A221740 a(n) = -4*((n-1)*(n+1)^(n+1)+1)/(((-1)^n-3)*n^3).

1, 7, 19, 293, 1493, 38127, 293479, 10593529, 109739369, 5135610071, 66987982331, 3856048810781, 60693710471869, 4149140360751583, 76519827268721103, 6058888636862818097, 128138108936443028945, 11533996620790579909159

Offset

1,2

Comments

Per exhaustive program, written for bases from 2 to 10, the number of permutations pairs, which have the same ratio, equal to A221740(n)/A221741(n) = (n^2*(n+1)^n - (n+1)^n + 1) / (-n^2 + n*(n+1)^n + (n+1)^n - n - 1), is: {2,2,3,3,5,3,7,5,7,...} for n >= 1 where n = r-1 and r is the base radix. Judging by above sequence it appears that the number of such permutations pairs is related to phi, which is the Euler totient function - according to A039649, A039650, A214288 (see bullet 1 of the analysis in the answer section of the Mathematics StackExchange link). - Alexander R. Povolotsky, Jan 26 2013

Links

G. C. Greubel, Table of n, a(n) for n = 1..385

Mathematics StackExchange, Permutations (with no duplicates) of decimal base digits 1,2,...,8,9,0.

NMBRTHRY, Number of specific permutations pairs relates to Euler Phi totient function?

Formula

a(n) = -4*A051846(n)/((-3 + (-1)^n)*n).

From Alexander R. Povolotsky, Oct 12 2022: (Start)

floor(a(n+1)/A221741(n+1)) = n.

Limit_{n->oo} (a(n)/A221741(n) - floor(a(n)/A221741(n))) = 0. (End)

Mathematica

Table[-4*((n - 1)*(n + 1)^(n + 1) + 1)/(((-1)^n - 3)*n^3), {n, 1, 50}] (* G. C. Greubel, Feb 19 2017 *)

Prog

(Maxima) makelist(-4*((n-1)*(n+1)^(n+1)+1)/(((-1)^n-3)*n^3), n, 1, 20); /* Martin Ettl, Jan 25 2013 */
(PARI) for(n=1, 25, print1(-4*((n - 1)*(n + 1)^(n + 1) + 1)/(((-1)^n - 3)*n^3), ", ")) \\ G. C. Greubel, Feb 19 2017

Crossrefs

Cf. A221741, A051846.

Sequence in context: A228150 A331404 A191624 * A364572 A335990 A359636

Adjacent sequences: A221737 A221738 A221739 * A221741 A221742 A221743

Keyword

nonn

Author

Alexander R. Povolotsky, Jan 23 2013

Status

approved