A259052 Sum of Pascal triples.

3, 4, 4, 4, 5, 6, 8, 6, 5, 6, 8, 13, 12, 13, 8, 6, 7, 10, 19, 20, 26, 20, 19, 10, 7, 8, 12, 26, 30, 45, 40, 45, 30, 26, 12, 8, 9, 14, 34, 42, 71, 70, 90, 70, 71, 42, 34, 14, 9, 10, 16, 43, 56, 105, 112, 161, 140, 161, 112, 105, 56, 43, 16, 10

Offset: 1

Author

Wolfdieter Lang, Jun 27 2015

Keywords

• nonn
• easy
• tabf

Comments

The sequence of row lengths of this irregular triangle is A005408(n-1) = 2*n - 1.

This entry is motivated by A258445 from Craig Knecht. There the minima of the Pascal triples are given.

A Pascal triple PT(n, k) for n >= 1, k = 1, 2, ..., 2*n-1 is defined for even k by (P(n-1, k/2-1), P(n-1, k/2), P(n, k/2)) with P(n, k) = A007318(n, k) = binomial(n, k), and for odd k by (P(n-1, (k-1)/2), P(n, (k-1)/2), P(n, (k+1)/2)).

The strip S_n between row n-1 and n of Pascal's triangle (written as symmetric equilateral triangle) is divided into 2*n-1 small equilateral up - down triangles connecting neighboring entries of Pascal's triangle. For odd k these triangles have their base on row n of Pascal's triangle (up triangles), and for even n their base is on row n-1 (down triangles). There are n up triangles and n-1 down triangles in strip S_n.

The present irregular triangle gives the sum of the Pascal triples.

This is motivated by the idea (see A258445) of considering equal touching cylinders (closed only with a bottom disk) with centers at the corners of the small up and down triangles and radius r/2 if the side of each triangle has length r. They are filled with a liquid to a height h with h/r given by the Pascal entry at the center of the bottom of the cylinder. If, for each of the three pairs from a triple of touching cylinders a hole on the bottom of the vertical touching line is opened, then the new height H of the liquid for such a triple will be the arithmetic mean of the three original heights of the three touching cylinders. The ratio H/r will be 1/3 of the corresponding irregular triangle entry for this Pascal triple.

The row sums of this irregular triangle give 3*A033484(n-1), n >= 1.

Examples

			The irregular triangle T(n, k) starts:
  n\k 1  2  3  4  5  6  7  8  9 10 11 12 13 ...
  1:  3
  2:  4  4  4
  3:  5  6  8  6  5
  4:  6  8 13 12 13  8  6
  5:  7 10 19 20 26 20 19 10  7
  6:  8 12 26 30 45 40 45 30 26 12  8
  7:  9 14 34 42 71 70 90 70 71 42 34 14  9
  ...
T(3, 2) = 6 from the sum of the Pascal triple (1, 2, 3) (from the first down triangle in Pascal's triangle strip S_3).
The height ratio H/r for this Pascal triple PT(3, 2) = (1, 2, 3) is (1 + 2 + 3)/3 = T(3, 2)/3 = 2.
		

Links

• Craig Knecht, Pascal triples
• Craig Knecht, Sum of triples table

Crossrefs

Cf. A007318, A258445, A005408, 3*A033484.

Programs

PARI

tabl(nn) = {for (n=1, nn, for (k=1, 2*n-1, kk = (k+1)\2; if (k%2, v = binomial(n-1, kk-1) + binomial(n, kk-1) + binomial(n, kk), v = binomial(n, kk) + binomial(n-1, kk-1) + binomial(n-1, kk)); print1(v, ", ");); print(););} \\ Michel Marcus, Jun 27 2015

Formula

T(n, 2*m) = P(n-1, m-1) + P(n-1, m) + P(n, m) with P(n, k) = A007318(n, k) = binomial(n, k), for m = 1..n-1, and

T(n, 2*m-1) = P(n-1, m-1) + P(n, m-1) + P(n, m) for m = 1..n.

T(n, 2*m-1) = A028263(n-1, m-1), n >= 1, m = 1..n, and T(n, 2*m) = 2*A014410(n, m), n >= 2, m = 1..n-1.