci[n, 2] -3 +(-1)^n)/4, {n,0,30}] (* G. C. Greubel, May 24 2021 *)

Vec((1+x+x^2)/((1-x^2)*(1-2*x-x^2)) + O(x^40)) \\ Colin Barker, Jun 24 2015

[(4*lucas_number1(n+2,2,-1) -2*lucas_number1(n+1,2,-1) -3 +(-1)^n)/4 for n in (0..30)] # G. C. Greubel, May 24 2021

a(n+2) - 2*a(n+1) + a(n) = A111955(n+2).
G.f.: (1+x+x^2)/((1-x)*(1+x)*(1-2*x-x^2)).
From Raphie Frank, Oct 01 2012: (Start)
a(2*n) = A216134(2*n+1).
a(2*n+1) = A006452(2*n+3)-1.
Lim_{n->infinity} a(n+1)/a(n) = A014176. (End)
a(n) = (2*A078343(n+2) - A010694(n))/4. - R. J. Mathar, Oct 02 2012
From Colin Barker, May 26 2016: (Start)
a(n) = ( 2*(-3 +(-1)^n) + (6-5*sqrt(2))*(1-sqrt(2))^n + (1+sqrt(2))^n*(6+5*sqrt(2)) )/8.
a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n>3. (End)
a(n) = (3*A002203(n) + 10*A000129(n) - 3 + (-1)^n)/4. - G. C. Greubel, May 24 2021

A129744 a(n) = -(u^n-1)*(v^n-1) with u = 1+sqrt(2), v = 1-sqrt(2).

Original entry on oeis.org

2, 4, 14, 32, 82, 196, 478, 1152, 2786, 6724, 16238, 39200, 94642, 228484, 551614, 1331712, 3215042, 7761796, 18738638, 45239072, 109216786, 263672644, 636562078, 1536796800, 3710155682, 8957108164, 21624372014, 52205852192

Offset: 1

N. J. A. Sloane, May 13 2007

Seiichi Manyama, Table of n, a(n) for n = 1..1000
G. Everest et al., Primes generated by recurrence sequences, Amer. Math. Monthly, 114 (No. 5, 2007), 417-431.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A002203, A000129, A001350, A113224.

u:=1+sqrt(2): v:=1-sqrt(2): a:=n->expand(-(u^n-1)*(v^n-1)): seq(a(n),n=1..33); # Emeric Deutsch, May 13 2007

Table[Simplify[ -((1 + Sqrt[2])^n - 1)*((1 - Sqrt[2])^n - 1)], {n, 1, 30}] (* Stefan Steinerberger, May 15 2007 *)

w = quadgen(8); vector(30, n, -((1+w)^n-1)*((1-w)^n-1)) \\ Michel Marcus, Mar 21 2015

Vec(2*x*(1+x^2)/((x^2+2*x-1)*(-1+x)*(1+x))+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2015

a(2n) = A002203(2n)-2. a(2n+1) = A002203(2n+1). - R. J. Mathar, corrected Dec 05 2007.
G.f.: 2*x*(1+x^2)/((x^2+2*x-1)*(-1+x)*(1+x)).
From Peter Bala, Mar 19 2015: (Start)
a(n) = -det(I - M^n) where I is the 2X2 identity matrix and M = [2, 1; 1, 0]. Cf. A001350.
a(n) = 2*A113224(n-1).
This is divisibility sequence, that is, if n | m then a(n) | a(m).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*Sum_{n >= 1} Pell(n) *x^n. (End)
a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n > 4. - Seiichi Manyama, Jun 07 2018

More terms from Emeric Deutsch and Stefan Steinerberger, May 13 2007

A113282 Logarithmic derivative of the g.f. of A113281.

Original entry on oeis.org

1, 5, 7, 17, 41, 101, 239, 577, 1393, 3365, 8119, 19601, 47321, 114245, 275807, 665857, 1607521, 3880901, 9369319, 22619537, 54608393, 131836325, 318281039, 768398401, 1855077841, 4478554085, 10812186007, 26102926097, 63018038201

Offset: 0

Paul D. Hanna, Oct 22 2005

nonn

Cf. A113281, A113283, A113284.

{a(n)=local(x=X+X*O(X^n)); polcoeff((1+x^2)/(1-x^2)/(1-2*x-x^2) + x*(3+x^2)/(1-x^4),n,X)}

G.f.: (1+x^2)/(1-x^2)/(1-2*x-x^2) + x*(3+x^2)/(1-x^4). a(2*n) = A113224(2*n), a(4*n+1) = 3 + A113224(4*n+1), a(4*n+3) = 1 + A113224(4*n+3), where A113224 is the self-convolution of A113281.

a(n) ~ (1 + sqrt(2))^(n+1) / 2. - Vaclav Kotesovec, Oct 18 2020

A114696 Expansion of (1+4x+x^2)/((1-x^2)(1-2*x-x^2)); a Pellian-related sequence.

Original entry on oeis.org

1, 6, 15, 40, 97, 238, 575, 1392, 3361, 8118, 19599, 47320, 114241, 275806, 665855, 1607520, 3880897, 9369318, 22619535, 54608392, 131836321, 318281038, 768398399, 1855077840, 4478554081, 10812186006, 26102926095, 63018038200, 152139002497, 367296043198

Offset: 0

Creighton Dement, Feb 18 2006

Elements of odd index give match to A065113: Sum of the squares of the n-th and the (n+1)st triangular numbers (A000217) is a perfect square.

Generating floretion: - 1.5'i + 'j + 'k - .5i' + j' + k' + .5'ii' - .5'jj' - .5'kk' - 'ij' + 'ik' - 'ji' + .5'jk' + 2'ki' - .5'kj' + .5e

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A000129, A002203, A005409, A100828, A111954, A113224.

Cf. A114647, A114688, A114689, A114695, A114697.

Q:= proc(n) option remember; # Q=A002203
    if n<2 then 2
  else 2*Q(n-1) + Q(n-2)
    fi; end:
seq((Q(n+2) -3 -(-1)^n)/2, n=0..40); # G. C. Greubel, May 24 2021

CoefficientList[Series[(1+4*x+x^2)/((1-x^2)*(1-2*x-x^2)), {x,0,30}], x] (* or *) LinearRecurrence[{2,2,-2,-1}, {1,6,15,40}, 30] (* Harvey P. Dale, Jan 23 2014 *)

Vec((1+4*x+x^2)/((1-x^2)*(1-2*x-x^2)) + O(x^30)) \\ Colin Barker, May 26 2016

[(lucas_number2(n+2,2,-1) -3 -(-1)^n)/2 for n in (0..30)] # G. C. Greubel, May 24 2021

G.f.: (1 +4*x +x^2)/((1-x)*(1+x)*(1-2*x-x^2)).
a(0)=1, a(1)=6, a(2)=15, a(3)=40, a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4). - Harvey P. Dale, Jan 23 2014
a(n) = (-3 - (-1)^n + (3-2*sqrt(2))*(1-sqrt(2))^n + (1+sqrt(2))^n*(3+2*sqrt(2)))/2. - Colin Barker, May 26 2016
From G. C. Greubel, May 24 2021: (Start)
a(n) = 3*A000129(n+1) + A000129(n) - (3 + (-1)^n)/2.
a(n) = (1/2)*(A002203(n+2) - 3 - (-1)^n). (End)

A113225 a(2n) = A011900(n), a(2n+1) = A001109(n+1).

Original entry on oeis.org

1, 1, 3, 6, 15, 35, 85, 204, 493, 1189, 2871, 6930, 16731, 40391, 97513, 235416, 568345, 1372105, 3312555, 7997214, 19306983, 46611179, 112529341, 271669860, 655869061, 1583407981, 3822685023, 9228778026, 22280241075, 53789260175

Offset: 0

Creighton Dement, Oct 18 2005

a(n+1) - a(n) = A097075(n+1), a(n) + a(n+1) = A024537(n+1), a(n+2) - a(n+1) - a(n) = A105635(n+1).
For n >= 1, a(n) is also the edge cover number and edge cut count of the n-Pell graph. - Eric W. Weisstein, Aug 01 2023
Also the independence number, Lovasz number, and Shannon capacity of the n-Pell graph. - Eric W. Weisstein, Aug 01 2023
Floretion Algebra Multiplication Program, FAMP Code: -2jbasejseq[B*C], B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'; C = + .5'i + .5i' + .5'ii' + .5e

C. Dement, Floretion Integer Sequences (work in progress).

Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See p. 19.
Eric Weisstein's World of Mathematics, Edge Cover Number.
Eric Weisstein's World of Mathematics, Edge Cut.
Eric Weisstein's World of Mathematics, Independence Number.
Eric Weisstein's World of Mathematics, Lovasz Number.
Eric Weisstein's World of Mathematics, Pell Graph.
Eric Weisstein's World of Mathematics, Shannon Capacity.
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A113224, A002315, A082639, A100828.

seq(iquo(fibonacci(n,2),1)-iquo(fibonacci(n,2),2),n=1..30); # Zerinvary Lajos, Apr 20 2008
with(combinat):seq(ceil(fibonacci(n,2)/2), n=1..30); # Zerinvary Lajos, Jan 12 2009

Ceiling[Fibonacci[Range[20], 2]/2]
Table[(1 + (-1)^n + 2 Fibonacci[n + 1, 2])/4, {n, 0, 20}] // Expand
CoefficientList[Series[-(-1 + x + x^2)/(1 - 2 x - 2 x^2 + 2 x^3 + x^4), {x, 0, 20}], x]
LinearRecurrence[{2, 2, -2, -1}, {1, 1, 3, 6}, 20]

{a(n)=local(y); if(n<0, 0, n++; y=x/(x^2+x-1)+x*O(x^n); polcoeff( y/(y^2-1), n))} /* Michael Somos, Sep 09 2006 */

G.f.: y/(y^2-1) where y=x/(x^2+x-1) if offset=1. - Michael Somos, Sep 09 2006
G.f.: (-1+x+x^2)/((1-x)*(x+1)*(x^2+2*x-1)).
Diagonal sums of A119468. - Paul Barry, May 21 2006
a(n) = (1 + (-1)^n + 2 A000129(n+1))/4. - Eric W. Weisstein, Aug 01 2023
a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4). - Eric W. Weisstein, Aug 01 2023

cp's OEIS Frontend

A113281 Self-convolution equals A113224.

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A002315 NSW numbers: a(n) = 6*a(n-1) - a(n-2); also a(n)^2 - 2*b(n)^2 = -1 with b(n) = A001653(n+1).

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A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

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A097783 Chebyshev polynomials S(n,11) + S(n-1,11) with Diophantine property.

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A114689 Expansion of (1 +4*x -x^2)/((1-x^2)*(1-2*x-x^2)); a Pellian-related sequence.

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A114697 Expansion of (1+x+x^2)/((1-x^2)*(1-2*x-x^2)); a Pellian-related sequence.

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A002315 NSW numbers: a(n) = 6a(n-1) - a(n-2); also a(n)^2 - 2b(n)^2 = -1 with b(n) = A001653(n+1).

A114689 Expansion of (1 +4x -x^2)/((1-x^2)(1-2*x-x^2)); a Pellian-related sequence.

A114697 Expansion of (1+x+x^2)/((1-x^2)(1-2x-x^2)); a Pellian-related sequence.

A114696 Expansion of (1+4x+x^2)/((1-x^2)(1-2*x-x^2)); a Pellian-related sequence.