A028263 -id:A028263 - cp's OEIS frontend

A014410 Elements in Pascal's triangle (by row) that are not 1.

2, 3, 3, 4, 6, 4, 5, 10, 10, 5, 6, 15, 20, 15, 6, 7, 21, 35, 35, 21, 7, 8, 28, 56, 70, 56, 28, 8, 9, 36, 84, 126, 126, 84, 36, 9, 10, 45, 120, 210, 252, 210, 120, 45, 10, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 13, 78

Offset: 2

Mohammad K. Azarian

Also, rows of triangle formed using Pascal's rule except begin and end n-th row with n+2. - Asher Auel.
Row sums are A000918. - Roger L. Bagula and Gary W. Adamson, Jan 15 2009
Given the triangle signed by rows (+ - + ...) = M, with V = a variant of the Bernoulli numbers starting [1/2, 1/6, 0, -1/30, 0, 1/42, ...]; M*V = [1, 1, 1, ...]. - Gary W. Adamson, Mar 05 2012
Also A014410 * [1/2, 1/6, 0, -1/30, 0, 1/42, 0, ...] = [1, 2, 3, 4, ...]. For an alternative way to derive the Bernoulli numbers from a modified version of Pascal's triangle see A135225. - Peter Bala, Dec 18 2014
T(n,k) mod n = A053201(n,k), k=1..n-1. - Reinhard Zumkeller, Aug 17 2013
From Wolfdieter Lang, May 22 2015: (Start)
This is Johannes Scheubel's (1494-1570) (also Scheybl, Schöblin) version of the arithmetical triangle from his 1545 book "De numeris et diversis rationibus". See the Kac reference, p. 396 and the Table 12.1 on p. 395.
The row sums give 2*A000225(n-1) = A000918(n) = 2*(2^n - 1), n >= 2. (See the second comment above).
The alternating row sums give repeat(2,0) = 2*A059841(n), n >= 2. (End)
T(n+1,k) is the number of k-facets of the n-simplex. - Jianing Song, Oct 22 2023

			The triangle T(n,k) begins:
n\k  1  2   3   4    5    6    7    8   9  10 11
2:   2
3:   3  3
4:   4  6   4
5:   5 10  10   5
6:   6 15  20  15    6
7:   7 21  35  35   21    7
8:   8 28  56  70   56   28    8
9:   9 36  84 126  126   84   36    9
10: 10 45 120 210  252  210  120   45  10
11: 11 55 165 330  462  462  330  165  55  11
12: 12 66 220 495  792  924  792  495 220  66 12
... reformatted. - _Wolfdieter Lang_, May 22 2015

Victor J. Kac, A History of Mathematics, third edition, Addison-Wesley, 2009, pp. 395, 396.

Reinhard Zumkeller, Rows n=2..150 of triangle, flattened
Carl McTague, On the Greatest Common Divisor of binomial(qn, q), binomial(qn,2q), ..., binomial(qn, qn-q), arXiv:1510.06696 [math.CO], 2015.
Wikipedia, Johannes Scheubel (in German).
Wikipedia, Simplex

Cf. A007318, A000918, A027641.
A180986 is the same sequence but regarded as a square array.
Cf. A000225,A059841, A257241 (Stifel's version).

a014410 n k = a014410_tabl !! (n-2) !! (k-1)
a014410_row n = a014410_tabl !! (n-2)
a014410_tabl = map (init . tail) $ drop 2 a007318_tabl
-- Reinhard Zumkeller, Mar 12 2012

for i from 0 to 12 do seq(binomial(i, j)*1^(i-j), j = 1 .. i-1) od; # Zerinvary Lajos, Dec 02 2007

Select[ Flatten[ Table[ Binomial[ n, i ], {n, 0, 13}, {i, 0, n} ] ], #>1& ]

T(n,k) = binomial(n,k) = A007318(n,k), n >= 2, k = 1, 2, ..., n-1.
a(n) = C(A003057(n),A002260(n)) = C(A003057(n),A004736(n)). - Lekraj Beedassy, Jul 29 2006
T(n,k) = A028263(n,k) - A007318(n,k). - Reinhard Zumkeller, Mar 12 2012
gcd_{k=1..n-1} T(n, k) = A014963(n), see Theorem 1 of McTague link. - Michel Marcus, Oct 23 2015

More terms from Erich Friedman

A259052 Sum of Pascal triples.

Original entry on oeis.org

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

Wolfdieter Lang, Jun 27 2015

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.

			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.

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

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

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

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.

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A051665 Duplicate of A028263.

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A014410 Elements in Pascal's triangle (by row) that are not 1.

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A259052 Sum of Pascal triples.

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