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x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.

The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013

From Peter Bala, Mar 23 2018: (Start)

Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have

2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.

a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

Let f = floor and c = ceiling. For x > 1, define four sequences as functions of x, as follows:

p1(0) = f(x), p1(n) = f(x*p1(n-1));

p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;

p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;

p4(0) = c(x), p4(n) = c(x*p4(n-1)).

The present sequence is given by a(n) = p3(n).

Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences. In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms. Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n).

x ......p1..... p2..... p3..... p4.......Limit

r.......A000045 A000045 A000045 A000045..r

r^2.....A001519 A001654 A061646 A001906..-1+S(5)

r^3.....A024551 A001076 A015448 A049652..-1+S(5)

r^4.....A049685 A157335 A214992 A004187..-19+9*S(5)

r^5.....A214993 A049666 A015457 A214994...(-9+5*S(5))/2

r^6.....A007805 A156085 A214995 A049660..-151+68*S(5)

1+S(2)..A024537 A000129 A001333 A048739..S(2)

2+S(2)..A007052 A214996 A214997 A007070..(1+S(2))/2

1+S(3)..A057960 A002605 A028859 A077846..(1+S(3))/2

2+S(3)..A001835 A109437 A214998 A001353..-4+3*S(3)

S(5)....A214999 A215091 A218982 A218983..1.26879683...

2+S(5)..A024551 A001076 A015448 A049652..-1+S(5)

2+S(6)..A218984 A090017 A123347 A218985..S(3/2)

2+S(7)..A218986 A015530 A126473 A218987..(1+S(7))/3

2+S(8)..A218988 A057087 A086347 A218989..(1+S(2))/2

3+S(8)..A001653 A084158 A218990 A001109..-13+10*S(2)

3+S(10).A218991 A005668 A015451 A218992..-2+S(10)

...

Properties of p1, p2, p3, p4:

(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.

(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.

(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer. If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences. Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?

(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent? If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?

(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=1 results shown here; y=-1 results are A188646.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/sqrt(n^2+1)) * T_{2*k+1}(sqrt(n^2+1)), with T the Chebyshev polynomial of the first kind. - Seiichi Manyama, Jan 02 2019

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:

A * c(n)+1 = a(n)^2,

(A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:

a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;

b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;

c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .

A157014 is the a(n) sequence for A=5.

For other A values the a(n), b(n) and c(n) sequences are in the OEIS:

A a-sequence b-sequence c-sequence

1 A001653 A002315(n-1) A078522

2 A072256 A054320(n-1) A045502(n-1)

3 A001570 A028230 A059989(n-1)

4 A007805 A049629(n-1) A157459

5 -> A157014 <- A133283 A157460

6 A153111 A157461 A157874

7 A157877 A157878 A157879

8 A077420(n-1) A046176 A157880

9 A097315(n-1) A097314(n-1) A157881

Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014

From Klaus Purath, Apr 22 2025: (Start)

Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.

(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.

a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.

The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,

with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.

Generic recurrences are:

A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.

B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.

k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).

x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;

Binet-type of solutions of these 2nd order recurrences are:

R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);

A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);

B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);

x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;

k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]

.C -A----- -B----- -k-----

01 A001519 A002878 A058038

02 A001653 A002315 A045899/2

03 A004253 A030221 A160695

04 A001653 A002315 A078522/4

05 A049685 A033890 A161582

06 A070997 A057080 A159683/2

07 A070998 A057081 A161585

08 A072256 A054320 A045502/4

09 A078922 A097783 A161586

10 A077417 A077416 A159681/2

11 A085260 A126816 A161588

12 A001570 A028230 A059989/4

13 A160682 A161591 A161584

14 A157456 A159678 A159679/2

15 A161595 A161599 A161583

16 A007805 A049629 A157459/4

For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]

Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Bisection of A001077; a(n) = A001077(2*n-1). - Greg Dresden, Jun 08 2021

From Peter Bala, Aug 25 2022: (Start)

The aerated sequence (b(n))n>=1 = [2, 0, 38, 0, 682, 0, 1238, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). The sequence (1/2)*(b(n))n>=1 is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. (End)

Gives those numbers which are Fibonacci numbers in A103135.

Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013

a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

From Tanya Khovanova, Jan 06 2023: (Start)

Fibonacci(6n+3) are divisible by 2 but not by 4.

These numbers are not divisible by 3. (End)