cp's OEIS Frontend

This is a fractal sequence: if the first instance of each number is deleted, the original sequence is recovered. - Franklin T. Adams-Watters, Dec 14 2013

Subsequences start at indices A000330 + 1. - Ralf Stephan, Dec 17 2013

When sequence fills a triangular array by rows, the main diagonal is A064865:

This triangle begins:

1

1 2

3 4 1

2 3 4 5

6 7 8 9 1

From Antti Karttunen, Feb 17 2014: (Start)

A more natural way of organizing this sequence is as an irregular table consisting of successively larger square matrices:

1;

1, 2;

3, 4;

1, 2, 3;

4, 5, 6;

7, 8, 9;

1, 2, 3, 4;

5, 6, 7, 8;

9,10,11,12;

13,14,15,16;

etc.

(End)

The n-th block consists of n subblocks, each of which counts from 1 to n.

This a fractal sequence: removing the first instance of each value leaves the original sequence.

The first comment implies that this gives the column index of the n-th element of a sequence whose terms are coefficients, read by rows, of a sequence of matrices of size 1 X 1, 2 X 2, 3 X 3, etc.; cf. example. The row index is given by A238013(n), and the size of the matrix by A074279(n). - M. F. Hasler, Feb 16 2014

Cases of enumeration in ascending order for the first m positive integers when one of them, j, is previously excluded (m=1,2,3,..., 1 <= j <= m).

Table of the k initial permutations, one per block, when all the k! permutations in lexicographic ascending order are split uniformly into k blocks. Such table read by rows for k=1,2,3,... .

These permutations might be considered the initial inputs for a parallel/distributed variant of the Narayana Pandita's algorithm. Such variant would deliver to each thread/core/host one or more of the mentioned inputs, then the remaining permutations can be obtained with (k-1)!-1 executions of the classic Narayana Pandita's algorithm for the next permutation in lexical order.

The terms of A237450 give the positions of rows of this table among the rows of A030298. The finite n X n square matrices converge towards the infinite square array A237447. Please see further comments there. - Antti Karttunen, Feb 10 2014

Alternative way to express this is that each row k=1..n of each n X n matrix contains the lexicographically earliest n-letter permutation beginning with number k, or equally, that each of the n X n square matrices contain in their n rows those n-letter permutations of the symmetric group S_n that correspond to the inverses of cycles (1), (1 2), (1 2 3), ..., (1 2 ... n). Please see the Example section. - Antti Karttunen, Feb 12 2014

Consider a sequence of matrices of size 1 X 1, 2 X 2, 3 X 3, etc., then list their elements, read by rows. The present sequence gives the row index of the n-th term of the sequence within "its" matrix, numbered by A074279(n); the column index being given by A121997.

The sequence could be seen as irregular table or triangle with row lengths equal to 1,2,2,3,3,3,4,4,4,4,... = A002024, each row being filled with k's, for the k-th row of given length; see Example.

With sequences constructed of successively larger square matrices (cf. A074279), a(n) will return the distance of n from the left edge of the matrix that n is located in, with 0 standing for the leftmost column (please see the Example section).

A237452 gives the corresponding row index.

A238013 and A121997 give these same row and column indices, starting the numbering with index 1. - M. F. Hasler, Feb 17 2014

With sequences constructed of successively larger kxk square matrices (cf. A074279), a(n) will return the distance of n from the top edge of the matrix that n is located in, with 0 standing for the topmost row in that matrix (please see the Example section).

A237451 gives the corresponding column index.

A238013 and A121997 give these same row and column indices, but starting the numbering with index 1. - M. F. Hasler, Feb 17 2014

From Jonathan Vos Post, Mar 18 2006: (Start)

The key to this sequence is: 1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+3+...+n)^2.

Since the last occurrence of n comes one before the first occurrence of n+1 and the former is at Sum_{i=0..n} i^3 = A000537(n) = (A000217(n))^2 = (n*(n+1)/2)^2 = (C(n+1,2))^2, have a(A000537(n)) = a((A000217(n))^2) = n and thus a(1+A000537(n)) = a(1+(A000217(n))^2) = n+1.

The current sequence is, loosely, the inverse function of the square of the triangular number sequence. (End)

In terms of the repetition convolution operator #, where (sequence A) # (sequence B) = the sequence consisting of A(n) copies of B(n), then this sequence is the repetition self-convolution (A000217(n)-0) # (A000217(n)-0). Over the set of positive infinite integer sequences, # gives a nonassociative noncommutative groupoid with a left identity (A000012) but no right identity, where the left identity is also a right nullifier and idempotent.

Triangle read by rows in which row n lists the first n^2 positive integers in decreasing order, n >= 1. - Omar E. Pol, Jan 20 2015