A078343 -id:A078343 - cp's OEIS frontend

A135138 a(n) = 5a(n-2) + 2a(n-3).

0, 0, 1, 0, 5, 2, 25, 20, 129, 150, 685, 1008, 3725, 6410, 20641, 39500, 116025, 238782, 659125, 1425960, 3773189, 8448050, 21717865, 49786628, 125485425, 292368870, 727000381, 1712815200, 4219739645, 10018076762, 24524328625, 58529863100

Offset: 0

Paul Curtz, Feb 13 2008

a(n+2), n>=0, is the (5,2)-Padovan sequence p(5,2;n)with o.g.f. 1/(1-5*x^2-2*x^3). See A000931(n+3) ((1,1)-Padovan), and the W. Lang link given there, also for a combinatorial interpretation. - Wolfdieter Lang, Jun 28 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,5,2).

Cf. A135139.

a = {0, 0, 1}; Do[AppendTo[a, 5*a[[ -2]] + 2*a[[ -3]]], {40}]; a (* Stefan Steinerberger, Feb 15 2008 *)
LinearRecurrence[{0, 5, 2}, {0, 0, 1}, 100] (* G. C. Greubel, Sep 28 2016 *)

From R. J. Mathar, Feb 15 2008: (Start)
O.g.f.: -x^2 / ( (2*x+1)*(x^2+2*x-1) ).
a(n) = [(-2)^n + A078343(n)]/7. (End)

More terms from R. J. Mathar and Stefan Steinerberger, Feb 15 2008

A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 57, 296, 1537, 7981, 41442, 215191, 1117397, 5802176, 30128277, 156443561, 812346082, 4218173971, 21903215937, 113734253656, 590574484217, 3066606674741, 15923607857922, 82684645964351, 429346837679677, 2229418834362736, 11576441009493357

Offset: 0

Vincenzo Librandi, Aug 17 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A000032, A001077, A052924, A078343, A179237, A180148, A329723.

[ n le 2 select (n) else 5*Self(n-1)+Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Sep 12 2013

LinearRecurrence[{5, 1}, {1, 2}, 40] (* or *) Rest[CoefficientList[Series [x (1 - 3 x) / (1 - 5 x - x^2), {x, 0, 40}], x]] (* Harvey P. Dale, May 02 2011 *)

Vec((1-3*x)/(1-5*x-x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

a(n) = 5*a(n-1)+a(n-2) = A052918(n)-3*A052918(n-1).
G.f.: (1-3*x)/(1-5*x-x^2).
a(n) = A052918(n) + A015449(n). - R. J. Mathar, Jul 06 2012
a(n) = (2^(-1-n)*((5-sqrt(29))^n*(1+sqrt(29))+(-1+sqrt(29))*(5+sqrt(29))^n))/sqrt(29). - Colin Barker, Oct 13 2015
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*5^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*5^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = A052918(n) -3*A052918(n-1). - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) - 7*(-1)^n/a(n)) = 3/10, since 1/(a(n) - 7*(-1)^n/a(n)) = b(n) - b(n+1), where b(n) = (1/5) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 7*(-1)^n/a(n)) = 1/10, since 1/(a(n) - 7*(-1)^n/a(n)) = c(n) + c(n+1), where c(n) = (1/5) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

A220414 a(n) = 6*a(n-1) - a(n-2), with a(1) = 13, a(2) = 73.

Original entry on oeis.org

13, 73, 425, 2477, 14437, 84145, 490433, 2858453, 16660285, 97103257, 565959257, 3298652285, 19225954453, 112057074433, 653116492145, 3806641878437, 22186734778477, 129313766792425, 753695865976073, 4392861429064013, 25603472708408005, 149227974821384017

Offset: 1

José Luis Ramírez Ramírez, Dec 13 2012

a(n) is the area of the 4-generalized Fibonacci snowflake.
a(n) is the area of the 5-generalized Fibonacci snowflake, for n >= 2.
From Wolfdieter Lang, Feb 07 2015: (Start)
This sequence gives one part of the positive proper (sometimes called primitive) solutions y of the Pell equation x^2 - 2*y^2 = - 7^2 based on the fundamental solution (x0, y0) = (-1, 5). The corresponding x solutions are given in A254757.
The other part of the proper solutions are given in (A254758(n), A254759(n)) for n >= 0.
The improper positive solutions come from 7*(x(n), y(n)) with the positive proper solutions of the Pell equation x^2 - 2*y^2 = -1 given in (A001653(n-1), A002315(n)), for n >= 1. (End)
The terms of this sequence are hypotenuses of Pythagorean triangles whose difference between legs is equal to 7. - César Aguilera, Sep 29 2023

			From _Wolfdieter Lang_, Feb 07 2015: (Start)
Pell equation x^2 - 2*y^2 = -7^2 instance:
A254757(3)^2 - 2*a(3)^2 = 601^2 - 2*425^2 = -49. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
César Aguilera, Notes on Perfect Numbers, OSF Preprints, 2023, p. 19.
R. De Castro, J. Ramírez, and G. Rubiano, Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv:1212.1368 [cs.DM], 2012-2014.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001653, A171587, A078343, A254757, A049310, A254758, A254759.

I:=[13, 73]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 01 2013

with(orthopoly): a := n -> `if`(n=1,13,13*U(n-1,3)-5*U(n-2,3)):
seq(a(n),n=1..22); # (after Wolfdieter Lang) Peter Luschny, Feb 07 2015

t = {13, 73}; Do[AppendTo[t, 6*t[[-1]] - t[[-2]]], {30}]; t (* T. D. Noe, Dec 20 2012 *)
LinearRecurrence[{6,-1},{13,73},40] (* Harvey P. Dale, Jan 26 2013 *)

a(n) = A078343(n)^2 + A078343(n+1)^2 = A060569(2*n-1).
G.f.: (13-5*x)/(x^2-6*x+1). - Harvey P. Dale, Jan 26 2013
From Wolfdieter Lang, Feb 07 2015: (Start)
a(n) = 13*S(n-1, 6) - 5*S(n-2, 6), n >= 1, with Chebyshev's S-polynomials evaluated at x = 6 (see A049310).
a(n) = 6*a(n-1) - a(n-2), n >= 2, with a(0) = 5 and a(1) = 13.
a(n) = irrational part of z(n), where z(n) = (-1+5*sqrt(2))*(3+2*sqrt(2))^n, n >= 1. (End)

A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

Original entry on oeis.org

2, 2, 1, 3, 4, 8, 9, 19, 22, 46, 53, 111, 128, 268, 309, 647, 746, 1562, 1801, 3771, 4348, 9104, 10497, 21979, 25342, 53062, 61181, 128103, 147704, 309268, 356589, 746639, 860882, 1802546, 2078353, 4351731, 5017588, 10506008, 12113529, 25363747, 29244646, 61233502

Offset: 0

Raphie Frank, Dec 30 2015

This sequence gives all x in N | 2*x^2 - 7(-1)^x = y^2. The companion sequence to this sequence, giving y values, is A266505.
A266505(n)/a(n) converges to sqrt(2).
Alternatively, 1/4*(3*A002203(floor[n/2]) - A002203(n-(-1)^n)), where A002203 gives the Companion Pell numbers, or, in Lucas sequence notation, V_n(2, -1).
Alternatively, bisection of A266506.
Alternatively, A048654(n -1) and A078343(n + 1) interlaced.
Alternatively, A100525(n-1), A266507(n), A038761(n) and A253811(n) interlaced.
Let b(n) = (a(n) - a(n)(mod 2))/2, that is b(n) = {1, 1, 0, 1, 2, 4, 4, 9, 11, 23, 26, 55, 64, ...}. Then:
A006452(n) = {b(4n+0) U b(4n+1)} gives n in N such that n^2 - 1 is triangular;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that n^2 + n + 1 is triangular (indices of Sophie Germain triangular numbers);
A216162(n) = {b(4n+0) U b(4n+2) U b(4n+1) U b(4n+3)}, sequences A006452 and A216134 interlaced.

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

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[2,2,1,3]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

LinearRecurrence[{0, 2, 0, 1}, {2, 2, 1, 3}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(1 - x) (2 + 4 x + x^2)/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 41}] (* Michael De Vlieger, Dec 31 2015 *)

Vec((1-x)*(2+4*x+x^2)/(1-2*x^2-x^4) + O(x^50)) \\ Colin Barker, Dec 31 2015

a(n) = 1/sqrt(8)*(+sqrt(2)*(1+sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n - 3*(1-sqrt(2))^(floor(n/2)-(-1)^n) + sqrt(2)*(1-sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n + 3*(1+sqrt(2))^(floor(n/2)-(-1)^n)).
a(n) = 1/4*((3*((1+sqrt(2))^floor(n/2)+(1-sqrt(2))^floor(n/2))) - (-1)^n*((1+sqrt(2))^(floor(n/2)-(-1)^n)+(1-sqrt(2))^(floor(n/2)-(-1)^n))).
a(2n) = (+sqrt(2)*(1+sqrt(2))^(n-1) - 3 *(1-sqrt(2))^(n-1) + sqrt(2)*(1-sqrt(2))^(n-1) + 3*(1 + sqrt(2))^(n-1))/sqrt(8) = A048654(n -1).
a(2n) = 1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) - ((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))) = A048654(n -1).
a(2n + 1) = (-sqrt(2)*(1+sqrt(2))^(n+1) - 3 *(1-sqrt(2))^(n+1) - sqrt(2)*(1-sqrt(2))^(n+1) + 3*(1+sqrt(2))^(n+1))/sqrt(8) = A078343(n + 1).
a(2n + 1) =1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) + ((1+sqrt(2))^(n+1)+(1-sqrt(2))^(n+1))) =  A078343(n + 1).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A100525(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A266507(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038761(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A253811(n).
sqrt(2*a(n)^2 - 7(-1)^a(n))*sgn(2*n - 1) = A266505(n).
(a(2n + 1) + a(2n))/2 = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 1) - a(2n))/2 = A000129(n), where A000129 gives the Pell numbers.
(a(2n+2) + a(2n+1))*2 = A002203(n+2)
(a(2n+2) - a(2n+1))*2 = A002203(n-1).
G.f.: (1-x)*(2+4*x+x^2) / (1-2*x^2-x^4). - Colin Barker, Dec 31 2015

A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

Original entry on oeis.org

-1, 1, 3, 5, 5, 11, 13, 27, 31, 65, 75, 157, 181, 379, 437, 915, 1055, 2209, 2547, 5333, 6149, 12875, 14845, 31083, 35839, 75041, 86523, 181165, 208885, 437371, 504293, 1055907, 1217471, 2549185, 2939235, 6154277, 7095941, 14857739, 17131117, 35869755, 41358175, 86597249, 99847467

Offset: 0

Raphie Frank, Dec 30 2015

a(n)/A266504(n) converges to sqrt(2).
Alternatively, bisection of A266506.
Alternatively, A135532(n) and A048655(n) interlaced.
Alternatively, A255236(n-1), A054490(n), A038762(n) and A101386(n) interlaced.
Let b(n) = (a(n) - (a(n) mod 2))/2, that is b(n) = {-1, 0, 1, 2, 2, 5, 6, 13, 15, 32, 37, 78, 90, ...}. Then:
A006451(n) = {b(4n+0) U b(4n+1)} gives n in N such that triangular(n) + 1 is square;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that triangular(n) follows form n^2 + n + 1 (twice a triangular number + 1).

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

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[-1,1,3,5]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

a:=proc(n) option remember; if n=0 then -1 elif n=1 then 1 elif n=2 then 3 elif n=3 then 5 else 2*a(n-2)+a(n-4); fi; end:  seq(a(n), n=0..50); # Wesley Ivan Hurt, Jan 01 2016

LinearRecurrence[{0, 2, 0, 1}, {-1, 1, 3, 5}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(-1 + 3 x) (1 + x)^2/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 42}] (* Michael De Vlieger, Dec 31 2015 *)

my(x='x+O('x^40)); Vec((-1+3*x)*(1+x)^2/(1-2*x^2-x^4)) \\ G. C. Greubel, Jul 26 2018

G.f.: (-1 + 3*x)*(1 + x)^2/(1 - 2*x^2 - x^4).
a(n) = (-(1+sqrt(2))^floor(n/2)*(-1)^n - sqrt(8)*(1-sqrt(2))^floor(n/2) - (1-sqrt(2))^floor(n/2)*(-1)^n + sqrt(8)*(1+sqrt(2))^floor(n/2))/2.
a(n) = 3*(((1+sqrt(2))^floor(n/2)-(1-sqrt(2))^floor(n/2))/sqrt(8)) - (-1)^n*(((1+sqrt(2))^(floor(n/2)-(-1)^n)-(1-sqrt(2))^(floor(n/2)-(-1)^n))/sqrt(8)).
a(n) = (3*A000129(floor(n/2)) - A000129(n-(-1)^n)), where A000129 gives the Pell numbers.
a(n) = sqrt(2*A266504(n)^2 - 7*(-1)^A266504(n))*sgn(2*n-1), where A266504 gives all x in N such that 2*x^2 - 7*(-1)^x = y^2. This sequence gives associated y values.
a(2n) = (-(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n - (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n) = A135532(n).
a(2n) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) - (((1+sqrt(2))^(n-1)-(1-sqrt(2))^(n-1))/sqrt(8)) = A135532(n).
a(2n+1) = (+(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n + (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n + 1) = A048655(n).
a(2n+1) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) + (((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/sqrt(8)) = A048655(n).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A255236(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A054490(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038762(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A101386(n).
(sqrt(2*(a(2n + 1) )^2 + 14*(-1)^floor(n/2)))/2 = A266504(n).
(a(2n + 1) + a(2n))/8 = A000129(n), where A000129 gives the Pell numbers.
a(2n + 1) - a(2n) = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 2) + a(2n + 1))/2 = A000129(n+2).
(a(2n + 2) - a(2n + 1))/2 = A000129(n-1).

A266507 a(n) = 6*a(n - 1) - a(n - 2) with a(0) = 2, a(1) = 8.

Original entry on oeis.org

2, 8, 46, 268, 1562, 9104, 53062, 309268, 1802546, 10506008, 61233502, 356895004, 2080136522, 12123924128, 70663408246, 411856525348, 2400475743842, 13990997937704, 81545511882382, 475282073356588, 2770146928257146, 16145599496186288, 94103450048860582

Offset: 0

Raphie Frank, Dec 30 2015

Bisection of A078343 = A078343(2*n + 1).
Quadrisection of A266504 = A266504(4*n + 1).
Octasection of A266506 = A266506(8*n + 2).

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

Bisection of A078343 = A078343(2n + 1).
Quadrisection of A266504 = A266504(4n + 1).
Octasection of A266506 = A266506(8n + 2).
Equals 2*A038723(n).

I:=[2,8]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

LinearRecurrence[{6, -1}, {2, 8}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[2 (1 - 2 x)/(1 - 6 x + x^2), {x, 0, n}], {n, 0, 22}] (* Michael De Vlieger, Dec 31 2015 *)

Vec(2*(1-2*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Dec 31 2015

a(n) = (-sqrt(2)*(1+sqrt(2))^(2*n+1) - 3 *(1-sqrt(2))^(2*n+1) - sqrt(2)*(1-sqrt(2))^(2*n+1) + 3*(1+sqrt(2))^(2*n+1))/sqrt(8).

G.f.: 2*(1-2*x) / (1-6*x+x^2). - Colin Barker, Dec 31 2015

A117894 Triangle, row sums = Pell numbers, A000129.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 3, 5, 1, 4, 4, 8, 12, 1, 5, 5, 11, 19, 29, 1, 6, 6, 14, 26, 46, 70, 1, 7, 7, 17, 33, 63, 111, 169, 1, 8, 8, 20, 40, 80, 152, 268, 408, 1, 9, 9, 23, 47, 97, 193, 367, 647, 985

Offset: 0

Gary W. Adamson, Mar 30 2006

Deleting the right border gives triangle A117895.

			First few rows of the triangle are:
  1;
  1, 1;
  1, 2, 2;
  1, 3, 3,  5;
  1, 4, 4,  8, 12;
  1, 5, 5, 11, 19, 29;
  1, 6, 6, 14, 26, 46,  70;
  1, 7, 7, 17, 33, 63, 111, 169;
...
Row 4 of A117584 = (1, 4, 7, 12). Difference terms (1, 3, 3, 5) = row 4 of A117894.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000129, A078343, A117584, A117895.

Pell:= func< n | Round(((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2))) >;
[k eq 0 select 1 else (k-n)*Pell(k+1) + (3*n-3*k+1)*Pell(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 27 2021

T[n_, k_]:= T[n, k]= If[k==0, 1, (k-n)*Fibonacci[k+1, 2] + (3*n-3*k+1)*Fibonacci[k, 2]]; Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 27 2021 *)

def P(n): return lucas_number1(n, 2, -1)
def A117894(n,k): return 1 if (k==0) else (k-n)*P(k+1) + (3*n-3*k+1)*P(k)
flatten([[A117894(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 27 2021

Rows are composed of difference terms of triangle A117584.
Rows sum to Pell numbers, A000129.
From G. C. Greubel, Sep 27 2021: (Start)
T(n, k) = (k-n)*A000129(k+1) + (3*n-3*k+1)*A000129(k) with T(n,0) = 1.
T(n, k) = A117584(n+1, k+1) - A117584(n+1, k) with T(n, 0) = 1.
T(n, 1) = n for n >= 1.
T(n, 2) = n for n >= 2.
T(n, n) = [n=0] + A000129(n).
T(n, n-1) = 2*[n=0] + A078343(n). (End)

A117895 Triangle T(n, k) = (k-n)A000129(k+1) + (3n-3k+1)A000129(k) with T(n,0) = 1, for 0 <= k <= n-1, read by rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 4, 8, 1, 5, 5, 11, 19, 1, 6, 6, 14, 26, 46, 1, 7, 7, 17, 33, 63, 111, 1, 8, 8, 20, 40, 80, 152, 268, 1, 9, 9, 23, 47, 97, 193, 367, 647, 1, 10, 10, 26, 54, 114, 234, 466, 886, 1562, 1, 11, 11, 29, 61, 131, 275, 565, 1125, 2139, 3771, 1, 12, 12, 32, 68, 148, 316, 664, 1364, 2716, 5164, 9104

Offset: 0

Gary W. Adamson, Mar 30 2006

Successive deletions of the right borders of triangle A117894 produces triangles whose row sums = generalized Pell sequences starting (1, 2...), (1, 3...), (1, 4...); etc. Row sums of A117894 = A000129: (1, 2, 5...). Row sums of A117895 = A001333: (1, 3, 7...). Deletion of the border of A117895 would produce a triangle with row sums of the Pell sequence A048654 (1, 4, 9...); and so on.

			First few rows of the triangle are:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 4,  8;
  1, 5, 5, 11, 19;
  1, 6, 6, 14, 26, 46;
  1, 7, 7, 17, 33, 63, 111;
  1, 8, 8, 20, 40, 80, 152, 268;
...
Row 4, (1, 4, 4, 8) is produced by adding (0, 1, 1, 3) to row 4 of A117894: (1, 3, 3, 5).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000129, A001333, A048654, A048655, A048693, A117894.

Pell:= func< n | Round(((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2))) >;
[k eq 0 select 1 else (k-n)*Pell(k+1) + (3*n-3*k+1)*Pell(k): k in [0..n-1], n in [0..12]]; // G. C. Greubel, Sep 27 2021

T[n_, k_]:= T[n, k]= If[k==0, 1, (k-n)*Fibonacci[k+1, 2] + (3*n-3*k +1)*Fibonacci[k, 2]]; Table[T[n, k], {n,0,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Sep 27 2021 *)

def P(n): return lucas_number1(n, 2, -1)
def A117895(n,k): return 1 if (k==0) else (k-n)*P(k+1) + (3*n-3*k+1)*P(k)
flatten([[A117895(n,k) for k in (0..n-1)] for n in (0..12)]) # G. C. Greubel, Sep 27 2021

Delete right border of triangle A117894. Alternatively, let row 1 = 1 and using the heading 0, 1, 1, 3, 7, 17, 41, 99, 239...(i.e. A001333 starting with 0, 1, 1, 3...); add the first n terms of the heading to n-th row of triangle A117894.
From G. C. Greubel, Sep 27 2021: (Start)
T(n, k) = (k-n)*A000129(k+1) + (3*n-3*k+1)*A000129(k) with T(n,0) = 1.
T(n, 1) = n+1 for n >= 1.
T(n, 2) = n+1 for n >= 2.
T(n, n) = 2*[n=0] + A078343(n). (End)

New name and more terms added by G. C. Greubel, Sep 27 2021

A135139 a(n) = 5a(n-2) + 2a(n-3).

Original entry on oeis.org

1, 2, 4, 12, 24, 68, 144, 388, 856, 2228, 5056, 12852, 29736, 74372, 174384, 431332, 1020664, 2505428, 5965984, 14568468, 34840776, 84774308, 203340816, 493553092, 1186252696, 2874447092, 6918369664, 16744740852, 40340742504, 97560443588

Offset: 0

Paul Curtz, Feb 13 2008

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 5, 2).

Cf. A135138.

a = {1, 2, 4}; Do[AppendTo[a, 5*a[[ -2]] + 2*a[[ -3]]], {40}]; a (* Stefan Steinerberger, Feb 15 2008 *)
LinearRecurrence[{0,5,2},{1,2,4},30] (* Harvey P. Dale, May 25 2012 *)

From R. J. Mathar, Feb 15 2008: (Start)
O.g.f.: 4*(x-2)/(7*(x^2+2*x-1)) - 1/(7*(1+2*x)).
a(n) = (4*A078343(n+1)-(-2)^n)/7.
a(n) = A135138(n+2) + 2*A135138(n+1) - A135138(n). (End)

More terms from R. J. Mathar and Stefan Steinerberger, Feb 15 2008

A217975 Integers k such that 2*k^2 - 7 is a square.

Original entry on oeis.org

2, 4, 8, 22, 46, 128, 268, 746, 1562, 4348, 9104, 25342, 53062, 147704, 309268, 860882, 1802546, 5017588, 10506008, 29244646, 61233502, 170450288, 356895004, 993457082, 2080136522, 5790292204, 12123924128, 33748296142, 70663408246, 196699484648

Offset: 1

Sture Sjöstedt, Oct 16 2012

a(n) gives y-values solving the Diophantine equation x^2 + 7 = 2*y^2. A077446(n) gives the x-values. - Sture Sjöstedt, Oct 16 2012

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 28 = 0. - Colin Barker, Feb 08 2014

			Since 2(4^2) - 7 = 25 = 5^2, and 4 is the second number with this property, a(2) = 4.

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

Cf. A077442 (2*n^2 + 7 is a square).

I:=[2, 4, 8, 22]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..31]]; // Vincenzo Librandi, Oct 16 2012

LinearRecurrence[{0, 6, 0, -1}, {2, 4, 8, 22}, 50] (* Sture Sjöstedt, Oct 16 2012 *)

Vec(2*x*(1-x)*(x^2+3*x+1)/(x^2-2*x-1)/(x^2+2*x-1)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

a(n) = 6*a(n - 2) - a(n - 4) with a(1)=2, a(2)=4, a(3)=8, a(4)=22. - Sture Sjöstedt, Oct 16 2012
a(n)*a(n+3)-a(n+1)*a(n+2) = 10-2*(-1)^n. - Bruno Berselli, Oct 25 2012
a(n) = 2*A006452(n). - R. J. Mathar, Oct 17 2012
G.f.: -2*x*(x - 1)*(x^2 + 3*x + 1)/((x^2 - 2*x - 1)*(x^2 + 2*x - 1)). - Colin Barker, Oct 24 2012
a(n) = a(-n+1) = ((4+sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))+(4-sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2)))/4. - Bruno Berselli, Oct 25 2012
a(2n-1) = A078343(2n-1), a(2n) = A100525(n-1). - Bruno Berselli, Oct 25 2012

cp's OEIS Frontend

A135138 a(n) = 5*a(n-2) + 2*a(n-3).

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A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

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A220414 a(n) = 6*a(n-1) - a(n-2), with a(1) = 13, a(2) = 73.

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A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

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A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

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A266507 a(n) = 6*a(n - 1) - a(n - 2) with a(0) = 2, a(1) = 8.

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A117894 Triangle, row sums = Pell numbers, A000129.

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A135138 a(n) = 5a(n-2) + 2a(n-3).

A117895 Triangle T(n, k) = (k-n)A000129(k+1) + (3n-3k+1)A000129(k) with T(n,0) = 1, for 0 <= k <= n-1, read by rows.

A135139 a(n) = 5a(n-2) + 2a(n-3).