A091602 Triangle: T(n,k) is the number of partitions of n such that some part is repeated k times and no part is repeated more than k times.

1, 1, 1, 2, 0, 1, 2, 2, 0, 1, 3, 2, 1, 0, 1, 4, 3, 2, 1, 0, 1, 5, 4, 3, 1, 1, 0, 1, 6, 7, 3, 3, 1, 1, 0, 1, 8, 8, 6, 3, 2, 1, 1, 0, 1, 10, 12, 7, 5, 3, 2, 1, 1, 0, 1, 12, 15, 11, 6, 5, 2, 2, 1, 1, 0, 1, 15, 21, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1, 18, 26, 20, 12, 9, 5, 4, 2, 2, 1, 1, 0, 1, 22, 35, 25, 18, 11, 8, 5, 4, 2, 2, 1, 1, 0, 1

Offset

1,4

Comments

From Gary W. Adamson, Mar 13 2010: (Start)

The triangle by rows = finite differences starting from the top, of an array in which row 1 = p(x)/p(x^2), row 2 = p(x)/p(x^3), ... row k = p(x)/p(x^k); such that p(x) = polcoeff A000041: (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...)

Note that p(x)/p(x^2) = polcoeff A000009: (1 + x + x^2 + 2x^3 + 2x^4 + ...).

Refer to the example. (End)

Links

Alois P. Heinz, Rows n = 1..141, flattened

Formula

G.f.: G = G(t,x) = sum(k>=1, t^k*(prod(j>=1, (1-x^((k+1)*j))/(1-x^j) ) -prod(j>=1, (1-x^(k*j))/(1-x^j) ) ) ). - Emeric Deutsch, Mar 30 2006

Sum_{k=1..n} k * T(n,k) = A264397(n). - Alois P. Heinz, Nov 20 2015

Example

Triangle starts:
   1:  1;
   2:  1,  1;
   3:  2,  0,  1;
   4:  2,  2,  0,  1;
   5:  3,  2,  1,  0,  1;
   6:  4,  3,  2,  1,  0,  1;
   7:  5,  4,  3,  1,  1,  0,  1;
   8:  6,  7,  3,  3,  1,  1,  0,  1;
   9:  8,  8,  6,  3,  2,  1,  1,  0,  1;
  10: 10, 12,  7,  5,  3,  2,  1,  1,  0,  1;
  11: 12, 15, 11,  6,  5,  2,  2,  1,  1,  0,  1;
  12: 15, 21, 14, 10,  5,  5,  2,  2,  1,  1,  0,  1;
  13: 18, 26, 20, 12,  9,  5,  4,  2,  2,  1,  1,  0,  1;
  14: 22, 35, 25, 18, 11,  8,  5,  4,  2,  2,  1,  1,  0,  1;
  ...
In the partition 5+2+2+2+1+1, 2 is repeated 3 times, no part is repeated more than 3 times.
From Gary W. Adamson, Mar 13 2010: (Start)
First few rows of the array =
  ...
  1, 1, 1, 2, 2, 3,  4,  5,  6,  8, 10, ... = p(x)/p(x^2) = A000009
  1, 1, 2, 2, 4, 5,  7,  9, 13, 16, 22, ... = p(x)/p(x^3)
  1, 1, 2, 3, 4, 6,  9, 12, 16, 22, 29, ... = p(x)/p(x^4)
  1, 1, 2, 3, 5, 6, 10, 13, 19, 25, 34, ... = p(x)/p(x^5)
  1, 1, 2, 3, 5, 7, 10, 14, 20, 27, 37, ... = p(x)/p(x^6)
  ...
Finally, taking finite differences from the top and deleting the first "1", we obtain triangle A091602 with row sums = A000041 starting with offset 1:
  1;
  1, 1;
  2, 0, 1;
  2, 2, 0, 1;
  3, 2, 1, 0, 1;
  4, 3, 2, 1, 0, 1;
  ...
(End)

Maple

g:=sum(t^k*(product((1-x^((k+1)*j))/(1-x^j), j=1..50)-product((1-x^(k*j))/(1-x^j), j=1..50)), k=1..50): gser:=simplify(series(g, x=0, 20)): for n from 1 to 13 do P[n]:=coeff(gser, x^n) od: for n from 1 to 13 do seq(coeff(P[n], t^j), j=1..n) od;
# yields sequence in triangular form - Emeric Deutsch, Mar 30 2006
b:= proc(n, i, k) option remember; `if`(n=0, 1,
      `if`(i>n, 0, add(b(n-i*j, i+1, min(k,
       iquo(n-i*j, i+1))), j=0..min(n/i, k))))
    end:
T:= (n, k)-> b(n, 1, k) -`if`(k=0, 0, b(n, 1, k-1)):
seq(seq(T(n, k), k=1..n), n=1..20);
# Alois P. Heinz, Nov 27 2013

Mathematica

b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i>n, 0, Sum[b[n-i*j, i+1, Min[k, Quotient[n-i*j, i+1]]], {j, 0, Min[n/i, k]}]]]; t[n_, k_] := b[n, 1, k] - If[k == 0, 0, b[n, 1, k-1]]; Table[t[n, k], {n, 1, 20}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 17 2014, after Alois P. Heinz's second Maple program *)

Crossrefs

Row sums: A000041. Inverse: A091603. Square: A091604.

Columns 1-6: A000009, A091605-A091609. Convergent of columns: A002865.

Cf. A000009. - Gary W. Adamson, Mar 13 2010

T(2n,n) gives: A232697.

Cf. A213177, A264397.

Sequence in context: A128187 A266477 A133121 * A336695 A035465 A096144

Adjacent sequences: A091599 A091600 A091601 * A091603 A091604 A091605

Keyword

nonn,tabl

Author

Christian G. Bower, Jan 23 2004

Status

approved