A046729 -id:A046729 - cp's OEIS frontend

A089499 a(0)=0; a(1)=1; a(2n) = 4*Sum_{k=0..n} a(2k-1); a(2n+1) = a(2n) + a(2n-1).

0, 1, 4, 5, 24, 29, 140, 169, 816, 985, 4756, 5741, 27720, 33461, 161564, 195025, 941664, 1136689, 5488420, 6625109, 31988856, 38613965, 186444716, 225058681, 1086679440, 1311738121, 6333631924, 7645370045, 36915112104, 44560482149

Offset: 0

Charlie Marion, Nov 11 2003

1, 4, 5, 24, 29, 140, ...= numerators in convergents to (sqrt(8) - 2) = continued fraction [0; 1, 4, 1, 4, 1, 4, ...]; where sqrt(8) - 2 = 0.828427124... = the inradius of a right triangle with hypotenuse 6, legs sqrt(32) and 2. Denominators of convergents to [0; 1, 4, 1, 4, 1, 4, ...] = A041011 starting (1, 5, 6, 29, 35, ...). - Gary W. Adamson, Dec 22 2007

This is a strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, May 12 2014

J. L. Ramirez, F. Sirvent, A q-Analogue of the Bi-Periodic Fibonacci Sequence, J. Int. Seq. 19 (2016) # 16.4.6, t_n at a=4, b=1.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A041011.

Numerator[NestList[(4/(4+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[0;1;4;5])[1,1] \\ Charles R Greathouse IV, Nov 13 2015

For n > 0, a(n) = A001333(n) + A084068(n-1)*(-1)^n.
a(n)*a(n+1) = A046729(n).
a(2n+1) = A001653(n); a(2n) = A005319(n).
a(1) = 1, a(2n) = 4*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Given the 2 X 2 matrix X = [1, 4; 1, 5], [a(2n-1), a(2n)] = top row of X^n. The sequence starting (1, 4, 5, 24, 29, ...) = numerators in continued fraction [0; 1, 4, 1, 4, 1, 4, ...] = (sqrt(8) - 2) = 0.828427124... E.g., X^3 = [29, 140; 35, 169], where 29/35, 140/169 are convergents to (sqrt(8)-2). - Gary W. Adamson, Dec 22 2007
From R. J. Mathar, Jul 08 2009: (Start)
a(n) = A000129(n)*A000034(n+1).
a(n) = 6*a(n-2) - a(n-4).
G.f.: -x*(-1-4*x+x^2)/((x^2-2*x-1)*(x^2+2*x-1)). (End)
From Peter Bala, May 12 2014: (Start)
a(2*n + 1) = A041011(2*n + 1); a(2*n) = 4*A041011(2*n).
For n odd, a(n) = (alpha^n - beta^n)/(alpha - beta), and for n even, a(n) = 4*(alpha^n - beta^n)/(alpha^2 - beta^2), where alpha = 1 + sqrt(2) and beta = 1 - sqrt(2).
a(n) = Product_{j = 1..floor(n/2)} ( 4 + 4*cos^2(j*Pi/n) ) for n >= 1. (End)

Corrected by T. D. Noe, Nov 08 2006

Definition corrected by Jonathan Sondow, Jun 06 2014

A114620 2*A084158 (twice Pell triangles).

Original entry on oeis.org

0, 2, 10, 60, 348, 2030, 11830, 68952, 401880, 2342330, 13652098, 79570260, 463769460, 2703046502, 15754509550, 91824010800, 535189555248, 3119313320690, 18180690368890, 105964828892652, 617608282987020

Offset: 0

Creighton Dement, Feb 17 2006

Cross-referenced sequences A116484, A001109, A108475, A090390 are also generated by A*B given in the following FAMP code.
Floretion Algebra Multiplication Program, FAMP Code: 1jesleftseq[A*B] with A = - .5'i + .5'j - .5i' + .5j' + 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj' and B = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'
Related to the reciprocals of the differences between successive convergents of the continued fraction of sqrt(2) (i.e., 1, 2, -10, 60, -348, 2030, -11830, 68952, ...). 1/1 + 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 + ... = sqrt(2). 2, 10, 60, ... are products of the denominators of two successive convergents of sqrt(2) (e.g., 11830 = 70*169, cf. A000129 (Pell numbers)). - Gerald McGarvey, Feb 28 2006
a(n) is half of the even leg (b(n)) of the ordered Pythagorean triple (x(n), y(n)=x(n)+1, z(n)). In fact b(n) = x(n) + (1-(-1)^n)/2: x(0)=0, b(0)=0, a(0)=0; x(1)=3, b(1)=4, a(1)=2. - George F. Johnson, Aug 13 2012
Given a square shape composed of A001110(n+1) elements, thinking of it graphically as a sum of layers, each layer having an odd number of elements (all layers together being a sum of consecutive odd numbers), a(n) is the number of last layers that we have to subtract from the square to get a square of squares that is made of A002965(2*(n+1))^4 elements. - Daniel Poveda Parrilla, Jul 17 2016
Also numbers m such that 8*m^2 - 4*m + 1 or 8*m^2 + 4*m + 1 is a perfect square (square roots are then A001653). - Lamine Ngom, Jul 25 2023

Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A116484, A001109, A108475, A090390.

Cf. A000129.

Table[Fibonacci[n, 2] Fibonacci[n + 1, 2], {n, 0, 20}] (* or *)
LinearRecurrence[{5, 5, -1}, {0, 2, 10}, 21] (* or *)
CoefficientList[Series[2 x/((x + 1) (x^2 - 6 x + 1)), {x, 0, 20}], x] (* Michael De Vlieger, Jul 17 2016 *)

G.f.: 2*x/((x+1)*(x^2-6*x+1)).
From George F. Johnson, Aug 13 2012: (Start)
a(n) = ((sqrt(2) + 1)^(2*n+1) - (sqrt(2) - 1)^(2*n+1) - 2*(-1)^n)/8. - corrected by Ilya Gutkovskiy, Jul 18 2016
4*a(n)*(2*a(n) + (-1)^n) + 1 = A000129(2*n+1)^2 is a perfect square.
For n >= 0, a(n+1) = 3*a(n) + (-1)^n + sqrt(4*a(n)*(2*a(n) + (-1)^n) + 1).
For n >  0, a(n-1) = 3*a(n) + (-1)^n - sqrt(4*a(n)*(2*a(n) + (-1)^n) + 1).
a(n+1) = 6*a(n) - a(n-1) + 2*(-1)^n.
a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
For n > 0, a(n+1)*a(n-1) = a(n)*(a(n) + 2*(-1)^n).
a(n) = A046729(n)/2. (End)
a(n) = A000129(n)*A000129(n+1). - Philippe Deléham, Apr 10 2013
a(n) = A002965(2*(n+1))*(A002965(2*(n+1)+1) - A002965(2*(n+1))). - Daniel Poveda Parrilla, Jul 17 2016

A105058 Expansion of g.f. (1+8x-x^2)/((1+x)(1-6*x+x^2)).

Original entry on oeis.org

1, 13, 69, 409, 2377, 13861, 80781, 470833, 2744209, 15994429, 93222357, 543339721, 3166815961, 18457556053, 107578520349, 627013566049, 3654502875937, 21300003689581, 124145519261541, 723573111879673

Offset: 0

Creighton Dement, Apr 04 2005

A floretion-generated sequence relating the squares of the numerators of continued fraction convergents to sqrt(2) to the squares of the denominators of continued fraction convergents to sqrt(2) (Pell numbers).
Floretion Algebra Multiplication Program, FAMP Code:
1dia[J]tesseq[ - .5'j + .5'k - .5j' + .5k' - 2'ii' + 'jj' - 'kk' + .5'ij' + .5'ik' + .5'ji' + 'jk' + .5'ki' + 'kj' + e ]. Identity used: dia[I]tes + dia[J]tes + dia[K]tes = jes + fam + 3tes.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,5,-1)

Cf. A000129, A001109, A001333, A046729, A077444.

Cf. A077445, A079291, A082639, A084158, A090390.

[Evaluate(DicksonSecond(2*n+1, -1), 2) -(-1)^n: n in [0..30]]; // G. C. Greubel, Aug 21 2022

CoefficientList[ Series[(1+8x-x^2)/((1+x)(1-6x+x^2)), {x,0,30}], x] (* Robert G. Wilson v, Apr 06 2005 *)
LinearRecurrence[{5,5,-1}, {1,13,69}, 30] (* Harvey P. Dale, Jun 03 2017 *)

[lucas_number1(2*n+2,2,-1) -(-1)^n for n in (0..30)] # G. C. Greubel, Aug 21 2022

a(n) = 2 * A001109(n+1) - (-1)^n.
G.f.: G(0)/(1-3*x) - 1/(1+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
From G. C. Greubel, Aug 21 2022: (Start)
a(n) = A000129(2*n+2) - (-1)^n.
E.g.f.: exp(3*x)*( 2*cosh(2*sqrt(2)*x) + (3/sqrt(2))*sinh(2*sqrt(2)*x)) - exp(-x). (End)

A111587 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.

Original entry on oeis.org

1, 4, 9, 20, 49, 120, 289, 696, 1681, 4060, 9801, 23660, 57121, 137904, 332929, 803760, 1940449, 4684660, 11309769, 27304196, 65918161, 159140520, 384199201, 927538920, 2239277041, 5406093004, 13051463049, 31509019100, 76069501249

Offset: 0

Creighton Dement, Aug 08 2005

Let (b(n)) be the p-INVERT of (1,2,2,2,2,2,...) using p(S) = 1 - S^2; then
b(0) = 0 and b(n) = a(n-1) for n >= 1; see A292400. - Clark Kimberling, Sep 30 2017
Floretion Algebra Multiplication Program, FAMP Code: 2kbasekseq[J+G] with J = + j' + k' + 'ii' and G = + .5'ii' + .5'jj' + .5'kk' + .5e

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

Cf. A019898, A046729, A090390, A111588.

I:=[1,4,9,20]; [n le 4 select I[n] else 2*Self(n-1) +2*Self(n-3)+Self(n-4): n in [1..35]]; // Vincenzo Librandi, Oct 01 2017

LinearRecurrence[{2,0,2,1},{1,4,9,20},30] (* Harvey P. Dale, Jul 26 2011 *)
CoefficientList[Series[(x + 1)^2 / ((x^2 + 1) (1 - 2 x - x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Oct 01 2017 *)

a(2n) = A090390(n+1), a(2n+1) = A046729(n+1);
G.f.: (x+1)^2/((x^2+1)*(1-2*x-x^2)). [sign flipped by R. J. Mathar, Nov 10 2009]
a(n) = A057077(n+1)/2 - A001333(n+2)/2. - R. J. Mathar, Nov 10 2009

cp's OEIS Frontend

A089499 a(0)=0; a(1)=1; a(2n) = 4*Sum_{k=0..n} a(2k-1); a(2n+1) = a(2n) + a(2n-1).

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A114620 2*A084158 (twice Pell triangles).

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A105058 Expansion of g.f. (1+8*x-x^2)/((1+x)*(1-6*x+x^2)).

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A111587 a(n) = 2*a(n-1) + 2*a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.

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A105058 Expansion of g.f. (1+8x-x^2)/((1+x)(1-6*x+x^2)).

A111587 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.