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Irregular table which maps each partition of n counted in A162932 to a binary number (converted to decimal).
1

%I #8 May 30 2013 13:11:40

%S 14,30,28,62,58,56,60,126,118,114,122,254,112,116,124,238,120,230,246,

%T 510,226,234,250,478,224,228,236,242,252,462,494,1022

%N Irregular table which maps each partition of n counted in A162932 to a binary number (converted to decimal).

%C The table encodes each partition of n which satisfies the requirements of A162932 into a binary number with run lengths of 1 determined by the frequency of the parts. As many most-significant-bits are set as determined by the run length of the largest part, each run of 1 is terminated by a 0, and missing parts in the partition are represented by consecutive zeros. Each part in the partition and its frequency pick one member of A079946, and the full binary number is a concatenation of these bits. - _R. J. Mathar_, Jul 13 2012

%e For n=17, the A162932(17)=4 solutions are 2+2+2+2+3+3+3, 2+3+3+3+3+3, 2+3+4+4+4, and 2+5+5+5. These are represented here by

%e 111011110=478 (3 threes give a run length of 3, and four 2's give a run length of 4),

%e 11111010=250 (five 3's give a run length of 5 and one 2 gives a run length of 1),

%e 11101010=234 (three 4's give a run length of 3, one 3 gives a run length of 1 and one 2 gives another run length of 1),

%e and 11100010=226 (three 5's give a run length of 3, missing 4 and 3 give two run lengths of zero, one 2 give one run length of 1),

%e then sorted into row 17. - _R. J. Mathar_, Jul 13 2012

%e A162932 begins 1 0 1 1 1 1 3 1 3 4 4 4 8 6 ... so the table begins

%e 14

%e .

%e 30

%e 28

%e 62

%e 58

%e 56,60,126

%e 118

%e 114,122,254

%e 112,116,124,238

%e 120,230,246,510

%e 226,234,250,478

%e 224,228,236,242,252,462,494,1022

%Y A162932 (number of entries per row). A079946

%K nonn,tabf,base,less

%O 6,1

%A _Alford Arnold_, Jul 17 2009