A238710 - OEIS

%I #4 Mar 10 2014 15:19:17

%S 1,1,1,2,1,1,1,3,1,1,3,3,2,1,1,1,6,3,2,1,1,3,6,6,2,2,1,1,2,10,6,5,2,2,

%T 1,1,3,11,11,6,4,2,2,1,1,1,16,13,10,5,4,2,2,1,1,5,17,19,12,9,4,4,2,2,

%U 1,1,1,24,24,18,11,8,4,4,2,2,1,1,3,27,34

%N Triangular array:  t(n,k) = number of partitions p = {x(1) >= x(2) >= ... >= x(k)} such that max(x(j) - x(j-1)) = k.

%C The first two columns are essentially A032741 and A237665.  Counting the top row as row 2, the sum of numbers in row n is A000041(n) - 1.

%H Clark Kimberling, <a href="/A238710/b238710.txt">Table of n, a(n) for n = 1..400</a>

%e row 2:  1

%e row 3:  1 ... 1

%e row 4:  2 ... 1 ... 1

%e row 5:  1 ... 3 ... 1 ... 1

%e row 6:  3 ... 3 ... 2 ... 1 ... 1

%e row 7:  1 ... 6 ... 3 ... 2 ... 1 ... 1

%e row 8:  3 ... 6 ... 6 ... 2 ... 2 ... 1 ... 1

%e row 9:  2 ... 10 .. 6 ... 5 ... 2 ... 2 ... 1 ... 1

%e Let m = max(x(j) - x(j-1)); then for row 5, the 1 partition with m = 0 is 11111; the 3 partitions with m = 1 are 32, 221, 2111; the 1 partition with m = 2 is 311, 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] = Max[-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, A238709, A238353.

%K nonn,tabl,easy

%O 1,4

%A _Clark Kimberling_, Mar 03 2014