A081018 -id:A081018 - cp's OEIS frontend

A033889 a(n) = Fibonacci(4*n + 1).

1, 5, 34, 233, 1597, 10946, 75025, 514229, 3524578, 24157817, 165580141, 1134903170, 7778742049, 53316291173, 365435296162, 2504730781961, 17167680177565, 117669030460994, 806515533049393, 5527939700884757, 37889062373143906, 259695496911122585, 1779979416004714189

Offset: 0

N. J. A. Sloane

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

Bruno Berselli, Table of n, a(n) for n = 0..500
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Kai Wan, Problem H-90, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 3 (2022), p. 282; Solution to Problem H-90, by Ángel Plaza, ibid., Vol. 62, No. 1 (2024), pp. 95-96.
Roman Wituła, Binomials Transformation Formulae of Scaled Lucas Numbers, Demonstratio Mathematica, Vol. XLVI, No. 1 (2013), pp. 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A014445, A016813, A122070, A167816, A094567.

Cf. A003881, A049684, A081018, A081016, A081068, A172968.

[Fibonacci(4*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Table[Fibonacci[4*n+1], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)

a(n)=fibonacci(4*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-2*x)/(1-7*x+x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = 7*a(n-1) - a(n-2) for n >= 2. - Floor van Lamoen, Dec 10 2001
From R. J. Mathar, Jan 17 2008: (Start)
O.g.f.: (1 - 2*x)/(1 - 7*x + x^2).
a(n) = A004187(n+1) - 2*A004187(n). (End); corrected by Klaus Purath, Jul 29 2020
a(n) = A167816(4*n+1). - Reinhard Zumkeller, Nov 13 2009
a(n) = sqrt(1 + 2 * Fibonacci(2*n) * Fibonacci(2*n + 1) + 5 * (Fibonacci(2*n) * Fibonacci(2*n + 1))^2). - Artur Jasinski, Feb 06 2010
a(n) = Sum_{k=0..n} A122070(n,k)*2^k. - Philippe Deléham, Mar 13 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n)*Fibonacci(2*n+2) + 1. - Gary Detlefs, Apr 18 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n+1)^2. - Bruno Berselli, Apr 19 2012
a(n) = Sum_{k = 0..n} A238731(n,k)*4^k. - Philippe Deléham, Mar 05 2014
a(n) = A000045(A016813(n)). - Michel Marcus, Mar 05 2014
2*a(n) = Fibonacci(4*n) + Lucas(4*n). - Bruno Berselli, Oct 13 2017
a(n) = A094567(n-1) + A094567(n), assuming A094567(-1) = 0. - Klaus Purath, Jul 29 2020
Sum_{n>=0} (-1)^n * arctan(3/a(n)) = Pi/4 (A003881) (Wan, 2022). - Amiram Eldar, Mar 01 2024
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A080239 Antidiagonal sums of triangle A035317.

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 15, 24, 40, 64, 104, 168, 273, 441, 714, 1155, 1870, 3025, 4895, 7920, 12816, 20736, 33552, 54288, 87841, 142129, 229970, 372099, 602070, 974169, 1576239, 2550408, 4126648, 6677056, 10803704, 17480760, 28284465, 45765225, 74049690

Offset: 1

Paul Barry, Feb 11 2003

Convolution of Fibonacci sequence with sequence (1, 0, 0, 0, 1, 0, 0, 0, 1, ...).
There is an interesting relation between a(n) and the Fibonacci sequence f(n). Sqrt(a(4n-2)) = f(2n). By using this fact we can calculate the value of a(n) by the following (1),(2),(3),(4) and (5). (1) a(1) = 1. (2) If n = 2 (mod 4), then a(n) = f((n+2)/2)^2. (3) If n = 3 (mod 4), then a(n) = (f((n+5)/2)^2-2f((n+1)/2)^2-1)/3. (4) If n = 0 (mod 4), then a(n) = (f((n+4)/2)^2+f(n/2)^2-1)/3. (5) If n = 1 (mod 4), then a(n) = (2f((n+3)/2)^2-f((n-1)/2)^2+1)/3. - Hiroshi Matsui and Ryohei Miyadera, Aug 08 2006
Sequences of the form s(0)=a, s(1)=b, s(n) = s(n-1) + s(n-2) + k if n mod m = p, else s(n) = s(n-1) + s(n-2) will have a form fib(n-1)*a + fib(n)*b + P(x)*k. a(n) is the P(x) sequence for m=4...s(n) = fib(n)*a + fib(n-1)*b + a(n-4-p)*k. - Gary Detlefs, Dec 05 2010
A different formula for a(n) as a function of the Fibonacci numbers f(n) may be conjectured. The pattern is of the form a(n) = f(p)*f(p-q) - 1 if n mod 4 = 3, else f(p)*f(p-q) where p = 2,2,4,4,4,4,6,6,6,6,8,8,8,8... and q = 0,1,3,2,0,1,3,2,0,1,3,2... p(n) = 2 * A002265(n+4) = 2*(floor((n+3)/2) - floor((n+3)/4)) (see comment by Jonathan Vos Post at A002265). A general formula for sequences having period 4 with terms a,b,c,d is given in A121262 (the discrete Fourier transform, as for all periodic sequences) and is a function of t(n)= 1/4*(2*cos(n*Pi/2) + 1 + (-1)^n). r4(a,b,c,d,n) = a*t(n+3) + b*t(n+2) + c*t(n+1) + d*t(n). This same formula may be used to subtract the 1 at n mod 4 = 3. a(n) = f(p(n))*f(p(n) - r4(1,0,3,2,n)) - r4(0,0,1,0,n). - Gary Detlefs, Dec 09 2010
This sequence is the sequence B4,1 on p. 34 of "Pascal-like triangles and Fibonacci-like sequences" in the references. In this article the authors treat more general sequences that have this sequence as an example. - Hiroshi Matsui and Ryohei Miyadera, Apr 11 2014
It is easy to see that a(n) = a(n-4) + f(n), where f(n) is the Fibonacci sequence. By using this repeatedly we have for a natural number m
a(4m) =a(4) + f(4m) + f(4m-4) + ... + f(8),
a(4m+1) = a(1) + f(4m) + f(4m-4) + ... + f(5),
a(4m+2) = a(2) + f(4m) + f(4m-4) + ... + f(6) and
a(4m+3) = a(3) + f(4m) + f(4m-4) + ... + f(7).
- Wataru Takeshita and Ryohei Miyadera, Apr 11 2014
a(n-1) counts partially ordered partitions of (n-1) into (1,2,3,4) where the position (order) of 2's is unimportant. E.g., a(5)=6 (n-1)=4 These are (4),(31),(13),(22),(211,121,112=one),(1111). - David Neil McGrath, May 12 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
H. Matsui et al., Problem B-1019, Fibonacci Quarterly, Vol. 45, Number 2; 2007; p. 182.
H. Matsui and R. Miyadera et al., Pascal-like triangles and Fibonacci-like sequences, The Mathematical Gazette, Vol. 94, Number 529; March 2010; pp. 27-41.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,1,-1,-1).

Cf. A000045, A004695, A026636, A026647, A035317.

List([1..40], n-> Sum([0..Int((n-1)/4)], k-> Fibonacci(n-4*k) )); # G. C. Greubel, Jul 13 2019

a080239 n = a080239_list !! (n-1)
a080239_list = 1 : 1 : zipWith (+)
   (tail a011765_list) (zipWith (+) a080239_list $ tail a080239_list)
-- Reinhard Zumkeller, Jan 06 2012

I:=[1,1,2,3,6,9]; [n le 6 select I[n] else Self(n-1)+Self(n-2)+Self(n-4)-Self(n-5)-Self(n-6): n in [1..50]]; // Vincenzo Librandi, Jun 07 2015

f:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 4 = 1 then t1:=1 else t1:=0; fi: f(n-1)+f(n-2)+t1; end; [seq(f(n), n=1..100)]; # N. J. A. Sloane, May 25 2008
with(combinat): f:=n-> fibonacci(n): p:=n-> 2*(floor((n+3)/2)-floor((n+3)/4)): t:=n-> 1/4*(2*cos(n*Pi/2)+1+(-1)^n): r4:=(a,b,c,d,n)-> a*t(n+3)+b*t(n+2)+c*t(n+1)+d*t(n): seq(f(p(n))*f(p(n)-r4(1,0,3,2,n))-r4(0,0,1,0,n), n = 1..33); # Gary Detlefs, Dec 09 2010
with(combinat): a:=proc(n); add(fibonacci(n-4*k),k=0..floor((n-1)/4)) end: seq(a(n), n = 1..33); # Johannes W. Meijer, Apr 19 2012

(*f[n] is the Fibonacci sequence and a[n] is the sequence of A080239*) f[n_]:= f[n] =f[n-1] +f[n-2]; f[1]=1; f[2]=1; a[n_]:= Which[n==1, 1, Mod[n, 4]==2, f[(n+2)/2]^2, Mod[n, 4]==3, (f[(n+5)/2]^2 - 2f[(n + 1)/2]^2 -1)/3, Mod[n, 4]==0, (f[(n+4)/2]^2 + f[n/2]^2 -1)/3, Mod[n, 4] == 1, (2f[(n+3)/2]^2 -f[(n-1)/2]^2 +1)/3] (* Hiroshi Matsui and Ryohei Miyadera, Aug 08 2006 *)
a=0; b=0; lst={a,b}; Do[z=a+b+1; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z,{n,4!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 16 2010 *)
(* Let f[n] be the Fibonacci sequence and a2[n] the sequence A080239 expressed by another formula discovered by Wataru Takeshita and Ryohei Miyadera *)
f=Fibonacci; a2[n_]:= Block[{m, s}, s = Mod[n, 4]; m = (n-s)/4;
Which[n==1, 1, n==2, 1, n==3, 2, s==0, 3 + Sum[f[4 i], {i, 2, m}], s == 1, 1 + Sum[f[4i+1], {i, 1, m}], s==2, 1 + Sum[f[4i+2], {i, 1, m}], s == 3, 2 + Sum[f[4i+3], {i, 1, m}]]]; Table[a2[n], {n, 1, 40}] (* Ryohei Miyadera, Apr 11 2014, minor update by Jean-François Alcover, Apr 29 2014 *)
LinearRecurrence[{1, 1, 0, 1, -1, -1}, {1, 1, 2, 3, 6, 9}, 41] (* Vincenzo Librandi, Jun 07 2015 *)

vector(40, n, f=fibonacci; sum(k=0,((n-1)\4), f(n-4*k))) \\ G. C. Greubel, Jul 13 2019

[sum(fibonacci(n-4*k) for k in (0..floor((n-1)/4))) for n in (1..40)] # G. C. Greubel, Jul 13 2019

G.f.: x/((1-x^4)(1 - x - x^2)) = x/(1 - x - x^2 - x^4 + x^5 + x^6).
a(n) = a(n-1) + a(n-2) + a(n-4) - a(n-5) - a(n-6).
a(n) = Sum_{j=0..floor(n/2)} Sum_{k=0..floor((n-j)/2)} binomial(n-j-2k, j-2k) for n>=0.
Another recurrence is given in the Maple code.
If n mod 4 = 1 then a(n) = a(n-1) + a(n-2) + 1, else a(n)= a(n-1) + a(n-2). - Gary Detlefs, Dec 05 2010
a(4n) = A058038(n) = Fibonacci(2n+2)*Fibonacci(2n).
a(4n+1) = A081016(n) = Fibonacci(2n+2)*Fibonacci(2n+1).
a(4n+2) = A049682(n+1) = Fibonacci(2n+2)^2.
a(4n+3) = A081018(n+1) = Fibonacci(2n+2)*Fibonacci(2n+3).
a(n) = 8*a(n-4) - 8*a(n-8) + a(n-12), n>12. - Gary Detlefs, Dec 10 2010
a(n+1) = a(n) + a(n-1) + A011765(n+1). - Reinhard Zumkeller, Jan 06 2012
a(n) = Sum_{k=0..floor((n-1)/4)} Fibonacci(n-4*k). - Johannes W. Meijer, Apr 19 2012

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

Original entry on oeis.org

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)

A089508 Solution to a binomial problem together with companion sequence A081016(n-1).

Original entry on oeis.org

1, 14, 103, 713, 4894, 33551, 229969, 1576238, 10803703, 74049689, 507544126, 3478759199, 23843770273, 163427632718, 1120149658759, 7677619978601, 52623190191454, 360684711361583, 2472169789339633, 16944503814015854

Offset: 1

Wolfdieter Lang, Dec 01 2003

a(n) and b(n) := A081016(n-1) are the solutions to the Diophantine equation binomial(a,b) = binomial(a+1,b-1). The first few binomials are given by A090162(n).

			n = 2: a(2) = 14, b(2) = A081016(1) = 6 satisfy binomial(14,6) = 3003 = binomial(15,5). 3003 = A090162(2).

A. I. Shirshov: On the equation binomial(n,m)=binomial(n+1,m-1), pp. 83-86, in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Wikipedia, Singmaster's conjecture
Index entries for linear recurrences with constant coefficients, signature (8, -8, 1).

Equals A081018 - 1.

[Fibonacci(2*n)*Fibonacci(2*n+1) - 1: n in [1..30]]; // G. C. Greubel, Dec 18 2017

Rest[CoefficientList[Series[x*(1 + 6*x - x^2)/((1 - x)*(1 - 7*x + x^2)), {x, 0, 50}], x]] (* G. C. Greubel, Dec 18 2017 *)

x='x+O('x^30); Vec(x*(1 + 6*x - x^2)/((1 - x)*(1 - 7*x + x^2))) \\ G. C. Greubel, Dec 18 2017

G.f.: x*(1+6*x-x^2)/((1-x)*(1-7*x+x^2)).

a(n) = A081018(n) - 1 = F(2*n)*F(2*n+1) - 1, n>=1; with F(n) := A000045(n) (Fibonacci).

A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.

Original entry on oeis.org

1, 3, 16, 105, 715, 4896, 33553, 229971, 1576240, 10803705, 74049691, 507544128, 3478759201, 23843770275, 163427632720, 1120149658761, 7677619978603, 52623190191456, 360684711361585, 2472169789339635

Offset: 1

James R. Buddenhagen, Feb 06 2012

A Pell sequence related to Heronian triangles (rational triangles), see A206334. The connection is this: consider the problem of finding triangles with area a positive integer n, and with sides (a, b, n) where a, b are rational. Note that n is both the area and one side. For many values of n this is not possible, and the sequence of such numbers n is quite erratic (see A206334). Nonetheless, each term in this sequence is such a value of n. For example, for n = 105 you can take the other two sides, a and b, to be 10817/104, and 233/104 and the area will equal n, i.e., 105.

			G.f. = x + 3*x^2 + 16*x^3 + 105*x^4 + 715*x^5 + 4896*x^6 + 33553*x^7 + ... - _Michael Somos_, Jun 26 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A003482, A089508, A094214.

Subsequence of A206334.

a206351 n = a206351_list !! (n-1)
a206351_list = 1 : 3 : map (subtract 4)
               (zipWith (-) (map (* 7) (tail a206351_list)) a206351_list)
-- Reinhard Zumkeller, Feb 08 2012

[Fibonacci(2*n)*Fibonacci(2*n-3): n in [1..30]]; // G. C. Greubel, Aug 12 2018

genZ := proc(n)
local start;
option remember;
    start := [1, 3];
    if n < 3 then start[n]
    else 7*genZ(n - 1) - genZ(n - 2) - 4
    end if
end proc:
seq(genZ(n),n=1..20);

LinearRecurrence[{8, -8, 1}, {1, 3, 16}, 50] (* Charles R Greathouse IV, Feb 07 2012 *)
RecurrenceTable[{a[1] == 1, a[2] == 3, a[n] == 7 a[n - 1] - a[n - 2] - 4}, a, {n, 20}] (* Bruno Berselli, Feb 07 2012 *)
a[ n_] := Fibonacci[2 n] Fibonacci[2 n - 3]; (* Michael Somos, Jun 26 2018 *)
nxt[{a_,b_}]:={b,7b-a-4}; NestList[nxt,{1,3},20][[;;,1]] (* Harvey P. Dale, Aug 29 2024 *)

Vec((1-5*x)/(1-8*x+8*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Feb 07 2012

{a(n) = fibonacci(2*n) * fibonacci(2*n - 3)}; /* Michael Somos, Jun 26 2018 */

From Bruno Berselli, Feb 07 2012: (Start)
G.f.: x*(1-5*x)/(1-8*x+8*x^2-x^3).
a(n) = A081018(n-1) + 1. (End)
a(n) = -A003482(-n) = Fibonacci(2*n)*Fibonacci(2*n-3). - Michael Somos, Jun 26 2018
a(n) = A089508(n-1) + 2 for n>1. - Bruno Berselli, Jun 20 2019 [Formula found by Umberto Cerruti]
Product_{n>=2} (1 - 1/a(n)) = 1/phi (A094214) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021
a(n) = (Fibonacci(2*n-2) + 1/Lucas(2*n-2))*(Fibonacci(2*n-1) + 1/Lucas(2*n-1)). - Peter Bala, Sep 03 2022

A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 5, 31, 209, 1429, 9791, 67105, 459941, 3152479, 21607409, 148099381, 1015088255, 6957518401, 47687540549, 326855265439, 2240299317521, 15355239957205, 105246380382911, 721369422723169, 4944339578679269, 33889007628031711, 232278713817542705

Offset: 0

XU Pingya, Sep 30 2020

			2*(F(5)^2)^3 + 2*(-F(4)^2)^3 + (-31)^3 = 2*(25)^3 + 2*(-9)^3 + (-31)^3 = 1, a(2) = 31.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A000578, A056573, A081018, A337929.

Table[(2*Fibonacci[2n+1]^6 - 2*Fibonacci[2n]^6 - 1)^(1/3), {n, 0, 21}]
Table[(Fibonacci[2n+1]*Fibonacci[2n+2]- Fibonacci[2n]^2), {n, 0, 21}] (* Wolfgang Berndt, May 26 2023 *)
LinearRecurrence[{8,-8,1},{1,5,31},30] (* Harvey P. Dale, Dec 17 2023 *)

Vec((1 - 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)) + O(x^20)) \\ Colin Barker, Oct 01 2020

a(n) = (2*F(2*n+1)^6 - 2*F(2*n)^6 - 1)^(1/3).
From Colin Barker, Oct 01 2020: (Start)
G.f.: (1 - 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>2.
(End)
a(n) = 2*A081018(n) + 1. - Hugo Pfoertner, Oct 01 2020
a(n) = A064170(n+2) + A033888(n). - Flávio V. Fernandes, Jan 10 2021
a(n) = F(2*n+1)*F(2*n+2) - F(2*n)^2. - Wolfgang Berndt, May 26 2023
a(2*n-1) = 5 + 6*Sum_{k=1..n-1} F(8*k+1), a(2*n) = 1 + 6*Sum_{k=1..n} F(8*k-3). - XU Pingya, Jun 09 2024

A081007 a(n) = Fibonacci(4n+1) - 1, or Fibonacci(2n)*Lucas(2n+1).

Original entry on oeis.org

0, 4, 33, 232, 1596, 10945, 75024, 514228, 3524577, 24157816, 165580140, 1134903169, 7778742048, 53316291172, 365435296161, 2504730781960, 17167680177564, 117669030460993, 806515533049392, 5527939700884756, 37889062373143905, 259695496911122584

Offset: 0

R. K. Guy, Mar 01 2003

Also the index of the first of two consecutive triangular numbers whose sum is equal to the sum of two consecutive heptagonal numbers. - Colin Barker, Dec 20 2014

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75

Nathaniel Johnston, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers).

Cf. A081018, A242671.

List([0..30], n-> Fibonacci(4*n+1)-1); # G. C. Greubel, Jul 14 2019

[Fibonacci(4*n+1) -1: n in [0..30]]; // Vincenzo Librandi, Apr 15 2011

with(combinat) for n from 0 to 30 do printf(`%d,`,fibonacci(4*n+1)-1) od # James Sellers, Mar 03 2003

Table[Fibonacci[4n+1] -1, {n,0,30}] (* Wesley Ivan Hurt, Oct 06 2013 *)
LinearRecurrence[{8,-8,1},{0,4,33},30] (* Harvey P. Dale, Jul 31 2018 *)
Table[Fibonacci[2n]LucasL[2n+1], {n,0,30}] (* Rigoberto Florez, Apr 19 2019 *)

A081007(n):=fib(4*n+1)-1$
makelist(A081007(n),n,0,30); /* Martin Ettl, Nov 12 2012 */

concat(0, Vec(x*(4+x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ Colin Barker, Dec 20 2014

vector(30, n, n--; fibonacci(4*n+1)-1) \\ G. C. Greubel, Jul 14 2019

[fibonacci(4*n+1)-1 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: x*(4+x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Jun 24 2012
a(n) = Sum_{i=1..2n} binomial(2n+i, 2n-i). - Wesley Ivan Hurt, Oct 06 2013
a(n) = Sum_{i=0..2n-1} F(i)*L(i+2), F(i) = A000045(i) and L(i) = A000032(i). - Rigoberto Florez, Apr 19 2019
Product_{n>=1} (1 - 1/a(n)) = (1 + 1/sqrt(5))/2 (A242671). - Amiram Eldar, Nov 28 2024

More terms from James Sellers, Mar 03 2003

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.

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)

cp's OEIS Frontend

A033889 a(n) = Fibonacci(4*n + 1).

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A080239 Antidiagonal sums of triangle A035317.

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

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A089508 Solution to a binomial problem together with companion sequence A081016(n-1).

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A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.

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A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

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A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where 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).