cp's OEIS Frontend

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

Numbers m such that n!*C(m,n) = C(m,n+1). - Lekraj Beedassy, Feb 18 2006

a(n) is also the maximum size for a deck of cards in the Communicating the Card magic trick. In this game Alice draws n+1 cards from the deck at random, without replacement, and passes n of them, one by one, to her accomplice Bob. If the deck has a(n) cards or fewer, there is an algorithm by which Alice can communicate to Bob the identity of the card she chooses to retain, using only the identity and the order of passing of the n passed cards. (One side of the proof, that no larger deck size will work, is easy: the retained card can be one of (n+1)! possibilities, since Bob knows that it is not one of the n passed cards. Alice has (n+1) ways to retain a card and n! ways to order the passing of the remaining cards, so she cannot communicate more than (n+1)! different possibilities.) - Lee A. Newberg, Jun 09 2010

There are 5! = 120 terms in this finite subsequence of A030299.

It would be interesting to conceive simple and/or efficient functions which yield (a) the n-th term of this sequence: f(n) = a(n), (b) for a given term, the subsequent one: f(a(n)) = a(1 + (n mod 5!)).

From Nathaniel Johnston, May 19 2011: (Start)

Individual terms a(n) can be computed efficiently via the following procedure: Define b(n,k) = 1 + floor(((n-1) mod (k+1)!)/k!) for k = 1, 2, 3, 4. The first digit of a(n) is b(n,4). The second digit of a(n) is the b(n,3)-th number not already used. The third digit of a(n) is the b(n,2)-th number not already used. The fourth digit of a(n) is the b(n,1)-th number not already used, and the final digit of a(n) is the only digit remaining. This procedure generalizes in the obvious way for related sequences such as A178476.

For example, if n = 38 then we compute b(38,1) = 2, b(38,2) = 1, b(38,3) = 3, b(38,4) = 2. Thus a(38) = 24153 (2, followed by the 3rd digit not yet used, followed by the 1st digit not yet used, followed by the 2nd digit not yet used, followed by the last remaining digit).

(End)

Also this is an irregular array where row n contains the n! consecutive multiples of n starting with n!.

For n >= 1, (a(n), A084555(n)) = (1,1), (2,3), (4,5), (6,8), (9,11), (12,14), ... gives the starting and ending offsets of the n-th permutation in the sequences like A030298 and A030496. Gives also the fixed points of A220662; we have A220662(a(n)) = a(n). - Antti Karttunen, Dec 18 2012

This finite sequence contains 6!=720 terms.

This is a subsequence of A030299, consisting of elements A030299(154)..A030299(873).

If individual digits are be split up into separate terms, we get a subsequence of A030298.

It would be interesting to conceive simple and/or efficient functions which yield (a) the n-th term of this sequence: f(n)=a(n), (b) for a given term, the subsequent one: f(a(n)) = a(1 + (n mod 6!)).

The expression a(n+6) - a(n) takes only 18 different values for n = 1..6!-6.

An efficient procedure for generating the n-th term of this sequence can be found at A178475. - Nathaniel Johnston, May 19 2011

From Hieronymus Fischer, Feb 13 2013: (Start)

The sum of all terms as decimal numbers is 279999720.

General formula for the sum of all terms (interpreted as decimal permutational numbers with exactly d different digits from the range 1..d < 10): sum = (d+1)!*(10^d-1)/18.

If the terms are interpreted as base-7 numbers the sum is 49412160.

General formula for the sum of all terms of the corresponding sequence of base-p permutational numbers (numbers with exactly p-1 different digits excluding the zero digit): sum = (p-2)!*(p^p-p)/2. (End)

There are 5!=120 terms in this finite sequence. Its origin is the fact that numbers whose decimal expansion is a permutation of 12345 are all of the form 9k+6.

The sequence is motivated by the fact that numbers whose decimal expansion is a permutation of 123456, are all of the form 9k+3.

There are 6!=720 terms in this finite sequence.

This table is otherwise similar to A030298, but lists permutations in the order given by the Steinhaus-Trotter-Johnson algorithm. - Antti Karttunen, Dec 28 2012

Used for computing A030298: a(n) tells the zero-based ranking of the n-th permutation in A030298 (A030299(n)) in the lexicographical ordering of all finite permutations of the same size.

It would be nice to have a simple explicit formula for the n-th term.

Contains A000142(7) = 5040 terms. - R. J. Mathar, Apr 08 2011

An efficient procedure for generating the n-th term of this sequence can be found at A178475. - Nathaniel Johnston, May 19 2011