cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A114619 a(n) = 2*A079291(n) (twice squares of Pell numbers).

Original entry on oeis.org

0, 2, 8, 50, 288, 1682, 9800, 57122, 332928, 1940450, 11309768, 65918162, 384199200, 2239277042, 13051463048, 76069501250, 443365544448, 2584123765442, 15061377048200, 87784138523762, 511643454094368
Offset: 0

Views

Author

Creighton Dement, Feb 17 2006

Keywords

Links

Crossrefs

Cf. A000129, A001109, A001333, A001542, A079291.
Cf. also A090390, A108475, A114619, A116484.

Programs

  • Magma
    [n le 3 select 2*(n-1)^2 else 5*Self(n-1) +5*Self(n-2) -Self(n-3): n in [1..31]]; // G. C. Greubel, Aug 18 2022
  • Mathematica
    2*Fibonacci[Range[0, 30], 2]^2 (* G. C. Greubel, Aug 18 2022 *)
  • SageMath
    [2*lucas_number1(n,2,-1)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

Formula

a(n) = 2*A000129(n)^2.
G.f.: 2*x*(1-x)/((1+x)*(1-6*x+x^2)).
a(n) = A001333(n)^2 - (-1)^n. - Antonio Pane (apane1(AT)spc.edu), Dec 15 2007

Extensions

Entry revised by N. J. A. Sloane, Mar 15 2024

A108477 A symmetric number triangle based on 2^n.

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 41, 41, 1, 1, 85, 321, 85, 1, 1, 145, 1289, 1289, 145, 1, 1, 221, 3649, 8989, 3649, 221, 1, 1, 313, 8361, 40081, 40081, 8361, 313, 1, 1, 421, 16641, 134245, 265729, 134245, 16641, 421, 1, 1, 545, 29961, 369305, 1256465, 1256465, 369305
Offset: 0

Views

Author

Paul Barry, Jun 04 2005

Keywords

Comments

Row sums are A108475. Diagonal sums are A108478.

Examples

			Rows begin
1;
1,1;
1,13,1;
1,41,41,1;
1,85,321,85,1;
1,145,1289,1289,145,1;

Formula

Number triangle T(n, k)=if(k<=n, sum{j=0..n, C(2(n-k), j)C(2k, j)2^j}, 0)

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

Views

Author

Creighton Dement, Feb 17 2006

Keywords

Comments

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

Links

Crossrefs

Cf. A116484, A001109, A108475, A090390.
Cf. A000129.

Programs

  • Mathematica
    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 *)

Formula

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
Showing 1-3 of 3 results.

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