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By definition a(1) is A045345(2).

This sequence has a very interesting behavior. If Mod(n, 2)(Mod(n, 20)-1)(Mod(n, 20)-9)(Mod(n, 20)-13)(Mod(n, 20)-17)!=0, a(n)=17, 23 or 25; in other cases a(n) may be too large. If Mod[n, 16] = 15, a(n) = 17. For example, a(n) = 17 for n = 15, 31, 47, 63, 79, 95, 111, 127, 143, 159, 175, 191, ...; also, a(n) = 23 for n = 1, 13, 23, 35, 45, 57, 67, 89, 101, 123, 133, 145, 155, 167, 177, 189, 199, ...; a(n) = 25 for n = 3, 5, 7, 11, 19, 25, 27, 39, 43, 51, 55, 59, 65, 71, 75, ..., . For a(n) = 19 for n = 2, 20, 38, 56, 74, 92, 110, 128, 146, 164, 182, 200, 218, ..., == 2 (mod 18).

From Alexander Adamchuk, Jul 20 2008: (Start)

Conjectures:

a(n) exists for all n; a(n) >= 17.

a(325)-a(575) = {25,19,25,5851,1843,61,23,821,89,301,17,37,131,455,25,1607,297,37,23,19,25,

325,25,37,353,47,17,1663,23,691,25,691,509269,155,25,269,105893,19,25,3971,

23,213215,17,26021,327,79,25,37,151,83,23,161,101,37,25,19,327,265,17,37,25,

43,23,41,169,61,25,113,21761,6289,25,47,23,19,17,4073,1137,565,25,527,25,

325,25,37,23,455,25,431,13195,37,17,19,53,155,23,37,89,455,25,18839,25,6221,

25,41,18597,229,17,811,623173,19,25,193,2079,673,25,881,23,47,25,37,25,97,

17,79,131,37,25,19,23,56501,25,37,299,455,25,167,2707,446963,17,157,25,325,

25,41,53,19,25,5917,103,1051,23,607,101,155,17,37,6233,455,25,9049,23,37,25,

19,327,5359,25,37,43,455,17,9187,23,193,25,1861,7923,301,25,113,25,19,23,41,

89,61,17,43,1785,131,25,37,1417,455,23,151,53,37,25,19,25,79,17,37,23,455,

25,289,59,47,25,511,47,83,25,739,23,19,17,301,25,269,25,41,707,2735,23,37,

299,43,25,283,69723,37,17,19,1785,479,23,37,25,455,25,1867,131,61,25,31799,

23,161,17}.

a(n) is currently unknown and a(n)>10^7 for n = {324, 576, ...}. (End)

All but one of the terms up to n=1000 are known and they are less than 10^8. Currently the only unknown term for n<=1000 is a(656)>10^8. - Alexander Adamchuk, May 24 2009

More terms: a(324) = 18642551, a(576) = 12824827. - Alexander Adamchuk, May 24 2009

a(656) > 23,491,000,000. - Robert Price, Apr 22 2014

a(656) > 10^12. - Paul W. Dyson, Nov 23 2024

From Paul W. Dyson, Jan 18 2025: (Start)

If n == 15 (mod 16), a(n) = 17; otherwise if n == 2 (mod 18), a(n) = 19; otherwise if n mod 22 = 1 or 13, a(n) = 23; otherwise if n mod 20 = 3, 5, 7, 11, 15 or 19, a(n) = 25; otherwise if n mod 36 = 12, 18 or 24, a(n) = 37; etc. These follow from the fact that a(n) will also be a divisor for a prime sum with power j when j == n (mod psi(a(n))) and both n and j are greater than or equal to the maximum exponent in the prime factorization of a(n), where psi is the reduced totient function (A002322). E.g. for n=15, a(n)=17 and psi(a(n)) = 16. So j = 31, 47, 63, ..., and a(31) = a(47) = a(63) = a(15) = 17. For proof, see the comment dated Dec 09 2022 in A111441.

If a(n) exists, a(n) >= 17. For k < 17, psi(k) <= 12 and the maximum exponent in a prime factorization is 4 (as 16=2^4). So any a(n) < 17 would appear with periodicity <= 12, and would be seen in the first 15 (=12+4-1) terms of the sequence. (End)

The primes correspond to indices n = 1, 4357, 230065, 32826947, 29578097627 = A125825.

a(12) > 3.7*10^16. - Paul W. Dyson, Jan 17 2025