A228208 -id:A228208 - cp's OEIS frontend

A054486 Expansion of (1+2x)/(1-3x+x^2).

1, 5, 14, 37, 97, 254, 665, 1741, 4558, 11933, 31241, 81790, 214129, 560597, 1467662, 3842389, 10059505, 26336126, 68948873, 180510493, 472582606, 1237237325, 3239129369, 8480150782, 22201322977, 58123818149, 152170131470, 398386576261, 1042989597313

Offset: 0

Barry E. Williams, May 06 2000

Binomial transform of A000285. - R. J. Mathar, Oct 26 2011

			G.f. = 1 + 5*x + 14*x^2 + 37*x^3 + 97*x^4 + 254*x^5 + 665*x^6 + 1741*x^7 + ...

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A000285, A002878, A100545, A104449, A228208.

F:=Fibonacci;; List([0..30], n-> F(2*n+2) +2*F(2*n) ); # G. C. Greubel, Nov 08 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/(1-3*x+x^2)) ); // Marius A. Burtea, Nov 05 2019

a:=[1,5]; [n le 2 select a[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // Marius A. Burtea, Nov 05 2019

with(combinat); f:=fibonacci; seq(f(2*n+2)+2*f(2*n), n=0..30); # G. C. Greubel, Nov 08 2019

CoefficientList[Series[(2*z+1)/(z^2-3*z+1), {z, 0, 30}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 15 2011 *)
a[ n_]:= 3 Fibonacci[2n] + Fibonacci[2n+1]; (* Michael Somos, Mar 17 2015 *)
LinearRecurrence[{3,-1},{1,5},40] (* Harvey P. Dale, Apr 24 2019 *)

Vec((1+2*x)/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jul 15 2011

{a(n) = 3*fibonacci(2*n) + fibonacci(2*n+1)}; /* Michael Somos, Mar 17 2015 */

f=fibonacci; [f(2*n+2) +2*f(2*n) for n in (0..30)] # G. C. Greubel, Nov 08 2019

a(n) = 3*a(n-1) - a(n-2), a(0)=1, a(1)=5.
a(n) = (5*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - (((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)))/sqrt(5).
a(n) + 7*A001519(n) = A005248(n). - Creighton Dement, Oct 30 2004
a(n) = Lucas(2*n+1) + Fibonacci(2*n) = A002878(n) + A001906(n) = A025169(n-1) + A001906(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-6)^k. - Philippe Deléham, Mar 05 2014
0 = -11 + a(n)^2 - 3*a(n)*a(n+1) + a(n+1)^2 for all n in Z. - Michael Somos, Mar 17 2015
a(n) = -2*F(n)^2 + 6*F(n)*F(n+1) + F(n+1)^2 for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = 3*F(2*n) + F(2*n+1) for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = -A100545(-2-n) for all n in Z. - Michael Somos, Mar 17 2015
a(n) = A000285(2*n) = A228208(2*n+1) = A104449(2*n+1) for all n in Z. - Michael Somos, Mar 17 2015
From Klaus Purath, Nov 05 2019: (Start)
a(n) = (a(n-m) + a(n+m))/Lucas(2*m), m <= n.
a(n) = sum of 2*m+1 consecutive terms starting with a(n-m) divided by Lucas(2*m+1), m <= n.
a(n) = alternating sum of 2*m+1 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+1), m <= n.
a(n) + a(n+1) = sum of 2*m+2 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+2), m <= n.
a(n) + a(n+1) = (a(n-m) + a(n+m+1))/Fibonacci(2*m+1), m <= n.
The following formulas are extended to negative indexes:
a(n) = 3*Fibonacci(2*n+1) - Fibonacci(2*n-3).
a(n) = (Fibonacci(2*n+5) - 3* Fibonacci(2*n-1))/2.
a(n) = (4*Lucas(2*n+2) - Lucas(2*n-4))/5.
a(n) = Fibonacci(2*n+5) - 4*Fibonacci(2*n+1).
a(n) = (5*Fibonacci(2*n+5) - Fibonacci(2*n-7))/12. (End)
E.g.f.: exp(-(1/2)*(-3+sqrt(5))*x)*(-7 + sqrt(5) + (7 + sqrt(5))*exp(sqrt(5)*x))/(2*sqrt(5)). - Stefano Spezia, Nov 19 2019
a(n) = 3*n + 1 + Sum_{k=1..n} k*a(n-k). - Yu Xiao, Jun 20 2020

"a(1)=5", not "a(0)=5" from Dan Nielsen (nielsed(AT)uah.edu), Sep 10 2009

A228207 x-values in the solution to x^2 - 20*y^2 = 176.

Original entry on oeis.org

14, 16, 26, 34, 64, 86, 166, 224, 434, 586, 1136, 1534, 2974, 4016, 7786, 10514, 20384, 27526, 53366, 72064, 139714, 188666, 365776, 493934, 957614, 1293136, 2507066, 3385474, 6563584, 8863286, 17183686, 23204384, 44987474, 60749866, 117778736, 159045214

Offset: 1

Colin Barker, Aug 16 2013

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A228208, A228210.

I:=[14,16,26,34]; [n le 4 select I[n] else 3*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Aug 17 2013

CoefficientList[Series[-2 (x - 1) (7 x^2 + 15 x + 7) / ((x^2 - x - 1) (x^2 + x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 17 2013 *)
LinearRecurrence[{0,3,0,-1},{14,16,26,34},50] (* Harvey P. Dale, May 25 2023 *)

Vec(-2*x*(x-1)*(7*x^2+15*x+7)/((x^2-x-1)*(x^2+x-1)) + O(x^100))

G.f.: -2*x*(x-1)*(7*x^2+15*x+7) / ((x^2-x-1)*(x^2+x-1)).
a(n) = 3*a(n-2)-a(n-4).
a(n) = 2*A228210(n). - Hugo Pfoertner, Feb 11 2024

A228210 x-values in the solutions to x^2 - 5y^2 = 44.

Original entry on oeis.org

7, 8, 13, 17, 32, 43, 83, 112, 217, 293, 568, 767, 1487, 2008, 3893, 5257, 10192, 13763, 26683, 36032, 69857, 94333, 182888, 246967, 478807, 646568, 1253533, 1692737, 3281792, 4431643, 8591843, 11602192, 22493737, 30374933, 58889368, 79522607, 154174367

Offset: 1

Colin Barker, Aug 16 2013

Also values of x (or y) in the solutions to x^2 - 3xy + y^2 + 55 = 0.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Project Euler, Problem 140: Modified Fibonacci golden nuggets
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A228208.

I:=[7,8,13,17]; [n le 4 select I[n] else 3*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Aug 17 2013

CoefficientList[Series[-(x - 1) (7 x^2 + 15 x + 7) / ((x^2 - x - 1) (x^2 + x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 17 2013 *)
LinearRecurrence[{0,3,0,-1},{7,8,13,17},40] (* Harvey P. Dale, Jun 01 2020 *)

Vec(-x*(x-1)*(7*x^2+15*x+7)/((x^2-x-1)*(x^2+x-1)) + O(x^100))

G.f.: -x*(x-1)*(7*x^2+15*x+7) / ((x^2-x-1)*(x^2+x-1)).

a(n) = 3*a(n-2)-a(n-4).

A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

5, 19, 31, 101, 179, 655, 1189, 4451, 8111, 30469, 55555, 208799, 380741, 1431091, 2609599, 9808805, 17886419, 67230511, 122595301, 460804739, 840280655, 3158402629, 5759369251, 21648013631, 39475304069, 148377692755, 270567759199, 1016995835621, 1854499010291

Offset: 1

XU Pingya, Aug 24 2022

Conjecture:
(i) For all k > 2, 2*x^3 + 2*y^3 + z^3 = A089270(k)^3 have primitive solutions form (c(n)^2, -d(n)^2, -w(n)) with d(n) = 3*d(n-2) - d(n-4), c(n) = d(n+2) - d(n) and w(n) = 8*w(n-2) - 8*w(n-4) + w(n-6).
(ii) This sequence is a subsequence of A089270.
From XU Pingya, Jun 07 2024: (Start)
Several positive examples of conjecture:
When A089270(4,5,6,7) = {19,29,31,41}, d(n) can be taken as:
(1/2) * (F(n+3) + (-1)^n * F(n-6));
((1-(-1)^n)/2) * (F(n+3) + F(n-4)) + ((1+(-1)^n)/2) * (F(n+3) - F(n-4));
((1-(-1)^n)/2) * (2*F(n-1) + 3*F(n-3)) + ((1+(-1)^n)/2) * (3*F(n-2) + 2*F(n-4));
and
((1-(-1)^n)/2) * (2*F(n+1) + F(n-5)) + ((1+(-1)^n)/2) * (F(n+2) + 2*F(n-4)).
When A089270(17) = 121, d(n) can be taken as d(1,2,3,4) = {-3,0,7,11}. (End)
From XU Pingya, Jul 17 2024: (Start)
Furthermore, we observe that if (x, y) (y < x/2) is the solution of the Diophantine equation x^2 + x * y - y^2 = A089270(k). Let
d(2*n-1) = x * F(2*n-2) - y * F(2*n-3), c(2*n-1) = d(2*n+1) - d(2*n-1);
d(2*n) = x * F(2*n-2) + y * F(2*n-1), c(2*n) = d(2*n+2) - d(2*n).
Then such c(n) and d(n) satisfy the conjecture. (End)

			For n=3, 2 * ((F(5) - F(0))^2)^3 + 2 * (-(F(4) - F(-1))^2)^3 + (-31)^3 = 2 * 25^3 - 2 * 4^3 - 31^3 = 1331, a(3) = 31.

Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-1,1).

Cf. A000045, A081016, A089270, A206351, A228208, A237132.

Cf. also A337928, A354336, A356717.

Table[(-1331+2*((Fibonacci[n+2]+(-1)^n*Fibonacci[n-3]))^6-2*(Fibonacci[n+1]+(-1)^n*Fibonacci[n-4])^6)^(1/3), {n,28}]

a(n) = (-1331 + 2 * A237132(n)^6 - 2 * A228208(n-1)^6)^(1/3).
a(n) = ((1-(-1)^n)/2) * (-1 + 6 * Sum_{k=0..n-1} Fibonacci(4*k-1) + 14 * Sum_{k=0..n-2} Fibonacci(4*k+1)) + ((1+(-1)^n)/2) * (-1 + 6 * Sum_{k=0..n-1} Fibonacci(4*k-1) + 14 * Sum_{k=0..n-1} Fibonacci(4*k+1)).
a(n) = ((1-(-1)^n)/2) * (-1 + 6*A206351(n) + 14*A081016(n-2)) + ((1+(-1)^n)/2) * (-1 + 6*A206351(n) + 14*A081016(n-1)).
From Stefano Spezia, Aug 25 2022: (Start)
G.f.: x*(5 + 14*x - 23*x^2 - 28*x^3 - x^4)/((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)
From XU Pingya, Jul 17 2024: (Start)
a(2*n-1) = (F(2*n) + F(2*n-2) + F(2*n-5))^2 + (F(2*n) + F(2*n-2) + F(2*n-5)) * (F(2*n-2) + F(2*n-4) + F(2*n-7)) - (F(2*n-2) + F(2*n-4) + F(2*n-7))^2;
a(2*n) = (F(2*n+2) + F(2*n-3))^2 + (F(2*n+2) + F(2*n-3)) * (F(2*n) + F(2*n-5)) - (F(2*n) + F(2*n-5))^2. (End)

A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 29, 59, 241, 445, 1691, 3089, 11629, 21211, 79745, 145421, 546619, 996769, 3746621, 6831995, 25679761, 46827229, 176011739, 320958641, 1206402445, 2199883291, 8268805409, 15078224429, 56675235451, 103347687745, 388457842781, 708355589819, 2662529664049

Offset: 1

XU Pingya, Aug 24 2022

			For n=3, 2 * ((F(5) - F(0))^2)^3 + 2 * (-(F(6) - F(1))^2)^3 + 59^3 = 2 * 25^3 - 2 * 49^3 + 59^3 = 1331, a(3) = 59.

Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-1,1).

Cf. A000045, A081016, A081018, A089270, A228208, A237132.

Cf. also A337929, A354337, A356716.

Table[(1331-2*((Fibonacci[n+2]+(-1)^n*Fibonacci[n-3]))^6+2*(Fibonacci[n+3]+(-1)^n*Fibonacci[n-2])^6)^(1/3), {n,28}]

a(n) = (1331 - 2 * A237132(n)^6 + 2 * A228208(n+1)^6)^(1/3).
a(n) = ((1-(-1)^n)/2) * (-5 + 14 * Sum_{k=1..n-1} Fibonacci(4*k-1) + 6 * Sum_{k=0..n-1} Fibonacci(4*k+1)) + ((1+(-1)^n)/2) * (-5 + 14 * Sum_{k=1..n} Fibonacci(4*k-1) + 6 * Sum_{k=0..n-1} Fibonacci(4*k+1)).
a(n) = ((1-(-1)^n)/2) * (-5 + 14 * A081018(n-1) + 6 * A081016(n-1)) + ((1+(-1)^n)/2) * (-5 + 14 * A081018(n) + 6 * A081016(n-1)).
From Stefano Spezia, Aug 25 2022: (Start)
G.f.: x*(1 + 28*x + 23*x^2 - 14*x^3 - 5*x^4)/((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)

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A054486 Expansion of (1+2*x)/(1-3*x+x^2).

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A228207 x-values in the solution to x^2 - 20*y^2 = 176.

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A228210 x-values in the solutions to x^2 - 5y^2 = 44.

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A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

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A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

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A054486 Expansion of (1+2x)/(1-3x+x^2).

A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).