cp's OEIS Frontend

Diagonal of triangles A062111, A152920.

Changing the formula by replacing T(2n,0)=T(n,2) by T(2n,0)=T(n,m) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058393, A058395, A057884 (and effectively A007318).

Binomial transform of [1,1,1,2,1,3,1,...]. [From Paul Barry, Mar 18 2009]

1, 9, 45, 161, 497, 1409, ... is the sequence of perimeters (sum of border elements) of the triangle.

1, 5, 80, 3520, 394240, 107233280, 68629299200, ... is the sequence of determinants of the triangle.

Only the first three terms are odd.

The array A(k, n) arises from the following Pascal-type triangles PTodd(k), k >= 0 based on the positive odd integers A005408.

For example, the Pascal-type triangle PTodd(k), for k = 3 is

1 3 5 7

4 8 12

12 20

32

Taken upside-down such triangles become so-called addition towers of height k+1 (Rechenturm in German elementary schools; thanks to my correspondent Bennet D.), starting with any k+1 numbers. Here the positive odd numbers are used.

The sequence s of the final number of these Pascal-type triangles PT(k), for k >= 0, begins 1, 4, 12, 32, ...; s(k) = (k+1)*2^k = A001787(k+1), for k >= 0.

For k -> infinity the left-aligned row sequences build the array A(k, n), with k >= 0 and n >= 0, namely A(k, n) = 2^k*(k + 2*n + 1); this array begins:

k\n 0 1 2 3 4 5 ...

-------------------------------

0: 1 3 5 7 9 11 ... {A005408(n)}

1: 4 8 12 16 20 24 ... {A008586(n+1)}

2: 12 20 28 36 44 52 ... {A017113(n+1)}

3: 32 48 64 80 96 112 ... {A008598(n+2)}

4: 80 112 144 176 208 240 ... {16*A005408(n+2)}

5: 192 256 320 384 448 512 ... {A152691(n+3)}

6: 448 576 704 832 960 1088 ... {64*A005408(n+3)}

...

The sequence s, the first (n=0) column of A, is always the binomial transform of the first (k=0) row in A.

A(k, n) = Sum_{j=0..k} binomial(k, j)*(2*(n+j)+1) = 2^k*(k + 1 + 2*n), for k >= 0 and n >= 0.

The corresponding antidiagonal-upwards read triangle is T(k, n) = A(k-n, n) = 2^(k-n)*(k + n + 1), n >= 0, k = 0..n.

If the nonnegative integers A001477 are used as k = 0 row of the array Anneg(k, n) = 2^(k-1)*(2*n + k), for k >= 0, n >= 0, with the triangle Tnneg(k, n) = Anneg(k-n, n) = (n + k)*2^(k-n-1), k >= 0, n = 0..k, then the s sequence is snneg(k) = Tnneg(k, 0) = k*2^{k-1} = A001787(k), the binomial transform of the sequence{A001477(n)}_{n>=0}. The triangle Tnneg begins [0], [1, 1], [4, 3, 2], [12, 8, 5, 3], [32, 20, 12, 7, 4], ... . See A062111 and the row-reversed triangle A152920 for other versions.

The ratio (count of ones)/(count of zeros) in the binary expansion of a(n) is > 1/2 and <= 5 for all n > 0, this is because the division by 9 adds a repeating pattern 111000... after some binary digits.

This sequence has in its "partial binomial transform" (see formula section) no other constants than 2 and 1 despite of its more complicated looking closed form expression. This transform has a deep connection to the Grünwald-Letnikov fractional derivative if we replace the order of the derivative with the variable x: D^x*f(x).