%I #38 Jan 19 2019 04:15:43

%S 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,

%T 30,45,40,45,30,26,12,8,9,14,34,42,71,70,90,70,71,42,34,14,9,10,16,43,

%U 56,105,112,161,140,161,112,105,56,43,16,10

%N Sum of Pascal triples.

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

%C This entry is motivated by A258445 from _Craig Knecht_. There the minima of the Pascal triples are given.

%C 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)).

%C 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.

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

%C 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.

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

%H Craig Knecht, <a href="/A259052/a259052_2.jpg">Pascal triples</a>

%H Craig Knecht, <a href="/A259052/a259052_4.jpg">Sum of triples table</a>

%F 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

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

%F 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.

%e The irregular triangle T(n, k) starts:

%e n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 ...

%e 1: 3

%e 2: 4 4 4

%e 3: 5 6 8 6 5

%e 4: 6 8 13 12 13 8 6

%e 5: 7 10 19 20 26 20 19 10 7

%e 6: 8 12 26 30 45 40 45 30 26 12 8

%e 7: 9 14 34 42 71 70 90 70 71 42 34 14 9

%e ...

%e 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).

%e 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.

%o (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

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

%K nonn,easy,tabf

%O 1,1

%A _Wolfdieter Lang_, Jun 27 2015