
COMMENTS

For n >= 2, a(n) = the nth positive integer such that a(n) (base n) has a block of exactly n consecutive zeros.
Comments from Alexander Adamchuk, Nov 12 2006 (Start)
(2n+1)^2 divides a(2n). a(2n)/(2n+1)^2 = {1,1,41,5713,1657009,826446281,633095889817,691413758034721,...} = A081215(2n).
p divides a(p1) for prime p. a(p1)/p = {1,3,205,39991,9090909091,8230246567621,...} = A081209(p1) = A076951(p1).
p^2 divides a(p1) for an odd prime p. a(p1)/p^2 = {1,41,5713,826446281,633095889817,1021273028302258913,1961870762757168078553, 14199269001914612973017444081,...} = A081215(p1).
Prime p divides a((p3)/2) for p = {13,17,19,23,37,41,43,47,61,67,71,89, 109,113,137,139,157,163,167,181,191,...}.
Prime p divides a((p5)/4) for p = {29,41,61,89,229,241,281,349,421,509,601,641,661,701,709,769,809,821,881,...} = A107218(n) Primes of the form 4x^2+25y^2.
Prime p divides a((p7)/6) for p = {79,109,127,151,313,421,541,601,613,751,757,787,...}.
Prime p divides a((p9)/8) for p = {41,337,401,521,569,577,601,857,929,937,953,977,...} A subset of A007519(n) Primes of form 8n+1.
Prime p divides a((p11)/10) for p = {41,181,331,601,761,1021,1151,1231,1801,...}.
Prime p divides a((p13)/12) for p = {313,337,433,1621,1873,1993,2161,2677,2833,...}. (End)
