A038761 -id:A038761 - cp's OEIS frontend

A164546 a(n) = 8a(n-1) - 8a(n-2) for n > 1; a(0) = 1, a(1) = 10.

1, 10, 72, 496, 3392, 23168, 158208, 1080320, 7376896, 50372608, 343965696, 2348744704, 16038232064, 109515898880, 747821334528, 5106443485184, 34868977205248, 238100269760512, 1625850340442112, 11102000565452800

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009

Binomial transform of A038761. Fourth binomial transform of A164640. Inverse binomial transform of A164547.

Vincenzo Librandi, Table of n, a(n) for n = 0..149
Index entries for linear recurrences with constant coefficients, signature (8,-8).

Cf. A038761, A164640, A164547.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((2+3*r)*(4+2*r)^n+(2-3*r)*(4-2*r)^n)/4: n in [0..19] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Aug 19 2009

LinearRecurrence[{8,-8}, {1,10}, 30] (* G. C. Greubel, Jul 17 2021 *)

[2*(2*sqrt(2))^(n-1)*(sqrt(2)*chebyshev_U(n, sqrt(2)) + chebyshev_U(n-1, sqrt(2))) for n in (0..30)] # G. C. Greubel, Jul 17 2021

a(n) = 8*a(n-1) - 8*a(n-2) for n > 1; a(0) = 1, a(1) = 10.
a(n) = ((2+3*sqrt(2))*(4+2*sqrt(2))^n + (2-3*sqrt(2))*(4-2*sqrt(2))^n)/4.
G.f.: (1 + 2*x)/(1 - 8*x + 8*x^2).
a(n) = 2*(2*sqrt(2))^(n-1)*(sqrt(2)*chebyshev_U(n, sqrt(2)) + chebyshev_U(n-1, sqrt(2))). - G. C. Greubel, Jul 17 2021

Edited and extended beyond a(5) by Klaus Brockhaus, Aug 19 2009

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

A111648 a(n) = A001541(n)^2 + A001653(n+1)^2 + A002315(n)^2.

Original entry on oeis.org

3, 83, 2811, 95483, 3243603, 110187011, 3743114763, 127155714923, 4319551192611, 146737584833843, 4984758333158043, 169335045742539611, 5752406796913188723, 195412496049305876963

Offset: 0

Charlie Marion, Aug 24 2005

			a(1) = 83 = 3^2+5^2+7^2.

Ray Chandler, Table of n, a(n) for n = 0..652
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001541, A001653, A002315, A111647, A111649.

LinearRecurrence[{35, -35, 1}, {3, 83, 2811}, 20] (* Paolo Xausa, Feb 06 2024 *)

a(n) = A038761(n)^2 + 2, e.g., 95483 = 309^2 + 2.
a(n) = A001652(2*n+1) - A001109(n+1)^2 - Sum_{k=1..n-1} A038723(2*n), e.g., 95483 = 137903 - 204^2 - (23 + 781).
For n > 0, 2*a(n) + A001652(2*n-1) = A001653(2*n+2), e.g., 2*2811 + 119 = 5741.
G.f.: -(11*x^2-22*x+3) / ((x-1)*(x^2-34*x+1)). - Colin Barker, Dec 14 2014 (Empirical g.f. confirmed for more terms and recurrence of source sequences. - Ray Chandler, Feb 05 2024)

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

Original entry on oeis.org

1, 1, 2, 1, 5, 2, 1, 9, 9, 2, 1, 14, 25, 13, 2, 1, 20, 55, 49, 17, 2, 1, 27, 105, 140, 81, 21, 2, 1, 35, 182, 336, 285, 121, 25, 2, 1, 44, 294, 714, 825, 506, 169, 29, 2, 1, 54, 450, 1386, 2079, 1716, 819, 225, 33, 2, 1, 65, 660, 2508, 4719, 5005, 3185, 1240, 289, 37, 2

Offset: 1

Clark Kimberling, Feb 19 2012

Subtriangle of the triangle T(n,k) given by (1, 0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 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;
  1,  2;
  1,  5,  2;
  1,  9,  9,  2;
  1, 14, 25, 13,  2;
Triangle (1, 0, 1/2, 1/2, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, ...) begins:
  1;
  1,  0;
  1,  2,  0;
  1,  5,  2,  0;
  1,  9,  9,  2,  0;
  1, 14, 25, 13,  2,  0;
  1, 20, 55, 49, 17,  2,  0;
  ...
1 = 2*1 - 1, 20 = 2*14 + 1 - 9, 55 = 2*25 + 14 - 9, 49 = 2*13 + 25 - 2, 17 = 2*2 + 1 - 0, 2 = 2*0 + 2 - 0. - _Philippe Deléham_, Mar 03 2012

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Cf. A207606.

A207607:= (n,k) -> `if`(k=1, 1, binomial(n+k-3, 2*k-2) + 2*binomial(n+k-3, 2*k-3) ); seq(seq(A207607(n, k), k = 1..n), n = 1..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 *)
Table[If[k==1, 1, Binomial[n+k-3, 2*k-2] + 2*Binomial[n+k-3, 2*k-3]], {n, 10}, {k, 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(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

def T(n, k):
    if k == 1: return 1
    else: return binomial(n+k-3, 2*k-2) + 2*binomial(n+k-3, 2*k-3)
[[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.
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Mar 03 2012
G.f.: (1-x+y*x)/(1-(y+2)*x+x^2). - Philippe Deléham, Mar 03 2012
For n >= 1, Sum{k=0..n} T(n,k)*x^k = A000012(n), A001906(n), A001834(n-1), A055271(n-1), A038761(n-1), A056914(n-1) for x = 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Mar 03 2012
T(n,k) = C(n+k-1,2*k) + 2*C(n+k-1,2*k-1). where C is binomial. - Yuchun Ji, May 23 2019
T(n,k) = T(n-1,k) + A207606(n,k-1). - Yuchun Ji, May 28 2019
Sum_{k=1..n} T(n, k)*x^k = { 4*(-1)^(n-1)*A016921(n-1) (x=-4), 3*(-1)^(n-1) * A130815(n-1) (x=-3), 2*(-1)^(n-1)*A010684(n-1) (x=-2), A057079(n+1) (x=-1), 0 (x=0), A001906(n) = Fibonacci(2*n) (x=1), 2*A001834(n-1) (x=2), 3*A055271(n-1) (x=3), 4*A038761(n-1) (x=4) }. - G. C. Greubel, Mar 15 2020

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).

A129346 a(2n) = A100525(n), a(2n+1) = A001653(n+1); a Pellian-related sequence.

Original entry on oeis.org

4, 5, 22, 29, 128, 169, 746, 985, 4348, 5741, 25342, 33461, 147704, 195025, 860882, 1136689, 5017588, 6625109, 29244646, 38613965, 170450288, 225058681, 993457082, 1311738121, 5790292204, 7645370045, 33748296142, 44560482149, 196699484648, 259717522849

Offset: 0

Creighton Dement, Apr 10 2007

Summation of -a(n) and A129345 returns twice Pell numbers A000129 (apart from signs; starting from 2nd term of A000129).

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

Cf. A129345, A000129, A001542, A038761.

LinearRecurrence[{0,6,0,-1},{4,5,22,29},30] (* Harvey P. Dale, Apr 08 2018 *)

Vec((4+5*x-2*x^2-x^3)/((x^2-2*x-1)*(x^2+2*x-1)) + O(x^40)) \\ Colin Barker, May 26 2016

O.g.f.: (4 + 5*x - 2*x^2 - x^3) / ((x^2 - 2*x - 1)*(x^2 + 2*x - 1)).
From Colin Barker, May 26 2016: (Start)
a(n) = (-(-1-sqrt(2))^(1+n)+(-1+sqrt(2))^(1+n)+(1-sqrt(2))^n*(-4+3*sqrt(2))+(1+sqrt(2))^n*(4+3*sqrt(2)))/(2*sqrt(2)).
a(n) = 6*a(n-2)-a(n-4) for n>3. (End)
E.g.f.: 2*cosh(sqrt(2)*x)*(sinh(x) + 2*cosh(x)) + (sinh(sqrt(2)*x)*(5*sinh(x) + 3*cosh(x)))/sqrt(2). - Ilya Gutkovskiy, May 26 2016

A154346 a(n) = 12a(n-1) - 28a(n-2) for n > 1, with a(0)=1, a(1)=12.

Original entry on oeis.org

1, 12, 116, 1056, 9424, 83520, 738368, 6521856, 57587968, 508443648, 4488860672, 39629905920, 349870772224, 3088811900928, 27269361188864, 240745601040384, 2125405099196416, 18763984361226240, 165656469557215232

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009

Binomial transform of A164547, second binomial transform of A164546, third binomial transform of A038761, fourth binomial transform of A164545, fifth binomial transform of A164544, sixth binomial transform of A164640.

Lim_{n -> infinity} a(n)/a(n-1) = 6 + 2*sqrt(2) = 8.8284271247....

R. J. Mathar, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (12, -28).

Cf. A002193 (decimal expansion of sqrt(2)), A164547, A164546, A038761, A164545, A164544, A164640.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((6+2*r)^n-(6-2*r)^n)/(4*r): n in [1..19] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jan 12 2009

Join[{a=1,b=12},Table[c=12*b-28*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 31 2011 *)
LinearRecurrence[{12,-28},{1,12},20] (* Harvey P. Dale, May 23 2012 *)
Rest@ CoefficientList[Series[x/(1 - 12 x + 28 x^2), {x, 0, 19}], x] (* Michael De Vlieger, Sep 13 2016 *)

a(n)=([0,1; -28,12]^(n-1)*[1;12])[1,1] \\ Charles R Greathouse IV, Sep 13 2016

[lucas_number1(n,12,28) for n in range(1, 20)] # Zerinvary Lajos, Apr 27 2009

a(n) = 12*a(n-1) - 28*a(n-2) for n > 1. - Philippe Deléham, Jan 12 2009
a(n) = ( (6 + 2*sqrt(2))^n - (6 - 2*sqrt(2))^n )/(4*sqrt(2)).
G.f.: x/(1 - 12*x + 28*x^2). - Klaus Brockhaus, Jan 12 2009, corrected Oct 08 2009
E.g.f.: (1/(2*sqrt(2)))*exp(6*x)*sinh(2*sqrt(2)*x). - G. C. Greubel, Sep 13 2016
a(n) =2^(n-1)*A081179(n). - R. J. Mathar, Feb 04 2021

Extended beyond a(7) by Klaus Brockhaus, Jan 12 2009
Edited by Klaus Brockhaus, Oct 08 2009
Offset corrected. - R. J. Mathar, Jun 19 2021

cp's OEIS Frontend

A164546 a(n) = 8*a(n-1) - 8*a(n-2) for n > 1; a(0) = 1, a(1) = 10.

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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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A111648 a(n) = A001541(n)^2 + A001653(n+1)^2 + A002315(n)^2.

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

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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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A129346 a(2n) = A100525(n), a(2n+1) = A001653(n+1); a Pellian-related sequence.

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A154346 a(n) = 12*a(n-1) - 28*a(n-2) for n > 1, with a(0)=1, a(1)=12.

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A164546 a(n) = 8a(n-1) - 8a(n-2) for n > 1; a(0) = 1, a(1) = 10.

A154346 a(n) = 12a(n-1) - 28a(n-2) for n > 1, with a(0)=1, a(1)=12.