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COMMENTS
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a(n) = 1 + sum of product of primes p_i from i = 1 to n such that p_i < p_(i+1) and a(n) is prime.
The corresponding primes p_i are:
2, 5, 7, 11, 41, 113, 149, 167, 173, 257, 281, 317, 431, 491, 839, 857, 953, 977, 1031, 1091, 2909, 3041, 3191, 3467, 4073, 4721, 5381, 6047, 6791, 7127,...
a(n) ==3 (mod 10) because p_1*p_2*...*p_n == 0 (mod 10).
Property of this sequence:
b(n) = (a(n) - 1)/2 = 6, 41, 426, 16211, 1799916,... is alternatively even == 6 mod 10 and odd == 1 mod 10.
An easy way to write and compute these numbers is the nested product 1 + p_1*(1 + p_2*(1 + p_3*...))). - T. D. Noe, May 03 2013
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