cp's OEIS Frontend

From Philippe Deléham, Apr 15 2007: (Start)

This triangle can be found in the Laisant reference in the following form:

.......................5...11..

...................4...9...20..

...............3...7..16...36..

...........2...5..12..28.......

.......1...3...8..20..48.......

...0...1...4..12..32..80....... (End)

Triangle A152920 reversed. - Philippe Deléham, Apr 21 2009

Diagonal of triangles A062111, A152920.

Last term in each row gives A001792. Difference between center term of row 2n-1 and row sum of row n, (A053220(n+4) - A053221(n+4)) gives A045618(n).

For all integers k >= 2, if a sequence k,k-1,k+2,k-3,k+4,...,2,2k-2,1,2k-1, b0(n) with offset 1, is written, the sequence b0(2)-b0(1), b0(3)-b0(2), b0(4)-b0(3), ..., b0(2k-1)-b0(2k-2), b1(n) with offset 1, is written under it, the sequence b1(2)-b1(1), b1(3)-b1(2), b1(4)-b1(3), ..., b1(2k-2)-b1(2k-3), b2(n) with offset 1, is written under this, and so on until the sequence b(2k-3)(2)-b(2k-3)(1), b(2k-2)(n) with offset 1 (which will contain only one term), is written, and then the sequence b1(1); b1(2),b2(1); b1(3),b2(2),b3(1); ...; b1(2k-2), b2(2k-3), b3(2k-4), ..., b(2k-2)(1) is obtained, then this sequence will be identical to the first 2k^2-3k+1 terms of a(n), except that the first term of this sequence will be negative, the next two terms will be positive, the next three will be negative, the next four positive, and so on.

Subtriangle of triangle in A152920. - Philippe Deléham, Nov 21 2011

Diagonal of triangles A062111, A152920.

Diagonal of triangles A062111, A152920.

Diagonal of triangles A062111, A152920.

Will every integer appear in the triangle?

After 0, is this the second column of A108284? [Bruno Berselli, Sep 13 2016 - this comment may be removed if the property is confirmed.]

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.