A206351 -id:A206351 - cp's OEIS frontend

A003482 a(n) = 7*a(n-1) - a(n-2) + 4, with a(0) = 0, a(1) = 5.

0, 5, 39, 272, 1869, 12815, 87840, 602069, 4126647, 28284464, 193864605, 1328767775, 9107509824, 62423800997, 427859097159, 2932589879120, 20100270056685, 137769300517679, 944284833567072, 6472224534451829, 44361286907595735, 304056783818718320

Offset: 0

N. J. A. Sloane

The values (a(n),x(n)), n >= 2, x(n)=Fibonacci(2*n+2)*Fibonacci(2*n+3)=A081018(n+1), are the integer solutions (a,x) of the equation binomial(x+1,a+1) + binomial(x+2,a+1) = binomial(x+3,a+1). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)

The values (a(n),x(n)), n >= 2 are also the integer solutions (a, x) of the equation x(a+1) = (x-a)(x-a-1) or, equivalently, binomial(x, a) = binomial(x-1, a+1). - Tomohiro Yamada, May 30 2018

			G.f. = 5*x + 39*x^2 + 272*x^3 + 1869*x^4 + 12815*x^5 + 87840*x^6 + ... - _Michael Somos_, Jun 26 2018

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Heiko Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose, California, August 1986, 1-5 (1988).
Sébastien Labbé and Jana Lepšová, A Fibonacci's complement numeration system, arXiv:2205.02574 [cs.FL], 2022.
Sébastien Labbé and Jana Lepšová, A Fibonacci analogue of the two's complement numeration system, RAIRO-Theor. Inf. Appl. (2023) Vol. 57, No. 12. See p. 16.
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart. 6(3) (1968), 86-93.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
John Riordan and N. J. A. Sloane, Correspondence, 1974
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart. 13 (1973), 295-298.
S. M. Tanny and M. Zuker, On a unimodal sequence of binomial coefficients, Discrete Math. 9 (1974), 79-89.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A001109, A001654, A081018, A206351.

A003482:=z*(-5+z)/(z-1)/(z**2-7*z+1); # conjectured by Simon Plouffe in his 1992 dissertation

LinearRecurrence[{8,-8,1},{0,5,39},30] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
nxt[{a_,b_}]:={b,7b-a+4}; NestList[nxt,{0,5},30][[;;,1]] (* Harvey P. Dale, Jun 09 2025 *)

a(n)=fibonacci(2*n)*fibonacci(2*n+3) \\ Charles R Greathouse IV, May 29 2013

a(n) = Fibonacci(2*n) * Fibonacci(2*n+3).
a(n) = Fibonacci(2*n+2)^2 - Fibonacci(2*n+1)^2. - Gary Detlefs, Oct 12 2011
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). - Vladimir Joseph Stephan Orlovsky and Vincenzo Librandi, Jan 22 2012
a(n) = -4/5 + (sqrt(5)/5 + 2/5)*(7/2 + 3*sqrt(5)/2)^n - (sqrt(5)/5 - 2/5)*(7/2 - 3*sqrt(5)/2)^n. - Antonio Alberto Olivares, May 29 2013
a(n) = -A206351(-n) for all n in Z. - Michael Somos, Jun 26 2018
From Sébastien Labbé, May 06 2022: (Start)
a(n) = Sum_{k=2..2*n+1} Fibonacci(k)^2.
a(n) = A001654(2*n+1)-1. (End)

A090162 Values of binomial(Fibonacci(2k)Fibonacci(2k+1),Fibonacci(2k-1)Fibonacci(2k)-1).

Original entry on oeis.org

1, 3003, 61218182743304701891431482520

Offset: 1

Eric W. Weisstein, Nov 23 2003 and Wolfdieter Lang, Dec 01 2003

These numbers are known to occur at least six times in Pascal's triangle.
The next term is approximately 3.537 * 10^204 and is in the b-file.
The numbers of digits in a(n), n >= 1, are given in A100022.

Hugo Pfoertner, Table of n, a(n) for n = 1..5
A. I. Shirshov, On the equation C(n, m) = C(n+1, m-1), chapter 10 in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am. Math. Soc., 1999, pp. 83-86
D. Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quarterly, 13 (1975), 295-298.
Eric Weisstein's World of Mathematics, Pascal's Triangle

Subsequence of A003015.

Cf. A081016, A089508, A062527, A206351.

a := proc(n) local a,b,s,p; s:= 1+sqrt(5); p:=16^n;
a := 4-2*p*s^(-4*n-1)+(s+2)*s^(4*n-1)/p:
b := 1+p*((s-2)^(1-4*n)/2-s^(-1-4*n)*(2+s)):
GAMMA(a/5)/(GAMMA(b/5)*GAMMA(1+(a-b)/5)) end:
digits := [1, 4, 29, 205, 1412]: A := n -> round(evalf(a(n),digits[n]+10)):
A(4); # Peter Luschny, Jul 15 2017

Table[Binomial[Fibonacci[2k]Fibonacci[2k+1],Fibonacci[2k-1] Fibonacci[2k]-1], {k,4}] (* Harvey P. Dale, Aug 18 2011 *)

A090162(n)=binomial(fibonacci(2*n+1)*fibonacci(2*n),fibonacci(2*n-1)*fibonacci(2*n)-1) \\ M. F. Hasler, Feb 17 2023

def A090162(n): return C(A000045(2*n+1)*A000045(2*n),A000045(2*n-1)*A000045(2*n)-1) # See A007318 for C(.,.). - M. F. Hasler, Feb 17 2023

a(n) = binomial(A089508(n), A081016(n-1)).
a(n) = binomial(A089508(n)+1, A081016(n-1)-1).
a(n) = Gamma(x)/(Gamma(y)*Gamma(1+x-y)) with x = A206351(n+1) and y = A081016(n-1). - Peter Luschny, Jul 15 2017

A206334 Numbers n such that there is a triangle with area n, side n, and the other two sides rational.

Original entry on oeis.org

3, 5, 7, 10, 12, 15, 16, 18, 19, 23, 25, 26, 27, 28, 29, 30, 33, 34, 36, 38, 39, 40, 41, 42, 43, 44, 46, 47, 51, 52, 55, 57, 58, 59, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 80, 83, 84, 85, 86, 87, 88, 89, 90, 91, 93, 95, 96, 97, 103, 104, 105, 106, 107, 109, 115, 119, 122, 123, 124, 125, 126

Offset: 1

James R. Buddenhagen, Feb 06 2012

nonn

n>3 is in the sequence just in case the elliptic curve y^2 = 4*x^4 + (n^2+8)*x^2 + 4 has positive rank.  Note that (0,2) is on that curve.
n is in the sequence just in case there are positive rational numbers x,y such that x*y>1 and x - 1/x + y - 1/y = n.
The triangle whose sides are [(4*k^6+8*k^5+8*k^4+4*k^3+2*k^2+2*k+1)/((k+1)*k*(2*k^2+2*k+1)), (4*k^6+16*k^5+28*k^4+28*k^3+18*k^2+6*k+1)/((k+1)*k*(2*k^2+2*k+1)), 4*k^2+4*k+4] has area equal to its third side.  Hence, starting with the second term, A112087 is a subsequence of the present sequence.
The triangle whose sides are [(k^6+2*k^4+k^2+1)/(k*(k^2+1)), (k^4+3*k^2+1)/(k*(k^2+1)), (k^2+2)*k] has area equal to its third side. Hence, starting with the first positive term, A054602 is a subsequence of the present sequence. [This subsequence found by Dragan K, see second link, below.]
The triangle whose sides are [(k^8+6*k^6+13*k^4+13*k^2+4)/(k*(k^2+2)*(k^2+1)), (k^6+3*k^4+5*k^2+4)/(k*(k^2+2)*(k^2+1)), k*(k^2+4)] has area equal to its third side. Hence A155965 is a subsequence of the present sequence.

			5 is in the sequence because the triangle with sides (37/6, 13/6, 5) has area 5, one side 5, and the other two sides rational.

James R. Buddenhagen, Table of triangles up to n = 145
Dragan K and Rita the dog, Question and answer [broken link]
Ian Connell, APECS elliptic curve software (which runs under old versions of Maple).
Eric Weisstein's World of Mathematics, Heronian Triangle.

Cf. A112087, A054602, A155965, and A206351 (subsequences, see comments).

A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

Original entry on oeis.org

-1, 2, 1, 7, 14, 41, 103, 274, 713, 1871, 4894, 12817, 33551, 87842, 229969, 602071, 1576238, 4126649, 10803703, 28284466, 74049689, 193864607, 507544126, 1328767777, 3478759199, 9107509826, 23843770273, 62423800999, 163427632718, 427859097161, 1120149658759

Offset: 0

Bruno Berselli, Oct 31 2015

Primes in sequence for n = 1, 3, 5, 6, 9, 24, 42, 48, 53, 71, 86, 102, 138, 182, 302, 438, 506, 926, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
A. Bremner, R. Høibakk, D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

First bisection of A111569.
Cf. A226205: numbers of the form F(n)*F(n+1)+(-1)^n.
Cf. A000045, A001654, A003482, A059929, A089508 (first bisection, without -1), A206351.

[Fibonacci(n)*Fibonacci(n+1)-(-1)^n: n in [0..30]];

with(combinat): A260259:=n->fibonacci(n)*fibonacci(n+1)-(-1)^n: seq(A260259(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017

Table[Fibonacci[n] Fibonacci[n + 1] - (-1)^n, {n, 0, 30}]

makelist(fib(n)*fib(n+1)-(-1)^n,n,0,30);

for(n=0, 30, print1(fibonacci(n)*fibonacci(n+1)-(-1)^n", "));

a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 29 2016

Vec(-(1-4*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

[fibonacci(n)*fibonacci(n+1)-(-1)^n for n in (0..30)]

G.f.: (-1 + 4*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = F(n+2)^2 - 2*F(n+1)^2.
a(n) = A059929(n) - A059929(n-1) with A059929(-1)=1.
a(n) = -A001654(n+1) + 4*A001654(n) - A001654(n-1).
a(n) = A206351((n+2)/2)-2 for even n; a(n) = A003482((n-1)/2)+2 for odd n.
Sum_{i>=0} 1/a(i) = .754301907697893871765121109686...
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=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).

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)

cp's OEIS Frontend

A003482 a(n) = 7*a(n-1) - a(n-2) + 4, with a(0) = 0, a(1) = 5.

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A090162 Values of binomial(Fibonacci(2k)*Fibonacci(2k+1),Fibonacci(2k-1)*Fibonacci(2k)-1).

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A206334 Numbers n such that there is a triangle with area n, side n, and the other two sides rational.

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A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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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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A090162 Values of binomial(Fibonacci(2k)Fibonacci(2k+1),Fibonacci(2k-1)Fibonacci(2k)-1).

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).