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

1,1

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.

Links

Table of n, a(n) for n=1..64.

Craig Knecht, Pascal triples

Craig Knecht, Sum of triples table

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, 2, ..., n-1, and

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

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

Example

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.

Prog

(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

Crossrefs

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

Sequence in context: A351911 A069812 A179842 * A268395 A304032 A022484

Adjacent sequences: A259049 A259050 A259051 * A259053 A259054 A259055

Keyword

nonn,easy,tabf

Author

Wolfdieter Lang, Jun 27 2015

Status

approved