A066633 - cp's OEIS frontend

1, 2, 1, 4, 1, 1, 7, 3, 1, 1, 12, 4, 2, 1, 1, 19, 8, 4, 2, 1, 1, 30, 11, 6, 3, 2, 1, 1, 45, 19, 9, 6, 3, 2, 1, 1, 67, 26, 15, 8, 5, 3, 2, 1, 1, 97, 41, 21, 13, 8, 5, 3, 2, 1, 1, 139, 56, 31, 18, 12, 7, 5, 3, 2, 1, 1, 195, 83, 45, 28, 17, 12, 7, 5, 3, 2, 1, 1, 272, 112, 63, 38, 25, 16, 11, 7, 5, 3, 2, 1, 1

Offset: 1

Naohiro Nomoto, Jan 09 2002

It appears that row n lists the first differences of the row n of triangle A181187 together with 1 (as the final term of the row n). - Omar E. Pol, Feb 26 2012
It appears that reversed rows converge to A000041. - Omar E. Pol, Mar 11 2012
Proof: For a partition of n with k>floor(n/2+1), k can only occur as the largest part; the other parts sum to n-k, so that T(n,n-k)=A000041(k). - George Beck, Jun 30 2019
T(n,k) is also the total number k's that are divisors of all positive integers in a sequence with n blocks where the m-th block consists of A000041(n-m) copies of m, with 1 <= m <= n. - Omar E. Pol, Feb 05 2021

			For n = 3, k = 1; 3 = 2+1 = 1+1+1. T(3,1) = (zero 1's) + (one 1's) + (three 1's), so T(3,1) = 4.
Triangle begins:
    1;
    2,   1;
    4,   1,  1;
    7,   3,  1,  1;
   12,   4,  2,  1,  1;
   19,   8,  4,  2,  1,  1;
   30,  11,  6,  3,  2,  1,  1;
   45,  19,  9,  6,  3,  2,  1, 1;
   67,  26, 15,  8,  5,  3,  2, 1, 1;
   97,  41, 21, 13,  8,  5,  3, 2, 1, 1;
  139,  56, 31, 18, 12,  7,  5, 3, 2, 1, 1;
  195,  83, 45, 28, 17, 12,  7, 5, 3, 2, 1, 1;
  272, 112, 63, 38, 25, 16, 11, 7, 5, 3, 2, 1, 1;
  ...

D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.2.1.5, Problem 73(b), pp. 415, 761. - N. J. A. Sloane, Dec 30 2018

Alois P. Heinz, Rows n = 1..141, flattened
Manosij Ghosh Dastidar and Sourav Sen Gupta, Generalization of a few results in Integer Partitions, arXiv preprint arXiv:1111.0094 [cs.DM], 2011.
Joseph Vandehey, Digital problems in the theory of partitions, Integers (2024) Vol. 24A, Art. No. A18. See p. 3.
Eric Weisstein's World of Mathematics, Elder's Theorem

Row sums give positive terms of A006128.

Columns (1-10): A000070, A024786-A024794.

b:= proc(n, i) option remember;
      `if`(n=0 or i=1, 1+n*x, b(n, i-1)+
      `if`(i>n, 0, (g->g+coeff(g, x, 0)*x^i)(b(n-i, i))))
    end:
T:= n-> (p->seq(coeff(p, x, i), i=1..n))(b(n, n)):
seq(T(n), n=1..14);  # Alois P. Heinz, Mar 21 2012

Table[Count[Flatten[IntegerPartitions[n]],k],
{n,1,20},{k,1,n}]
TableForm[% ] (* as a triangle *)
Flatten[%%]   (* as a sequence *)
(* Clark Kimberling, Mar 03 2010 *)
T[n_, n_] = 1; T[n_, k_] /; k, ] = 0; Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 21 2015, after Omar E. Pol *)

from math import isqrt, comb
from sympy import partition
def A066633(n):
    a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
    b = n-comb(a,2)
    return sum(partition(j) for j in range(a%b,a,b)) # Chai Wah Wu, Nov 13 2024

G.f. for the number of k's in all partitions of n is x^k/(1-x^k)* Product_{m>=1} 1/(1-x^m). - Vladeta Jovovic, Jan 15 2002

T(n, k) = Sum_{j

Equals triangle A027293 * A051731 as infinite lower triangular matrices. - Gary W. Adamson Mar 21 2011

It appears that T(n+k,k) = T(n,k) + A000041(n). - Omar E. Pol, Feb 04 2012. This was proved in the Dastidar-Gupta paper in Lemma 1. - George Beck, Jun 26 2019

It appears that T(n,k) = A206563(n,k) - A206563(n,k+2). - Omar E. Pol, Feb 26 2012

T(n,k) = Sum_{j=1..n} A182703(j,k). - Omar E. Pol, May 02 2012

More terms from Vladeta Jovovic, Jan 11 2002

cp's OEIS Frontend

A066633 Triangle T(n,k), n >= 1, 1 <= k <= n, giving number of k's in all partitions of n.

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