A276468 Irregular triangular array: T(n,i) = number of partitions of n having crossover index k; see Comments.
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, 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, 12, 3, 2, 1, 1, 67, 108, 36, 13, 3, 2, 1, 1, 67, 132, 64, 18, 9
Offset: 1
Examples
First 15 rows (indexed by column 1): 1... 1 2... 2 3... 2 1 4... 4 1 5... 4 2 1 6... 7 3 1 7... 7 6 1 1 8... 12 8 1 1 9... 12 12 4 1 1 10.. 19 16 5 1 1 11... 19 25 8 2 1 1 12.. 30 34 9 2 1 1 13.. 30 44 17 6 2 1 1 14.. 45 59 20 7 2 1 1 15.. 45 81 31 12 3 2 1 1
Links
- Clark Kimberling, Table of n, a(n) for n = 1..500
Crossrefs
Programs
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Mathematica
p[n_] := p[n] = IntegerPartitions[n]; t[n_, k_] := t[n, k] = p[n][[k]]; q[n_, k_] := q[n, k] = Select[Range[50], Sum[t[n, k][[i]], {i, 1, #}] >= n/2 &, 1]; u[n_] := u[n] = Flatten[Table[q[n, k], {k, 1, Length[p[n]]}]]; c[n_, k_] := c[n, k] = Count[u[n], k]; v = Table[c[n, k], {n, 1, 25}, {k, 1, Ceiling[n/2]}]; TableForm[v] (* A276468 array *) Flatten[v] (* A276468 sequence *)
Comments