OFFSET
0,2
COMMENTS
a(n) exists for all n: by Dirichlet's theorem on arithmetic progressions, there exists primes of the form 2*c_1 + 1, 2*c_2 + 1, ..., 2*c_(n+1) + 1 where c_i are pairwise coprime odd numbers, then the multiplicative group of integers modulo (2*c_1 + 1)(2*c_2 + 1)*...*(2*c_(n+1) + 1) is isomorphic to (C_2)^n x C_(2*c_1*c_2*...*c_(n+1)).
Conjecture: (a) a(j) is divisible by a(i) if i < j; (b) a(n)/4 is squarefree for all n.
LINKS
Jianing Song, Table of n, a(n) for n = 0..61 [This now agrees with the list of factorized terms. - N. J. A. Sloane, Jan 19 2019]
Jianing Song, Factorizations of a(0) to a(61) [Corrected by Jianing Song, Jan 19 2019]
EXAMPLE
The multiplicative group of integers modulo 212520 is isomorphic to (C_2)^6 x C_660 and 212520 is the smallest number N such that the multiplicative group of integers modulo N contains six C_2 as the product of cyclic groups, so a(6) = 212520.
a(17) = 2^3 * 3 * 5 * 7 * 11 * 17 * 19 * 23 * 47 * 59 * 71 * 83 * 107 * 167 * 179 * 227 * 239 * 263, and the multiplicative group of integers modulo a(17) is isomorphic to (C_2)^17 x C_(4*3*5*7) x C_(16*9*5*7*11*17*23*29*41*53*83*113*131).
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Apr 01 2018
STATUS
approved