cp's OEIS Frontend

For a graph G, pick a permutation of its vertices that minimizes the bitstring obtained by reading the lower triangular part of the corresponding adjacency matrix by rows. The code of G is that bitstring interpreted as a binary number plus 2^(v*(v-1)/2), where v is the number of vertices of G; see example. As a special case, the code of the null graph is 0. The sequence consists of all such minimal codes.

For n >= 1, the numbers of vertices and edges of the graph with code a(n) are A002024(A000523(a(n))+1) and A000120(a(n))-1 = A382758(n), respectively.

This sequence can be used to define sequences for:

- graph invariants (examples: A382758, A382759, A382760);

- graph operators, either by code (A382763) or by index (A382764);

- lists of subsets of graphs, either by code (A382761) or by index (A382762).

From Davis Smith, May 09 2021: (Start)

For n > 2, a(n) cannot be a power of 2.

If A007088(n) (the binary expansion of n) contains a string of k zeros, then it contains A007088(2^m), where 0 <= m <= k, as a substring. Similarly, if A007088(n) contains a string of k ones, then it contains A007088(2^m - 1), where 1 <= m <= k. Strings of zeros and ones are the most compact way to have powers of 2 and powers of 2 minus 1 (respectively) as substrings in a binary expansion. This means that A007088(a(n)) will contain a string of A000523(n) ones and a string of A000523(n) zeros. The binary expansion of a(2^k - 1) will contain a string of k ones and a string of k - 1 zeros.

Conjecture: a(n) == 0 (mod A053644(n)), i.e., A007088(a(n)) ends with the longest string of zeros. It follows from this that a(2^k) = 2*a(2^k - 1). A conjecture related to this is that a(2^k - 1) = 2*a(2^k - 2) + 2^(k - 1), i.e., A007088(a(2^k - 1)) ends with the longest string of ones followed by the longest string of zeros. Ending with the longest string of ones followed by the longest string of zeros is not true for all A007088(a(n)), as some have a hiccup before starting their string of zeros, e.g., a(10), a(18), a(22), and a(34).

Conjecture: a(2^k + 1) = 2^(k + floor(log_2(a(2^k)))) + a(2^k), i.e., concatenate the binary expansion of 2^(k - 1) to the front of the binary expansion of a(2^k) in order to get the binary expansion of a(2^k + 1).

(End)

All terms belong to A261467. - Rémy Sigrist, May 11 2021

From Jon E. Schoenfield, Jun 03 2021: (Start)

Conjecture: the binary expansion of a(n) contains exactly ceiling(n/2) 1's iff 2^m - 7 <= n <= 2^m + 6 for some integer m >= 3. (See Links.)

Conjecture: for n > 1, the binary expansion of a(n) begins with that of 2^floor(log_2(n-1)) + 1.(End)

From Davis Smith, Jun 05 2021: (Start)

For a proof that a(n) == 2^floor(log_2(n)) (mod 2^(floor(log_2(n)) + 1)), see my second link (not the b-file). This also proves the conjecture from May 09 2021 which states that it is congruent to 0 (mod A053644(n)). A proof for the related conjecture would likely rely on an explanation of values of n such that a(n) is not congruent to (2^floor(log_2(n)) - 1)*2^floor(log_2(n)) (mod 2^(2*floor(log_2(n)))), i.e. the values of n such that A007088(a(n)) does not end with a string of floor(log_2(n)) ones followed immediately by a string of floor(log_2(n)) zeros. A proof for Jon E. Schoenfield's second conjecture on Jun 03 2021 would satisfy my more restricted second conjecture and it may follow necessarily from my proof, assuming that A007088(a(n)) must begin with either A007088(2^floor(log_2(n - 1)) + 1) or A007088(2^floor(log_2(n))). (End)

Does not satisfy Benford's law [Whyman et al., 2016]. - N. J. A. Sloane, Feb 12 2017

In an increasing rooted graph, nodes are numbered and the numbers increase as you move away from the root.

(a(n+1)/a(n))/n tends to 1/A073003 = 1.676875... (same limit as A029768). - Vaclav Kotesovec, Jul 26 2014

A measure of the growth rate of the primorials.

A023416(a(n)) = A023416(n), A000120(a(n)) = A000120(n).

Fixed points are { 0 } union { A099627 }. - Alois P. Heinz, Jan 30 2025