cp's OEIS Frontend

Relates squares of numerators and denominators of continued fraction convergents to sqrt(5).

Sequence is generated by the floretion A*B*C with A = + 'i - 'k + i' - k' - 'jj' - 'ij' - 'ji' - 'jk' - 'kj' ; B = - 'i + 'j - i' + j' - 'kk' - 'ik' - 'jk' - 'ki' - 'kj' ; C = - 'j + 'k - j' + k' - 'ii' - 'ij' - 'ik' - 'ji' - 'ki' (apart from a factor (-1)^n)

Floretion Algebra Multiplication Program, FAMP Code: tesseq[A*B*C].

The sequence a(n-1), n >= 0, with a(-1) = 1, is also the curvature of circles inscribed in a special way in the larger segment of a circle of radius 5/4 (in some units) cut by a chord of length 2. For the smaller segment, the analogous curvature sequence is given in A240926. For more details see comments on A240926. See also the illustration in the present sequence, and the proof of the coincidence of the curvatures with a(n-1) in part I of the W. Lang link. - Kival Ngaokrajang, Aug 23 2014

(11*b(n))^2 - 5*(5*a(n))^2 = -4 with b(n)=A097842(n) give all positive solutions of this Pell equation.

B is of the form B(i) = 26*B(i-1) - B(i-2) for B(0) = 1, B(1) = 25 (this sequence).

A is of the form A(i) = 26*A(i-1) - A(i-2) for A(0) = 1, A(1) = 27.

In general a Pell-like equation of the form 1 + X*A*A = (X + 1)*B*B has the solution A(i) = (4*X + 2)*A(i-1) - A(i-2), for A(0) = 1 and A(1) = (4*X + 3), and B(i) = (4*X + 2)*B(i-1) - B(i-2) for B(0) = 1 and B(1) = (4*X + 1).

Examples in the OEIS:

X = 1 gives A002315 for A(i) and A001653 for B(i);

X = 2 gives A054320 for A(i) and A072256 for B(i);

X = 3 gives A028230 for A(i) and A001570 for B(i);

X = 4 gives A049629 for A(i) and A007805 for B(i);

X = 5 gives A133283 for A(i) and A157014 for B(i);

X = 6 gives A157461 for A(i) and this sequence for B(i).

Positive values of x (or y) satisfying x^2 - 26*x*y + y^2 + 24 = 0. - Colin Barker, Feb 20 2014

This sequence is part of a solution of a more general problem involving two 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; for details see comment in A157014.

A157459 is the c(n) sequence for A=4.

See A195500 for a discussion and references.

See A195500 for discussion and list of related sequences; see A195614 for Mathematica program.

The centered square numbers are n^2 + (n+1)^2 while the centered pentagonal numbers are (5*r^2 + 5*r + 2)/2. A number has both properties iff 5*(2*r+1)^2 = (4*n+2)^2 + 1. We solve the equation 5*Y^2 - 1 = X^2 whose solutions in positive integers are given by A075796 and A007805 respectively. The r values are 0,8,..., i.e., A053606. The n values define A119032.

See A214992 for a discussion of power ceiling-floor sequence and the power ceiling-floor function, p3(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = (golden ratio)^6, and the limit p3(r) = 17.94722275971790615684809...

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.

Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.

The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).

Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

This sequence gives all solutions of one of two classes of positive proper solutions a(n) = a1(n) of the Pell equation a(n)^2 - 5*b(n) = -4 with b(n) = b1(n) = Fibonacci(6*n+1) = A134493(n). These solutions are obtained from the fundamental positive solution [1, 1] (to be read as a column vector) by application of positive powers of the automorphic matrix A = matrix([9, 20], [4, 9]) with determinant +1.

The other class of positive proper solutions is obtained similarly from the fundamental solution [11,5] and is given by [a5(n), b5(n)], with a5(n) = A305316(n) and b5(n) = A134497(n) = F(6*n+5), with the Fibonacci numbers F = A000045.

The remaining positive solutions are improper and are obtained by application of the same matrix A on the fundamental improper solution [4, 2]. They are given by [a3(n), b3(n)], with a3(n) = 4*A049629(n) and b3(n) = A134495(n) = 2*A007805(n).

Via the Cayley-Hamilton theorem the powers of the automorphic matrix A are: A^n = matrix([S(n) - 9*S(n-1), 20*S(n-1)], [4*S(n-1), S(n) - 9*S(n-1)]) with the Chebyshev polynomials S(n-1) = S(n-1, x=18) = A049660(n), n >= 0.

This shows that ordered Markoff (Markov) triples [1, y, m], with 1 <= y <= m, have for m from the union of sets {m1(k)}{k>=0} U {m5(k)}{k>=0) U {m3(k)}_{k>=0)}, with mj(k) = F(6*k+j), for j = 1, 5, and 3, the unique solutions yj(k) = (3*F(6*k+j) - aj(k))/2 < mj(k), namely y1(k) = F(6*k-1) = A134497(k-1) with F(-1) = 1, y5(k) = F(6*k+3) = A134495(k) and y3(k) = F(6*k+1) = A134493. The solutions with the + sign are excluded because they are > mj(k). This trisection of the odd-indexed Fibonacci numbers as m numbers shows again the well known fact that each of them appears as largest member in a Markoff triple if the smallest member is x = 1. The positions of the odd-indexed Fibonacci numbers in the Markoff sequence A002559 are given in A158381. The conjecture in this case is that the odd-indexed Fibonacci numbers appear as largest numbers only in the ordered Markov triples with x = 1. See, e.g., the Aigner reference for the general Frobenius-Markoff conjecture.

Also Lucas numbers that are congruent to 1 mod 4. - Fred Patrick Doty, Aug 03 2020