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%I #4 Mar 10 2014 15:18:41
%S 1,1,1,3,0,1,4,1,0,1,7,1,1,0,1,10,2,0,1,0,1,16,2,1,0,1,0,1,22,3,1,1,0,
%T 1,0,1,32,4,2,0,1,0,1,0,1,44,5,2,1,0,1,0,1,0,1,62,6,3,1,1,0,1,0,1,0,1,
%U 83,8,3,2,0,1,0,1,0,1,0,1,113,10,4,2,1
%N Triangular array: t(n,k) = number of partitions p = {x(1) >= x(2) >= ... >= x(k)} such that min(x(j) - x(j-1)) = k.
%C The first two columns are essentially A047967 and A238708. Counting the top row as row 2, the sum of numbers in row n is A000041(n) - 1.
%H Clark Kimberling, <a href="/A238709/b238709.txt">Table of n, a(n) for n = 1..400</a>
%e row 2: 1
%e row 3: 1 ... 1
%e row 4: 3 ... 0 ... 1
%e row 5: 4 ... 1 ... 0 ... 1
%e row 6: 7 ... 1 ... 1 ... 0 ... 1
%e row 7: 10 .. 2 ... 0 ... 1 ... 0 ... 1
%e row 8: 16 .. 2 ... 1 ... 0 ... 1 ... 0 ... 1
%e row 9: 22 .. 3 ... 1 ... 1 ... 0 ... 1 ... 0 ... 1
%e Let m = min(x(j) - x(j-1)); then for row 5, the 4 partitions with m = 0 are 311, 221, 2111, 11111; the 1 partition with m = 1 is 32, and the 1 partition with m = 3 is 41.
%t z = 25; p[n_, k_] := p[n, k] = IntegerPartitions[n][[k]]; m[n_, k_] := m[n, k] = Min[-Differences[p[n, k]]]; c[n_] := Table[m[n, h], {h, 1, PartitionsP[n]}]; v = Table[Count[c[n], h], {n, 2, z}, {h, 0, n - 2}]; Flatten[v]
%t TableForm[v]
%Y Cf. A238710, A238708.
%K nonn,tabl,easy
%O 1,4
%A _Clark Kimberling_, Mar 03 2014
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