1,1

Are all the terms in this sequence divisible by 6?

Let's look at the sequence in base 12 with X for ten and E for eleven. Recall that all prmes greater than three end in a 1, 5, 7, or E. The sequence [n,(23*n^2+1)mod 12], 0<=n<=11, is [0, 1], [1, 0], [2, 9], [3, 4], [4, 9], [5, 0], [6, 1], [7, 0], [8, 9], [9, 4], [10, 9], [11, 0]. Thus the only possible primes are in 0 or 6 mod 12, that is, all multiples of 6 and all such primes end in 1. The sequence in base 12 is [6,591],[10,1E01], [20,7801], [56,49E91], [60,59001], [70,79E01], [76,8E991], [80,X2801]. - Walter A. Kehowski (wkehowski(AT)cox.net), Oct 05 2005

If n=144 then (23*n^2) + 1 = 476929 (prime).

select(proc(z) isprime(z[2]) end, [seq([n, 23*n^2 + 1], n=0..9*12)]); (Kehowski)

Sequence in context: A091629 A089529 A001766 this_sequence A065106 A030775 A057029

Adjacent sequences: A110956 A110957 A110958 this_sequence A110960 A110961 A110962

nonn

Parthasarathy Nambi (PachaNambi(AT)yahoo.com), Sep 26 2005