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Set b(n) = A290040(n). Then a(n) is the least divisor d > 1 of b(n) with binomial(b(n)+d, d) == 1 mod b(n).
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%I #31 Dec 20 2018 02:01:41

%S 10,264,10,10,55,18,20,10,52,10,18,34,10,20,34,18,10,10,18,10,2525,10,

%T 20,10,38,18,10,10,10,20,10,55,272,10,26,10,57,10,68,18,10,10,68,18,

%U 10,18,20,10,36,10,22,10,18,33,10,38,10,18,10,10,20,10,18,10,33,10,10,10,10,18,10,34,10,50,10,20,10,33,18,10,10

%N Set b(n) = A290040(n). Then a(n) is the least divisor d > 1 of b(n) with binomial(b(n)+d, d) == 1 mod b(n).

%C Sondow (2017) shows that if d is any divisor of m > 0 with d > 1 and binomial(m+d, d) == 1 mod m, then d is composite. Can d be a prime power?

%C The first term of A290040 for which more than one such d exists is A290040(165) = 101000, where d = 20 or d = 100.

%C d cannot be a prime power p^r | m if p^(r+1) does not divide m. Can d = 6? - _Jonathan Sondow_, Dec 26 2017

%H C. Babbage, <a href="http://books.google.com/books?id=KrA-AAAAYAAJ&amp;pg=PA46">Demonstration of a theorem relating to prime numbers</a>, Edinburgh Philosophical Journal, 1 (1819), 46-49.

%H J. Sondow, Extending Babbage's (non-)primality tests, in <a href="https://doi.org/10.1007/978-3-319-68032-3_19">Combinatorial and Additive Number Theory II</a>, Springer Proc. in Math. & Stat., Vol. 220, 269-277, CANT 2015 and 2016, New York, 2017; <a href="http://arxiv.org/abs/1812.07650">arXiv:1812.07650 [math.NT]</a>, 2018.

%H J. Sondow, <a href="https://doi.org/10.1080/00029890.2018.1424478">Problem 12030</a>, Amer. Math. Monthly, 125 (2018), 276.

%F binomial(A290040(n) + a(n), a(n)) == 1 mod A290040(n).

%e b(1) = 260 and binomial(260+10,10) = 479322759878148681 == 1 mod 260 gives the least d = 10, so a(1) = 10.

%t M = Select[ Range[ 30000], (d = Divisors[#]; Product[ Mod[ Binomial[# + d[[k]], d[[k]]], #] - 1, {k, 2, Length[d]}] == 0) &]; Table[ Select[ Divisors[ M[[k]]], # > 1 && Mod[ Binomial[# + M[[k]], #], M[[k]]] == 1 &], {k, 1, Length[M]}] // Flatten

%o (PARI) terms(n) = my(i=0); for(k=1, oo, my(d=divisors(k)); for(x=2, #d, if(Mod(binomial(k+d[x], d[x]), k)==1, print1(d[x], ", "))); if(i==n, break))

%o /* Print initial 40 terms as follows */

%o terms(40) \\ _Felix Fröhlich_, Dec 26 2017

%Y Cf. A290040, A299115.

%K nonn

%O 1,1

%A _Jonathan Sondow_, Jul 23 2017