OFFSET
1,3
COMMENTS
Row n has floor(sqrt(n)) terms. Row sums yield A000009. T(n,1)=A001227(n) (n>=1). Sum(k*T(n,k),k>=1)=A038348(n-1) (n>=1).
Conjecture: Also the number of strict integer partitions of n with k maximal runs of consecutive parts decreasing by 1. - Gus Wiseman, Jun 24 2025
LINKS
Alois P. Heinz, Rows n = 1..1000, flattened
FORMULA
G.f.: product(1+tx^(2j-1)/(1-x^(2j-1)), j=1..infinity).
EXAMPLE
From Gus Wiseman, Jun 24 2025: (Start)
Triangle begins:
1: 1
2: 1
3: 2
4: 1 1
5: 2 1
6: 2 2
7: 2 3
8: 1 5
9: 3 4 1
10: 2 7 1
11: 2 8 2
12: 2 10 3
13: 2 11 5
14: 2 13 7
15: 4 12 11
16: 1 19 11 1
17: 2 18 17 1
18: 3 20 21 2
19: 2 22 27 3
20: 2 25 32 5
Row n = 9 counts the following partitions into odd parts by number of distinct parts:
(9) (7,1,1) (5,3,1)
(3,3,3) (3,3,1,1,1)
(1,1,1,1,1,1,1,1,1) (5,1,1,1,1)
(3,1,1,1,1,1,1)
Row n = 9 counts the following strict partitions by number of maximal runs:
(9) (6,3) (5,3,1)
(5,4) (7,2)
(4,3,2) (8,1)
(6,2,1)
(End)
MAPLE
g:=product(1+t*x^(2*j-1)/(1-x^(2*j-1)), j=1..35): gser:=simplify(series(g, x=0, 34)): for n from 1 to 29 do P[n]:=coeff(gser, x^n) od: for n from 1 to 29 do seq(coeff(P[n], t, j), j=1..floor(sqrt(n))) od; # yields sequence in triangular form
# Alternative:
with(numtheory):
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, add(b(n-i*j, i-2)*`if`(j=0, 1, x), j=0..n/i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(
b(n, iquo(n+1, 2)*2-1)):
seq(T(n), n=1..30); # Alois P. Heinz, Mar 08 2015
MATHEMATICA
b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i<1, 0, Sum[b[n-i*j, i-2]*If[j == 0, 1, x], {j, 0, n/i}]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][b[n, Quotient[n+1, 2]*2-1]]; Table[T[n], {n, 1, 30}] // Flatten (* Jean-François Alcover, May 22 2015, after Alois P. Heinz *)
Table[Length[Select[IntegerPartitions[n], OddQ[Times@@#]&&Length[Union[#]]==k&]], {n, 1, 12}, {k, 1, Floor[Sqrt[n]]}] (* Gus Wiseman, Jun 24 2025 *)
CROSSREFS
KEYWORD
AUTHOR
Emeric Deutsch, Feb 22 2006
STATUS
approved
