cp's OEIS Frontend

The subset {a(1),...,a(2k)} together with a(2k+2) is the set of proper divisors of 5*2^k.

For a(n)>4: number of vertices of complete graphs that can be properly edge-colored in such a way that the edges can be partitioned into edge disjoint multicolored isomorphic spanning trees.

(Editor's note: The following 3 comments are equivalent.)

From Wouter Meeussen, Apr 10 2005: This appears to be the same sequence as "Numbers n such that n^2 is not the sum of three nonzero squares". Don Reble and Paul Pollack respond: Yes, that is correct.

Also numbers k such that k^2=a^2+b^2+c^2 has no solutions in the positive integers a, b and c. - Wouter Meeussen, Apr 20 2005

The only natural numbers which cannot be the lengths of an interior diagonal of a cuboid with natural edges. - Michael Somos, Mar 02 2004

Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j).

Equivalently, the number of walks of length n-1 on the path graph P_k. - Andrew Howroyd, Apr 17 2017

Interleaving of A000079 without initial 1 and A007283.

Agrees from a(2) onward with A145751 for all terms listed there (up to 65536). Apparently equal to 2, 3 followed by A090989. Equals 2 followed by A163978.

Binomial transform is A000129 without first two terms, second binomial transform is A020727, third binomial transform is A164033, fourth binomial transform is A164034, fifth binomial transform is A164035.

Number of achiral necklaces or bracelets with n beads using up to 2 colors. For n=5, the eight achiral necklaces or bracelets are AAAAA, AAAAB, AAABB, AABAB, AABBB, ABABB, ABBBB, and BBBBB. - Robert A. Russell, Sep 22 2018

The left column, A(n,0), is A000129(n), Pell Numbers.

The top row, A(0,k), is A077957(k) plus an initial 0, which is the inverse binomial transform of A000129.

These may be considered initializing values, or results, depending the perspective taken, since there are several ways to generate the array. See Formula section for details.

The columns of the array hold all values, in sequential order, of numbers m such that 2m^2 + 2^k or 2m^2 - 2^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for k+1.

For example A(n,0) are numbers such that 2m^2 +- 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 2m^2 +- 2 are squares, with those square roots in A(n,2), etc.

A(n, k)/A(n,k-2) = 2; A(n,k)/A(n,k-1) converges to sqrt(2) for large n.

A(n,k)/A(n-1,k) converges to 1 + sqrt(2) for large n.

The other columns of this array hold current OEIS sequences as follows:

A(n,1) = A001333(n); A(n,2) = A163271(n); A(n,3) = A002203(n);

Bisections of these column-oriented sequences also appear in the OEIS, corresponding to the even and odd rows of the array, which in turn correspond to the two different recursive square root equations in the formula section below.

Farey fraction denominators interleave columns 0 and 1, and the corresponding numerators interleave columns 1 and 2, for approximating sqrt(2). See A002965 and A119016, respectively.

The other rows of this array hold current OEIS sequences as follows:

A(1,k) = A016116(k); A(2,k) = A029744(k) less the initial 1;

A(3,k) = A070875(k); A(4,k) = A091523(k) less the initial 8.

The Pell Numbers (A000219) are the only initializing set of numbers where the two recursive square root equations (see below) produce exclusively integer values, for all iterations of k. For any other initial values only even iterations (at k = 2, 4, ...) produce integers.

The numbers in this array, especially the first three columns, are also integer square roots of these expressions: floor(m^2/2), floor(m^2/2 + 1), floor (m^2/2 - 1). See A227972 for specific arrangements and relationships. Also: ceiling(m^2/2), ceiling(m^2/2 + 1), ceiling (m^2/2 -1), m^2+1, m^2-1, m^2*(m^2-1)/2, m^2*(m^2-1)/2, in various different arrangements. Many of these involve: A000129(2n)/2 = A001109(n).

A001109 also appears when multiplying adjacent columns: A(n,k) * A(n,k+1) = (k+1) * A001109(n), for all k.

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple incremental action, bundle existing output as an integrated product (which requires p=-1 steps), or implement the current bundled action (which requires q=5 steps). The first use of a novel bundle erases (or makes obsolete) all prior actions. How large an output can be generated in n time steps?"

1. A production scheme with replacement R(p,q) eliminates existing output following a bundling action, while an additive scheme A(p,q) retains the output. The schemes correspond according to A(p,q)=R(p-q,q), with the replacement scheme serving as the default presentation.

2. This problem is structurally similar to the Copy and Paste Keyboard problem: Existing sequences (A178715 and A193286) should be regarded as Paradigm-Shift Sequences with production schemes R(1,1) and R(2,1) with replacement, respectively.

3. The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.

4. All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

Apparently identical to A111386! Is this a theorem? - Klaus Brockhaus, Jul 21 2009

For n > 1, number of necklaces with n-1 beads and 5 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

Number of chiral bracelets with n beads and two colors.

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple incremental action, bundle existing output as an integrated product (which requires p=-3 steps), or implement the current bundled action (which requires q=5 steps). The first use of a novel bundle erases (or makes obsolete) all prior actions. How large an output can be generated in n time steps?"

1. A production scheme with replacement R(p,q) eliminates existing output followinging a bundling action, while an additive scheme A(p,q) retains the output. The schemes correspond according to A(p,q)=R(p-q,q), with the replacement scheme serving as the default presentation.

2. This problem is structurally similar to the Copy and Paste Keyboard problem: Existing sequences (A178715 and A193286) should be regarded as Paradigm-Shift Sequences with production schemes R(1,1) and R(2,1) with replacement, respectively.

3. The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 2.

4. All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

5. For large n, the sequence is recursively defined.