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a(n) = the n-th instance of b(k)/2 such that b(k-1) and b(k-2) are both odd, where b(n) = A359804(n).
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%I #11 Jul 06 2023 20:56:29

%S 2,4,8,13,16,17,19,23,26,29,31,32,34,37,38,41,43,46,47,53,58,59,61,62,

%T 64,67,68,71,73,74,76,79,82,83,86,89,92,94,97,101,103,106,107,109,113,

%U 116,118,122,124,127,128,131,134,136,137,139,142,146,148,149,151,152,157,158,163,164,166,167,172

%N a(n) = the n-th instance of b(k)/2 such that b(k-1) and b(k-2) are both odd, where b(n) = A359804(n).

%C The sequence strictly increases, a consequence of definition of A359804.

%C Conjecture: { A000079 \ {1} } U { A000040 \ {3, 5, 7, 11} } is a subset. In other words, this sequence is the union of powers of 2 greater than 1, and primes greater than 11.

%C This sequence is conjectured to be infinite. It tracks all occurrences of consecutive odd terms in A359804, which are (by definition) always followed by an even term, from which a(n) is derived. - _David James Sycamore_, Jun 21 2023

%H Michael De Vlieger, <a href="/A363594/b363594.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A359804(A363593(n)+2)/2 = A361639(A363593(n)+1).

%e a(1) = 2 since b(3..5) = {3, 5, 4}; 4/2 = 2.

%e a(2) = 4 since b(8..10) = {7, 9, 8}; 8/2 = 4.

%e a(3) = 8 since b(22..24) = {33, 35, 16}; 16/2 = 8.

%e a(4) = 13 since b(29..31) = {45, 49, 26}; 26/2 = 13.

%e a(5) = 16 since b(36..38) = {55, 63, 32}; 32/2 = 16, etc.

%t nn = 500; c[_] = False; q[_] = 1;

%t Set[{i, j}, {1, 2}]; c[1] = c[2] = True; q[2] = 2; u = 3;

%t Reap[Do[

%t (k = q[#]; While[c[k #], k++]; k *= #;

%t While[c[# q[#]], q[#]++]) &[(p = 2;

%t While[Divisible[i j, p], p = NextPrime[p]]; p)];

%t If[OddQ[i j], Sow[k/2]];

%t Set[{c[k], i, j}, {True, j, k}];

%t If[k == u, While[c[u], u++]], {n, 3, nn}] ][[-1, -1]]

%Y Cf. A000040, A000079, A100484, A359804, A361639, A363593.

%K nonn

%O 1,1

%A _Michael De Vlieger_ and _David James Sycamore_, Jun 12 2023