A081016 -id:A081016 - cp's OEIS frontend

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

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).

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

Original entry on oeis.org

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

A104763 Triangle read by rows: Fibonacci(1), Fibonacci(2), ..., Fibonacci(n) in row n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 8, 1, 1, 2, 3, 5, 8, 13, 1, 1, 2, 3, 5, 8, 13, 21, 1, 1, 2, 3, 5, 8, 13, 21, 34, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233

Offset: 1

Gary W. Adamson, Mar 23 2005

Triangle of A104762, Fibonacci sequence in each row starts from the right.
The triangle or chess sums, see A180662 for their definitions, link the Fibonacci(n) triangle to sixteen different sequences, see the crossrefs. The knight sums Kn14 - Kn18 have been added. As could be expected all sums are related to the Fibonacci numbers. - Johannes W. Meijer, Sep 22 2010
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A104763 is reluctant sequence of Fibonacci numbers (A000045), except 0. - Boris Putievskiy, Dec 13 2012

			First few rows of the triangle are:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 2, 3;
  1, 1, 2, 3, 5;
  1, 1, 2, 3, 5, 8;
  1, 1, 2, 3, 5, 8, 13; ...

Reinhard Zumkeller, Rows n = 1..100 of table, flattened
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A104762, A000045.
Cf. A000071 (row sums). - R. J. Mathar, Jul 22 2009
Triangle sums (see the comments): A000071 (Row1; Kn4 & Ca1 & Ca4 & Gi1 & Gi4); A008346 (Row2); A131524 (Kn11); A001911 (Kn12); A006327 (Kn13); A167616 (Kn14); A180671 (Kn15); A180672 (Kn16); A180673 (Kn17); A180674 (Kn18); A052952 (Kn21 & Kn22 & Kn23 & Fi2 & Ze2); A001906 (Kn3 &Fi1 & Ze3); A004695 (Ca2 & Ze4); A001076 (Ca3 & Ze1); A080239 (Gi2); A081016 (Gi3). - Johannes W. Meijer, Sep 22 2010

Flat(List([1..15], n -> List([1..n], k -> Fibonacci(k)))); # G. C. Greubel, Jul 13 2019

a104763 n k = a104763_tabl !! (n-1) !! (k-1)
a104763_row n = a104763_tabl !! (n-1)
a104763_tabl = map (flip take $ tail a000045_list) [1..]
-- Reinhard Zumkeller, Aug 15 2013

[Fibonacci(k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 13 2019

Table[Fibonacci[k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jul 13 2019 *)

for(n=1,15, for(k=1,n, print1(fibonacci(k), ", "))) \\ G. C. Greubel, Jul 13 2019

[[fibonacci(k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Jul 13 2019

F(1) through F(n) starting from the left in n-th row.
T(n,k) = A000045(k), 1<=k<=n. - R. J. Mathar, May 02 2008
a(n) = A000045(m), where m= n-t(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 13 2012
G.f.: (x*y)/((x-1)*(x^2*y^2+x*y-1)). - Vladimir Kruchinin, Jun 21 2025

Edited by R. J. Mathar, May 02 2008

Extended by R. J. Mathar, Aug 27 2008

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

A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

Original entry on oeis.org

0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855, 116139356908771352, 796030994547383610

Offset: 0

R. K. Guy, Mar 01 2003

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010
Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014
Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015
The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

			G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Colin Barker, Table of n, a(n) for n = 0..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.
Dae S. Hong, When is the Generating Function of the Fibonacci Numbers an Integer?, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Cf. A001622, A081007, A081016, A178898.

List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # G. C. Greubel, Jul 14 2019

[(Lucas(4*n+1)-1)/5: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # James Sellers, Mar 03 2003

(LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* G. C. Greubel, Aug 24 2015, modified Jul 14 2019 *)

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

[(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Mar 30 2012
E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 24 2015
From - Michael Somos, Aug 27 2015: (Start)
a(n) = -A081016(-1-n) for all n in Z.
0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - Stefano Spezia, Mar 11 2021
Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021

More terms from James Sellers, Mar 03 2003

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

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604

Offset: 0

Bruno Berselli, Jun 19 2017

a(n) mod 4 gives A101000.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A101000.
Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

[Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017

LucasL[4 Range[0, 21] + 3]
LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)

Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

from sympy import lucas
def a(n):  return lucas(4*n + 3)
print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021

def L():
    x, y = -1, 4
    while True:
        yield y
        x, y = y, 7*y - x
r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

A094565 Triangle read by rows: binary products of Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 13, 15, 16, 21, 34, 39, 40, 42, 55, 89, 102, 104, 105, 110, 144, 233, 267, 272, 273, 275, 288, 377, 610, 699, 712, 714, 715, 720, 754, 987, 1597, 1830, 1864, 1869, 1870, 1872, 1885, 1974, 2584, 4181, 4791, 4880, 4893, 4895, 4896, 4901, 4935, 5168, 6765

Offset: 1

Clark Kimberling, May 12 2004

Row n consists of n numbers, first F(2n-1) and last F(2n).
Central numbers: (1,6,40,273,...) = A081016.
Row sums: A001870.
Alternating row sums: 1,1,7,7,48,48,329,329; the sequence b=(1,7,48,329,...) is A004187, given by b(n)=F(4n+2)-b(n-1) for n>=2, with b(1)=1.
In each row, the difference between neighboring terms is a Fibonacci number.

			Triangle begins:
   1;
   2,   3;
   5,   6    8;
  13,  15,  16,  21;
  34,  39,  40,  42,  55;
  89, 102, 104, 105, 110, 144; ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Cf. A000045, A094566, A094568.

Flat(List([1..12], n-> List([1..n], k-> Fibonacci(2*k)*Fibonacci(2*n-2*k+1) ))); # G. C. Greubel, Jul 15 2019

[Fibonacci(2*k)*Fibonacci(2*n-2*k+1): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 15 2019

Table[Fibonacci[2*k]*Fibonacci[2*n-2*k+1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 15 2019 *)

row(n) = vector(n, k, fibonacci(2*k)*fibonacci(2*n-2*k+1));
tabl(nn) = for(n=1, nn, print(row(n))); \\ Michel Marcus, May 03 2016

[[fibonacci(2*k)*fibonacci(2*n-2*k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 15 2019

Row n: F(2)F(2n-1), F(4)F(2n-3), ..., F(2n)F(1).

A094566 Triangle of binary products of Fibonacci numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 9, 10, 13, 21, 24, 25, 26, 34, 55, 63, 64, 65, 68, 89, 144, 165, 168, 169, 170, 178, 233, 377, 432, 440, 441, 442, 445, 466, 610, 987, 1131, 1152, 1155, 1156, 1157, 1165, 1220, 1597, 2584, 2961, 3016, 3024, 3025, 3026, 3029, 3050, 3194, 4181

Offset: 1

Clark Kimberling, May 12 2004

For n>1, row n consists of n numbers, first F(2n-2) and last F(2n-1).
Central numbers: (1,4,9,25,64,...), essentially A081016.
Row sums: A027991. Alternating row sums: 1,1,4,4,30,30,203,203; the sequence b=(1,4,30,203,1394,...) is A094567.
In each row, the difference between neighboring terms is a Fibonacci number.

			Rows 1 to 4:
1
1 2
3 4 5
8 9 10 13

Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Cf. A000045, A094565, A094567, A094568.

pef(k, n) = fibonacci(2*k)*fibonacci(2*n-2*k);
pof(k, n) = fibonacci(2*n-2*k+1)*fibonacci(2*k-1);
tabl(nn) = {for (n=1, nn, if (n==1, print1(1, ", "), if (n % 2 == 0, for (k=1, n/2, print1(pef(k,n), ", ");); forstep (k=n/2, 1, -1, print1(pof(k,n), ", "););, for (k=1, n\2, print1(pef(k,n), ", ");); forstep (k=n\2+1, 1, -1, print1(pof(k,n), ", ");););); print(););} \\ Michel Marcus, May 04 2016

Row 1 is the single number 1. For m>=1, Row 2m: F(2)F(4m-2), F(4)F(4m-4), ..., F(2m)F(2m), F(2m+1)F(2m-1), F(2m+3)F(2m-3), ..., F(4m-1)F(1) Row 2m+1: F(2)F(4m), F(4)F(4m-2), ..., F(2m+1)F(2m+1), F(2m+3)F(2m-1), F(2m+5)F(2m-3), ..., F(4m+1)F(1)

A145995 a(n) = 8 - 12n + 5n^2.

Original entry on oeis.org

1, 4, 17, 40, 73, 116, 169, 232, 305, 388, 481, 584, 697, 820, 953, 1096, 1249, 1412, 1585, 1768, 1961, 2164, 2377, 2600, 2833, 3076, 3329, 3592, 3865, 4148, 4441, 4744, 5057, 5380, 5713, 6056, 6409, 6772, 7145, 7528, 7921, 8324, 8737, 9160, 9593, 10036

Offset: 1

Artur Jasinski, Oct 26 2008

For n > 1, a(n) is square if and only if n-1 is in A081016.

a(n) and a(-n) give all numbers m such that 5*m-4 is a square. - Bruno Berselli, Feb 03 2016

			A081016(0) = 1 and a(2) = 2^2 = 4; A081016(1) = 6 and a(7) = 13^2 = 169; A081016(2) = 40 and a(41) = 89^2 = 7921; A081016(3) = 273 and a(274) = 610^2 = 372100; A081016(4) = 1870 and a(1871) = 4181^2 = 17480761. - _Klaus Brockhaus_, Oct 29 2008

G. C. Greubel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A081016, A000217, A080855.

Cf. A195162 (numbers m such that 5*m+4 is a square).

List([1..50], n-> 8-12*n+5*n^2); # G. C. Greubel, Jul 15 2019

[8-12*n+5*n^2: n in [1..50]]; // G. C. Greubel, Jul 15 2019

Table[8 -12x +5x^2, {x,50}]
s = 1; lst = {s}; Do[s += n + 2; AppendTo[lst, s], {n, 1, 450, 10}]; lst (* Zerinvary Lajos, Jul 11 2009 *)
LinearRecurrence[{3, -3, 1}, {1, 4, 17}, 51] (* G. C. Greubel, Jan 30 2016 *)

for(n=1, 50, print1(8-12*n+5*n^2, ",")) \\ Klaus Brockhaus, Oct 29 2008

[8-12*n+5*n^2 for n in (1..50)] # G. C. Greubel, Jul 15 2019

a(n) = a(n-1) + 10*n - 17, with a(1)=1. - Vincenzo Librandi, Nov 26 2010
From G. C. Greubel, Jan 30 2016 (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3, a(1)=1, a(2)=4, a(3)=17.
G.f.: x*(1 + x + 8*x^2)/(1-x)^3.
E.g.f.: (5*x^2 - 7*x + 8)*exp(x) - 8. (End)

Corrected definition; corrected comment; added keyword. - Klaus Brockhaus, Oct 29 2008

cp's OEIS Frontend

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

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

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A104763 Triangle read by rows: Fibonacci(1), Fibonacci(2), ..., Fibonacci(n) in row n.

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A080239 Antidiagonal sums of triangle A035317.

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A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

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

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

A145995 a(n) = 8 - 12n + 5n^2.