This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A259052 #45 May 18 2025 19:03:29
%S A259052 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 A259052 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 A259052 56,105,112,161,140,161,112,105,56,43,16,10
%N A259052 Sum of Pascal triples.
%C A259052 The sequence of row lengths of this irregular triangle is A005408(n-1) = 2*n - 1.
%C A259052 This entry is motivated by A258445 from _Craig Knecht_. There the minima of the Pascal triples are given.
%C A259052 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 A259052 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 A259052 The present irregular triangle gives the sum of the Pascal triples.
%C A259052 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 A259052 The row sums of this irregular triangle give 3*A033484(n-1), n >= 1.
%H A259052 Craig Knecht, <a href="/A259052/a259052_2.jpg">Pascal triples</a>
%H A259052 Craig Knecht, <a href="/A259052/a259052_4.jpg">Sum of triples table</a>
%F A259052 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
%F A259052 T(n, 2*m-1) = P(n-1, m-1) + P(n, m-1) + P(n, m) for m = 1..n.
%F A259052 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.
%e A259052 The irregular triangle T(n, k) starts:
%e A259052 n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
%e A259052 1: 3
%e A259052 2: 4 4 4
%e A259052 3: 5 6 8 6 5
%e A259052 4: 6 8 13 12 13 8 6
%e A259052 5: 7 10 19 20 26 20 19 10 7
%e A259052 6: 8 12 26 30 45 40 45 30 26 12 8
%e A259052 7: 9 14 34 42 71 70 90 70 71 42 34 14 9
%e A259052 ...
%e A259052 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 A259052 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 A259052 (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 A259052 Cf. A007318, A258445, A005408, 3*A033484.
%K A259052 nonn,easy,tabf
%O A259052 1,1
%A A259052 _Wolfdieter Lang_, Jun 27 2015