A178476 Permutations of 123456: Numbers having each of the decimal digits 1,...,6 exactly once, and no other digit.

123456, 123465, 123546, 123564, 123645, 123654, 124356, 124365, 124536, 124563, 124635, 124653, 125346, 125364, 125436, 125463, 125634, 125643, 126345, 126354, 126435, 126453, 126534, 126543, 132456, 132465, 132546, 132564, 132645, 132654

Offset: 1

Author

M. F. Hasler, May 28 2010

Keywords

• fini
• full
• easy
• nonn
• base

Comments

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)

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..720 (full sequence)

Crossrefs

Cf. A030298, A030299, A055089, A060117.

Programs

Mathematica

Take[FromDigits/@Permutations[Range[6]],40] (* Harvey P. Dale, Jun 05 2012 *)

PARI

v=vector(6,i,10^(i-1))~; A178476=vecsort(vector(6!,i,numtoperm(6,i)*v));
is_A178476(x)= { vecsort(Vec(Str(x)))==Vec("123456") }
forstep( m=123456,654321,9, is_A178476(m) & print1(m","))

Formula

a(n) + a(6! + 1 - n) = 777777.

floor( a(n) / 10^5 ) = ceiling( n / 5! ).

a(n) = A030299(n+153).

a(n) == 3 (mod 9).

a(n) = 3 + 9*A178486(n).