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%I #9 Dec 05 2016 11:34:10
%S 1,2,2,1,4,1,4,2,1,7,3,1,7,6,1,1,12,8,1,1,12,12,4,1,1,19,16,5,1,1,19,
%T 25,8,2,1,1,30,34,9,2,1,1,30,44,17,6,2,1,1,45,59,20,7,2,1,1,45,81,31,
%U 12,3,2,1,1,67,108,36,13,3,2,1,1,67,132,64,18,9
%N Irregular triangular array: T(n,i) = number of partitions of n having crossover index k; see Comments.
%C Suppose that P = [p(1),p(2),...,p(k)] is a partition of n, where p(1) >= p(2) >= ... >= p(k). The crossover index of P is the least h such that p(1)+...+p(h) > = n/2. Equivalently for k > 1, p(1)+...+p(h) >= p(h+1)+...+p(k). The n-th row sum is the number of partitions of n, A000041. The bisections of column 1 are given by A000070. The limit of the reversal of row n is given by A000041.
%H Clark Kimberling, <a href="/A276468/b276468.txt">Table of n, a(n) for n = 1..500</a>
%e First 15 rows (indexed by column 1):
%e 1... 1
%e 2... 2
%e 3... 2 1
%e 4... 4 1
%e 5... 4 2 1
%e 6... 7 3 1
%e 7... 7 6 1 1
%e 8... 12 8 1 1
%e 9... 12 12 4 1 1
%e 10.. 19 16 5 1 1
%e 11... 19 25 8 2 1 1
%e 12.. 30 34 9 2 1 1
%e 13.. 30 44 17 6 2 1 1
%e 14.. 45 59 20 7 2 1 1
%e 15.. 45 81 31 12 3 2 1 1
%t p[n_] := p[n] = IntegerPartitions[n]; t[n_, k_] := t[n, k] = p[n][[k]];
%t q[n_, k_] := q[n, k] = Select[Range[50], Sum[t[n, k][[i]], {i, 1, #}] >= n/2 &, 1];
%t u[n_] := u[n] = Flatten[Table[q[n, k], {k, 1, Length[p[n]]}]];
%t c[n_, k_] := c[n, k] = Count[u[n], k];
%t v = Table[c[n, k], {n, 1, 25}, {k, 1, Ceiling[n/2]}];
%t TableForm[v] (* A276468 array *)
%t Flatten[v] (* A276468 sequence *)
%Y Cf. A000041, A000070 (bisections of column 1), A279033 (crossover index for strict partitions), A279044 (crossover parts).
%K nonn,easy,tabf
%O 1,2
%A _Clark Kimberling_, Dec 03 2016
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