cp's OEIS Frontend

Terms do not depend on the choice of m, provided that m!>n (the index of the considered term), and the numbers associated to a permutation s of {0,...,m-1} are N(s) = Sum_{i=1..m} s(i)*10^(m-i). This defines the present sequence for any arbitrarily large index, not limited to n <= 10!, for example.

Similar sequences might be built in another base b, they would always start (1, b-1, 2, b-1, 1, ...). The partial sums of this kind of sequence would yield the analog of A215940 in the corresponding base.

There are at least two palindromic patterns which are repeated throughout this sequence: one of them is "1,b-1,2,b-1,1" (It is optional here whether or not to include the 1's), another is built from the first 4!-1 terms (See the corresponding link for details).

Also, for 1<=n<=(9!)-1: The repeating parts in the first differences of A030299 divided by nine, i.e. a(n) = A219664(n)/9. - Antti Karttunen, Dec 18 2012. Edited by: R. J. Cano, May 09 2017

There are more palindromic patterns than those mentioned above: Similar to the first 3!-1 and the first 4!-1 terms, the first k!-1 terms are repeated for all other k>4. Frequent are also multiples of these, e.g., k*[1,9,2,9,1] = [2,18,4,18,2], [3,27,6,27,3], ...), [1, 19, 3, 8, 2, 67, 1, 29, 4, 7, 3, 176, 3, 7, 4, 29, 1, 67, 2, 8, 3, 19, 1], and others. The "middle part" of roughly half the length (e.g., [9,2,9] or [67,...,67] in the last example), is repeated even more frequently. - M. F. Hasler, Jan 14 2013

From R. J. Cano, Apr 04 2016: (Start)

Conjecture 1: Given 1A217626 and so on).

Lemma: Let P be an arbitrary set consisting of m integers; let x[i] be an element in P (with 1<=i<=m); let y[j] = x[j+1] - x[j] (with 1 <= j <= m-1) be the 1st differences of P. These differences are symmetric if y[j]=y[m-j] which for P implies the condition x[j]+x[m-j+1]=x[j+1]+x[m-j];

Consequence: When m=n! and P is a set with all the permutations for the letters 0..n-1, the preceding lemma implies P has associated at least a set Q such that 1st differences in Q are symmetric.

Generating algorithm: Such Q can be built based upon P and the condition given by the preceding lemma if it is removed from P (until P becomes empty) its 1st element tau, inserting them both in Q tau and its arithmetic complement to repdigit (n-1)*111...1 (n times 1) removing the mentioned complement from P.

Conjecture 2: The autosimilarity shown by a(n) is a consequence of the fact that the corresponding P is the set of the n! permutations in increasing sequence for the letters 0..n-1, and Q=P (it holds if they are replaced "a(n)" and "increasing" respectively with "-1*a(n)" and "decreasing").

Note: "Q=P" is a necessary but not sufficient condition for observing the autosimilarity in a(n).

Application: The "generating algorithm" described previously might be potentially useful for parallel computing. In combination with the partition scheme proposed at links in A237265, and multiple indirection. For example notice that in such sense an algorithm for generating k! permutations with an increasing sequence would require only k!/2 iterations because the other half would be already determined by symmetry.

Conjecture 3: For n>2, given P the set of permutations in increasing sequence for the letters 0..n-1, there are distributed with a symmetric pattern among its (n!)! permutations all those A000165(n!\2) of them such that their 1st differences are symmetric. Moreover by setting to zero the other elements whose 1st differences are not symmetric, we obtain an antisymmetric sequence.

(End)

Conjecture 4: If 2<=mR. J. Cano, Apr 19 2017

Consider the first y!-1 terms for even y; The central term a(y!/2) is determined by the difference between the (y/2+1)th row from the y-th matrix defining the irregular table in A237265 and the consecutive permutation preceding it in lexicographic order (See EXAMPLE). - R. J. Cano, May 09 2017

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

Row n is the lexicographically earliest permutation of positive integers beginning with n. This also holds for the reverse colexicographic order, thus A007489(n-1) gives the position of n-th row of this array (which is one-based) in zero-based arrays A195663 & A055089.

The finite n X n square matrices in sequence A237265 converge towards this infinite square array.

Rows can be constructed also simply as follows: The first row is A000027 (natural numbers, also known as positive integers). For the n-th row, n=2, ..., pick n out from the terms of A000027 and move it to the front. This will create a permutation with one cycle of length n, in cycle notation: (1 n n-1 n-2 ... 3 2), which is the inverse of (1 2 ... n-1 n).

There are A000110(n) ways to choose n permutations from the n first rows of this table so that their composition is identity (counting all the different composition orders). This comment is essentially the same as my May 01 2006 comment on A000110, please see there for more information. - Antti Karttunen, Feb 10 2014

Also, for n > 1, the whole symmetric group S_n can be generated with just two rows, row 2, which is transposition (1 2), and row n, which is the inverse of cycle (1 ... n). See Rotman, p. 24, Exercise 2.9 (iii).

This is transpose of A237447, please see comments there.

When organized as a triangular table

1;

2, 3;

4, 6, 8;

10, 16, 22, 28;

34, 58, 82, 106, 130;

...

the k-th term of row n gives the position of the first n-letter permutation beginning with number k among all the lexicographically ordered permutations A030299. Thus the terms give the positions of rows of irregular table A237265 among the rows of A030298.

Note: the notation !n stands for the left factorial, A003422(n).