A095664 -id:A095664 - 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.

A225624 Triangle read by rows: T(n,k) is the number of descent sequences of length n with exactly k-1 descents, n>=1, 1<=k<=n.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 5, 0, 0, 5, 15, 3, 0, 0, 6, 35, 25, 1, 0, 0, 7, 70, 117, 28, 0, 0, 0, 8, 126, 405, 271, 22, 0, 0, 0, 9, 210, 1155, 1631, 483, 13, 0, 0, 0, 10, 330, 2871, 7359, 5126, 711, 5, 0, 0, 0, 11, 495, 6435, 27223, 36526, 13482, 889, 1, 0, 0, 0, 12, 715, 13299, 86919, 199924, 151276, 30906, 962, 0, 0, 0, 0

Offset: 1

Joerg Arndt, May 11 2013

A descent sequence is a sequence [d(1), d(2), ..., d(n)] where d(1)=0, d(k)>=0, and d(k) <= 1 + desc([d(1), d(2), ..., d(k-1)]) where desc(.) gives the number of descents of its argument, see example.
Row sums are A225588 (number of descent sequences).
First column is C(n,1)=n, second column is C(n+1,4) = A000332(n+1), third column appears to be A095664(n-5) for n>=5.

			Triangle begins:
01:  1,
02:  2, 0,
03:  3, 1, 0,
04:  4, 5, 0, 0,
05:  5, 15, 3, 0, 0,
06:  6, 35, 25, 1, 0, 0,
07:  7, 70, 117, 28, 0, 0, 0,
08:  8, 126, 405, 271, 22, 0, 0, 0,
09:  9, 210, 1155, 1631, 483, 13, 0, 0, 0,
10:  10, 330, 2871, 7359, 5126, 711, 5, 0, 0, 0,
11:  11, 495, 6435, 27223, 36526, 13482, 889, 1, 0, 0, 0,
12:  12, 715, 13299, 86919, 199924, 151276, 30906, 962, 0, 0, 0, 0,
13:  13, 1001, 25740, 247508, 903511, 1216203, 546001, 63462, 903, 0, 0, 0, 0,
...
The number of descents for the A225588(5)=23 descent sequences of length 5 are (dots for zeros):
.#:  descent seq.   no. of descents
01:  [ . . . . . ]    0
02:  [ . . . . 1 ]    0
03:  [ . . . 1 . ]    1
04:  [ . . . 1 1 ]    0
05:  [ . . 1 . . ]    1
06:  [ . . 1 . 1 ]    1
07:  [ . . 1 . 2 ]    1
08:  [ . . 1 1 . ]    1
09:  [ . . 1 1 1 ]    0
10:  [ . 1 . . . ]    1
11:  [ . 1 . . 1 ]    1
12:  [ . 1 . . 2 ]    1
13:  [ . 1 . 1 . ]    2
14:  [ . 1 . 1 1 ]    1
15:  [ . 1 . 1 2 ]    1
16:  [ . 1 . 2 . ]    2
17:  [ . 1 . 2 1 ]    2
18:  [ . 1 . 2 2 ]    1
19:  [ . 1 1 . . ]    1
20:  [ . 1 1 . 1 ]    1
21:  [ . 1 1 . 2 ]    1
22:  [ . 1 1 1 . ]    1
23:  [ . 1 1 1 1 ]    0
There are 5 sequences with 0 descents, 15 with 1 descents, 3 with 2 descents, and 0 for 3 or 5 descents. Therefore row 5 is [5, 15, 3, 0, 0].

Joerg Arndt and Alois P. Heinz, Rows n = 1..100, flattened (Rows n = 1..18 from Joerg Arndt)

b:= proc(n, i, t) option remember; local j; if n<1 then [0$t, 1]
      else []; for j from 0 to t+1 do zip((x, y)->x+y, %,
      b(n-1, j, t+`if`(jAlois P. Heinz, May 18 2013

b[n_, i_, t_] :=  b[n, i, t] =  Module[{j, pc}, If[n<1, Append[Array[0 &, t], 1], pc = {}; For[j = 0, j <= t+1, j++, pc = Plus @@ PadRight[ {pc, b[n-1, j, t+If[jJean-François Alcover, Feb 27 2014, after Alois P. Heinz *)

# After Alois P. Heinz.
@CachedFunction
def b(n, i, t, N):
    B = [0 for x in range(N)]
    if n < 1: B[t] = 1; return B
    for j in (0..t+1):
        B = map(operator.add, B, b(n-1, j, t+int(jPeter Luschny, May 20 2013; updated May 21 2013

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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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A225624 Triangle read by rows: T(n,k) is the number of descent sequences of length n with exactly k-1 descents, n>=1, 1<=k<=n.

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