A173238 Triangle by columns, A000041 in every column shifted down twice for columns > 0.

1, 1, 2, 1, 3, 1, 5, 2, 1, 7, 3, 1, 11, 5, 2, 1, 15, 7, 3, 1, 22, 11, 5, 2, 1, 30, 15, 7, 3, 1, 42, 22, 11, 5, 2, 1, 56, 30, 15, 7, 3, 1, 77, 42, 22, 11, 5, 2, 1, 101, 56, 30, 15, 7, 3, 1, 135, 77, 42, 22, 11, 5, 2, 1, 176, 101, 56, 30, 15, 7, 3, 1

Offset: 0

Author

Gary W. Adamson, Feb 13 2010

Keywords

• nonn
• tabf
• easy

Comments

Row sums = A024786 starting with A024786(2): (1, 1, 3, 4, 8, 11, 19, 26, ...) = number of 2's in all partitions of n.

A173238 as an infinite lower triangular matrix * [1, 2, 3, ...] = A103650.

Let the triangle = M. Then Lim_{n->inf} M^n = (1, 1, 3, 4, 10, 13, 26, ...), the left-shifted vector considered as a sequence = A092119, the Euler transform of the ruler sequence, A001511.

Given P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x)/A(x^2), where A(x) = polcoeff A092119: (1 + x + 3x^2 + 4x^3 + ...).

Conjectures: Given the infinite set of triangles with A000041 in every column shifted down 0, 1, 2, ... n times, row sums of n-th triangle (where A173238 = 2nd in the set) = the numbers of n's in all partitions of n. E.g., row sums = of A173238 = A024786, the numbers of 2's in all partitions of n.

Similarly, row sums of triangle A173239 with columns >0 shifted down thrice = numbers of 3's in all partitions of n, and so on. Refer to comments in A000041 regarding the numbers of 1's in partitions of n.

...

Second conjecture: Given the infinite set of analogous triangles with columns shifted down 2, 3, 4, ..., k times, we let such triangles = T(k) and perform lim_{n->inf} T^n(k), obtaining the left-shifted vectors considered as sequences. The conjecture states that the infinite set of such left-shifted vectors = the Euler transform of the infinite set of Ruler functions starting with the ruler function for k=2 = A001511: (1, 2, 1, 3, 1, 2, 1, ...)

To obtain the k-th ruler functions, begin with the natural numbers, 1,...2,...3,...4,...5,...6,...7,...8,...9,...; then for k = 2 we get A001511: 1,...2,...1,...3,...1,...2,...1,...4,...1,...; by finding the highest exponent of k dividing n, then adding 1. Similarly, for k = 2, we obtain A051064: 1,...1,...2,...1,...1,...2,...1,...1,...3,...

Next, we obtain the Euler transforms of the ruler functions (e.g., Euler transform of A001511 = A092119: (1, 1, 3, 4, 10, 13, 26, ...), noting that A092119 is lim_{n->inf} A173238^n, the left-shifted vector.

...

Third conjecture: Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + ...), and A(k)(x) = the Euler transform of k-th ruler sequence, (k=2,3,...). Then P(x) = A(k)(x) / A(k)(x^k).

Examples: for k=2, A(x) = A092119: (1, 1, 3, 4, 10, 13, ...), then P(x) = (1 + x + 2x^2 + 3x^3 + ...) = (1 + x + 3x^2 + 4x^3 + ...) / (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...). For k=3 relating to triangle A173238, the left-shifted vector = the Euler transform of A051064 = A(x) for k=3, then P(x) = A(x) / A(x^3).

The conjecture extends the analogous conclusions to all k.

From Gary W. Adamson, Feb 25 2010: (Start)

Proof of second conjecture received from Helmut Prodinger 02/28/10 with a summary by R. J. Mathar:

Consider Product_{n>=0} z^(t^n)/(1-z^(t^n)) = Sum_{k>=1} (1+v_t(k))z^k where v_t(n) is the number of trailing zeros in the t-ary expansion of n, and its Euler transform A(z) = product_{k >= 1} 1/(1-z^k)^{1+v_t(k)}, then A(z)/A(z^t) = product_{k >= 1} 1/(1-z^k) is the partition generating function.

Here is the proof: A(z)/A(z^t) = Product_{k>=1} (1-z^(tk))^(1+v_t(k))/(1-z^k)^(1+v_t(k))

= Product_{k>=1} (1-z^(tk))^(v_t(tk))/(1-z^k)^(1+v_t(k))

= Product_{k>=1} (1-z^k)^(v_t(k))/(1-z^k)^(1+v_t(k)) (*)

= Product_{k>=1} 1/(1-z^k)

as desired. Notice that for (*), that v_t(n)=0 if n is not divisible by t. [Helmut Prodinger, hproding(AT)sun.ac.za, Feb 28 2010] (End)

Examples

			First few rows of the triangle:
    1;
    1;
    2,   1;
    3,   1;
    5,   2,  1;
    7,   3,  1;
   11,   5,  2,  1;
   15,   7,  3,  1;
   22,  11,  5,  2,  1;
   30,  15,  7,  3,  1;
   42,  22, 11,  5,  2, 1;
   56,  30, 15,  7,  3, 1;
   77,  42, 22, 11,  5, 2, 1;
  101,  56, 30, 15,  7, 3, 1;
  135,  77, 42, 22, 11, 5, 2, 1;
  176, 101, 56, 30, 15, 7, 3, 1;
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..9999 (rows 0 <= n <= 199, flattened)

Crossrefs

Cf. A000041, A001511, A092119, A173239.

Programs

Mathematica

Table[PartitionsP[n - 2 k], {n, 17}, {k, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 23 2021 *)

Formula

T(n,k) = A000041(n-2*k) for k=0..floor(n/2).

Extensions

a(24), a(25) corrected by Georg Fischer, Nov 23 2021