A007854 -id:A007854 - cp's OEIS frontend

A116395 Riordan array (1/sqrt(1-4x), (1/sqrt(1-4x)-1)/2).

1, 2, 1, 6, 5, 1, 20, 22, 8, 1, 70, 93, 47, 11, 1, 252, 386, 244, 81, 14, 1, 924, 1586, 1186, 500, 124, 17, 1, 3432, 6476, 5536, 2794, 888, 176, 20, 1, 12870, 26333, 25147, 14649, 5615, 1435, 237, 23, 1, 48620, 106762, 112028, 73489, 32714, 10135, 2168, 307, 26, 1

Offset: 0

Paul Barry, Feb 12 2006

Row sums are A007854. Diagonal sums are A116396.
Triangle T(n,k), 0 <= k <= n, read by rows given by [2,1,1,1,1,1,1,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 05 2007
Inverse of Riordan array (1/(1+2*x), x*(1+x)/(1+2*x)^2) (see A123876). - Philippe Deléham, Oct 25 2007

			Triangle begins:
    1;
    2,   1;
    6,   5,   1;
   20,  22,   8,  1;
   70,  93,  47, 11,  1;
  252, 386, 244, 81, 14, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Joseph Pappe, Digjoy Paul and Anne Schilling, An area-depth symmetric q,t-Catalan polynomial, arXiv:2109.06300 [math.CO], 2021. See Remark 2.4 p. 4.

[[ Round((4^n/2^k)*(&+[ (-1)^(k-j)*Binomial(k, j)*Gamma(n+(j+1)/2)/(Factorial(n)*Gamma((j+1)/2)) : j in [0..k]])) : k in [0..n]]: n in [0..12]]; // G. C. Greubel, May 28 2019

T[n_, k_]:= (4^n/2^k)*Sum[(-1)^(k-j)*Binomial[k, j]*Binomial[n+(j-1)/2, n], {j, 0, k}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 28 2019 *)

{T(n,k) = (4^n/2^k)*sum(j=0, k, (-1)^(k-j)*binomial(k, j)* binomial(n+(j-1)/2, n))}; \\ G. C. Greubel, May 28 2019

[[(4^n/2^k)*sum( (-1)^(k-j)*binomial(k, j)* binomial(n+(j-1)/2, n) for j in (0..k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, May 28 2019

Number triangle T(n,k) = (4^n/2^k)*Sum_{j=0..k} C(k,j)*C(n+(j-1)/2,n)*(-1)^(k-j).
Sum_{k=0..n} (-1)^k*T(n,k) = A000108(n), Catalan numbers. - Philippe Deléham, Nov 07 2006
T(n,k) = Sum_{j>=0} A039599(n,j)*binomial(j,k). - Philippe Deléham, Mar 30 2007
Sum_{k=0..n} T(n,k)*x^k = A127053(n), A126985(n), A127016(n), A127017(n), A126987(n), A126986(n), A126982(n), A126984(n), A126983(n), A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Oct 25 2007

A067347 Square array read by antidiagonals: T(n,k)=(T(n,k-1)n^2-Catalan(k-1)n)/(n-1) with a(n,0)=1 and a(1,k)=Catalan(k) where Catalan(k)=C(2k,k)/(k+1)=A000108(k).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 6, 3, 1, 0, 14, 20, 12, 4, 1, 0, 42, 70, 51, 20, 5, 1, 0, 132, 252, 222, 104, 30, 6, 1, 0, 429, 924, 978, 548, 185, 42, 7, 1, 0, 1430, 3432, 4338, 2904, 1150, 300, 56, 8, 1, 0, 4862, 12870, 19323, 15432, 7170, 2154, 455, 72, 9, 1, 0, 16796

Offset: 0

Henry Bottomley, Jan 16 2002

			Array begins
1 0 0 0 0 0 0 0 ... k=0
1 1 2 5 14 42 132 429 ... k=1
1 2 6 20 70 252 924 3432 ... k=2
1 3 12 51 222 978 4338 19323 ... k=3

Rows give A000007, A000108, A000984, A007854, A076035, A076036. Columns give A000012, A001700, A002378, A062158.

T(n, k) =A067345(n, k)*n =A067346(n, k)*n/(n-1)

A112657 A Motzkin transform of Jacobsthal numbers.

Original entry on oeis.org

1, 2, 7, 23, 79, 272, 943, 3278, 11419, 39830, 139057, 485795, 1697905, 5936348, 20760271, 72615143, 254028355, 888758030, 3109714117, 10881403229, 38077702909, 133251869648, 466325356273, 1631981113112, 5711490384901

Offset: 0

Paul Barry, Jan 11 2006

Binomial transform of A100098.

Inverse binomial transform of A007854. The Hankel transform of this sequence is 3^n (see A000244). - Philippe Deléham, Nov 25 2007

Cf. A000244, A007854, A026300, A089942, A100098.

a(n) = Sum_{k=0..n} A026300(n, k)*(2^(k+1) + (-1)^k)/3, where A026300 is the Motzkin triangle; a(n) = Sum_{k=0..n} ((k+1)/(n+1))*Sum_{j=0..n+1} C(n+1, j)*C(j, 2j-n+k)*(2^(k+1) + (-1)^k)/3.

a(n) = Sum_{k=0..n} A089942(n,k)*2^k = Sum_{k=0..n} A071947(n,k)*2^(n-k). - Philippe Deléham, Mar 31 2007

A127543 Triangle T(n,k), 0<=k<=n, read by rows given by :[ -1,1,1,1,1,1,1,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, -1, 1, -1, 1, -2, 0, 2, -1, 1, -6, 2, 1, 3, -1, 1, -18, 5, 7, 2, 4, -1, 1, -57, 17, 19, 13, 3, 5, -1, 1, -186, 56, 64, 36, 20, 4, 6, -1, 1, -622, 190, 212, 124, 56, 28, 5, 7, -1, 1, -2120, 654, 722, 416, 198, 79, 37, 6, 8, -1, 1, -7338, 2282, 2494, 1434, 673, 287, 105, 47, 7, 9, -1, 1

Offset: 0

Philippe Deléham, Apr 01 2007

Riordan array (2/(3-sqrt(1-4*x)), (1-sqrt(1-4*x))/(3-sqrt(1-4*x))). - Philippe Deléham, Jan 27 2014

			Triangle begins:
    1;
   -1,  1;
    0, -1,  1;
   -1,  1, -1,  1;
   -2,  0,  2, -1,  1;
   -6,  2,  1,  3, -1,  1;
  -18,  5,  7,  2,  4, -1,  1;
  -57, 17, 19, 13,  3,  5, -1, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

A065600[n_, k_]:= If[k==n, 1, Sum[j*Binomial[k+j, j]*Binomial[2*(n-k-j), n-k]/(n-k-j), {j,0, Floor[(n-k)/2]}]];
A127543[n_, k_]:= A065600[n-1,k-1] - A065600[n-1,k];
Table[A127543[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 17 2021 *)

def A065600(n,k): return 1 if (k==n) else sum( j*binomial(k+j, j)*binomial(2*(n-k-j), n-k)/(n-k-j) for j in (0..(n-k)//2) )
def A127543(n,k): return A065600(n-1, k-1) - A065600(n-1, k)
flatten([[A127543(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 17 2021

T(n,k) = A065600(n-1,k-1) - A065600(n-1,k).
Sum_{k=0..n} T(n,k)*x^k = A127053(n), A126985(n), A127016(n), A127017(n), A126987(n), A126986(n), A126982(n), A126984(n), A126983(n), A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for n= -8,-7,...,8,9 respectively.
Sum_{j>=0} T(n,j)*A007318(j,k) = A106566(n,k).
Sum_{j>=0} T(n,j)*A038207(j,k) = A039599(n,k).
Sum_{j>=0} T(n,j)*A027465(j,k) = A116395(n,k).

A117375 Riordan array (1/(1-3x*c(x)),xc(x)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 3, 1, 12, 4, 1, 51, 17, 5, 1, 222, 74, 23, 6, 1, 978, 326, 104, 30, 7, 1, 4338, 1446, 468, 142, 38, 8, 1, 19323, 6441, 2103, 657, 189, 47, 9, 1, 86310, 28770, 9447, 3006, 903, 246, 57, 10, 1, 386250, 128750, 42440, 13670, 4223, 1217, 314, 68, 11, 1, 1730832

Offset: 0

Paul Barry, Mar 10 2006

Triangle factors as (1,xc(x))*(1/(1-3x),x). First row is A007854. Second row is A049027(n)-0^n. Row sums are A049027(n+1). Diagonal sums are A117376.

			Triangle begins
1,
3, 1,
12, 4, 1,
51, 17, 5, 1,
222, 74, 23, 6, 1,
978, 326, 104, 30, 7, 1,
4338, 1446, 468, 142, 38, 8, 1
Production array begins
3, 1
3, 1, 1
3, 1, 1, 1
3, 1, 1, 1, 1
3, 1, 1, 1, 1, 1
3, 1, 1, 1, 1, 1, 1
3, 1, 1, 1, 1, 1, 1, 1
3, 1, 1, 1, 1, 1, 1, 1, 1
... - _Philippe Deléham_, Mar 05 2013

Number triangle T(0,0)=1, T(n,k)=[k<=n]*sum{j=0..n, (j/(n-j))*C(2n-j,n-j)[k<=j]*3^(j-k)}

A165992 Triangle read by rows: T(n,0) = 3^n, T(n,k) = T(n,k-1) + T(n-1,k) for 0 < k < n, and T(n,n) = T(n,n-1).

Original entry on oeis.org

1, 3, 3, 9, 12, 12, 27, 39, 51, 51, 81, 120, 171, 222, 222, 243, 363, 534, 756, 978, 978, 729, 1092, 1626, 2382, 3360, 4338, 4338, 2187, 3279, 4905, 7287, 10647, 14985, 19323, 19323, 6561, 9840, 14745, 22032, 32679, 47664, 66987, 86310, 86310, 19683

Offset: 0

Gerald McGarvey, Oct 03 2009

A007854 (main diagonal)

s=10;M=matrix(s,s);for(n=1,s,M[n,1]=3^(n-1)); for(n=2,s,for(k=2,n,M[n,k]=M[n,k-1]+M[n-1,k])); for(n=1,10,for(k=1,n,print1(M[n,k],", ")))

cp's OEIS Frontend

A116395 Riordan array (1/sqrt(1-4*x), (1/sqrt(1-4*x)-1)/2).

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A067347 Square array read by antidiagonals: T(n,k)=(T(n,k-1)*n^2-Catalan(k-1)*n)/(n-1) with a(n,0)=1 and a(1,k)=Catalan(k) where Catalan(k)=C(2k,k)/(k+1)=A000108(k).

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A112657 A Motzkin transform of Jacobsthal numbers.

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A127543 Triangle T(n,k), 0<=k<=n, read by rows given by :[ -1,1,1,1,1,1,1,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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A117375 Riordan array (1/(1-3x*c(x)),xc(x)), c(x) the g.f. of A000108.

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Formula

A165992 Triangle read by rows: T(n,0) = 3^n, T(n,k) = T(n,k-1) + T(n-1,k) for 0 < k < n, and T(n,n) = T(n,n-1).

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A116395 Riordan array (1/sqrt(1-4x), (1/sqrt(1-4x)-1)/2).

A067347 Square array read by antidiagonals: T(n,k)=(T(n,k-1)n^2-Catalan(k-1)n)/(n-1) with a(n,0)=1 and a(1,k)=Catalan(k) where Catalan(k)=C(2k,k)/(k+1)=A000108(k).