A060073 - OEIS

A060073

a(n) = (n^(n-1)-1)/(n-1)^2.

1, 2, 7, 39, 311, 3268, 42799, 672605, 12345679, 259374246, 6140565047, 161792257795, 4696537119847, 148943500129544, 5124095576030431, 190082780764323705, 7563707819165039903, 321380710796022350410, 14523213296398891966759, 695546073617378871592991

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OFFSET

2,2

COMMENTS

Written in base n, a(n) has n-2 digits and looks like 12345... except that the final digit is n-1 rather than n-2.

Note that 2^m-1 divides a(m+1) = ((m+1)^m-1)/m^2 if and only if m = 2^k-1 with gcd(k,m) = 1. Mersenne numbers M = 2^p-1 such that a(M+1)/(2^M-1) is prime are Mersenne primes 2^3-1 = 7 and 2^7-1 = 127. - Thomas Ordowski, Sep 19 2021

LINKS

Harry J. Smith, Table of n, a(n) for n = 2..200

Index entries for sequences related to final digits of numbers

FORMULA

a(n) = A037205(n-1)/(n-1)^2 = A060072(n)/(n-1) = A058128(n)/n = A059522(n)/A000142(n).

EXAMPLE

a(10) = 999999999/81 = 111111111/9 = 12345679.

MATHEMATICA

Table[(n^(n - 1) - 1)/(n - 1)^2, {n, 2, 20}] (* Michael De Vlieger, Oct 28 2021 *)

PROG

(PARI) a(n) = { (n^(n - 1) - 1)/(n - 1)^2 } \\ Harry J. Smith, Jul 01 2009

CROSSREFS