A363594 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).
2, 4, 8, 13, 16, 17, 19, 23, 26, 29, 31, 32, 34, 37, 38, 41, 43, 46, 47, 53, 58, 59, 61, 62, 64, 67, 68, 71, 73, 74, 76, 79, 82, 83, 86, 89, 92, 94, 97, 101, 103, 106, 107, 109, 113, 116, 118, 122, 124, 127, 128, 131, 134, 136, 137, 139, 142, 146, 148, 149, 151, 152, 157, 158, 163, 164, 166, 167, 172
Offset: 1
Keywords
Examples
a(1) = 2 since b(3..5) = {3, 5, 4}; 4/2 = 2. a(2) = 4 since b(8..10) = {7, 9, 8}; 8/2 = 4. a(3) = 8 since b(22..24) = {33, 35, 16}; 16/2 = 8. a(4) = 13 since b(29..31) = {45, 49, 26}; 26/2 = 13. a(5) = 16 since b(36..38) = {55, 63, 32}; 32/2 = 16, etc.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
nn = 500; c[] = False; q[] = 1; Set[{i, j}, {1, 2}]; c[1] = c[2] = True; q[2] = 2; u = 3; Reap[Do[ (k = q[#]; While[c[k #], k++]; k *= #; While[c[# q[#]], q[#]++]) &[(p = 2; While[Divisible[i j, p], p = NextPrime[p]]; p)]; If[OddQ[i j], Sow[k/2]]; Set[{c[k], i, j}, {True, j, k}]; If[k == u, While[c[u], u++]], {n, 3, nn}] ][[-1, -1]]
Comments