cp's OEIS Frontend

Note: 1 might be a more natural starting offset for this sequence, although the identities concerning A053645 and A161511 would have to be changed. - Antti Karttunen, Oct 12 2009.

This can be regarded as an arrangement of the partitions, indexed by position in A125106. The partitions in a given row all have the same remaining partition when the largest part is removed; specifically, the partition indexed by the row number in A125106 (with row 0 having the empty partition remaining).

The first value on row n is A004760(n+1). The second value on each row is A004760(n+1) plus A062383(n); subsequent values increase by ever enlarging powers of two. Or equivalently, each subsequent value on the row after the first nonzero value is given by A004754(previous value on the same row).

A055941(r) tells how many terms the row r (>= 0) has been shifted rightward from its "natural position", i.e. with how many zeros that row has been prepended.

The number of (nonzero) entries in column k is A000041(k).

Let A_n be the upper triangular matrix in the group GL(n,3) of invertible n X n matrices over GF(3) that has zero entries below the diagonal and 1 elsewhere. For example for n=4 the matrix is / 1,1,1,1 / 0,1,1,1 / 0,0,1,1 / 0,0,0,1 /. a(n) is the order of this matrix as an element of GL(n,3).

For n>1 a(n) is the smallest integer such that gcd(a(n),2^a(n)+1) >= n. - Benoit Cloitre, Apr 21 2002

From Jianing Song, Jul 05 2025: (Start)

a(n+1) is the period of {binomial(N,n) mod 3: N in Z}. For the general result, see A349593.

Since the modulus (3) is a prime, the remainder of binomial(N,n) is given by Lucas's theorem. (End)

Consider the symbols as positive integers. By the greedy way we mean to fill the grid row by row from left to right always with the least possible positive integer such that the three constraints (on rows, columns and rectangular blocks) are satisfied. In contrast to the sudoku case, the m X m rectangles have "floating" borders, so the constraint is actually equivalent to say that any element must be different from all neighbors in a Moore neighborhood of range m-1 (having up to (2m-1)^2 grid points). See A292673 for examples.

One can consider the infinite square array A(m,n) defined in the same way, but the lower triangular part of it is uninteresting: for m>n one has A(m,n) = n^2, i.e., the columns continue below the diagonal indefinitely with the same value, n^2.

Either a(n) = a(n-1) or a(n) = a(n-1) + 1. Proof: Suppose n is a power of 2, then a(n+1) = a(n) + a(0) = a(n). Otherwise let 2m be the largest power of 2 greater than n, so a(n) = a(m) + a(n-1-m) and a(n+1) = a(m) + a(n-m) and then proceed by induction. - Charles R Greathouse IV, Aug 27 2014

It appears that this sequence gives the partial sums of A164349. - Arie Groeneveld, Aug 27 2014

Each term other than zero appears at least twice. Suppose m is a power of 2, then a(2m) and a(4m) appear at least twice by my above comment. Otherwise suppose 3 <= k+2 <= 2m, then a(2m+k) = a(m) + a(m+k-1), a(2m+k+1) = a(m) + a(2m+k), and a(2m+k+2) = a(m) + a(m) + a(m+k+1), so a(2m+k+2) - a(2m+k) = a(m+k+1) - a(m+k-1). So if each term from a(m) to a(2m) appears at least twice then so will each term in a(2m) to a(4m). - Charles R Greathouse IV, Sep 10 2014

a(n) = Theta(n), see link. - Benoit Jubin, Sep 16 2014

The position of where n first appears: 0, 1, 4, 6, 10, 13, 15, 18, 21, 23, 27, 30, 32, 34, 37, 39, 43, 46, 48, 51, 54, 56, 60, 63, 66, 69, ... - Robert G. Wilson v, Sep 19 2014

The (10^k)-th term: 0, 3, 36, 355, 3549, 35494, 354942, ... - Robert G. Wilson v, Sep 19 2014

Is this a permutation of the positive integers?

The smallest number not in {a(n) | n<=8000000} is 5083527. It appears that the quotient (a(1)+...+a(n))/n^2 meanders around between 1/2 (=perfect permutation) and 2/3: at n=8000000 the value is approximately 0.5866 (does it converge? 1/2? Golden ratio?).

The scatterplot of the first 100000 terms (see "graph") has some remarkable features which have not yet been explained. - Leroy Quet, Jul 05 2009

The lines that appear in the scatterplot seem to be related to the position of n in the sum of the first n terms; see colorized scatterplots in the Links section. - Rémy Sigrist, May 08 2017

From Michael De Vlieger, May 09 2017: (Start)

Starting positions of n in Sum_{k=1..n} a(k) written in binary: {1, 1, 1, 2, 1, 1, 1, 3, 2, 4, 3, 1, 1, 1, 5, 3, 2, 4, 3, 5, 4, 5, ...}.

Running total of a(n) in binary: {1, 100, 110, 1100, 10100, 11000, 11101, 101000, 110010, 1001010, 1010110, 1100011, 1101010, 1110011, ...}.

(End)

Usually successive powers of 2, but "stutters" when n is power of 2. (G.f. must satisfy some interesting functional equations!). Empty partition of 0 defined as 1.

This is an instance of what I like to call "numbral theory": whenever you have a set of indexed objects that you can do some kind of arithmetic on, then the indices act as "shadows" of the objects and you can generally talk about lots of analogs, such as partitions, primes, even generating functions, etc. It would be worthwhile to systematically "fill out" the entries for as many of these systems as possible in the OEIS.

The "AND" version is just the all-ones sequence. - Christian G. Bower, Jun 07 2005

a(n) is the number of orbits of the FlipAfter1 map on integers with n+1 binary digits. The FlipAfter1 map on an integer in binary form is: flip each bit that is immediately preceded by a "1". For example, the orbits on 4-bit numbers are 1000 -> 1100 -> 1010 -> 1111 and 1001 -> 1101 -> 1011 -> 1110. The orbits on n-bit numbers are all of length 2^floor(log_2(n-1)+1) (for n >= 2), A062383. There is precisely one member of each orbit in the following set: integers in binary form such that each bit at distance a power of two from the leading "1" is 0. This set of orbit representatives begins 1, 10, 100, 1000, 1001, 10000, 10010, 100000, 100001, 100100, 100101. - David Callan, Oct 13 2012

The EG1 matrix coefficients are defined by EG1[2m-1,1] = 2*eta(2m-1) and the recurrence relation EG1[2m-1,n] = EG1[2m-1,n-1] - EG1[2m-3,n-1]/(n-1)^2 with m = .., -2, -1, 0, 1, 2, ... and n = 1, 2, 3, ... . As usual, eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. For the EG2 matrix, the even counterpart of the EG1 matrix, see A008955.

The coefficients in the columns of the EG1 matrix, for m >= 1 and n >= 2, can be generated with GFE(z;n) = ((-1)^(n-1)*r(n)*CFN1(z,n)*GFE(z;n=1) + ETA(z,n))/pg(n) for n >= 2.

The CFN1(z,n) polynomials depend on the central factorial numbers A008955 and the ETA(z,n) are the Eta polynomials which led to the Eta triangle, see for both A160464.

The pg(n) sequence can be generated with the first Maple program and the EG1[2m-1,n] matrix coefficients can be generated with the second Maple program.

The EG1 matrix is related to the ES1 matrix, see A160464 and the formulas below.

Permutation of the natural numbers with inverse A180077;

if a(n-1) > a(n) then a(n) < a(n+1) = 3*a(n)+1;

see A180110 for m with a(m-2) < a(m-1) < a(m);

A180078(n) = a(a(n));

a(A180079(n)) = A180079(a(n)) = A180077(n);

A180080 and A180081 give record values and where they occur.