cp's OEIS Frontend

Original definition: "Quotients of the polynomial remainder theorem for Diophantine equations among permutations."

The set built from the first terms of this sequence {0, 1, 10, 12, 21, 22, ....} contains the general solutions for a class of Diophantine linear equations among permutations, if one takes into account the unlimited number of distinct bases where these may be read.

From M. F. Hasler, Jan 12 2013, edited by R. J. Cano, May 08 2017: (Start)

Let P be the sequence of permutations of (0,...,m-1) interpreted as decimal numbers, P(n) = Sum_{i=1..m} 10^(m-i)*s(i) where s=(s(1),...,s(m)) is the n-th permutation (in lexicographical order), n <= m!. Then the difference P(n)-P(1) is independent of the choice of m, and divisible by 9. (Since 10 == 1 (mod 9), the numbers P(n) are all congruent (mod 9) to the sum s(1)+...+s(m).) This yields well-defined terms a(n)=(P(n)-P(1))/9.

For n>10!, P(n) will no longer be the concatenation of the "digits" (some of which will exceed 9). The pattern present in the decimal representation of the first terms will also be lost since there will be digits as large as d+1.

Note that the same a(n) is obtained independently of the chosen base b, provided that (i) 10 and 9 in the above are replaced with b and b-1, (ii) the result is (and can be) written in base b. (This implies the restriction to terms which can be written using the digits 0-9 to which the OEIS is limited.) See EXAMPLES for an illustration. (End)

We have P(n)-P(1)=a(n)*g(n), with g(n) = 9 = 10-1. Considering this and P(n) as polynomials in x=10, one can see an analogy with the polynomial remainder theorem. [Given as "formula" by R. J. Cano, rephrased by M. F. Hasler, Jan 12 2013]

Contribution by R. J. Cano, Feb 09 2013, (Start)

The maximum of the first m! terms of this sequence is given in base R by the explicit formula (please see A211869): max(m,R)=Sum_{k=1..m} k*(m-k)*R^(m-k-1);

If the first m! terms of this sequence were computed reading the permutations in base A033638(m), dividing their differences by A033638(m)-1, the resulting quotients would be written in the same way (with the same digits) in every base > A033638(m).

(End)

From R. J. Cano, Apr 29 2016, (Start)

Although in the sequence name it reads: "permutation of (0,...,m-1)", the most general statement that could replace it is: "permutation of any m-tuple of integers all of them in arithmetic progression", obtaining a multiple of this sequence, lambda*a(n), where lambda is the common difference for the progression. It works in such way because only the differences is what matters here.

Given x>1 and k>=0, if a polynomial G(x) of degree k is divided by x-1 then the remainder will be the sum of all the coefficients in G. Let us consider the case in which those coefficients are the differences among the letters ("digits") of two permutations for the same set of letters (0..x-1): The sum of all those differences must vanish. This explains how the difference between two of such permutations expressed in base x is 0 mod x-1, particularly why differences for pairs of permutations are divisible by 9.

Another way of introducing this sequence takes advantage of the fact that for n>1, n! is even. Consider for n>1 to obtain only the first n! terms. This can be done by subtracting the last permutation from the first, the penultimate permutation from the second, and so on by following the pattern (P(k)-P(n!-k+1))/9 with 1<=k<=n!. Such procedure generates an antisymmetric sequence f(k) from which a(k)=(f(k)+f(n!))/2. This partially explains why A217626 is symmetric. Also a base-independent treatment is possible using linear algebra: Column vectors and the strictly lower triangular matrix instead of the division by (r-1) where r is the base (and r=10 here for this sequence). This approach leads one to conclude that terms in this sequence are the differences between pairs of vectors made from the first n-1 partial sums of letters ("digits") taken from permutations for n consecutive letters, when components in these vectors are viewed as coefficients for a power series in r=10.

(End)

We can produce similar sequences of length n!-1 from all the n-set permutations (1,...,n), starting from n=2 up to n=9. The next larger sequence contains always the preceding sequence as its proper prefix. See A219664 for the largest such sequence. - Antti Karttunen, Dec 18 2012

See A209280 for the extension of this sequence to 9!-1 terms, and for comments and formulas which apply to this subsequence. - M. F. Hasler, Jan 15 2013

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)

a(9) is a zeroless pandigital number in base 10, with 9 digits such that every k-digit substring ( 1 <= k <= 9 ) taken from the left, is divisible by k (see A163574). - Michel Marcus, Dec 01 2013

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.

Correspondingly, gives the positions of those terms in A030299 whose first digit is not 1, as long as the decimal encoding system employed is valid.

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.

A non-averaging permutation avoids any 3-term arithmetic progression.

The encoding of the empty permutation () is 0. For positive n each element in the permutation is encoded using 1+floor(log_10(n)) = A055642(n) digits with leading 0's if necessary. Then all elements are concatenated.

All terms are in increasing order.