A062344 -id:A062344 - cp's OEIS frontend

A193601 Augmentation of the triangle A062344. See Comments.

1, 1, 2, 1, 6, 10, 1, 12, 49, 76, 1, 20, 149, 508, 756, 1, 30, 354, 2082, 6353, 9192, 1, 42, 720, 6484, 32852, 92750, 131406, 1, 56, 1315, 16820, 127365, 580606, 1545757, 2153912, 1, 72, 2219, 38256, 404559, 2706150, 11385058, 28931758, 39768798

Offset: 0

Clark Kimberling, Jul 31 2011

For an introduction to the unary operation "augmentation" as applied to triangular arrays or sequences of polynomials, see A193091.

			First 5 rows of A193601:
1
1...2
1...6....10
1...12...49...76
1...20...149..508...756

Cf. A193091, A062344.

p[n_, k_] := Binomial[2 n, k] (* A062344 *)
Table[p[n, k], {n, 0, 5}, {k, 0, n}]
m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
TableForm[m[4]]
w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
v[n_] := v[n - 1].m[n]
TableForm[Table[v[n], {n, 0, 6}]] (* A193601 *)
Flatten[Table[v[n], {n, 0, 8}]]

A094527 Triangle T(n,k), read by rows, defined by T(n,k) = binomial(2*n,n-k).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 20, 15, 6, 1, 70, 56, 28, 8, 1, 252, 210, 120, 45, 10, 1, 924, 792, 495, 220, 66, 12, 1, 3432, 3003, 2002, 1001, 364, 91, 14, 1, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1, 48620, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1, 184756, 167960

Offset: 0

Paul Barry, May 07 2004

Right-hand side of even-numbered rows of Pascal's triangle.
Row sums are A032443. Reverse of A062344. Right-hand side of A034870. Binomial transform of trinomial triangle A094531.
Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 2*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 14 2007
Central coefficients T(2n,n) are binomial(4n,n) (A005810).
The A- and Z-sequence for this Riordan triangle is [1,2,1] and [2,2], respectively. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. See also the Philippe Deléham comment above. - Wolfdieter Lang, Nov 22 2012

			The triangle T(n,k) begins:
  n\k      0      1      2     3     4     5    6    7   8  9 10
  0:       1
  1:       2      1
  2:       6      4      1
  3:      20     15      6     1
  4:      70     56     28     8     1
  5:     252    210    120    45    10     1
  6:     924    792    495   220    66    12    1
  7:    3432   3003   2002  1001   364    91   14    1
  8:   12870  11440   8008  4368  1820   560  120   16   1
  9:   48620  43758  31824 18564  8568  3060  816  153  18  1
  10: 184756 167960 125970 77520 38760 15504 4845 1140 190 20  1
  ... Reformatted ad extended by _Wolfdieter Lang_, Nov 22 2012
From _Paul Barry_, Sep 07 2009: (Start)
Production array is
  2, 1,
  2, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Nov 22 2012: (Start)
Recurrence from the Riordan A-sequence [1,2,1]: T(4,1) = 56 = 1*T(3,0) + 2*T(3,1) + 1*T(3,2) = 1*20 + 2*15 + 1*6.
Recurrence from the Riordan Z-sequence [2,2]: T(7,0) = 3432 = 2*T(6,0) + 2*T(6,1) = 2*924 + 2*792. See the _Philippe Deléham_ comment above. (End)

Indranil Ghosh, Rows 0..100 of triangle, flattened
Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 19.
A. Luzón, D. Merlini, M. A. Morón, R. Sprugnoli, Complementary Riordan arrays, Discrete Applied Mathematics, 172 (2014) 75-87.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012. - From _N. J. A. Sloane_, Sep 16 2012
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263
T. M. Richardson, The Reciprocal Pascal Matrix, arXiv preprint arXiv:1405.6315 [math.CO], 2014.

Cf. A007318, A032443, A039599, A039598.

A094527 := proc(n,k)
    binomial(2*n,n-k) ;
end proc: # R. J. Mathar, Jun 04 2013

T[n_, k_] := Binomial[2*n, n - k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017 *)

Riordan array (1/sqrt(1-4*x), (1-2*x-sqrt(1-4*x))/(2*x)). Column k has e.g.f. exp(2*x)*Bessel_I(k, 2*x). - Paul Barry, Jul 14 2005
Product of Riordan arrays (1/(1-x), x/(1-x)) (Pascal's triangle, A007318) and (1/sqrt(1-2x-3x^2), (1-x-sqrt(1-2x-3x^2))/(2x)) (A094531). Inverse is A110162. - Paul Barry, Jul 14 2005
T(n,k) = Sum_{j=0..n} C(n,j)*C(n,j-k). - Paul Barry, Mar 07 2006
T(n,k) = Sum_{h>=k} A039599(n,h). Sum_{k=0..n} T(n,k) = A032443(n). - Philippe Deléham, May 01 2006
Sum_{k=0..n} T(n,k)^2 = A036910(n). - Philippe Deléham, May 07 2006
Sum_{k=0..n} T(n,k)*(-1)^k = A088218(n). - Philippe Deléham, Mar 14 2007
From Wolfdieter Lang, Nov 22 2012: (Start)
The o.g.f. for the row polynomials P(n,x) := Sum_{k=0..n} T(n,k)*x^k is G(z,x) = (-x + (1+x)*z + x*z*c(z))/(sqrt(1-4*z)*((1+x)^2*z -x)) with c the o.g.f. of A000108 (Catalan). This follows from the Riordan property.
The o.g.f. for column no. k is (c(x)-1)^k/sqrt(1-4*x) (from the Riordan property). (End)
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - 2*x - sqrt(1 - 4*x) )/(2*x) and so belongs to the hitting time subgroup of the Riordan group (see Peart and Woan, Example 5.1).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 + x)^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
From Peter Bala, Jul 21 2015: (Start)
n-th row polynomial R(n,t) = [x^n] ( (1 + (1 + t)*x)^2/(1 + t*x) )^n.
exp ( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (2 + t)*x + (5 + 4*t + t^2)*x^2 + ... is the o.g.f. for A039598. (End)

Entry revised by N. J. A. Sloane, Mar 23 2007

A034868 Left half of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 4, 6, 1, 5, 10, 1, 6, 15, 20, 1, 7, 21, 35, 1, 8, 28, 56, 70, 1, 9, 36, 84, 126, 1, 10, 45, 120, 210, 252, 1, 11, 55, 165, 330, 462, 1, 12, 66, 220, 495, 792, 924, 1, 13, 78, 286, 715, 1287, 1716, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 15

Offset: 0

N. J. A. Sloane

			1;
1;
1, 2;
1, 3;
1, 4,  6;
1, 5, 10;
1, 6, 15, 20;
...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A107430, A062344, A122366, A027306 (row sums).
Cf. A008619.
Cf. A225860.
Cf. A126257.
Cf. A034869 (right half), A014413, A014462, A265848.

a034868 n k = a034868_tabf !! n !! k
a034868_row n = a034868_tabf !! n
a034868_tabf = map reverse a034869_tabf
-- Reinhard Zumkeller, improved Dec 20 2015, Jul 27 2012

Flatten[ Table[ Binomial[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}]] (* Robert G. Wilson v, May 28 2005 *)

for(n=0, 14, for(k=0, floor(n/2), print1(binomial(n, k),", ");); print();) \\ Indranil Ghosh, Mar 31 2017

import math
from sympy import binomial
for n in range(15):
    print([binomial(n, k) for k in range(int(math.floor(n/2)) + 1)]) # Indranil Ghosh, Mar 31 2017

from itertools import count, islice
def A034868_gen(): # generator of terms
    yield from (s:=(1,))
    for i in count(0):
        yield from (s:=(1,)+tuple(s[j]+s[j+1] for j in range(len(s)-1)) + ((s[-1]<<1,) if i&1 else ()))
A034868_list = list(islice(A034868_gen(),30)) # Chai Wah Wu, Oct 17 2023

T(n,k) = A034869(n,floor(n/2)-k), k = 0..floor(n/2). - Reinhard Zumkeller, Jul 27 2012

A122366 Triangle read by rows: T(n,k) = binomial(2*n+1,k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 5, 10, 1, 7, 21, 35, 1, 9, 36, 84, 126, 1, 11, 55, 165, 330, 462, 1, 13, 78, 286, 715, 1287, 1716, 1, 15, 105, 455, 1365, 3003, 5005, 6435, 1, 17, 136, 680, 2380, 6188, 12376, 19448, 24310, 1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 1, 21

Offset: 0

Reinhard Zumkeller, Aug 30 2006

Sum of n-th row = A000302(n) = 4^n.
Central terms give A052203.
Reversal of A111418. - Philippe Deléham, Mar 22 2007
Coefficient triangle for the expansion of one half of odd powers of 2*x in terms of Chebyshev's T-polynomials: ((2*x)^(2*n+1))/2 = Sum_{k=0..n} a(n,k)*T(2*(n-k)+1,x) with Chebyshev's T-polynomials. See A053120. - Wolfdieter Lang, Mar 07 2007
The signed triangle T(n,k)*(-1)^(n-k) appears in the formula ((2*sin(phi))^(2*n+1))/2 = Sum_{k=0..n} ((-1)^(n-k))*a(n,k)*sin((2*(n-k)+1)*phi). - Wolfdieter Lang, Mar 07 2007
The signed triangle T(n,k)*(-1)^(n-k) appears therefore in the formula (4-x^2)^n = Sum_{k=0..n} ((-1)^(n-k))*a(n,k)*S(2*(n-k),x) with Chebyshev's S-polynomials. See A049310 for S(n,x). - Wolfdieter Lang, Mar 07 2007
From Wolfdieter Lang, Sep 18 2012: (Start)
The triangle T(n,k) appears also in the formula F(2*l+1)^(2*n+1) = (1/5^n)*Sum_{k=0..n} T(n,k)*F((2*(n-k)+1)*(2*l+1)), l >= 0, n >= 0, with F=A000045 (Fibonacci).
The signed triangle Ts(n,k):=T(n,k)*(-1)^k appears also in the formula
  F(2*l)^(2*n+1) = (1/5^n)*Sum_{k=0..n} Ts(n,k)*F((2(n-k)+1)*2*l), l >= 0, n >= 0, with F=A000045 (Fibonacci).
This is Lemma 2 of the K. Ozeki reference, p. 108, written for odd and even indices separately.
(End)

			.......... / 1 \ .......... =A062344(0,0)=A034868(0,0),
......... / 1 . \ ......... =T(0,0)=A034868(1,0),
........ / 1 2 . \ ........ =A062344(1,0..1)=A034868(2,0..1),
....... / 1 3 ... \ ....... =T(1,0..1)=A034868(3,0..1),
...... / 1 4 6 ... \ ...... =A062344(2,0..2)=A034868(4,0..2),
..... / 1 5 10 .... \ ..... =T(2,0..2)=A034868(5,0..2),
.... / 1 6 15 20 ... \ .... =A062344(3,0..3)=A034868(6,0..3),
... / 1 7 21 35 ..... \ ... =T(3,0..3)=A034868(7,0..3),
.. / 1 8 28 56 70 .... \ .. =A062344(4,0..4)=A034868(8,0..4),
. / 1 9 36 84 126 ..... \ . =T(4,0..4)=A034868(9,0..4).
Row n=2:[1,5,10] appears in the expansion ((2*x)^5)/2 = T(5,x)+5*T(3,x)+10*T(1,x).
Row n=2:[1,5,10] appears in the expansion ((2*cos(phi))^5)/2 = cos(5*phi)+5*cos(3*phi)+10*cos(1*phi).
The signed row n=2:[1,-5,10] appears in the expansion ((2*sin(phi))^5)/2 = sin(5*phi)-5*sin(3*phi)+10*sin(phi).
The signed row n=2:[1,-5,10] appears therefore in the expansion (4-x^2)^2 = S(4,x)-5*S(2,x)+10*S(0,x).
Triangle T(n,k) starts:
  n\k 0  1   2   3    4     5     6     7     8     9  ...
  0   1
  1   1  3
  2   1  5  10
  3   1  7  21  35
  4   1  9  36  84  126
  5   1 11  55 165  330   462
  6   1 13  78 286  715  1287  1716
  7   1 15 105 455 1365  3003  5005  6435
  8   1 17 136 680 2380  6188 12376 19448 24310
  9   1 19 171 969 3876 11628 27132 50388 75582 92378
  ...  - _Wolfdieter Lang_, Sep 18 2012
Row n=2, with F(n)=A000045(n) (Fibonacci number), l >= 0, see a comment above:
F(2*l)^5   = (1*F(10*l) - 5*F(6*l) + 10*F(2*l))/25,
F(2*l+1)^5 = (1*F(10*l+5) + 5*F(6*l+3) + 10*F(2*l+1))/25.
- _Wolfdieter Lang_, Sep 19 2012

T. J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2nd ed., Wiley, New York, 1990, pp. 54-55, Ex. 1.5.31.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 795.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
Index entries for sequences related to Chebyshev polynomials.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A062344.

Odd numbered rows of A008314. Even numbered rows of A008314 are A127673.

a122366 n k = a122366_tabl !! n !! k
a122366_row n = a122366_tabl !! n
a122366_tabl = f 1 a007318_tabl where
   f x (_:bs:pss) = (take x bs) : f (x + 1) pss
-- Reinhard Zumkeller, Mar 14 2014

T[_, 0] = 1;
T[n_, k_] := T[n, k] = T[n-1, k-1] 2n(2n+1)/(k(2n-k+1));
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 01 2021 *)

T(n,0)=1; T(n,k) = T(n-1,k-1)*2*n*(2*n+1)/(k*(2*n-k+1)) for k > 0.
T(n,0)=1; for n > 0: T(n,1)=n+2; for n > 1: T(n,n) = T(n-1,n-2) + 3*T(n-1,n-1), T(n,k) = T(n-1,k-2) + 2*T(n-1,k-1) + T(n-1,k), 1 < k < n.
T(n,n) = A001700(n).
T(n,k) = A034868(2*n+1,k) = A007318(2*n+1,k), 0 <= k <= n;
G.f.: (2*y)/((y-1)*sqrt(1-4*x*y)-4*x*y^2+(1-4*x)*y+1). - Vladimir Kruchinin, Oct 30 2020

Chebyshev and trigonometric comments from Wolfdieter Lang, Mar 07 2007.

Typo in comments fixed, thanks to Philippe Deléham, who indicated this.

A225419 Triangle read by rows: T(n,k) (0 <= k <= n) = binomial(2*n+2,k).

Original entry on oeis.org

1, 1, 4, 1, 6, 15, 1, 8, 28, 56, 1, 10, 45, 120, 210, 1, 12, 66, 220, 495, 792, 1, 14, 91, 364, 1001, 2002, 3003, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 1, 20, 190

Offset: 0

Roger L. Bagula, May 07 2013

Row sums are A000346.

			Triangle begins:
1,
1, 4,
1, 6, 15,
1, 8, 28, 56,
1, 10, 45, 120, 210,
1, 12, 66, 220, 495, 792,
1, 14, 91, 364, 1001, 2002, 3003,
...

Roger L. Bagula, a225419nb.txt

Cf. A127673, A062344, A034870, A000346.

Flatten[Table[Table[DifferenceRoot[Function[{y, n}, {(-(2*m + 1) + n) y[n] + n y[1 + n] == 0, y[1] == 1}]][k], {k, 1, m}], {m, 1, 10}]]
Flatten[Table[Binomial[2n+2,k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Apr 13 2014 *)

Edited by N. J. A. Sloane, May 11 2013

A142243 Triangle T(n,k) = binomial(2n,k) binomial(2n-2k,n-k), read by rows; 0<=k<=n.

Original entry on oeis.org

1, 2, 2, 6, 8, 6, 20, 36, 30, 20, 70, 160, 168, 112, 70, 252, 700, 900, 720, 420, 252, 924, 3024, 4620, 4400, 2970, 1584, 924, 3432, 12936, 22932, 25480, 20020, 12012, 6006, 3432, 12870, 54912, 110880, 141120, 127400, 87360, 48048, 22880, 12870, 48620

Offset: 0

Roger L. Bagula and Gary W. Adamson, Sep 17 2008

Row sums are s(n) = 1, 4, 20, 106, 580, 3244,,...

			1;
2, 2;
6, 8, 6;
20, 36, 30, 20;
70, 160, 168, 112, 70;
252, 700, 900, 720, 420, 252;
924, 3024, 4620, 4400, 2970, 1584, 924;
3432, 12936, 22932, 25480, 20020, 12012, 6006, 3432;
12870, 54912, 110880, 141120, 127400, 87360, 48048, 22880, 12870;
48620, 231660, 525096, 753984, 771120, 599760, 371280, 190944, 87516, 48620';
184756, 972400, 2445300, 3912480, 4476780, 3907008, 2713200, 1550400, 755820, 335920, 184756;

Cf. A062344.

t[n_, m_] = (Binomial[2*n, m]*Binomial[2*(n - m), (n - m)]); Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%]

Conjecture for row sums: 2*(n+1)*(2*n+1)*s(n) +(-81*n^2+19*n-8)*s(n-1) +10*(51*n^2-77*n+30)*s(n-2) -500*(n-1)*(2*n-3)*s(n-3)=0. - R. J. Mathar, Sep 13 2013

cp's OEIS Frontend

A193601 Augmentation of the triangle A062344. See Comments.

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A094527 Triangle T(n,k), read by rows, defined by T(n,k) = binomial(2*n,n-k).

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A034868 Left half of Pascal's triangle.

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A122366 Triangle read by rows: T(n,k) = binomial(2*n+1,k), 0 <= k <= n.

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A225419 Triangle read by rows: T(n,k) (0 <= k <= n) = binomial(2*n+2,k).

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A142243 Triangle T(n,k) = binomial(2*n,k) *binomial(2*n-2*k,n-k), read by rows; 0<=k<=n.

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A142243 Triangle T(n,k) = binomial(2n,k) binomial(2n-2k,n-k), read by rows; 0<=k<=n.