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COMMENTS
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p divides a(p+1) for all prime p except 3. p^2 divides a(p+1) for prime p in A123374.
2 divides a(n) for n = {2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, 18, 19, 20, 22, 23, 24, 26, 27, 28, 30, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, ...}.
2^2 divides a(n) for n = {2, 3, 6, 7, 8, 10, 11, 14, 15, 16, 18, 19, 22, 23, 24, 26, 27, 30, 31, 32, 34, 35, 38, 39, 40, 42, 43, 46, 47, 48, 50, ...}.
2^3 divides a(n) for n = {2, 3, 6, 7, 10, 11, 14, 15, 16, 18, 19, 22, 23, 26, 27, 30, 31, 32, 34, 35, 38, 39, 42, 43, 46, 47, 48, 50, ...}.
2^4 divides a(n) for n = {7, 14, 15, 18, 23, 30, 31, 32, 34, 39, 46, 47, 50, ...}.
2^5 divides a(n) for n = {15, 30, 31, 34, 47, 62, 63, 64, 66, 79, 94, 95, 98, ...}.
2^6 divides a(n) for n = {31, 62, 63, 66, 95, ...}.
2^7 divides a(n) for n = {63, 126, 127, 130, ...}.
It appears that for k > 2 the least few n such that a(n) is divisible by 2^(k+1) are n = {(2^k-1), 2*(2^k-1), 2*(2^k-1)+1, 2*(2^k-1)+3, 3*(2^k-1)+2, 4*(2^k-1)+2, 4*(2^k-1)+3, 4*(2^k-1)+4, 4*(2^k-1)+6, 5*(2^k-1)+4, 6*(2^k-1)+4, 6*(2^k-1)+5, 6*(2^k-1)+8, ...}. - Alexander Adamchuk, Oct 08 2006
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