A094570 -id:A094570 - cp's OEIS frontend

A006355 Number of binary vectors of length n containing no singletons.

1, 0, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634

Offset: 0

David M. Bloom

Number of cvtemplates at n-2 letters given <= 2 consecutive consonants or vowels (n >= 4).
Number of (n,2) Freiman-Wyner sequences.
Diagonal sums of the Riordan array ((1-x+x^2)/(1-x), x/(1-x)), A072405 (where this begins 1,0,1,1,1,1,...). - Paul Barry, May 04 2005
Central terms of the triangle in A094570. - Reinhard Zumkeller, Mar 22 2011
Pisano period lengths: 1, 1, 8, 3, 20, 8, 16, 6, 24, 20, 10, 24, 28, 16, 40, 12, 36, 24, 18, 60, ... . - R. J. Mathar, Aug 10 2012
Also the number of matchings in the (n-2)-pan graph for n >= 5. - Eric W. Weisstein, Oct 03 2017
a(n) is the number of bimultus bitstrings of length n. A bitstring is bimultus if each of its 1's possess at least one neighboring 1 and each of its 0's possess at least one neighboring 0. - Steven Finch, May 26 2020

			a(6)=10 because we have: 000000, 000011, 000111, 001100, 001111, 110000, 110011, 111000, 111100, 111111. - _Geoffrey Critzer_, Jan 26 2014

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 16, 51.

Charles R Greathouse IV, Table of n, a(n) for n = 0..4786 (next term has 1001 digits)
Kassie Archer and Aaron Geary, Powers of permutations that avoid chains of patterns, arXiv:2312.14351 [math.CO], 2023. See p. 15.
J. Berman and P. Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), 103-124. [Annotated scanned copy]
Ian F. Blake, The enumeration of certain run length sequences, Information and Control, 55 (1982), 222-237.
Alexander Burstein, Sergey Kitaev, and Toufik Mansour, Partially ordered patterns and their combinatorial interpretations, PU. M. A. Vol. 19 (2008), No. 2-3, pp. 27-38.
Steven Finch, Variance of longest run duration in a random bitstring, arXiv:2005.12185 [math.CO], 2020.
Enoch Haga, Room for Expansion, Word Ways, 33 (No. 2, 2000), pp. 106-113 (see p. 110).
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 898.
Sergey Kitaev and Jeffrey Remmel, (a,b)-rectangle patterns in permutations and words, arXiv:1304.4286 [math.CO], 2013.
Thor Martinsen, Non-Fisherian generalized Fibonacci numbers, Notes Num. Theor. Disc. Math. (2025) Vol. 31, No. 2, 370-389. See p. 389.
Noriaki Sannomiya, Hosho Katsura, and Yu Nakayama, Supersymmetry breaking and Nambu-Goldstone fermions with cubic dispersion, arXiv preprint arXiv:1612.02285 [cond-mat.str-el], 2016-2017. See Table II, line 2.
Eric Weisstein's World of Mathematics, Independent Edge Set, Matching and Pan Graph.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Except for initial term, = 2*Fibonacci numbers (A000045).
Essentially the same as A047992, A054886, A055389, A068922, and A090991.
Column 2 in A265584.
Cf. A094570, A097925, A097926, A118658, A119457.

a006355 n = a006355_list !! n
a006355_list = 1 : fib2s where
   fib2s = 0 : map (+ 1) (scanl (+) 1 fib2s)
-- Reinhard Zumkeller, Mar 20 2013

[1] cat [Lucas(n) - Fibonacci(n): n in [1..50]]; // Vincenzo Librandi, Aug 02 2014

a:= n-> if n=0 then 1 else (Matrix([[2,-2]]). Matrix([[1,1], [1,0]])^n) [1,1] fi: seq(a(n), n=0..38); # Alois P. Heinz, Aug 18 2008
a := n -> ifelse(n=0, 1, -2*I^n*ChebyshevU(n-2, -I/2)):
seq(simplify(a(n)), n = 0..38);  # Peter Luschny, Dec 03 2023

Join[{1}, Last[#] - First[#] & /@ Partition[Fibonacci[Range[-1, 40]], 4, 1]] (* Harvey P. Dale, Sep 30 2011 *)
Join[{1}, LinearRecurrence[{1, 1}, {0, 2}, 38]] (* Jean-François Alcover, Sep 23 2017 *)
(* Programs from Eric W. Weisstein, Oct 03 2017 *)
Join[{1}, Table[2 Fibonacci[n], {n, 0, 40}]]
Join[{1}, 2 Fibonacci[Range[0, 40]]]
CoefficientList[Series[(1-x+x^2)/(1-x-x^2), {x, 0, 40}], x] (* End *)

a(n)=if(n,2*fibonacci(n-1),1) \\ Charles R Greathouse IV, Mar 14 2012

my(x='x+O('x^50)); Vec((1-x+x^2)/(1-x-x^2)) \\ Altug Alkan, Nov 01 2015

def A006355(n): return 2*fibonacci(n-1) - int(n==0)
print([A006355(n) for n in range(51)]) # G. C. Greubel, Apr 18 2025

a(n+2) = F(n-1) + F(n+2), for n > 0.
G.f.: (1-x+x^2)/(1-x-x^2). - Paul Barry, May 04 2005
a(n) = A119457(n-1,n-2) for n > 2. - Reinhard Zumkeller, May 20 2006
a(n) = 2*F(n-1) for n > 0, F(n)=A000045(n) and a(0)=1. - Mircea Merca, Jun 28 2012
G.f.: 1 - x + x*Q(0), where Q(k) = 1 + x^2 + (2*k+3)*x - x*(2*k+1 + x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 05 2013
a(n) = A118658(n) - 0^n. - M. F. Hasler, Nov 05 2014
a(n) = 2^(-n)*((1+r)*(1-r)^n - (1-r)*(1+r)^n)/r for n > 0, where r=sqrt(5). - Colin Barker, Jan 28 2017
a(n) = a(n-1) + a(n-2) for n >= 3. - Armend Shabani, Nov 25 2020
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) - sqrt(5)*sinh(sqrt(5)*x/2))/5 - 1. - Stefano Spezia, Apr 18 2022
a(n) = F(n-3) + F(n-2) + F(n-1) for n >= 3, where F(n)=A000045(n). - Gergely Földvári, Aug 03 2024

Corrected by T. D. Noe, Oct 31 2006

A141169 Triangle of Fibonacci numbers, read by rows: T(n,k) = A000045(k), 0<=k<=n.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 3, 0, 1, 1, 2, 3, 5, 0, 1, 1, 2, 3, 5, 8, 0, 1, 1, 2, 3, 5, 8, 13, 0, 1, 1, 2, 3, 5, 8, 13, 21, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 0, 1, 1, 2, 3, 5, 8, 13, 21

Offset: 0

Reinhard Zumkeller, Mar 21 2011

T(n,0) = A000004(n); T(n,n) = A000045(n);
central terms: T(2*n,n) = A000045(n);
sums of rows: Sum(T(n,k): 0<=k<=n) = A000071(n+2);
alternating sums of rows: Sum(T(n,k)*(-1)^k: 0<=k<=n) = A119282(n);
T(n,k) + T(n,n-k) = A094570(n,k).

Reinhard Zumkeller, Rows n=0..125 of triangle, flattened

Cf. A000045, A000004, A000071, A094570, A119282.

import Data.List (inits)
a141169 n k = a141169_tabl !! n !! k
a141169_row n = a141169_tabl !! n
a141169_tabl = tail $ inits a000045_list
a141169_list = concat $ a141169_tabl
-- Reinhard Zumkeller, Aug 24 2015, Mar 21 2011

A088060 For each Fibonacci number F(m) = A000045(m), m >= 1, look for the smallest Fibonacci number F(k) > F(m) such that F(m) + F(k) is a square. If it exists, append F(k) to the sequence.

Original entry on oeis.org

3, 3, 34, 13, 89, 14930352

Offset: 1

Amarnath Murthy, Sep 21 2003

Conjecture: sequence is finite.
m=1 ==> k=4 or 6, m=2 ==> k=4 or 6, m=3 ==> k=3 or 9, m=4 ==> k=1 or 2 or 7 or 17, m=6 ==> k=1 or 2 or 6, m=7 ==> k=4, m=9 ==> k=3, m=10 ==> k=11, m=11 ==> k=10, m=12 ==> k=36, m=17 ==> k=4, m=36 ==> k=12, for other m there is no solution. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 14 2004
Joe Silverman wrote: "Here's a general finiteness result: fib(n) + c = m^2 has only finitely many solutions, regardless of the value of c. Write fib(n) = a(u^n - u^(-n)), where a=1/sqrt(5) and u=(1+sqrt(5))/2. So generalizing further, can look at the equation au^n + bu^(-n) + c = dm^2, where a,b,c,d are fixed integers (a,b,d nonzero) and u is a unit, or even just an algebraic integer, in some number field.
"Rewrite this as au^(2n) + b + cu^n = d m^2 u^n. Using standard methods, it is not hard to prove that any such equation has only finitely many solutions. One way is to consider the congruence class of n modulo (say) 2. Thus write n = 2*n1+n2. We can assume that n2 is fixed (either 0 or 1). Now let x = u^n1 and y = m be the unknowns and let e=u^n2 be another fixed value.
"Then we need to solve (ae^2)x^4 + b + (ce)x^2 = dx^2y^2. This is the equation of an elliptic curve (albeit a singular one), so has only finitely many solutions in the ring of integers of any field (by a theorem of Siegel). Or one can consider n modulo a higher congruence and then the resulting curve will have genus larger than one, in which case one can quote Faltings' theorem that there are only finitely many points with coordinates in any number field."

			F(10) = 55, F(11) = 89, 55 + 89 = 144 is a square, and the first of the form F(10) + F(k), k > 10. F(10) is the 5th Fibonacci term that has at least one qualifying value, k. Hence a(5) = 89. - _Peter Munn_, Mar 21 2021

Cf. A000045, A094570, A182576.

a(5) and a(6) from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 14 2004

Name clarified by Peter Munn, Mar 22 2021

cp's OEIS Frontend

A006355 Number of binary vectors of length n containing no singletons.

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A141169 Triangle of Fibonacci numbers, read by rows: T(n,k) = A000045(k), 0<=k<=n.

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A088060 For each Fibonacci number F(m) = A000045(m), m >= 1, look for the smallest Fibonacci number F(k) > F(m) such that F(m) + F(k) is a square. If it exists, append F(k) to the sequence.

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