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For n >= 2, a(n) = the n-th 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(p-1) for prime p. a(p-1)/p = {1,3,205,39991,9090909091,8230246567621,...} = A081209(p-1) = A076951(p-1).

p^2 divides a(p-1) for an odd prime p. a(p-1)/p^2 = {1,41,5713,826446281,633095889817,1021273028302258913,1961870762757168078553, 14199269001914612973017444081,...} = A081215(p-1).

Prime p divides a((p-3)/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((p-5)/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((p-7)/6) for p = {79,109,127,151,313,421,541,601,613,751,757,787,...}.

Prime p divides a((p-9)/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((p-11)/10) for p = {41,181,331,601,761,1021,1151,1231,1801,...}.

Prime p divides a((p-13)/12) for p = {313,337,433,1621,1873,1993,2161,2677,2833,...}. (End)

From Mathew Englander, Oct 19 2020: (Start)

The sum of two adjacent terms of the sequence cannot be prime.

In base n, a(n) has n-1 digits, which are (beginning from the left): n-2, 2, n-4, 4, and so on, except that if n is even the rightmost digit is 1 instead of 0. In that case, the other digits form a palindrome with every even digit from 2 to n-2 appearing twice. For example, a(14) in base 14 is c2a486684a2c1. If n is odd, then all digits from 1 to n-1 occur exactly once. For example, a(15) in base 15 is d2b496785a3c1e.

For any positive integer k, any prime p, and any positive integer h such that h*p > 2, a(h*p^k - 2) == (-1)^h * (1 - 2^(h-1)) (mod p). For example, a(7*p^k - 2) == 63 (mod p); a(10*p^k - 2) == -511 (mod p).

Suppose k and m are positive integers. If k is even, then a(k*m) == 1, a(k*m+1) == 0, and a(k*m-1) == -1 (all mod m). If k is odd, then a(k*m) == (-1)^m and a(k*m+1) == ceiling(m/2) (both mod m), while a(k*m-1) == m/2 - 1 for m even, and a(k*m-1) == 1 for m odd (mod m).

For proofs of the above, see the Englander link. (End)

p divides a(p) for odd primes p. a(p) = A081209(p-1) = p*A081215(p-1) for odd primes p. a(p)/p = A081215(p-1) = {1,41,5713,826446281,633095889817,1021273028302258913,...} - Alexander Adamchuk, Nov 12 2006