cp's OEIS Frontend

The sequences A001607, A077020, A107920, A167433, A169998 are all essentially the same except for signs.

This is an example of a sequence of Lehmer numbers. In this case, the two parameters, alpha and beta, are (1 +- i*sqrt(7))/2. Bilu, Hanrot, Voutier and Mignotte show that all terms of a Lehmer sequence a(n) have a primitive factor for n > 30. Note that for this sequence, a(30) = 24475 = 5*5*11*89 has no primitive factors. - T. D. Noe, Oct 29 2003

Row sums of Riordan array (1/(1+2*x^2), x/(1+2*x^2)). - Paul Barry, Sep 10 2005

Pisano period lengths: 1, 1, 8, 2, 24, 8, 21, 2, 24, 24, 10, 8, 168, 21, 24, 4, 144, 24, 360, 24, ... - R. J. Mathar, Aug 10 2012

This is the Lucas Sequence U_n(P, Q) = U_n(1, 2). V_n(1, 2) = A002249(n). - Raphie Frank, Dec 25 2013

Note that (A002249(n)/2)^2 + 7*(a(n)/2)^2 = 2^n for all n in N. This is a specific case of the Lucas sequence identity (V_n/2)^2 - D*(U_n/2)^2 = Q^n where V_n = (a^n + b^n), U_n = (a^n - b^n)/(a - b), Q = (a*b) = 2 and D = (a - b)^2 = -7; a = (1 + sqrt(-7))/2 and b = (1 - sqrt(-7))/2. - Raphie Frank, Nov 26 2015

For the special case where |a(n)| = 1, true for n if and only if n is in {1, 2, 3, 5, 13} = {A215795(n) + 1} = {A060728(n) - 2}, then, additionally, by the Lucas sequence identity (U_2n = U_n*V_n), we have (a(2n)/2)^2 + 7*(a(n)/2)^2 = 2^n. - Raphie Frank, Nov 26 2015

See A038198 for corresponding x. - Lekraj Beedassy, Sep 07 2004

Also numbers such that 2^(n-3)-1 is in A000217, i.e., a triangular number. - M. F. Hasler, Feb 23 2009

With respect to M. F. Hasler's comment above, all terms 2^(n-3) - 1 are known as the Ramanujan-Nagell triangular numbers (A076046). - Raphie Frank, Mar 31 2013

Interestingly enough, all the solutions correspond to noncomposite x, i.e., x = 1 for the first term, and primes 3, 5, 11, 181 for the following terms. - M. F. Hasler, Mar 11 2024

a(n) = (2*x - 1)^2 = (sqrt(2)*sqrt(sqrt(6*y^2 - 5) + 1) - 1)^2 = 2^(z + 3) - 7 for x, y, z are the solutions to two Diophantine equations noted by R. K. Guy: 2*x^2*(x^2 - 1) = 3*(y^2 - 1) & x*(x - 1)/2 = 2^z - 1 (see A180445). x = {1, 2, 3, 6, 91} = A180445(n), y = {1, 3, 7, 29, 6761} = A227077(n), and z = {0, 1, 2, 4, 12} = A215795(n).

2*(x^2)*((x^2)-1) = 3*((y^2)-1) has only these five positive solutions.

x*(x-1)/2 = (2^z)-1 has only these five positive solutions.

Richard K. Guy notes, as Example 29: "True, but why the coincidence?"

Algebraically, y solutions = {1, 3, 7, 29, 6761} can be derived from x solutions as follows: y = sqrt(((2*x^2 - 1)^2 + 5)/6). From this relationship it becomes clear that the form (((2*x^2 - 1)^2 + 5)/6) can only be an integer square for x is in {1, 2, 3, 6, 91}. Thus, x and y solutions are also unique integer solutions to the following equivalency: (2x^2 - 1)^2 = 6y^2 - 5. From this relationship the following statement naturally follows: ((sqrt(6*y^2 - 5) + 1)/2 - sqrt((sqrt(6*(y^2) - 5) + 1)/2))/2 = (2^z - 1) = {0, 1, 3, 15, 4095} = A076046(n), the Ramanujan-Nagell triangular numbers; z = {0, 1, 2, 4, 12} = (A060728(n) - 3). - Raphie Frank, Jun 26 2013

Start with a point marked on an axis at unit distance from the origin. How many distinct points can be marked on the axis, if the offset from the origin of each additional point is a positive integer multiple or submultiple of that of a point already marked, and no less than unit distance nor greater than a given maximum, d? The possible finite answers, which occur only if d < 27/5, are the numbers in this sequence.

If we think of an interval with lower bound 1 as growing in time, equivalence classes of cardinality a(n) appear when the interval reaches [1, r_min(n)] for a rational r_min(n) whose numerator and denominator are 5-smooth, and disappear when the interval reaches [1, r_max(n)) where r_max(n) = (r_min(n+1))^2/r_min(n).

The initially known sequence terms (n <= 20) have the following property: a(n) = 2 * a(n-1) or, for some i < n, a(n) = 2 * a(n-1) - a(i); the former case occurring if r_min(n) equals a prime number, the latter occurring if and only if r_min(n) = r_max(i). Having identified underlying reasons that look like the basis of a proof, the author conjectures this is true for all n > 1.

Infinite equivalence classes occur if the interval is a scaled image of an interval containing (1, 27/5). Intervals that are scaled images of an interval containing [1, 50/9) or (1, 50/9] have no finite equivalence classes; for other intervals, the smallest equivalence class has cardinality 1, 2, 4 or 12 (cf. A215795).

The 5-smooth rationals in the critical semi-open interval [1, 50/9) are therefore partitioned into infinite equivalence classes. I believe these classes can be numbered, ordered by the power(s), k, of 5, for which 5^k*2^j is in the class for at least one integer j. 8/5, 1/1 and 5/1 are in class 0; 25/8 and 125/64 are in class 1; 32/25 and 128/125 are in class -1. If we close the critical interval, the inclusion of 50/9 unites classes 0 and 1 as defined above. If an interval, I, (or any scaled image of I) is a proper superset of the closed interval [1, 50/9], I am confident all 5-smooth rationals in I are in the same equivalence class. - Peter Munn, Jun 29 2022

From Peter Munn, Sep 16 2022: (Start)

The linked figure, "The disappearance of finite classes", shows various interval parts of the interval [1, 27/5] as rectangles. The arrowed lines between the rectangles represent the relation R. It can be seen as summarizing a graph between uncountably many points.

There are 12 rectangles around the circumference annotated as representing points that are on a finite path. These correspond to the points (equivalently real numbers) that are in equivalence classes of cardinality 12 -- the only finite equivalence classes when the interval is large enough to have infinite equivalence classes.

If we animate this figure to show the upper bound of the interval increasing from 27/5, the 12 outer rectangles would shrink. Consider, in particular, the one on the right marked with bounds 27/25 and 10/9. 10/9 derives from (((1 * 5) / 3) * 2) / 3, a progression that can followed around the inner edge of the circuit in the figure. 10/9 stays fixed. 27/25 grows as 1/5 of the upper bound, and the finite classes/paths disappear when the upper bound reaches 50/9 = 5 * 10/9.

(End)