OFFSET
0,3
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.
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.
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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)
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..9999 (rows 0 <= n <= 199, flattened)
FORMULA
T(n,k) = A000041(n-2*k) for k=0..floor(n/2).
EXAMPLE
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;
...
MATHEMATICA
Table[PartitionsP[n - 2 k], {n, 17}, {k, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 23 2021 *)
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Gary W. Adamson, Feb 13 2010
EXTENSIONS
a(24), a(25) corrected by Georg Fischer, Nov 23 2021
STATUS
approved