cp's OEIS Frontend

Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. - Benoit Cloitre, Aug 15 2002

Binomial transformation yields A081908, with A081908(0)=1 dropped. - R. J. Mathar, Oct 05 2008

1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). - Wolfdieter Lang, Mar 01 2013

a(n) is the number of election results for an election with n+2 candidates, say C1, C2, ..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. - Dennis P. Walsh, May 08 2013

This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) = A005248; at n=2, b(k) = 2*A001541; at n=3, b(k)= A057076; at n=4, b(k) = 2*A023039. This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) = A005248. The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also see A000032 for a related family of sequences. - Richard R. Forberg, Nov 22 2014

Also the number of maximum matchings in the n-gear graph. - Eric W. Weisstein, Dec 31 2017

Also the Wiener index of the n-dipyramidal graph. - Eric W. Weisstein, Jun 14 2018

Numbers of the form n^2+2 have no factors that are congruent to 7 (mod 8). - Gordon E. Michaels, Sep 12 2019

For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {n, 2n}]. - Magus K. Chu, Sep 10 2022

2/a(n) = R(n)/r, n >= 0, with R(n) the n-th radius of the counterclockwise Pappus chain of the arbelos with semicircle radii r, r1 = 2r/3, r2 = r - r1 = r/3. See the MathWorld link for such a Pappus chain. The clockwise chain companion has circle radii R'(n)/r = 2/A222465(n), n >= 0. - Wolfdieter Lang, Mar 01 2013

Old name was: "Numbers of the form x^2 + 14".

x^2 + 14 != y^n for all x,y and n > 1.

Conjecture: n^2 + 10 != x^k for all n,x, and k > 1.

The conjecture is true: See Cohn. - James Rayman, Feb 14 2023

3/a(n) = R(n)/r, n >= 0, with R(n) the n-th radius of the counterclockwise Pappus chain of the arbelos with semicircle radii r, r1 = 3*r/4, r2 = r - r1 = r/4. See a comment on A114949 also for the MathWorld Pappus chain link. - Wolfdieter Lang, Jun 29 2015

Refer to sequential curvatures from Wikipedia. For any integer k > 0, there exists an Apollonian gasket defined by the following curvatures:

(-k, k+1, k*(k+1), k*(k+1)+1).

For example, the gaskets defined by (-1, 2, 2, 3), (-2, 3, 6, 7), (-3, 4, 12, 13), ..., all follow this pattern (all curvatures are integral). Because every interior circle that is defined by k+1 can become the bounding circle (defined by -k) in another gasket, these gaskets can be nested. When one considers only circles that contact both circles -k and k+1, the pattern will be nested Pappus chains. T(n,k) is the curvature when n = 0 is the circle at the center and n > 0 is in the clockwise direction, k >= 1 for each nested iteration. See illustration in links.

Conjecture: n^2 + 34 != x^k for all n,x and k > 1.

The conjecture is true: See Cohn. - James Rayman, Feb 14 2023