cp's OEIS Frontend

From M. F. Hasler, Jan 12 2013: (Start)

Note [updated Mar 03 2013]: The definition of sequence A030299 has been slightly modified in Jan. 2013, and as a consequence the following properties remain valid beyond the first A007489(9)-1 = 409112 terms, which had not been the case before, when A030299 had been defined through concatenation of the lexicographically ordered permutations, which in case of elements >= 10 broke up the nice mathematical properties (esp. of the sequence A219664 = 9*A217626 cited below).

This sequence taken modulo 9 is zero except (possibly) at indices where a run of permutations ends in A030299. (These indices are given by A007489(n), n>0.) There it equals (mod 9) the "n" of the following run. E.g., a(1)=2 (mod 9), and A030299(1+1)=12 is the start of the run for n=2; a(3)=3 (mod 9) and A030299(3+1)=123 is the start of the run for n=3, a(9)=4 (mod 9) and A030299(9+1)=1234 is the start of the run for n=4, etc.

The subsequence between these indices (A007489(n)+1,...,A007489(n+1)-1), always starts with the same terms, listed in A219664 = 9*A217626 (= A209280 = A107346 where the latter are defined). (End)

Conjecture: there are infinitely many terms.

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).

Contains every finite sequence of distinct numbers, infinitely many times.

Length of n-th row = sum of n-th row = n!; number of zeros in n-th row = A000166(n); number of positive terms in n-th row = A002467(n). [Reinhard Zumkeller, Mar 29 2012]

A preferential arrangement is like a permutation, except that ties are allowed. Preferential arrangements are also called ordered partitions. See A000670.

There are A000670(n) terms of length n.

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

First 5!-1 terms are identical to A107346, and the 9!-1 terms are identical to A209280. (Updated by M. F. Hasler, Jan 12 2013, Mar 03 2013)

Because of the self-similar nature of A220664, this sequence can be also constructed by picking appropriate terms from there with the auxiliary sequence A220655, cf. formula.

Similarly, differences between successive permutations of {1,2,...,k} in lexicographic order interpreted as decimal numbers, for any k=2..9, produce the first (k!)-1 terms of this sequence. But for k=10 the result is ill-defined, so we can consider the sequence finite, well-defined only for n=1..362879. [See however the following comment. - Editor's note]

In sequence A030299 it is clearly defined how it extends beyond index n = 1!+2!+...+9! = A007489(9), so the sequence A220664 of its first differences is well-defined up to infinity. (The "result" mentioned above is ill defined because the meaning of "interpreted" is not clear.) But the preceding comment is misleading by speaking of "self similar nature", and the sequence definition as "repeating part" is also misleading: If the sequence is defined to be of finite length 9!-1 (thus equal to A209280), then it is indeed infinitely often repeated as a subsequence (of consecutive terms) in A220664 (even when the latter was defined using concatenation for permutations of more than 9 elements, but then not as differences of the terms following 12345678910 where it was expected, but, e.g., as differences of the terms following 10123456789, etc.).

Since A030299 has been defined through a ("simpler") sum rather than concatenation, the nice mathematical properties of A220664, and even more this sequence A219664, persist beyond n=9!. - M. F. Hasler, Jan 12 2013

This sequence is the natural extension of A107346 (and others, see below) from 5!-1 to 9!-1 terms, which is the natural (since maximal) length, given that OEIS sequence data are stored as decimal numbers. On the other hand, it is quite different from A219664 in many aspects, not only for the reason that the other sequence is infinite and therefore differs from this one in all terms beyond n = 9!-1.

The sequence is finite, with 9!-1 terms, and symmetric: a(n)=a(9!-n).

All terms are multiples of 9, cf. formula.

The subsequence of the first n!-1 terms (n=2,...,9) yields the first differences of the sequence of numbers whose digits are a permutation of (1,...,n):

The first 8!-1 terms yield the first differences of A178478: numbers whose digits are a permutation of 12345678.

The first 7!-1 terms yield the first differences of A178477: numbers whose digits are a permutation of 1234567.

The first 6!-1 terms yield the first differences of A178476: numbers whose digits are a permutation of 123456.

The first 5!-1 terms yield A107346, the first differences of A178475: numbers whose digits are a permutation of 12345.