1,1

Numbers of the form (b^n-1)/(b-1) for n>2 and b>1. - T. D. Noe (noe(AT)sspectra.com), Jun 07 2006

Numbers m which are nontrivial repunits for any base b >= 2. For k = 2 (I use k for the exponent since n is used as the index in a(n)) we get (b^k-1)/(b-1) = (b^2-1)/(b-1) = b+1, so every integer m >= 3 is a 2 digit repunit in base b = m-1. And for n = 1 (the 1 digit degenerate repunit) we get (b-1)/(b-1) = 1 for any base b >= 2. If we considered all k >= 1 we would get the sequence of all positive integers except 2 since it is the smallest uniform base used in positional representation (2 might be seen as the "repunit" in a non positional base representation such as the Roman numbers were 2 is expressed as II.) [From Daniel Forgues (squid(AT)zensearch.com), Mar 01 2009]

T. D. Noe, Table of n, a(n) for n = 1..1172

a(n) ~ n^2 + n + 1 or lim n -> inf {a(n)/(n^2 + n + 1)} = 1 since as n grows the density of repunits of degree 2 among all the repunits tends to 1. [From Daniel Forgues (squid(AT)zensearch.com), Dec 09 2008]

a(5)=31 because 31 can be written as 111 base 5 (or indeed 11111 base 2)

Cf. A119598 (numbers that are repunits in four or more bases).

Sequence in context: A076701 A076196 A167782 this_sequence A090503 A059520 A102797

Adjacent sequences: A053693 A053694 A053695 this_sequence A053697 A053698 A053699

nonn

Henry Bottomley (se16(AT)btinternet.com), Mar 23 2000