OFFSET
0,6
COMMENTS
It is conjectured that the triangle is an integer triangle. The triangle and its inverse both appear to have row sums equal to the all 1's sequence.
The triangle is equivalent to the lower semi-matrix T = e_{1,1} + Sum_{i>=2} Sum_{p>=0} ( e_{2^p i, i} ceiling(i/2) - e_{2^p (i+1), i} ceiling(i/2) ) , where e_{i,j} is the matrix unit. The conjecture above is true, deduced from the formula of the matrix. - FUNG Cheok Yin, Sep 12 2022
All the conjectures above are true (see Fried link). - Sela Fried, Nov 14 2025
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows n = 0..150, flattened).
Sela Fried, On the inverse of a certain triangular matrix and its connection to the largest odd divisor, Notes Num. Theor. Disc. Math. 32(2) (2026), 255-262. See pp. 1-2, 5.
Sela Fried, On the inverse of a certain triangular matrix and its connection to the largest odd divisor, 2025. [Cached copy]
FUNG Cheok Yin, the triangle with the first 61 rows.
FORMULA
From Sela Fried, Nov 14 2025: (Start)
Denote the triangle by T and its elements by T(n,k).
For n >= 1, write n = 2^v * u with u odd and integer v >= 0.
For each j in {0,1,...,v}, set theta(j) = 2^(v-j) * u.
Then for k >= 1, T(n,k) = Sum_{p>=0} ( ceiling(theta(p)/2) * [k = theta(p)] - floor(theta(p)/2) * [k = theta(p) - 1] ) where [] is the Iverson bracket.
G.f.: t(x,y) = ((1 + x*y)/(x*(1 - y))) * log((1 - x^2*y^2)/(1 - x^2*y)).
G.f.: T(x,y) = Sum_{p >= 0} y*x^(2^p)*(1 - x^(2^p)) / ((1 - y*x^(2^p))^2 * (1 + y*x^(2^p))). (End)
EXAMPLE
Triangle begins
1;
0, 1;
0, -1, 2;
0, 1, -2, 2;
0, 0, 0, -2, 3;
0, -1, 2, 0, -3, 3;
0, 0, 0, 0, 0, -3, 4;
0, 1, -2, 2, 0, 0, -4, 4;
0, 0, 0, 0, 0, 0, 0, -4, 5;
0, 0, 0, -2, 3, 0, 0, 0, -5, 5;
0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 6;
0, -1, 2, 0, -3, 3, 0, 0, 0, 0, -6, 6;
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -6, 7;
Inverse of the triangle begins
1;
0, 1;
0, 1/2, 1/2;
0, 0, 1/2, 1/2;
0, 0, 1/3, 1/3, 1/3;
0, 0, 0, 1/3, 1/3, 1/3;
0, 0, 0, 1/4, 1/4, 1/4, 1/4;
0, 0, 0, 0, 1/4, 1/4, 1/4, 1/4;
0, 0, 0, 0, 1/5, 1/5, 1/5, 1/5, 1/5;
0, 0, 0, 0, 0, 1/5, 1/5, 1/5, 1/5, 1/5;
0, 0, 0, 0, 0, 1/6, 1/6, 1/6, 1/6, 1/6, 1/6;
MATHEMATICA
rows = 11;
t[n_, k_] := If[k <= n, If[n <= 2 k, 1/Floor[(n+2)/2] , 0], 0];
T = Table[t[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse;
Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Stefano Spezia, Sep 30 2018 *)
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Paul Barry, Jan 29 2007
STATUS
approved