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COMMENTS
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For every n, all sufficiently large primes p such that n*p+1 is a perfect power are of the form ((n+1)^q-1)/n with q prime.
a(34) = (35^313-1)/34 is too large to include; it has 482 decimal digits.
a(35) - a(37) = {37, 61, 1483}.
a(38) = (39^349-1)/38 is too large to include; it has 554 decimal digits.
a(39) - a(100) = {5, 2, 43, 3500201, 5, 71, 43, 3851, 178481, 11, 47, 3221, 5, 178250690949465223, 2971, 127, 53, 3, 7, 3541, 61, 2, 59, 2, 61, 17, 3, 751410597400064602523400427092397, 21700501, 4831, 7, 19, 73, 5, 7, 5701, 73, 6007, 79, 39449441, 6481, 19, 79, 48037081, 6218272796370530483675222621221, 2, 3, 438668366137, 89, 5, 23, 331, 89, 654022685443, 11, 1001523179, 97, 3, 792806586866086631668831, 9901, 97, 10303}.
If n*p+1 = m^k, then n*p = m^k-1 = (m-1)*(m^(k-1) + m^(k-2) + ... + m + 1). If p >= n, then m^k = n*p+1 >= n^2+1 > n^2, and we have these three cases: Case 1: m-1 > n, then p can't be prime. Case 2: m-1 = n, this is A084738. Case 3: m-1 < n. If gcd(n, m-1) != m-1, then because m^(k-1) + m^(k-2) + ... + m + 1 > n, p can't be prime. This implies m-1 | n. The three cases means that we only need to check p < n and numbers m such that m-1 | n.
The first numbers n such that a(n) = 0 are {124, 215, 224, 242, ...}. a(268) is unknown; it is the smallest prime of the form (269^q - 1)/268 with prime q if such a prime exists (in which case it must be greater than (269^63659-1)/268), otherwise 0.
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