A141072 Sum of diagonal numbers in a Pascal-like triangle with index of asymmetry y = 3 and index of obliquity z = 0 (going upwards, left to right).

1, 1, 2, 3, 6, 11, 22, 42, 83, 162, 319, 626, 1231, 2419, 4756, 9349, 18380, 36133, 71036, 139652, 274549, 539748, 1061117, 2086100, 4101165, 8062677, 15850806, 31161863, 61262610, 120439119, 236777074, 465491470, 915132135, 1799102406, 3536942203, 6953445286

Offset: 1

Author

Juri-Stepan Gerasimov, Jul 16 2008

Keywords

• nonn

Comments

For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for n >= 0 and k = 1..(n+1). (This is array A140996.)

For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and for k = 4..(n+4). (This is array A140995.)

From Petros Hadjicostas, Jun 13 2019: (Start)

In the example below the author uses array A140996 to create this sequence. If we use array A140995, which is the mirror image of A140996, and follow the same process, we get a different sequence: 1, 1, 2, 3, 4, 7, 8, 15, 16, 31, 33, 63, 68, 127, 140, 255, 288, 512, 592, ...

Even though array A140996 starts at row n = 0, the offset of the current sequence was set at n = 1. In other words, a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) = G(n-1, 0) + G(n-2, 1) + ... + G(ceiling((n-1)/2), floor((n-1)/2)) for n >= 1, where G(n, k) = A140996(n, k).

To get the g.f. of this sequence, we take the bivariate g.f. of sequence A140996, and set x = y. We multiply the result by x because the offset here was set at n = 1.

Finally, we mention that in the attached photograph about Stepan's triangle, the index of asymmetry is denoted by s (rather than y) and the index of obliqueness is denoted by e (rather than z). For the Pascal triangle, s = y = 0.

(End)

Examples

			Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
  1, so a(1) = 1.
  1   1, so a(2) = 1.
  1   2   1, so a(3) = 1 + 1 = 2.
  1   4   2   1, so a(4) = 1 + 2 = 3.
  1   8   4   2  1, so a(5) = 1 + 4 + 1 = 6.
  1  16   8   4  2  1, so a(6) = 1 + 8 + 2 = 11.
  1  31  17   8  4  2  1, so a(7) = 1 + 16 + 4 + 1 = 22.
  1  60  35  17  8  4  2 1, so a(8) = 1 + 31 + 8 + 2 = 42.
  1 116  72  35 17  8  4 2 1, so a(9) = 1 + 60 + 17 + 4 + 1 = 83.
  1 224 148  72 35 17  8 4 2 1, so a(10) = 1 +  116 + 35 + 8 + 2 = 162.
  1 432 303 149 72 35 17 8 4 2 1, so a(11) = 1 + 224 + 72 + 17 + 4 + 1 = 319.
... [edited by _Petros Hadjicostas_, Jun 13 2019]
		

Links

• Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Crossrefs

Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141073.

Formula

From Petros Hadjicostas, Jun 13 2019: (Start)

a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) for n >= 1, where G(n, k) = A140996(n, k) for 0 <= k <= n.

G.f.: x*(1 - x^2 - x^3 - x^4 - x^5)/((1 - x)*(1 + x)*(1 - x - x^2 - x^3 - x^4 - x^5)) = x*(1 - x^2 - x^3 - x^4 - x^5)/(1 - x - 2*x^2 + x^6 + x^7).

Recurrence: a(n) = -(3 + (-1)^n)/2 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 6 with a(1) = a(2) = 1, a(3) = 2, a(4) = 3, and a(5) = 6.

(End)

Extensions

Partially edited by N. J. A. Sloane, Jul 18 2008

Name edited by and more terms from Petros Hadjicostas, Jun 13 2019