A095662 -id:A095662 - cp's OEIS frontend

A095660 Pascal (1,3) triangle.

3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3

Offset: 0

Wolfdieter Lang, May 21 2004

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012
Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

			Triangle starts:
  3;
  1,  3;
  1,  4,  3;
  1,  5,  7,   3;
  1,  6, 12,  10,   3;
  1,  7, 18,  22,  13,   3;
  1,  8, 25,  40,  35,  16,   3;
  1,  9, 33,  65,  75,  51,  19,   3;
  1, 10, 42,  98, 140, 126,  70,  22,   3;
  1, 11, 52, 140, 238, 266, 196,  92,  25,   3;
  1, 12, 63, 192, 378, 504, 462, 288, 117,  28,  3;
  1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.
Cf. A097805.

a095660 n k = a095660_tabl !! n !! k
a095660_row n = a095660_tabl !! n
a095660_tabl = [3] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3]
-- Reinhard Zumkeller, Apr 08 2012

A095660:= func< n,k | n eq 0 select 3 else (1+2*k/n)*Binomial(n,k) >;
[A095660(n,k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 02 2021

T(n,k):=piecewise(n=0,3,0Mircea Merca, Apr 08 2012

{3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 14 2015 *)

def A095660(n,k): return 3 if n==0 else (1+2*k/n)*binomial(n,k)
flatten([[A095660(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).
G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

A095663 Eighth column (m=7) of (1,3)-Pascal triangle A095660.

Original entry on oeis.org

3, 22, 92, 288, 750, 1716, 3564, 6864, 12441, 21450, 35464, 56576, 87516, 131784, 193800, 279072, 394383, 547998, 749892, 1012000, 1348490, 1776060, 2314260, 2985840, 3817125, 4838418, 6084432, 7594752, 9414328, 11594000, 14191056

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 3-subset of an n-set X then, for n>=9, a(n-9) is the number of 7-subsets of X having at most one element in common with Y. - Milan Janjic, Nov 23 2007

Seventh column: A095662. Ninth column: A095664.

G.f.: (3-2*x)/(1-x)^8.

a(n)= binomial(n+6, 6)*(n+21)/7 = 3*b(n)-2*b(n-1), with b(n):=binomial(n+7, 7); cf. A000580.

A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

Original entry on oeis.org

1, 2, 1, 1, 3, 3, 4, 2, 1, 4, 6, 10, 9, 7, 3, 1, 5, 10, 20, 25, 26, 19, 11, 4, 1, 6, 15, 35, 55, 71, 70, 56, 34, 16, 5, 1, 7, 21, 56, 105, 161, 196, 197, 160, 106, 55, 22, 6, 1, 8, 28, 84, 182, 322, 462, 554, 553, 463, 321, 183, 83, 29, 7, 1, 9, 36, 120, 294, 588, 966, 1338, 1569, 1570, 1337, 967, 587, 295, 119, 37, 8, 1, 10, 45, 165, 450, 1002, 1848, 2892, 3873, 4477, 4476, 3874

Offset: 0

Werner Schulte, Nov 08 2015

The triangle is related to the triangle of trinomial coefficients.

			The irregular triangle T(n,k) begins:
n\k:  0   1   2    3    4    5    6    7    8    9   10  11  12  13  14  ...
0     1;
1     2   1   1;
2     3   3   4    2    1;
3     4   6  10    9    7    3    1;
4     5  10  20   25   26   19   11    4    1;
5     6  15  35   55   71   70   56   34   16    5    1;
6     7  21  56  105  161  196  197  160  106   55   22   6   1;
7     8  28  84  182  322  462  554  553  463  321  183  83  29   7   1;
etc.
The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A000027 (col 0), A000217 (col 1), A000292 (col 2), A001590, A002426, A004524, A005582 (col 3), A008937, A027907, A095662 (col 5), A113682, A246437.

A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n,0,10}, {k, 0, 2*n} ] // Flatten (* G. C. Greubel, Mar 07 2017 *)

T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
T(n,n-1) = A246437(n+1), and T(n,n-1)+T(n,n) = A002426(n+1) for n > 0.
If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.

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A095660 Pascal (1,3) triangle.

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A095663 Eighth column (m=7) of (1,3)-Pascal triangle A095660.

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A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

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