A116685 Triangle read by rows: T(n,k) is number of partitions of n that have k parts smaller than the largest part (n>=1, k>=0).

1, 2, 2, 1, 3, 1, 1, 2, 3, 1, 1, 4, 2, 3, 1, 1, 2, 5, 3, 3, 1, 1, 4, 4, 6, 3, 3, 1, 1, 3, 6, 6, 7, 3, 3, 1, 1, 4, 6, 10, 7, 7, 3, 3, 1, 1, 2, 9, 10, 12, 8, 7, 3, 3, 1, 1, 6, 6, 15, 14, 13, 8, 7, 3, 3, 1, 1, 2, 11, 15, 20, 16, 14, 8, 7, 3, 3, 1, 1, 4, 10, 21, 22, 24, 17, 14, 8, 7, 3, 3, 1, 1, 4, 11, 21

Offset: 1

Author

Emeric Deutsch, Feb 23 2006

Keywords

• nonn
• tabf

Comments

Same as A097364 without the 0's.

Also number of partitions of n such that the difference between the largest and smallest parts is k (see A097364). Example: T(6,2)=3 because we have [4,2],[3,2,1] and [3,1,1,1].

Row 1 has one term; row n (n>=2) has n-1 terms.

Row sums yield the partition numbers (A000041).

T(n,0)=A000005(n) (number of divisors of n).

T(n,1)=A049820(n) (n minus number of divisors of n).

T(n,2)=A008805(n-4) for n>=4.

Sum(k*T(n,k),k=0..n-2)=A116686

Examples

			Triangle starts:
01:  1
02:  2
03:  2  1
04:  3  1  1
05:  2  3  1  1
06:  4  2  3  1  1
07:  2  5  3  3  1  1
08:  4  4  6  3  3  1 1
09:  3  6  6  7  3  3 1 1
10:  4  6 10  7  7  3 3 1 1
11:  2  9 10 12  8  7 3 3 1 1
12:  6  6 15 14 13  8 7 3 3 1 1
13:  2 11 15 20 16 14 8 7 3 3 1 1
14:  4 10 21 22 24 17 ...
T(6,2)=3 because we have [4,1,1],[3,2,1] and [2,2,1,1].
		

Links

• Alois P. Heinz, Rows n = 1..142, flattened
• G. E. Andrews, M. Beck and N. Robbins, Partitions with fixed differences between largest and smallest parts, arXiv:1406.3374 [math.NT], 2014.
• Bernard L. S. Lin, Saisai Zheng, k-regular partitions and overpartitions with bounded part differences, The Raman. J. 56 (2021) 685-695

Crossrefs

Cf. A000041, A000005, A049820, A008805, A116686, A097364.

Columns k=3-10 give: A128508, A218567, A218568, A218569, A218570, A218571, A218572, A218573. T(2*n,n) = A117989(n). - Alois P. Heinz, Nov 02 2012

Programs

Maple

g:=sum(x^i/(1-x^i)/product(1-t*x^j,j=1..i-1),i=1..50): gser:=simplify(series(g,x=0,18)): for n from 1 to 15 do P[n]:=coeff(gser,x^n) od: 1; for n from 2 to 15 do seq(coeff(P[n],t,j),j=0..n-2) od;
# yields sequence in triangular form

Mathematica

rows = 15; max = rows + 2; col[k0_ /; k0 > 0] := col[k0] = Sum[x^(2*k + k0)/Product[ (1 - x^(k + j)), {j, 0, k0}], {k, 1, Ceiling[max/2]}] + O[x]^max // CoefficientList[#, x] &; col[0] := Table[Switch[n, 1, 0, 2, 1, , n - 1 - col[1][[n]]], {n, 1, Length[col[1]]}]; Join[{1}, Table[ col[k][[n+2]], {n, 0, rows-1}, {k, 0, n-1}] // Flatten] (* _Jean-François Alcover, Sep 11 2017, after Alois P. Heinz *)

Formula

G.f.: sum(i>=1, x^i/(1-x^i)/prod(j=1..i-1, 1-t*x^j) ).