A070550 -id:A070550 - cp's OEIS frontend

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760

Offset: 0

N. J. A. Sloane

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with aa(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Ömer Egecioglu, Elif Saygi, and Zülfükar Saygi, The Mostar index of Fibonacci and Lucas cubes, arXiv:2101.04740 [math.CO], 2021. Mentions this sequence.
Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048".
Jonny Griffiths and Martin Griffiths, Fibonacci-related sequences via iterated QRT maps, Fib. Q., 51 (2013), 218-227.
James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Lemma 4.1 p. 34.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
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
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
M. Renault, Dissertation
Wikipedia, Illustration of 273 as a golden rectangle number.
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1993
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013

I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007

LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)

A001654(n)=fibonacci(n)*fibonacci(n+1);

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016

from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017

from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extended by Wolfdieter Lang, Jun 27 2000

A035607 Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 8, 2, 1, 8, 18, 12, 2, 1, 10, 32, 38, 16, 2, 1, 12, 50, 88, 66, 20, 2, 1, 14, 72, 170, 192, 102, 24, 2, 1, 16, 98, 292, 450, 360, 146, 28, 2, 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2, 1, 20, 162, 688, 1666, 2364, 1970, 952, 258, 36, 2, 1, 22, 200, 978, 2816

Offset: 0

N. J. A. Sloane

Table also gives coordination sequences of same lattices.
Rows sums are given by A001333. Rising and falling diagonals are the tribonacci numbers A000213, A001590. - Paul Barry, Feb 13 2003
a(d,m) also gives the number of ways to choose m squares from a 2 X (d-1) grid so that no two squares in the selection are (horizontally or vertically) adjacent. - Jacob A. Siehler, May 13 2006
Mirror image of triangle A113413. - Philippe Deléham, Oct 15 2006
The Ca1 sums lead to A126116 and the Ca2 sums lead to A070550, see A180662 for the definitions of these triangle sums. - Johannes W. Meijer, Aug 05 2011
A035607 is jointly generated with the Delannoy triangle A008288 as an array of coefficients of polynomials v(n,x):  initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x).  See the Mathematica section. - Clark Kimberling, Mar 05 2012
Also, the polynomial v(n,x) above is x + (x + 1)*f(n-1,x), where f(0,x) = 1. - Clark Kimberling, Oct 24 2014
Rows also give the coefficients of the independence polynomial of the n-ladder graph. - Eric W. Weisstein, Dec 29 2017
Considering both sequences as square arrays (offset by one row), the rows of A035607 are the first differences of the rows of A008288, and the rows of A008288 are the partial sums of the rows of A035607. - Shel Kaphan, Feb 23 2023
Considering only points with nonnegative coordinates, the number of points at L1 distance = m in d dimensions is the same as the number of ways of putting m indistinguishable balls into d distinguishable urns, binomial(m+d-1, d-1). This is one facet of the cross-polytope. Allowing for + and - coordinates, there are binomial(d,i)*2^i facets containing points with up to i nonzero coordinates. Eliminating double counting of points with any coordinates = 0, there are Sum_{i=1..d} (-1)^(d-i)*binomial(m+i-1,i-1)*binomial(d,i)*2^i points at distance m in d dimensions. One may avoid the alternating sum by using binomial(m-1,i-1) to count only the points per facet with exactly i nonzero coordinates, avoiding any double counting, but the result is the same. - Shel Kaphan, Mar 04 2023

			From _Clark Kimberling_, Oct 24 2014: (Start)
As a triangle of coefficients in polynomials v(n,x) in Comments, the first 6 rows are
  1
  1   2
  1   4   2
  1   6   8   2
  1   8  18  12   2
  1  10  32  38  16   2
  ... (End)
From _Shel Kaphan_, Mar 04 2023: (Start)
For d=3, m=4:
There are binomial(3,1)*2^1 = 6 facets (vertices) of binomial(4+1-1,1-1) = 1 point with <= one nonzero coordinate.
There are binomial(3,2)*2^2 = 12 facets (edges) of binomial(4+2-1,2-1) = 5 points with <= two nonzero coordinates.
There are binomial(3,3)*2^3 = 8 facets (faces) of binomial(4+3-1,3-1) = 15 points with <= three nonzero coordinates.
a(3,4) = 8*15 - 12*5 + 6*1 = 120 - 60 + 6 = 66. (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Sect. 2.3.
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 392.
R. J. Mathar, Bivariate Generating Functions for Non-attacking Wazirs on Rectangular Boards (2024), Table 2.
Emanuele Munarini, Combinatorial properties of the antichains of a garland, Integers, 9 (2009), 353-374.
Joan Serra-Sagrista, Enumeration of lattice points in l_1 norm, Inf. Proc. Lett. 76 (1-2) (2000) 39-44.
J. Siehler, Adjacency-free selections from a 2xN grid (Mathematica notebook) [broken link]
Jacob A. Siehler, Selections Without Adjacency on a Rectangular Grid, arXiv:1409.3869 [math.CO], 2014.
Eric Weisstein's World of Mathematics, Independence Polynomial
Eric Weisstein's World of Mathematics, Ladder Graph

Other versions: A113413, A119800, A122542, A266213.
Cf. A008288, which has g.f. 1/(1-x-x*y-x^2*y).
Cf. A078057 (row sums), A050146 (central terms).
Cf. A050146.

a035607 n k = a035607_tabl !! n !! k
a035607_row n = a035607_tabl !! n
a035607_tabl = map fst $ iterate
   (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                      zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 2])
-- Reinhard Zumkeller, Jul 20 2013

A035607 := proc(d,m) local j: add(binomial(floor((d-1+j)/2),d-m-1)*binomial(d-m-1, floor((d-1-j)/2)),j=0..d-1) end: seq(seq(A035607(d,m),m=0..d-1),d=1..11); # d=dimension, m=norm # Johannes W. Meijer, Aug 05 2011

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A008288 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A035607 *)
(* Clark Kimberling, Mar 09 2012 *)
Reverse /@ CoefficientList[CoefficientList[Series[(1 + x)/(1 - x - x y - x^2 y), {x, 0, 10}], x], y] // Flatten (* Eric W. Weisstein, Dec 29 2017 *)

T(n, k) = if (k==0, 1, sum(i=0, k-1, binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1)));
tabl(nn) = for (n=1, nn, for (k=0, n-1, print1(T(n, k), ", ")); print); \\ as a triangle; Michel Marcus, Feb 27 2018

def A035607_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return prec(n-1,k-1)+2*sum(prec(n-i,k-1) for i in (2..n-k+1))
    return [prec(n, n-k) for k in (0..n-1)]
for n in (1..10): print(A035607_row(n)) # Peter Luschny, Mar 16 2016

From Johannes W. Meijer, Aug 05 2011: (Start)
f(d,m) = Sum_{j=0..d-1} binomial(floor((d-1+j)/2), d-m-1)*binomial(d-m-1, floor((d-1-j)/2)), d >= 1 and 0 <= m <= d-1.
f(d,m) = f(d-1,m-1) + f(d-1,m) + f(d-2,m-1) (d >= 3 and 1 <= m <= d-1) with f(d,0) = 1 (d >= 1) and f(d,d-1) = 2 (d>=2). (End)
From Roger Cuculière, Apr 10 2006: (Start)
The generating function G(x,y) of this double sequence is the sum of a(n,p)*x^n*y^p, n=1..oo, p=0..oo, which is G(x,y) = x*(1+y)/(1-x-y-x*y).
The horizontal generating function H_n(y), which generates the rows of the table: (1, 2, 2, 2, 2, ...), (1, 4, 8, 12, 16, ...), (1, 6, 18, 38, 66, ...), is the sum of a(n,p)*y^p, p=0..oo, for each fixed n. This is H_n(y) = ((1+y)^n)/((1-y)^n).
The vertical generating function V_p(x), which generates the columns of the table: (1, 1, 1, 1, 1, ...), (2, 4, 6, 8, 10, ...), (2, 8, 18, 32, 50, ...), is the sum of a(n,p)*x^n, n=1..oo, for each fixed p. This is V_p(x) = 2*((1+x)^(p-1))/((1-x)^(p+1)) for p >= 1 and V_0(x) = x/(1-x). (End)
G.f.: (1+x)/(1-x-x*y-x^2*y). - Vladeta Jovovic, Apr 02 2002 (But see previous lines!)
T(2*n,n) = A050146(n+1). - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle read by rows: T(n,0) = 1, for n > 1: T(n,n-1) = 2, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-1), 0 < k < n. - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle T(n,k) with 0 <= k < n read by rows: T(n,0)=1 for n > 0 and T(n,k) = Sum_{i=0..k-1} binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1) for k > 0. - Werner Schulte, Feb 22 2018
With p >= 1 and q >= 0, as a square array a(p,q) = T(p+q-1,q) = 2*p*Hypergeometric2F1[1-p, 1-q, 2, 2] for q >= 1. Consequently, a(p,q) = a(q,p)*p/q. - Shel Kaphan, Feb 14 2023
For n >= 1, T(2*n,n) = A002003(n), T(3*n,2*n) = A103885(n) and T(4*n,3*n) = A333715(n). - Peter Bala, Jun 15 2023

More terms from David W. Wilson

Maple program corrected and information added by Johannes W. Meijer, Aug 05 2011

A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

Original entry on oeis.org

0, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 20737, 54288, 142130, 372099, 974170, 2550408, 6677057, 17480760, 45765226, 119814915, 313679522, 821223648, 2149991425, 5628750624, 14736260450, 38580030723, 101003831722, 264431464440, 692290561601, 1812440220360

Offset: 0

Henry Bottomley, Feb 09 2001

Expansion of golden ratio (1+sqrt(5))/2 as an infinite product: phi = Product_{i>=0} (1+1/(Fibonacci(2*i+1) * Fibonacci(2*i+3)-1)) * (1-1/(Fibonacci(2*i+2) * Fibonacci(2i+4)+1)). - Thomas Baruchel, Nov 11 2003
Each of these is one short of or one over the square of a Fibonacci number (A007598). This means that a rectangle sized F(n) by F(n + 2) units can't be converted into a square with sides of length F(n + 1) units unless one square unit of material is added or removed. - Alonso del Arte, May 03 2011
These are the integer parts of the numerators of the numbers with continued fraction representations [1, 2, 2, 2, ...], [1, 1, 2, 2, 2, ...], [1, 1, 1, 2, 2, 2, ...], etc., that is, sqrt(2), (2+sqrt(2))/2, 3-sqrt(2), (10+sqrt(2))/7, (24-sqrt(2))/14, etc. - Geoffrey Caveney, May 03 2014
a(n) appears also as the third component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), A079472(n+1), a(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
Numbers with a continued fraction expansion with the repeating sequence of length n [1, 1, ..., 1, 2], n-1 ones followed by a single two, for n > = 1, appear to be equal to (F(n) + sqrt(a(n)))/F(n+1), where F(n) = A000045(n). - R. James Evans, Nov 21 2018
The preceding conjecture is true. Proof: For n >= 1 let c(n) := confrac(repeat(1^{n-1}, 2)) where 1^{k} denotes 1 taken k times. This can be computed, e.g. from [Perron, third and fourth eq. on p. 62], as c(n) = (F(n) + sqrt(F(n+1)^2 - (-1)^n)) / F(n+1), which is the conjectured formula because F(n+1)^2 - (-1)^n = a(n). - Wolfdieter Lang, Jan 05 2019

			G.f. = 2*x + 3*x^2 + 10*x^3 + 24*x^4 + 65*x^5 + 168*x^6 + ... - _Michael Somos_, Mar 18 2022

Oskar Perron, Die Lehre von den Kettenbrüchen, Band I, 3. Auflage, B. G. Teubner, Stuttgart, 1954, pp. 61-61.

Muniru A Asiru, Table of n, a(n) for n = 0..2374 (first 501 terms from Harry J. Smith)
Tim Jones (producer), Engineering Bits & Bytes: The Fibonacci Puzzle, Wayne State University College of Engineering (2011).
Eric H. Kuo, Applications of graphical condensation for enumerating matchings and tilings, arXiv:math/0304090 [math.CO], 2003.
Marc Renault, Properties of the Fibonacci Sequence Under Various Moduli, Master's Thesis, Wake Forest University, 1996.
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Bisection of A070550.
First differences of A059840.
Cf. A000045, A001519, A001654, A007598, A033999, A035513, A079472, A121801, A126358, A147973, A248161.

a:=List([0..30],n->Fibonacci(n)*Fibonacci(n+2));; Print(a); # Muniru A Asiru, Jan 05 2019

[Fibonacci(n)*Fibonacci(n+2): n in [0..30]]; // Vincenzo Librandi, Jul 02 2014

with(combinat): a:=n->fibonacci(n)*fibonacci(n+2): seq(a(n), n=0..26); # Zerinvary Lajos, Oct 07 2007

Table[Fibonacci[n]*Fibonacci[n+2],{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)

a(n) = fibonacci(n)*fibonacci(n + 2) \\ Harry J. Smith, Jun 30 2009

from sympy import fibonacci
[fibonacci(n)*fibonacci(n+2) for n in range(30)] # Stefano Spezia, Jan 05 2019

[fibonacci(n)*fibonacci(n+2) for n in range(30)] # G. C. Greubel, Nov 21 2018

a(n) = Fibonacci(n+1)^2 - (-1)^n = A007598(n+1) + A033999(n+1) = A000045(n+1)^2 - A033999(n).
G.f.: (2*x-x^2) / ((1+x)*(1-3*x+x^2)).
Sum_{n>=1} 1/a(n) = 1.
Sum_{n>=1} (-1)^n/a(n) = 2 - sqrt(5).
Sum_{n>=1} 1/a(2n-1) = 1/phi = (sqrt(5) - 1)/2. - Franz Vrabec, Sep 15 2005
Sum_{n>=1} 1/a(2n) = (3 - sqrt(5))/2. - Franz Vrabec, Nov 30 2009
a(n) = ((7+3*sqrt(5))/10)*((3+sqrt(5))/2)^(n-1) + ((7-3*sqrt(5))/10)*((3-sqrt(5))/2)^(n-1) + (3/5)*(-1)^(n-1). - Tim Monahan, Aug 09 2011
a(n) = (Lucas(n+1)^2 - Fibonacci(n+1)^2)/4. - Vincenzo Librandi, Aug 02 2014
a(n) = F(n-2)*F(n) + F(n-1)*F(n) + F(n-2)*F(n+1) + F(n-1)*F(n+1), where F=A000045, F(-2)=-1, F(-1)=1. - Bruno Berselli, Nov 03 2015
a(n) = A035513(1,n-1)*A035513(3,n-1)/2 = A035513(1,n-1)*A035513(4,n-1)/3. - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A001519(n+2). - R. J. Mathar, Oct 19 2021
a(n) = 2*A001654(n) - A001654(n-1). - R. J. Mathar, Oct 19 2021
a(n)+a(n+3) = 2*F(2n+5) = A126358(n+2). - Andrés Ventas, Oct 25 2021
Sum_{n>=1} Fibonacci(n+1)/a(n) = 2. - Amiram Eldar, Jan 11 2022
a(n) = a(-2-n) and a(n) + a(n+3) = 2*(a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Mar 18 2022

A171861 Expansion of x(1+x+x^2) / ( (x-1)(x^3+x^2-1) ).

Original entry on oeis.org

1, 2, 4, 6, 9, 13, 18, 25, 34, 46, 62, 83, 111, 148, 197, 262, 348, 462, 613, 813, 1078, 1429, 1894, 2510, 3326, 4407, 5839, 7736, 10249, 13578, 17988, 23830, 31569, 41821, 55402, 73393, 97226, 128798, 170622, 226027, 299423, 396652, 525453, 696078, 922108

Offset: 1

Ed Pegg Jr, Oct 16 2010

Number of wins in Penney's game if the two players start HHT and TTT and HHT beats TTT.

HHT beats TTT 70% of the time. - Geoffrey Critzer, Mar 01 2014

			a(n) enumerates length n+2 sequences on {H,T} that end in HHT but do not contain the contiguous subsequence TTT.
a(3)=4 because we have: TTHHT, THHHT, HTHHT, HHHHT.
a(4)=6 because we have: TTHHHT, THTHHT, THHHHT, HTTHHT, HTHHHT, HHHHHT. - _Geoffrey Critzer_, Mar 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..1000
Wikipedia, Penney's game
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1).

Related sequences are A000045 (HHH beats HHT, HTT beats TTH), A006498 (HHH beats HTH), A023434 (HHH beats HTT), A000930 (HHH beats THT, HTH beats HHT), A000931 (HHH beats TTH), A077868 (HHT beats HTH), A002620 (HHT beats HTT), A000012 (HHT beats THH), A004277 (HHT beats THT), A070550 (HTH beats HHH), A000027 (HTH beats HTT), A097333 (HTH beats THH), A040000 (HTH beats TTH), A068921 (HTH beats TTT), A054405 (HTT beats HHH), A008619 (HTT beats HHT), A038718 (HTT beats THT), A128588 (HTT beats TTT).

Cf. A164315 (essentially the same sequence).

A171861 := proc(n) option remember; if n <=4 then op(n,[1,2,4,6]); else procname(n-1)+procname(n-2)-procname(n-4) ; end if; end proc:

nn=44;CoefficientList[Series[x(1+x+x^2)/(1-x-x^2+x^4),{x,0,nn}],x] (* Geoffrey Critzer, Mar 01 2014 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,1,1]^(n-1)*[1;2;4;6])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

a(n) = a(n-1) +a(n-2) -a(n-4) = A000931(n+10)-3 = A134816(n+6)-3 = A078027(n+12)-3.

a(n) = A164315(n-1). - Alois P. Heinz, Oct 12 2017

A126116 a(n) = a(n-1) + a(n-3) + a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 7, 11, 19, 31, 49, 79, 129, 209, 337, 545, 883, 1429, 2311, 3739, 6051, 9791, 15841, 25631, 41473, 67105, 108577, 175681, 284259, 459941, 744199, 1204139, 1948339, 3152479, 5100817, 8253295, 13354113, 21607409, 34961521

Offset: 0

Luis A Restrepo (luisiii(AT)mac.com), Mar 05 2007

nonn

This sequence has the same growth rate as the Fibonacci sequence, since x^4 - x^3 - x - 1 has the real roots phi and -1/phi.

The Ca1 sums, see A180662 for the definition of these sums, of triangle A035607 equal the terms of this sequence without the first term. - Johannes W. Meijer, Aug 05 2011

			G.f. = 1 + x + x^2 + x^3 + 3*x^4 + 5*x^5 + 7*x^6 + 11*x^7 + 19*x^8 + 31*x^9 + ...

S. Wolfram, A New Kind of Science. Champaign, IL: Wolfram Media, pp. 82-92, 2002

Seiichi Manyama, Table of n, a(n) for n = 0..4786
K. T. Atanassov, D. R. Deford, A. G. Shannon, Pulsated Fibonacci recurrences, Fibonacci Quarterly, Vol. 52, No. 5, Dec. 2014, pp. 22-27.
Kelley L. Ross, The Golden Ratio and The Fibonacci Numbers
Eric Weisstein's World of Mathematics, Golden Ratio
Wikipedia, Golden Ratio
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. Fibonacci numbers A000045; Lucas numbers A000032; tribonacci numbers A000213; tetranacci numbers A000288; pentanacci numbers A000322; hexanacci numbers A000383; 7th-order Fibonacci numbers A060455; octanacci numbers A079262; 9th-order Fibonacci sequence A127193; 10th-order Fibonacci sequence A127194; 11th-order Fibonacci sequence A127624, A128429.

a:=[1,1,1,1];; for n in [5..50] do a[n]:=a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jul 15 2019

[n le 4 select 1 else Self(n-1) + Self(n-3) + Self(n-4): n in [1..50]]; // Vincenzo Librandi, Dec 25 2015

# From R. J. Mathar, Jul 22 2010: (Start)
A010684 := proc(n) 1+2*(n mod 2) ; end proc:
A000032 := proc(n) coeftayl((2-x)/(1-x-x^2),x=0,n) ; end proc:
A126116 := proc(n) ((-1)^floor(n/2)*A010684(n)+2*A000032(n))/5 ; end proc: seq(A126116(n),n=0..80) ; # (End)
with(combinat): A126116 := proc(n): fibonacci(n-1) + fibonacci(floor((n-4)/2)+1)* fibonacci(ceil((n-4)/2)+2) end: seq(A126116(n), n=0..38); # Johannes W. Meijer, Aug 05 2011

LinearRecurrence[{1,0,1,1},{1,1,1,1},50] (* Harvey P. Dale, Nov 08 2011 *)

Vec((x-1)*(1+x+x^2)/((x^2+x-1)*(x^2+1)) + O(x^50)) \\ Altug Alkan, Dec 25 2015

((1-x)*(1+x+x^2)/((1-x-x^2)*(1+x^2))).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jul 15 2019

From R. J. Mathar, Jul 22 2010: (Start)
G.f.: (1-x)*(1+x+x^2)/((1-x-x^2)*(1+x^2)).
a(n) = ( (-1)^floor(n/2) * A010684(n) + 2*A000032(n))/5.
a(2*n) = A061646(n). (End)
From Johannes W. Meijer, Aug 05 2011: (Start)
a(n) = F(n-1) + A070550(n-4) with F(n) = A000045(n).
a(n) = F(n-1) + F(floor((n-4)/2) + 1)*F(ceiling((n-4)/2) + 2). (End)
a(n) = (1/5)*((sqrt(5)-1)*(1/2*(1+sqrt(5)))^n - (1+sqrt(5))*(1/2*(1-sqrt(5)))^n + sin((Pi*n)/2) - 3*cos((Pi*n)/2)). - Harvey P. Dale, Nov 08 2011
(-1)^n * a(-n) = a(n) = F(n) - A070550(n - 6). - Michael Somos, Feb 05 2012
a(n)^2 + 3*a(n-2)^2 + 6*a(n-5)^2 + 3*a(n-7)^2 = a(n-8)^2 + 3*a(n-6)^2 + 6*a(n-3)^2 + 3*a(n-1)^2. - Greg Dresden, Jul 07 2021
a(n) = A293411(n)-A293411(n-1). - R. J. Mathar, Jul 20 2025

Edited by Don Reble, Mar 09 2007

A049853 a(n) = a(n-1) + Sum_{k=0..n-3} a(k) for n >= 2, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 2, 3, 6, 11, 19, 33, 58, 102, 179, 314, 551, 967, 1697, 2978, 5226, 9171, 16094, 28243, 49563, 86977, 152634, 267854, 470051, 824882, 1447567, 2540303, 4457921, 7823106, 13728594, 24092003, 42278518, 74193627

Offset: 0

Clark Kimberling

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1).

Cf. A070550, A180662 (Ca2).

a049853 n = a049853_list !! n
a049853_list = 1 : 2 : 2 : 3 :
   zipWith (+) a049853_list
               (zipWith (+) (drop 2 a049853_list) (drop 3 a049853_list))
-- Reinhard Zumkeller, Aug 06 2011

a := proc(n) option remember: if n<2 then n+1 else a(n-1) + add(a(k), k=0..n-3) fi end: seq(a(n), n=0..33); # Johannes W. Meijer, Jun 18 2018

LinearRecurrence[{2,-1,1},{1,2,2},40] (* Harvey P. Dale, May 12 2022 *)

Vec((1 - x)*(1 + x) / (1 - 2*x + x^2 - x^3) + O(x^40)) \\ Colin Barker, Jun 17 2018

a(n) = 2*a(n-1) - a(n-2) + a(n-3); 3 initial terms required.
a(n) = a(n-1) + a(n-2) + a(n-4) for n > 3. - Reinhard Zumkeller, Aug 06 2011
Empirical: a(n) = Sum_{k=0..floor(n/3)} A084534(n-2*k, n-3*k). - Johannes W. Meijer, Jun 17 2018
G.f.: (1 - x)*(1 + x) / (1 - 2*x + x^2 - x^3). - Colin Barker, Jun 17 2018

A128494 Coefficient table for sums of Chebyshev's S-Polynomials.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, -1, 1, 1, 1, -1, -2, 1, 1, 1, 2, -2, -3, 1, 1, 0, 2, 4, -3, -4, 1, 1, 0, -2, 4, 7, -4, -5, 1, 1, 1, -2, -6, 7, 11, -5, -6, 1, 1, 1, 3, -6, -13, 11, 16, -6, -7, 1, 1, 0, 3, 9, -13, -24, 16, 22, -7, -8, 1, 1, 0, -3, 9, 22, -24, -40, 22, 29, -8, -9, 1, 1, 1, -3, -12, 22, 46, -40, -62, 29, 37, -9, -10, 1, 1, 1, 4, -12

Offset: 0

Wolfdieter Lang, Apr 04 2007

See A049310 for the coefficient table of Chebyshev's S(n,x)=U(n,x/2) polynomials.

This is a 'repetition triangle' based on a signed version of triangle A059260: a(2*p,2*k) = a(2*p+1,2*k) = A059260(p+k,2*k)*(-1)^(p+k) and a(2*p+1,2*k+1) = a(2*p+2,2*k+1) = A059260(p+k+1,2*k+1)*(-1)^(p+k), k >= 0.

			The triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6   7   8   9  10
   0:  1
   1:  1   1
   2:  0   1   1
   3:  0  -1   1   1
   4:  1  -1  -2   1   1
   5:  1   2  -2  -3   1   1
   6:  0   2   4  -3  -4   1   1
   7:  0  -2   4   7  -4  -5   1   1
   8:  1  -2  -6   7  11  -5  -6   1   1
   9:  1   3  -6 -13  11  16  -6  -7   1   1
  10:  0   3   9 -13 -24  16  22  -7  -8   1   1
... reformatted by _Wolfdieter Lang_, Oct 16 2012
Row polynomial S(1;4,x) = 1 - x - 2*x^2 + x^3 + x^4 = Sum_{k=0..4} S(k,x).
S(4,y)*S(5,y)/y = 3 - 13*y^2 + 16*y^4 - 7*y^6 + y^8, with y=sqrt(2+x) this becomes S(1;4,x).
From _Wolfdieter Lang_, Oct 16 2012: (Start)
S(1;4,x) = (1 - (S(5,x) - S(4,x)))/(2-x) = (1-x)*(2-x)*(1+x)*(1-x-x^2)/(2-x) = (1-x)*(1+x)*(1-x-x^2).
S(5,x) - S(4,x) = R(11,sqrt(2+x))/sqrt(2+x) = -1 + 3*x + 3*x^2 - 4*x^3 - x^4 + x^5. (End)

Wolfdieter Lang, First 15 rows
Per Alexandersson, Luis Angel González-Serrano, and Egor A. Maximenko, Mario Alberto Moctezuma-Salazar, Symmetric polynomials in the symplectic alphabet and their expression via Dickson-Zhukovsky variables, arXiv:1912.12725 [math.CO], 2019.
Per Alexandersson, Luis Angel González-Serrano, Egor A. Maximenko, and Mario Alberto Moctezuma-Salazar, Symmetric Polynomials in the Symplectic Alphabet and the Change of Variables z_j = x_j + x_j^(-1), The Elec. J. of Combinatorics (2021) Vol. 28, No. 1, #P1.56.

Row sums (signed): A021823(n+2). Row sums (unsigned): A070550(n).
Cf. A128495 for S(2; n, x) coefficient table.
The column sequences (unsigned) are, for m=0..4: A021923, A002265, A008642, A128498, A128499.
For m >= 1 the column sequences (without leading zeros) are of the form a(m, 2*k) = a(m, 2*k+1) = ((-1)^k)*b(m, k) with the sequences b(m, k), given for m=1..11 by A008619, A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808.

S(1;n,x) = Sum_{k=0..n} S(k,x) = Sum_{m=0..n} a(n,m)*x^m, n >= 0.
a(n,m) = [x^m](S(n,y)*S(n+1,y)/y) with y:=sqrt(2+x).
G.f. for column m: (x^m)/((1-x)*(1+x^2)^(m+1)), which shows that this is a lower diagonal matrix of the Riordan type, named (1/((1+x^2)*(1-x)), x/(1+x^2)).
From Wolfdieter Lang, Oct 16 2012: (Start)
a(n,m) = [x^m](1- (S(n+1,x) - S(n,x)))/(2-x). From the Binet - de Moivre formula for S and use of the geometric sum.
a(n,m) = [x^m](1- R(2*n+3,sqrt(2+x))/sqrt(2+x))/(2-x) with the monic integer T-polynomials R with coefficient triangle given in A127672. From the odd part of the bisection of the T-polynomials. (End)

A191373 Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 5i+j=n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 4, 1, 2, 2, 4, 2, 3, 3, 5, 1, 3, 2, 5, 2, 3, 4, 6, 1, 3, 2, 6, 2, 3, 4, 6, 2, 4, 3, 7, 3, 5, 5, 8, 1, 4, 3, 8, 2, 3, 5, 8, 2, 4, 3, 8, 4, 6, 6, 9, 1, 5, 3, 9, 2, 3, 6, 9, 2, 4, 3, 9

Offset: 0

Johannes W. Meijer, Jun 05 2011

nonn

The Le1{1,5} and Le2{5,1} triangle sums of Sierpinski’s triangle A047999 equal this sequence; see the formulas for their definitions. The Le1{1,5} and Le2{5,1} triangle sums are similar to the Kn11 and Kn21 sums, the Ca1 and Ca2 sums, and the Gi1 and Gi2 sums, see A180662.

Some A191373(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs.

Sam Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.

Cf. A001316 (1,1), A002487 (2,1), A120562 (3,1), A112970 (4,1), A191373 (5,1)

Cf. A000012 (p=0), A006498 (p=1, p=2, p=4, p=8, p=16, p=32), A070550 (p=3, p=6, p=12, p=24), A000071 (p=15, p=30), A115008 (p=23).

A191373:=proc(n) option remember; if n <0 then A191373(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A191373(n/2) else A191373((n-1)/2) + A191373(((n-1)/2)-2); fi; end: seq(A191373(n),n=0..75);

a(2*n) = a(n) and a(2*n+1) = a(n) + a(n-2) with a(0) = 1, a(1) = 1 and a(n)=0 for n<=-1.
a(n) = Le1{1,5}(n) = add(T(n-4*k,k),k=0..floor(n/5))
a(n) = Le1{1,5}(n) = sum(binomial(i + j, i) mod 2 | (i + 5*j) = n)
a(n) = Le2{5,1}(n) = add(T(n-4*k,n-5*k),k=0..floor(n/5))
a(n) = Le2{5,1}(n) = sum(binomial(i + j, i) mod 2 | (5*i + j) = n)
G.f.: Product_{n>=0} (1+x^(2^n)+x^(5*2^n)).
G.f. A(x) satisfies: A(x) = (1 + x + x^5) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019

A262342 Area of Lewis Carroll's paradoxical F(2n+1) X F(2n+3) rectangle.

Original entry on oeis.org

10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290, 3372041405099481410

Offset: 1

Jonathan Sondow, Oct 16 2015

Warren Weaver (1938): "In a familiar geometrical paradox a square of area 8 X 8 = 64 square units is cut into four parts which may be refitted to form a rectangle of apparent area 5 X 13 = 65 square units.... Lewis Carroll generalized this paradox...."
Carroll cuts a F(2n+2) X F(2n+2) square into four parts, where F(n) is the n-th Fibonacci number. Two parts are right triangles with legs F(2n) and F(2n+2); two are right trapezoids three of whose sides are F(2n), F(2n+1), and F(2n+1). (Thus n > 0.) The paradox (or dissection fallacy) depends on Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1.
For an extension of the paradox to a F(2n+1) X F(2n+1) square using Cassini's identity F(2n) * F(2n+2) = F(2n+1)^2 - 1, see Dudeney (1970), Gardner (1956), Horadam (1962), Knott (2014), Kumar (1964), and Sillke (2004). Sillke also has many additional references and links.

			F(3) * F(5) = 2 * 5 = 10 = 3^2 + 1 = F(4)^2 + 1, so a(1) = 10.
G.f. = 10*x + 65*x^2 + 442*x^3 + 3026*x^4 + 20737*x^5 + 142130*x^6 + 974170*x^7 + ...

W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th edition, Dover, 1987, p. 85.
Henry E. Dudeney, 536 Puzzles and Curious Problems, Scribner, reprinted 1970, Problems 352-353 and their answers.
Martin Gardner, Mathematics, Magic and Mystery, Dover, 1956, Chap. 8.
Edward Wakeling, Rediscovered Lewis Carroll Puzzles, Dover, 1995, p. 12.
David Wells, The Penguin Book of Curious and Interesting Puzzles, Penguin, 1997, Puzzle 143.

Colin Barker, Table of n, a(n) for n = 1..1000
Margherita Barile, Dissection Fallacy, MathWorld.
A. F. Horadam, Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 35, No. 1 (1962), pp. 1-11.
Ron Knott, Fibonacci jigsaw puzzles, 2014.
Santosh Kumar, On Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 37, No. 4 (1964), pp. 221-223.
Oskar Schlömilch, Ein geometrisches Paradoxon, Zeitschrift für Mathematik und Physik, Vol. 13 (1868), p. 162.
Torsten Sillke, Jigsaw paradox, 2004.
David Singmaster, Vanishing area puzzles, Recreational Math. Mag., Vol. 1 (2014), pp. 10-21.
Warren Weaver, Lewis Carroll and a Geometrical Paradox, American Math. Monthly, Vol. 45, No. 4 (1938), pp. 234-236.
Wikipedia, Cassini and Catalan identities.
Wikipedia, Fibonacci number.
Wikipedia, Missing square puzzle; also see External Links.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A001519, A059929, A064170, A070550, A166516, A176055, A190018, A236165, A245306.

[Fibonacci(2*n+1)*Fibonacci(2*n+3) : n in [1..30]]; // Wesley Ivan Hurt, Oct 16 2015

with(combinat): A262342:=n->fibonacci(2*n+1)*fibonacci(2*n+3): seq(A262342(n), n=1..30); # Wesley Ivan Hurt, Oct 16 2015

Table[Fibonacci[2 n + 1] Fibonacci[2 n + 3], {n, 22}]
LinearRecurrence[{8,-8,1},{10,65,442},30] (* Harvey P. Dale, Aug 06 2024 *)

Vec(-x*(2*x^2-15*x+10)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ Colin Barker, Oct 17 2015

a(n) = fibonacci(2*n+1) * fibonacci(2*n+3) \\ Altug Alkan, Oct 17 2015

a(n) = Fibonacci(2n+1) * Fibonacci(2n+3) = Fibonacci(2n+2)^2 + 1 for n > 0.
From Colin Barker, Oct 17 2015: (Start)
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: -x*(2*x^2-15*x+10) / ((x-1)*(x^2-7*x+1)).
(End)
a(3*k-2) mod 2 = 0; a(3*k-1) mod 2 = 1; a(3*k) mod 2 = 0, k > 0. - Altug Alkan, Oct 17 2015
a(n) = A059929(2*n+1) = A070550(4*n+1) = A166516(2*n+2) = A190018(8*n) = A236165(4*n+4) = A245306(2*n+2). - Bruno Berselli, Oct 17 2015
a(n) = A064170(n+3). - Alois P. Heinz, Oct 17 2015
E.g.f.: (1/5)*((1/phi*r)*exp(b*x) + (phi^4/r)*exp(a*x) + 3*exp(x) - 10), where r = 2*phi+1, 2*a=7+3*sqrt(5), 2*b=7-3*sqrt(5). - G. C. Greubel, Oct 17 2015
a(n) = (A337928(n+1) - A337929(n+1)) / 2. - Flávio V. Fernandes, Feb 06 2021
Sum_{n>=1} 1/a(n) = sqrt(5)/2 - 1 = A176055 - 2. - Amiram Eldar, Mar 04 2021

Showing 1-9 of 9 results.

cp's OEIS Frontend

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

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A035607 Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).

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A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

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A171861 Expansion of x*(1+x+x^2) / ( (x-1)*(x^3+x^2-1) ).

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A126116 a(n) = a(n-1) + a(n-3) + a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

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A171861 Expansion of x(1+x+x^2) / ( (x-1)(x^3+x^2-1) ).