A238709
Triangular array: t(n,k) = number of partitions p = {x(1) >= x(2) >= ... >= x(k)} such that min(x(j) - x(j-1)) = k.
Original entry on oeis.org
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, 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, 83, 8, 3, 2, 0, 1, 0, 1, 0, 1, 0, 1, 113, 10, 4, 2, 1
Offset: 1
row 2: 1
row 3: 1 ... 1
row 4: 3 ... 0 ... 1
row 5: 4 ... 1 ... 0 ... 1
row 6: 7 ... 1 ... 1 ... 0 ... 1
row 7: 10 .. 2 ... 0 ... 1 ... 0 ... 1
row 8: 16 .. 2 ... 1 ... 0 ... 1 ... 0 ... 1
row 9: 22 .. 3 ... 1 ... 1 ... 0 ... 1 ... 0 ... 1
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.
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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]
TableForm[v]
A238354
Triangle T(n,k) read by rows: T(n,k) is the number of partitions of n (as weakly ascending list of parts) with minimal ascent k, n >= 0, 0 <= k <= n.
Original entry on oeis.org
1, 1, 0, 2, 0, 0, 2, 1, 0, 0, 4, 0, 1, 0, 0, 5, 1, 0, 1, 0, 0, 8, 1, 1, 0, 1, 0, 0, 11, 2, 0, 1, 0, 1, 0, 0, 17, 2, 1, 0, 1, 0, 1, 0, 0, 23, 3, 1, 1, 0, 1, 0, 1, 0, 0, 33, 4, 2, 0, 1, 0, 1, 0, 1, 0, 0, 45, 5, 2, 1, 0, 1, 0, 1, 0, 1, 0, 0, 63, 6, 3, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 84, 8, 3, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 114, 10, 4, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0
Offset: 0
Triangle starts:
00: 1;
01: 1, 0;
02: 2, 0, 0;
03: 2, 1, 0, 0;
04: 4, 0, 1, 0, 0;
05: 5, 1, 0, 1, 0, 0;
06: 8, 1, 1, 0, 1, 0, 0;
07: 11, 2, 0, 1, 0, 1, 0, 0;
08: 17, 2, 1, 0, 1, 0, 1, 0, 0;
09: 23, 3, 1, 1, 0, 1, 0, 1, 0, 0;
10: 33, 4, 2, 0, 1, 0, 1, 0, 1, 0, 0;
11: 45, 5, 2, 1, 0, 1, 0, 1, 0, 1, 0, 0;
12: 63, 6, 3, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0;
13: 84, 8, 3, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0;
14: 114, 10, 4, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0;
15: 150, 13, 4, 3, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0;
...
The 11 partitions of 6 together with their minimal ascents are:
01: [ 1 1 1 1 1 1 ] 0
02: [ 1 1 1 1 2 ] 0
03: [ 1 1 1 3 ] 0
04: [ 1 1 2 2 ] 0
05: [ 1 1 4 ] 0
06: [ 1 2 3 ] 1
07: [ 1 5 ] 4
08: [ 2 2 2 ] 0
09: [ 2 4 ] 2
10: [ 3 3 ] 0
11: [ 6 ] 0
There are 8 partitions of 6 with min ascent 0, 1 with min ascents 1, 2, and 4, giving row 6 of the triangle: 8, 1, 1, 0, 1, 0, 0.
Cf.
A238353 (partitions by maximal ascent).
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b:= proc(n, i, t) option remember; `if`(n=0, 1/x, `if`(i<1, 0,
b(n, i-1, t)+`if`(i>n, 0, (p->`if`(t=0, p, add(coeff(
p, x, j)*x^`if`(j<0, t-i, min(j, t-i)),
j=-1..degree(p))))(b(n-i, i, i)))))
end:
T:= n->(p->seq(coeff(p, x, k)+`if`(k=0, 1, 0), k=0..n))(b(n$2, 0)):
seq(T(n), n=0..15);
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b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1/x, If[i<1, 0, b[n, i-1, t]+If[i>n, 0, Function[{p}, If[t == 0, p, Sum[Coefficient[p, x, j]*x^If[j<0, t-i, Min[j, t-i]], {j, -1, Exponent[p, x]}]]][b[n-i, i, i]]]]]; T[n_] := Function[{p}, Table[ Coefficient[p, x, k]+If[k == 0, 1, 0], {k, 0, n}]][b[n, n, 0]]; Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Jan 12 2015, translated from Maple *)
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