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A259052
Sum of Pascal triples.
4
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.
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
KEYWORD
nonn,easy,tabf
AUTHOR
Wolfdieter Lang, Jun 27 2015
STATUS
approved