Steinar H. Gunderson: A286874(16) >= 48

Following up on the previous post, here are some heuristic results: First, if restricting oneself to 5-uniform values (all values have exactly five bits set), the best 15-bit code one can make is indeed 42 elements, and there are two distinct solutions: {31, 227, 364, 692, 1240, 1577, 1606, 2353, 3008, 3205, 3338, 4434, 4746, 4869, 5536, 6182, 6217, 7696, 8582, 8984, 9266, 9537, 10324, 10408, 10755, 12433, 12896, 13324, 16777, 16977, 17186, 17684, 18578, 18956, 19552, 20536, 20676, 21507, 24613, 24650, 26240, 30976} and {31, 227, 364, 692, 849, 906, 1240, 2354, 3206, 3337, 3680, 4485, 5169, 5442, 5644, 6228, 6312, 6659, 8745, 9285, 9632, 9746, 10314, 10385, 11012, 12326, 12568, 12992, 16966, 17450, 17684, 18049, 18469, 18880, 18968, 20553, 20626, 21280, 24688, 24716, 24835, 31744}. This supports, but does not prove, the conjecture that A286874(15) = 42. Second, A286874(16) >= 48 (the best previously known bound was 45), since this is a valid 48-element solution: 0000000000011111 0000000011100011 0000000101101100 0000001010110100 0000010011011000 0000011100000011 0000100100110001 0000101000101010 0000101111000000 0001000110001001 0001010000110010 0001011000001100 0001100100000110 0001110001000001 0010000110010010 0010010010000101 0010011001100000 0010100001010100 0010110100001000 0011000001001010 0011001000010001 0011100010100000 0100001001001001 0100010001000110 0100010110100000 0100100010001100 0100111000010000 0101000000100101 0101000101010000 0101001010000010 0110000000111000 0110001100000100 0110100000000011 1000001001010010 1000010000101001 1000010100010100 1000101000000101 1000110010000010 1001000011000100 1001001100100000 1001100000011000 1010000000100110 1010000101000001 1010001010001000 1100000010010001 1100000100001010 1100100001100000 1111010000000000 I won't be sweeping all of the 15- or 16-bit spaces.