A022121 -id:A022121 - cp's OEIS frontend

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803

Offset: 0

N. J. A. Sloane, May 24 1994

Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd  b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd  b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3)  = 2*L(n - 2);
L(n) - L(n - 4)  = 5*F(n - 2);
L(n) - L(n - 6)  = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...

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Index entries for sequences related to Chebyshev polynomials..
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for two-way infinite sequences.
Index entries for "core" sequences.
Index entries for sequences related to Benford's law.

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.

a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011

Magma
```
[Lucas(n): n in [0..120]];
```

with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)

a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)

{a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};

{a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};

{a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};

a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011

polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011

def A000032_gen(): # generator of terms
    a, b = 2, 1
    while True:
        yield a
        a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for  in range(50)] # _Cole Dykstra, Aug 02 2022

from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023

[(i:=3)+(j:=-1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025

[lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) ==  - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(n+2) = 1 + A001610(n+1) = 1 + Sum_{k=0..n} L(k). - Tom Edgar, Apr 15 2015
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 2022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
L(n) = ((Sum_{i=0..n-1} L(i)^2) - 2)/L(n-1). - Jules Beauchamp, May 03 2025
From Peter Bala, Jul 09 2025: (Start)
The following series telescope:
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651

Offset: 0

Patrick De Geest

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, World!Of Numbers, Palindromes of the form n+(n+1)^2.
Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Nandini Nilakantan and Anurag Singh, Homotopy type of neighborhood complexes of Kneser graphs, KG_{2,k}, Proceeding-Mathematical Sciences, 128, Article number: 53(2018).
Yanni Pei and Jiang Zeng, Counting signed derangements with right-to-left minima and excedances, arXiv:2206.11236 [math.CO], 2022.
Popular Computing (Calabasas, CA), The CSR Function, Vol. 4 (No. 34, Jan 1976), pages PC34-10 to PC34-11. Annotated and scanned copy.
Zdzislaw Skupień and Andrzej Żak, Pair-sums packing and rainbow cliques, in Topics In Graph Theory, A tribute to A. A. and T. E. Zykovs on the occasion of A. A. Zykov's 90th birthday, ed. R. Tyshkevich, Univ. Illinois, 2013, pages 131-144, (in English and Russian).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).
Cf. also A135223, A176271, A214604, A105728, A005408, A002378, A084990, A253607, A260910, A089270.

a028387 n = n + (n + 1) ^ 2  -- Reinhard Zumkeller, Jul 17 2014

[n + (n+1)^2: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

FoldList[## + 2 &, 1, 2 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Table[n + (n + 1)^2, {n, 0, 100}] (* Vincenzo Librandi, Oct 17 2012 *)
Table[ FrobeniusNumber[{n, n + 1}], {n, 2, 30}] (* Zak Seidov, Jan 14 2015 *)

a(n)=n^2+3*n+1 \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return (n**2+3*n+1) # Torlach Rush, May 07 2024

[n+(n+1)^2 for n in range(0,48)] # Zerinvary Lajos, Jul 03 2008

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A035513 Wythoff array read by falling antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 7, 6, 5, 11, 10, 9, 8, 18, 16, 15, 12, 13, 29, 26, 24, 20, 14, 21, 47, 42, 39, 32, 23, 17, 34, 76, 68, 63, 52, 37, 28, 19, 55, 123, 110, 102, 84, 60, 45, 31, 22, 89, 199, 178, 165, 136, 97, 73, 50, 36, 25, 144, 322, 288, 267, 220, 157, 118, 81, 58, 41, 27, 233, 521

Offset: 1

N. J. A. Sloane

T(0,0)=1, T(0,1)=2,...; y^2-x^2-xy

Inverse of sequence A064274 considered as a permutation of the nonnegative integers. - Howard A. Landman, Sep 25 2001

The Wythoff array W consists of all the Wythoff pairs (x(n),y(n)), where x=A000201 and y=A001950, so that W contains every positive integer exactly once. The differences T(i,2j+1)-T(i,2j) form the Wythoff difference array, A080164, which also contains every positive integer exactly once. - Clark Kimberling, Feb 08 2003

For n>2 the determinant of any n X n contiguous subarray of A035513 (as a square array) is 0. - Gerald McGarvey, Sep 18 2004

From Clark Kimberling, Nov 14 2007: (Start)

Except for initial terms in some cases:

(Row 1) = A000045

(Row 2) = A000032

(Row 3) = A006355

(Row 4) = A022086

(Row 5) = A022087

(Row 6) = A000285

(Row 7) = A022095

(Row 8) = A013655 (sum of Fibonacci and Lucas numbers)

(Row 9) = A022112

(Row 10-19) = A022113, A022120, A022121, A022379, A022130, A022382, A022088, A022136, A022137, A022089

(Row 20-28) = A022388, A022096, A022090, A022389, A022097, A022091, A022390, A022098, A022092

(Column 1) = A003622 = AA Wythoff sequence

(Column 2) = A035336 = BA Wythoff sequence

(Column 3) = A035337 = ABA Wythoff sequence

(Column 4) = A035338 = BBA Wythoff sequence

(Column 5) = A035339 = ABBA Wythoff sequence

(Column 6) = A035340 = BBBA Wythoff sequence

Main diagonal = A020941. (End)

The Wythoff array is the dispersion of the sequence given by floor(n*x+x-1), where x=(golden ratio). See A191426 for a discussion of dispersions. - Clark Kimberling, Jun 03 2011

If u and v are finite sets of numbers in a row of the Wythoff array such that (product of all the numbers in u) = (product of all the numbers in v), then u = v. See A160009 (row 1 products), A274286 (row 2), A274287 (row 3), A274288 (row 4). - Clark Kimberling, Jun 17 2016

All columns of the Wythoff array are compound Wythoff sequences. This follows from the main theorem in the 1972 paper by Carlitz, Scoville and Hoggatt. For an explicit expression see Theorem 10 in Kimberling's paper from 2008 in JIS. - Michel Dekking, Aug 31 2017

The Wythoff array can be viewed as an infinite graph over the set of nonnegative integers, built as follows: start with an empty graph; for all n = 0, 1, ..., create an edge between n and the sum of the degrees of all i < n. Finally, remove vertex 0. In the resulting graph, the connected components are chains and correspond to the rows of the Wythoff array. - Luc Rousseau, Sep 28 2017

Suppose that h < k are consecutive terms in a row of the Wythoff array. If k is in an even numbered column, then h = floor(k/tau); otherwise, h = -1 + floor(k/tau). - Clark Kimberling, Mar 05 2020

From Clark Kimberling, May 26 2020: (Start)

For k > = 0, column k shows the numbers m having F(k+1) as least term in the Zeckendorf representation of m. For n >= 1, let r(n,k) be the number of terms in column k that are <= n. Then n/r(n,k) = n/(F(k+1)*tau + F(k)*(n-1)), by Bottomley's formula, so that the limiting ratio is 1/(F(k+1)*tau + F(k)). Summing over all k gives Sum_{k>=0} 1/(F(k+1)*tau + F(k)) = 1. Thus, in the limiting sense:

38.19...% of the numbers m have least term 1;

23.60...% have least term 2;

14.58...% have least term 3;

9.01...% have least term 5, etc. (End)

Named after the Dutch mathematician Willem Abraham Wythoff (1865-1939). - Amiram Eldar, Jun 11 2021

From Clark Kimberling, Jun 04 2025: (Start)

Let u(n) = (T(n,1),T(n,2)) mod 2. The positive integers (A000027) are partitioned into 4 sets (sequences):

{n : u(n) = (0,0)} = (3, 5, 9, 15, 19, 25, 29,...) = 1 + 2*A190429

{n: u(n) = (0,1)} = (2, 6, 12, 16, 18, 22, 28,...) = A191331

{n : u(n) = (1,0)} = (1, 7, 11, 13, 17, 21, 23,...) = A086843

{n: u(n) = (1,1)} = (4, 8, 10, 14, 20, 24, 26,...) = A191330.

Let v(n) = (T(n,1),T(n,2)) mod 3. The positive integers are partitioned into 9 sets (sequences):

{n : v(n) = (0,0)} = (4, 13, 19, 28, 43, 52,...) = 1 + 3*A190434

{n: v(n) = (0,1)} = (3, 12, 27, 36, 42, 51,...) = 3*A140399

{n : v(n) = (0,2)} = (5, 11, 20, 35, 44, 50,...) = 2 + 3*A190439

{n: v(n) = (1,0)} = (9, 18, 24, 33, 48, 57,...) = 3*A140400

{n: v(n) = (1,1)} = (2, 8, 17, 26, 32, 41,...) = A384601

{n : v(n) = (1,2)} = (1, 10, 16, 25, 34, 40,...) = A384602

{n: v(n) = (2,0)} = (14, 23, 29, 38, 47, 53,...) = 2 + 3*A190438

{n: v(n) = (2,1)} = (7, 22, 31, 37, 46, 61,...) = 1 + 3*A190433

{n : v(n) = (2,2)} = (6, 15, 21, 30, 39, 45,...) = 3*A140398.

Conjecture: If m >= 2, then {(T(n,1), T(n,2)) mod m} has cardinality m^2. (End)

			The Wythoff array begins:
   1    2    3    5    8   13   21   34   55   89  144 ...
   4    7   11   18   29   47   76  123  199  322  521 ...
   6   10   16   26   42   68  110  178  288  466  754 ...
   9   15   24   39   63  102  165  267  432  699 1131 ...
  12   20   32   52   84  136  220  356  576  932 1508 ...
  14   23   37   60   97  157  254  411  665 1076 1741 ...
  17   28   45   73  118  191  309  500  809 1309 2118 ...
  19   31   50   81  131  212  343  555  898 1453 2351 ...
  22   36   58   94  152  246  398  644 1042 1686 2728 ...
  25   41   66  107  173  280  453  733 1186 1919 3105 ...
  27   44   71  115  186  301  487  788 1275 2063 3338 ...
  ...
The extended Wythoff array has two extra columns, giving the row number n and A000201(n), separated from the main array by a vertical bar:
0     1  |   1    2    3    5    8   13   21   34   55   89  144   ...
1     3  |   4    7   11   18   29   47   76  123  199  322  521   ...
2     4  |   6   10   16   26   42   68  110  178  288  466  754   ...
3     6  |   9   15   24   39   63  102  165  267  432  699 1131   ...
4     8  |  12   20   32   52   84  136  220  356  576  932 1508   ...
5     9  |  14   23   37   60   97  157  254  411  665 1076 1741   ...
6    11  |  17   28   45   73  118  191  309  500  809 1309 2118   ...
7    12  |  19   31   50   81  131  212  343  555  898 1453 2351   ...
8    14  |  22   36   58   94  152  246  398  644 1042 1686 2728   ...
9    16  |  25   41   66  107  173  280  453  733 1186 1919 3105   ...
10   17  |  27   44   71  115  186  301  487  788 1275 2063 3338   ...
11   19  |  30   49   79   ...
12   21  |  33   54   87   ...
13   22  |  35   57   92   ...
14   24  |  38   62   ...
15   25  |  40   65   ...
16   27  |  43   70   ...
17   29  |  46   75   ...
18   30  |  48   78   ...
19   32  |  51   83   ...
20   33  |  53   86   ...
21   35  |  56   91   ...
22   37  |  59   96   ...
23   38  |  61   99   ...
24   40  |  64   ...
25   42  |  67   ...
26   43  |  69   ...
27   45  |  72   ...
28   46  |  74   ...
29   48  |  77   ...
30   50  |  80   ...
31   51  |  82   ...
32   53  |  85   ...
33   55  |  88   ...
34   56  |  90   ...
35   58  |  93   ...
36   59  |  95   ...
37   61  |  98   ...
38   63  |     ...
   ...
Each row of the extended Wythoff array also satisfies the Fibonacci recurrence, and may be extended to the left using this recurrence backwards.
From _Peter Munn_, Jun 11 2021: (Start)
The Wythoff array appears to have the following relationship to the traditional Fibonacci rabbit breeding story, modified for simplicity to be a story of asexual reproduction.
Give each rabbit a number, 0 for the initial rabbit.
When a new round of rabbits is born, allocate consecutive numbers according to 2 rules (the opposite of many cultural rules for inheritance precedence): (1) newly born child of Rabbit 0 gets the next available number; (2) the descendants of a younger child of any given rabbit precede the descendants of an older child of the same rabbit.
Row n of the Wythoff array lists the children of Rabbit n (so Rabbit 0's children have the Fibonacci numbers: 1, 2, 3, 5, ...). The generation tree below shows rabbits 0 to 20. It is modified so that each round of births appears on a row.
                                                                 0
                                                                 :
                                       ,-------------------------:
                                       :                         :
                       ,---------------:                         1
                       :               :                         :
              ,--------:               2               ,---------:
              :        :               :               :         :
        ,-----:        3         ,-----:         ,-----:         4
        :     :        :         :     :         :     :         :
     ,--:     5     ,--:     ,---:     6     ,---:     7     ,---:
     :  :     :     :  :     :   :     :     :   :     :     :   :
  ,--:  8  ,--:  ,--:  9  ,--:  10  ,--:  ,--:  11  ,--:  ,--:  12
  :  :  :  :  :  :  :  :  :  :   :  :  :  :  :   :  :  :  :  :   :
  : 13  :  : 14  : 15  :  : 16   :  : 17  : 18   :  : 19  : 20   :
The extended array's nontrivial extra column (A000201) gives the number that would have been allocated to the first child of Rabbit n, if Rabbit n (and only Rabbit n) had started breeding one round early.
(End)

John H. Conway, Posting to Math Fun Mailing List, Nov 25 1996.
Clark Kimberling, "Stolarsky interspersions," Ars Combinatoria 39 (1995) 129-138.

Alois P. Heinz, Table of n, a(n) for n = 1..5151
Peter G. Anderson, More Properties of the Zeckendorf Array, Fib. Quart. 52-5 (2014), 15-21.
L. Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Fibonacci representations, Fib. Quart., Vol. 10, No. 1 (1972), pp. 1-28.
Code Golf Stack Exchange, New Order #5: where Fibonacci and Beatty meet at Wythoff.
J. H. Conway, Allan Wechsler, Marc LeBrun, Dan Hoey, and N. J. A. Sloane, On Kimberling Sums and Para-Fibonacci Sequences, Correspondence and Postings to Math-Fun Mailing List, Nov 1996 to Jan 1997.
John Conway and Alex Ryba, The extra Fibonacci series and the Empire State Building, Math. Intelligencer, Vol. 38, No. 1 (2016), pp. 41-48.
Larry Ericksen and Peter G. Anderson, Patterns in differences between rows in k-Zeckendorf arrays, The Fibonacci Quarterly, Vol. 50, No. 1 (February 2012), pp. 11-18. - _N. J. A. Sloane_, Jun 10 2012
Clark Kimberling, Interspersions.
Clark Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 3-8.
Clark Kimberling, Complementary equations and Wythoff Sequences, JIS, Vol. 11 (2008), Article 08.3.3.
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
Clark Kimberling and Kenneth B. Stolarsky, Slow Beatty sequences, devious convergence, and partitional divergence, Amer. Math. Monthly, 123, No. 2 (2016), pp. 267-273.
Stéphane Legendre, Labeled Fibonacci Trees, Fibonacci Quart. 53 (2015), no. 2, 152-167.
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See pages 1-2.
Casey Mongoven, Sonification of multiple Fibonacci-related sequences, Annales Mathematicae et Informaticae, Vol. 41 (2013), pp. 175-192.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Classic Sequences
Sam Vandervelde, On the divisibility of Fibonacci sequences by primes of index two, The Fibonacci Quarterly, Vol. 50, No. 3 (2012), pp. 207-216. See Figure 1.
Eric Weisstein's World of Mathematics, Wythoff Array.
Wikipedia, Wythoff array.
Pedro Zanetti, C++ code snippet, for generating this sequence.
Index entries for sequences that are permutations of the natural numbers

See comments above for more cross-references.
Cf. A003622, A064274 (inverse), A083412 (transpose), A000201, A001950, A080164, A003603, A265650, A019586 (row that contains n).
For two versions of the extended Wythoff array, see A287869, A287870.

W:= proc(n,k) Digits:= 100; (Matrix([n, floor((1+sqrt(5))/2* (n+1))]). Matrix([[0,1], [1,1]])^(k+1))[1,2] end: seq(seq(W(n, d-n), n=0..d), d=0..10); # Alois P. Heinz, Aug 18 2008
A035513 := proc(r, c)
    option remember;
    if c = 1 then
        A003622(r) ;
    else
        A022342(1+procname(r, c-1)) ;
    end if;
end proc:
seq(seq(A035513(r,d-r),r=1..d-1),d=2..15) ; # R. J. Mathar, Jan 25 2015

W[n_, k_] := Fibonacci[k + 1] Floor[n*GoldenRatio] + (n - 1) Fibonacci[k]; Table[ W[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten

T(n,k)=(n+sqrtint(5*n^2))\2*fibonacci(k+1) + (n-1)*fibonacci(k)
for(k=0,9,for(n=1,k, print1(T(n,k+1-n)", "))) \\ Charles R Greathouse IV, Mar 09 2016

from sympy import fibonacci as F, sqrt
import math
tau = (sqrt(5) + 1)/2
def T(n, k): return F(k + 1)*int(math.floor(n*tau)) + F(k)*(n - 1)
for n in range(1, 11): print([T(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Apr 23 2017

from math import isqrt, comb
from gmpy2 import fib2
def A035513(n):
    a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
    x = n-comb(a,2)
    b, c = fib2(a-x+2)
    return b*(x+isqrt(5*x*x)>>1)+c*(x-1) # Chai Wah Wu, Jun 26 2025

T(n, k) = Fib(k+1)*floor[n*tau]+Fib(k)*(n-1) where tau = (sqrt(5)+1)/2 = A001622 and Fib(n) = A000045(n). - Henry Bottomley, Dec 10 2001

T(n,-1) = n-1. T(n,0) = floor(n*tau). T(n,k) = T(n,k-1) + T(n,k-2) for k>=1. - R. J. Mathar, Sep 03 2016

Comments about the extended Wythoff array added by N. J. A. Sloane, Mar 07 2016

A127830 a(n) = Sum_{k=0..n} (binomial(floor(k/2),n-k) mod 2).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 3, 2, 3, 5, 5, 4, 4, 5, 4, 3, 3, 3, 4, 4, 3, 4, 5, 3, 5, 8, 8, 7, 6, 7, 7, 5, 6, 8, 7, 6, 5, 5, 5, 4, 4, 5, 6, 5, 5, 7, 6, 4, 5, 6, 7, 7, 5, 6, 8, 5, 8, 13, 13, 11, 10, 12, 11, 8, 9, 11, 11, 10, 8, 9, 10, 7, 9, 13, 12

Offset: 0

Paul Barry, Feb 01 2007

Row sums of number triangle A127829.
From Johannes W. Meijer, Jun 05 2011: (Start)
The Ze3 and Ze4 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence.
The sequences A127830(2^n-p), p>=0, are apparently all Fibonacci like sequences, i.e., the next term is the sum of the two nonzero terms that precede it; see the crossrefs. (End)

Cf.: A000045 (p=0), A000204 (p=7), A001060 (p=13), A000285 (p=14), A022095 (p=16), A022120 (p=24), A022121 (p=25), A022113 (p=28), A022096 (p=30), A022097 (p=31), A022098 (p=32), A022130 (p=44), A022137 (p=48), A022138 (p=49), A022122 (p=52), A022114 (p=53), A022123 (p=56), A022115 (p=60), A022100 (p=62), A022101 (p=63), A022103 (p=64), A022136 (p=79), A022388 (p=80), A022389 (p=88). - Johannes W. Meijer, Jun 05 2011

A127830 := proc(n) local k: option remember: add(binomial(floor(k/2), n-k) mod 2, k=0..n) end: seq(A127830(n), n=0..80); # Johannes W. Meijer, Jun 05 2011

Table[Sum[Mod[Binomial[Floor[k/2],n-k],2],{k,0,n}],{n,0,80}] (* James C. McMahon, Jan 04 2025 *)

def A127830(n): return sum(not ~(k>>1)&n-k for k in range(n+1)) # Chai Wah Wu, Jul 29 2025

a(2^n) = F(n); a(2^(n+1)+1) = L(n).

a(n) mod 2 = A000931(n+5) mod 2 = A011656(n+4).

A199535 Clark Kimberling's even first column Stolarsky array read by antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 7, 6, 5, 11, 9, 10, 8, 18, 15, 17, 12, 13, 29, 24, 27, 19, 14, 21, 47, 39, 44, 31, 23, 16, 34, 76, 63, 71, 50, 37, 25, 20, 55, 123, 102, 115, 81, 60, 41, 33, 22, 89, 199, 165, 186, 131, 97, 66, 53, 35, 26, 144, 322, 267, 301, 212, 157, 107, 86, 57, 43, 28

Offset: 1

Casey Mongoven, Nov 07 2011

The rows of the array can be seen to have the form A(n, k) = p(n)*Fibonacci(k) + q(n)*Fibonacci(k+1) where p(n) is the sequence {0, 1, 3, 3, 3, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, ...}{n >= 1} and q(n) is the sequence {1, 3, 3, 7, 2, 9, 9, 13, 13, 17, 17, 19, 19, 23, 23, 25, ...}{n >= 1}. - G. C. Greubel, Jun 23 2022

			The even first column stolarsky array (EFC array), northwest corner:
  1......2.....3.....5.....8....13....21....34....55....89...144 ... A000045;
  4......7....11....18....29....47....76...123...199...322...521 ... A000032;
  6......9....15....24....39....63...102...165...267...432...699 ... A022086;
  10....17....27....44....71...115...186...301...487...788..1275 ... A022120;
  12....19....31....50....81...131...212...343...555...898..1453 ... A013655;
  14....23....37....60....97...157...254...411...665..1076..1741 ... A000285;
  16....25....41....66...107...173...280...453...733..1186..1919 ... A022113;
  20....33....53....86...139...225...364...589...953..1542..2495 ... A022096;
  22....35....57....92...149...241...390...631..1021..1652..2673 ... A022130;
Antidiagonal rows (T(n, k)):
   1;
   2,   4;
   3,   7,   6;
   5,  11,   9,  10;
   8,  18,  15,  17, 12;
  13,  29,  24,  27, 19, 14;
  21,  47,  39,  44, 31, 23, 16;
  34,  76,  63,  71, 50, 37, 25, 20;
  55, 123, 102, 115, 81, 60, 41, 33, 22;

Clark Kimberling, The first column of an interspersion, Fibonacci Quarterly 32 (1994), pp. 301-314.

Cf. A000032, A000045, A000285, A013655, A022086, A022088, A022095.
Cf. A022096, A022097, A022113, A022120, A022121, A022122, A022130.
Cf. A022388, A022390, A035506, A035513, A097135, A199536, A199537.

From G. C. Greubel, Jun 23 2022: (Start)
T(n, 1) = A000045(n+1).
T(n, 2) = A000032(n+1), n >= 2.
T(n, 3) = A022086(n) = A097135(n), n >= 3.
T(n, 4) = A022120(n-2), n >= 4.
T(n, 5) = A013655(n-1), n >= 5.
T(n, 6) = A000285(n-2), n >= 6.
T(n, 7) = A022113(n-4), n >= 7.
T(n, 8) = A022096(n-4), n >= 8.
T(n, 9) = A022130(n-6), n >= 9.
T(n, 10) = A022098(n-5), n >= 10.
T(n, 11) = A022095(n-7), n >= 11.
T(n, 12) = A022121(n-8), n >= 12.
T(n, 13) = A022388(n-10), n >= 13.
T(n, 14) = A022122(n-10), n >= 14.
T(n, 15) = A022097(n-10), n >= 15.
T(n, 16) = A022088(n-10), n >= 16.
T(n, 17) = A022390(n-14), n >= 17.
T(n, n) = A199536(n).
T(n, n-1) = A199537(n-1), n >= 2. (End)

More terms added by G. C. Greubel, Jun 23 2022

A271315 Array T(n,k) read by diagonals: T(n,k) = T(n,k-1) + T(n,k-2) where T(n,0) = F(n+1), T(n,1) = F(n); F(n) = Fibonacci(n) = A000045(n).

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 3, 2, 5, 3, 5, 4, 5, 3, 8, 5, 8, 7, 7, 8, 5, 13, 8, 13, 11, 12, 11, 13, 8, 21, 13, 21, 18, 19, 19, 18, 21, 13, 34, 21, 34, 29, 31, 30, 31, 29, 34, 21, 55, 34, 55, 47, 50, 49, 49, 50, 47, 55, 34, 89

Offset: 0

Bob Selcoe, Apr 03 2016

The array is built by treating rows as Fibonacci-type sequences with seed values being two consecutive Fibonacci numbers (A000045(n) = F(n)) in reverse order: For row n, a(0) = F(n+1), a(1) = F(n). As a result, columns are Fibonacci-type sequences with seed values b(0) = F(k-1), b(1) = F(k+1); so starting with T(n,1), Row n == Column k=n+1.
Therefore, an alternative title is: Array T(n,k) read by diagonals: T(n,k) = T(n-1,k) + T(n-2,k) where T(0,k) = F(k-1) and T(1,k) = F(k+1), k>=1.
Patterns exist for certain generalized (a,b)-Pascal triangle transforms of row sequences. Definitions, explanation and examples: (Start)
Define (a,b)-Pascal triangles as having conditions T(0,0) = 1, a = left boundary and b = right boundary.
Let R_n be Row n, and R_n(k) be terms k in sequence R_n.
Let Tr_(k) be the (a,b)-Pascal triangle transform of R_n; define Tr_n(k) as when a = R_n(1) and b = R_n(0). Then Tr_n(k) = R_n(n+2k-2), k>=1. (Trivially, Tr_n(0) = R_n(0)).
For example, n=4: R_4 = {5, 3, 8, 11, 19, 30, 49, 79, 128, 207, 335, 542...}; a=3, b=5.
(3,5)-Pascal triangle is:
  1
  3    5
  3    8    5
  3   11   13   5
  3   14   24   18   5
  etc.
Transform Tr_4(k) is:
  Tr_4(0) = 5*1 = 5 = R_4(0).
  Tr_4(1) = 5*3 + 3*5 = 30 = R_4(5).
  Tr_4(2) = 5*3 + 3*8 + 8*5 = 79 = R_4(7).
  Tr_4(3) = 5*3 + 3*11 + 8*13 + 11*5 = 207 = R_4(9).
  Tr_4(4) = 5*3 + 3*14 + 8*24 + 11*18 + 19*5 = 542 = R_4(11).
  etc.
Examples of sequences where such transforms apply:
  Tr_0 = A001906 starting A001906(0)=0.
  Tr_1 = A001519 starting A001519(2)=2.
  Tr_2 = A002878 starting A002878(1)=4.
  Tr_4 = A167375 starting A167375(3)=30.
(End)

			Array Starts:
  n/k   0   1   2    3    4    5    6    7     8     9     10
  0     1   0   1    1    2    3    5    8     13    21    34
  1     1   1   2    3    5    8    13   21    34    55    89
  2     2   1   3    4    7    11   18   29    47    76    123
  3     3   2   5    7    12   19   31   50    81    131   212
  4     5   3   8    11   19   30   49   79    128   207   335
  5     8   5   13   18   31   49   80   129   209   338   547
  6     13  8   21   29   50   79   129  208   337   545   882
  7     21  13  34   47   81   128  209  337   546   883   1429
  8     34  21  55   76   131  207  338  545   883   1428  2311
  9     55  34  89   123  212  335  547  882   1429  2311  3740
  10    89  55  144  199  343  542  885  1427  2312  3739  6051
Row 7 starts {21,13} because A000045(8)=21 and A000045(7)=13.
T(9,2)=89 + T(9,3)=123 = T(9,4)=212; alternatively, T(7,4)=81 + T(8,4)=131 = T(9,4)=212.

Cf. A000045 (Fibonacci numbers).
Cf. additional sequences related to rows and columns: A000032 (Lucas numbers), A013655, A022121, A022138, A206610.
Cf. sequences related to falling diagonals: A061646, A079472.
Cf. sequences related to (a,b)-Pascal triangle transforms of rows: A001906, A001519, A002878, A167375.

{T(n, k) = fibonacci(n) * fibonacci(k) + fibonacci(n+1) * fibonacci(k-1)}; /* Michael Somos, Apr 03 2016 */

T(n,k) = T(n,k-1) + T(n,k-2) = T(n-1,k) + T(n-2,k).
T(n,n) = T(n-1,n+1) = A061646(n).
T(n,n+1) = A079472(n+1). Omitting T(n,0), the array is symmetric about this falling diagonal.
Treating rows and columns as individual sequences, let R_n be Row n and C_k be Column k; let R_n(k) and C_k(n) be terms k and n, respectively, in these sequences:
C_0(n) = A000045(n+1).
R_0(k) = A000045(k-1); C_1(n) = A000045(n).
R_1(k) = A000045(k+1); C_2(n) = A000045(n+2).
R_2(k) = A000032(k); C_3(n) = A000032(n+1) .
R_3(k) = A013655(k); C_4(n) = A013655(n+1).
R_4(k) = A022121(k-1); C_5(n) = A022121(n).
R_5(k) = A022138(k-1); C_6(n) = A022138(n).
R_6(k) = A206610(k+1); C_7(n) = A206610(n+2).

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A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

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A028387 a(n) = n + (n+1)^2.

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A035513 Wythoff array read by falling antidiagonals.

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A127830 a(n) = Sum_{k=0..n} (binomial(floor(k/2),n-k) mod 2).

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A199535 Clark Kimberling's even first column Stolarsky array read by antidiagonals.

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