A106270 -id:A106270 - cp's OEIS frontend

A106271 Row sums of number triangle A106270.

1, 0, -2, -7, -21, -63, -195, -624, -2054, -6916, -23712, -82498, -290510, -1033410, -3707850, -13402695, -48760365, -178405155, -656043855, -2423307045, -8987427465, -33453694485, -124936258125, -467995871775, -1757900019099, -6619846420551, -24987199492703

Offset: 0

Paul Barry, Apr 28 2005

From Petros Hadjicostas, Jul 15 2019: (Start)
To prove R. J. Mathar's conjecture, let A(x) be the g.f. of the current sequence. We note first that
Sum_{n >= 2} (n+1)*a(n)*x^n = (x*A(x))' - 1,
Sum_{n >= 2} (1-5*n)*a(n-1)*x^n = x*A(x) - 5*x*(x*A(x))' + 4*x, and
Sum_{n >= 2} 2*(2*n-1)*a(n-2)*x^n = 4*x*(x^2*A(x))' - 2*x^2*A(x).
Adding these equations (side by side), we get
Sum_{n >= 2} ((n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2))*x^n = 0,
which proves the conjecture. (End)

Michael De Vlieger, Table of n, a(n) for n = 0..1669

Cf. A014138, A014137 (partial sums of Catalan numbers (A000108)).

Cf. A106270.

Table[1 - Sum[(2n)!/n!/(n+1)!,{n,1,k}],{k,0,30}] (* Alexander Adamchuk, Feb 23 2007 *)

G.f.: c(x)*sqrt(1 - 4x)/(1 - x), where c(x) is the g.f. of A000108.
a(n) = Sum_{k = 0..n} 2*0^(n-k) - C(n-k), where C(m) = A000108(m) (Catalan numbers).
a(n) = 2 - A014137(n) for n >= 0 and a(n) = 1 - A014138(n) for n >= 0. - Alexander Adamchuk, Feb 23 2007, corrected by Vaclav Kotesovec, Jul 22 2019
Conjecture: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ -2^(2*n + 2) / (3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 22 2019

More terms from Alexander Adamchuk, Feb 23 2007

A106272 Antidiagonal sums of number triangle A106270.

Original entry on oeis.org

1, -1, -1, -6, -15, -48, -147, -477, -1577, -5339, -18373, -64125, -226385, -807025, -2900825, -10501870, -38258495, -140146660, -515897195, -1907409850, -7080017615, -26373676870, -98562581255, -369433290520, -1388466728579, -5231379691972

Offset: 0

Paul Barry, Apr 28 2005

To prove R. J. Mathar's conjecture, let A(x) be the g.f. of the current sequence. We note first that
Sum_{n >= 3} (n+1)*a(n)*x^n = (x*A(x))' + (-1 + 2*x + 3*x^2),
Sum_{n >= 3} 2*(1-2*n)*a(n-1)*x^n = 2*x*A(x) - 4*x*(x*A(x))' + (2*x - 6*x^2),
Sum_{n >= 3} -(n+1)*a(n-2)*x^n = -(x^3*A(x))' + 3*x^2, and
Sum_{n >= 3} 2*(2*n-1)*a(n-3)*x^n = 4*x*(x^3*A(x))' - 2*x^3*A(x).
Adding these equations (side by side), we get
Sum_{n >= 3} ((n+1)*a(n) + 2*(1-2*n)*a(n-1) - (n+1)*a(n-2) + 2*(2*n-1)*a(n-3))*x^n = 0,
which proves the conjecture. - Petros Hadjicostas, Jul 15 2019

Cf. A000108, A106270.

c(x) = (1-sqrt(1-4*x))/(2*x);
my(x='x+O('x^35)); Vec(c(x)*sqrt(1 - 4*x)/(1 - x^2)) \\ Michel Marcus, Jul 16 2019

G.f.: c(x)*sqrt(1 - 4*x)/(1 - x^2), where c(x) is the g.f. of A000108.
a(n) = Sum_{k = 0..floor(n/2)} 2*0^(n-2k) - C(n-2k).
Conjecture: (n+1)*a(n) + 2*(1-2*n)*a(n-1) - (n+1)*a(n-2) + 2*(2*n-1)*a(n-3) = 0. - R. J. Mathar, Nov 09 2012

A014318 Convolution of Catalan numbers and powers of 2.

Original entry on oeis.org

1, 3, 8, 21, 56, 154, 440, 1309, 4048, 12958, 42712, 144210, 496432, 1735764, 6145968, 21986781, 79331232, 288307254, 1054253208, 3875769606, 14315659632, 53097586284, 197677736208, 738415086066

Offset: 0

N. J. A. Sloane

nonn

Binomial transform of A097332: (1, 2, 3, 5, 9, 18, 39, ...). - Gary W. Adamson, Aug 01 2011

Hankel transform is A087960. - Wathek Chammam, Dec 02 2011

Fung Lam, Table of n, a(n) for n = 0..1600
W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra and its Applications, Volume 436, Issue 7, 1 April 2012, Pages 2105-2116.

Cf. A000079, A000108, A087960, A097332, A106270.

A014318:= func< n | (&+[2^(n-j)*Catalan(j): j in [0..n]]) >;
[A014318(n): n in [0..40]]; // G. C. Greubel, Jan 09 2023

a:=proc(n) options operator, arrow: sum(2^(n-j)*binomial(2*j,j)/(j+1), j=0..n) end proc: seq(a(n), n=0..23); # Emeric Deutsch, Oct 16 2008

a[n_]:= a[n]= Sum[2^(n-j)*CatalanNumber[j], {j,0,n}];
Table[a[n], {n,0,40}] (* G. C. Greubel, Jan 09 2023 *)

def A014318(n): return sum(2^(n-j)*catalan_number(j) for j in range(n+1))
[A014318(n) for n in range(41)] # G. C. Greubel, Jan 09 2023

From Emeric Deutsch, Oct 16 2008: (Start)
G.f.: (1-sqrt(1-4*z))/(2*z*(1-2*z)).
a(n) = Sum_{j=0..n} (2^(n-j) * binomial(2*j,j)/(j+1)). (End)
a(n) = Sum_{j=0..n} abs(A106270(n, j)) * A000079(j). - Gary W. Adamson, Apr 02 2009
Recurrence: (n+1)*a(n) = 32*(2*n-7)*a(n-5) + 48*(8-3*n)*a(n-4) + 8*(16*n-29)*a(n-3) + 4*(13-14*n)*a(n-2) + 12*n*a(n-1), n>=5. - Fung Lam, Mar 09 2014
Asymptotics: a(n) ~ 2^(2n+1)/n^(3/2)/sqrt(Pi). - Fung Lam, Mar 21 2014
G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x) * A(x)^2. - Ilya Gutkovskiy, Nov 21 2021

A014140 Apply partial sum operator twice to Catalan numbers.

Original entry on oeis.org

1, 3, 7, 16, 39, 104, 301, 927, 2983, 9901, 33615, 116115, 406627, 1440039, 5147891, 18550588, 67310955, 245716112, 901759969, 3325067016, 12312494483, 45766188970, 170702447097, 638698318874, 2396598337975, 9016444758528, 34003644251233, 128524394659942, 486793096819011

Offset: 0

N. J. A. Sloane

nonn

From Alexander Adamchuk, Jul 04 2006: (Start)
p divides a(p-1) and a((p-3)/2) for primes in A002476.
p divides a((p-5)/2) for primes in A068228.
p^2 divides a(p^2-1) for all primes p > 3. (End)
Equals triangle A106270(unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Apr 02 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Partial sums of A014137.

Cf. A000108, A005043, A106270.

b:= proc(n) option remember; `if`(n<0, [0$2], (q->(f->
      [f[2]+q, q]+f)(b(n-1)))(binomial(2*n, n)/(n+1)))
    end:
a:= n-> b(n)[1]:
seq(a(n), n=0..28);  # Alois P. Heinz, Feb 13 2022

Table[Sum[Sum[(2k)!/k!/(k+1)!,{k,0,m}],{m,0,n}],{n,0,50}] Table[Sum[(n+1-k)*(2k)!/k!/(k+1)!,{k,0,n}],{n,0,50}] (* Alexander Adamchuk, Jul 04 2006 *)

sm(v)={my(s=vector(#v)); s[1]=v[1]; for(n=2, #v, s[n]=v[n]+s[n-1]); s; }
C(n)=binomial(2*n, n)/(n+1);
sm(sm(vector(66, n, C(n-1))))
/* Joerg Arndt, May 04 2013 */

1*C(n) + 2*C(n-1) + 3*C(n-2) + ... + (n+1-k)*C(k) + ... + n*C(1) + (n+1)*C(0), where C(k) = (2k)!/(k!*(k+1)!) is Catalan Number A000108(k). - Alexander Adamchuk, Jul 04 2006
From Alexander Adamchuk, Jul 04 2006: (Start)
a(n) = Sum_{m=0..n} Sum_{k=0..m} (2k)!/(k!*(k+1)!).
a(n) = Sum_{k=0..n} (n+1-k)*(2k)!/(k!*(k+1)!). (End)
G.f.: 1/(1-x)^2*(1-sqrt(1-4*x))/(2*x). - Vladimir Kruchinin, Oct 14 2016
a(n) = Sum_{k=0..n} binomial(n+2,k+2)*r(k), where r(k) are the Riordan numbers A005043. - Vladimir Kruchinin, Oct 14 2016
a(n) ~ 2^(2*n+4) / (9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2016

More terms from Alexander Adamchuk, Jul 04 2006

A106268 Number triangle T(n,k) = (-1)^(n-k)binomial(k-n, n-k) = (0^(n-k) + binomial(2(n-k), n-k))/2 if k <= n, 0 otherwise; Riordan array (1/(2-C(x)), x) where C(x) is g.f. for Catalan numbers A000108.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 10, 3, 1, 1, 35, 10, 3, 1, 1, 126, 35, 10, 3, 1, 1, 462, 126, 35, 10, 3, 1, 1, 1716, 462, 126, 35, 10, 3, 1, 1, 6435, 1716, 462, 126, 35, 10, 3, 1, 1, 24310, 6435, 1716, 462, 126, 35, 10, 3, 1, 1, 92378, 24310, 6435, 1716, 462, 126, 35, 10, 3, 1, 1

Offset: 0

Paul Barry, Apr 28 2005

Triangle includes A088218.

Inverse is A106270.

			Triangle (with rows n >= 0 and columns k >= 0) begins as follows:
    1;
    1,  1;
    3,  1,  1;
   10,  3,  1, 1;
   35, 10,  3, 1, 1;
  126, 35, 10, 3, 1, 1;
  ...
Production matrix begins:
    1, 1;
    2, 0, 1;
    5, 0, 0, 1;
   14, 0, 0, 0, 1;
   42, 0, 0, 0, 0, 1;
  132, 0, 0, 0, 0, 0, 1;
  429, 0, 0, 0, 0, 0, 0, 1;
  ... - _Philippe Deléham_, Oct 02 2014

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

Cf. A000108, A024718 (row sums), A088218, A106269 (diagonal sums), A106270.

A106268:= func< n,k | k eq n select 1 else (n-k+1)*Catalan(n-k)/2 >;
[A106268(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 10 2023

T[n_, k_]:= (-1)^(n-k)*Binomial[k-n, n-k];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 10 2023 *)

trg(nn) = {for (n=1, nn, for (k=1, n, print1(binomial(k-n,n-k)*(-1)^(n-k), ", ");); print(););} \\ Michel Marcus, Oct 03 2014

def A106268(n,k): return (1/2)*(0^(n-k) + (n-k+1)*catalan_number(n-k))
flatten([[A106268(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 10 2023

T(n, k) = (-1)^(n-k)*binomial(k-n, n-k).
T(n, k) = (1/2)*(0^(n-k) + binomial(2*(n-k), n-k)).
Sum_{k=0..n} T(n, k) = A024718(n) (row sums).
Sum_{k=0..floor(n/2)} T(n-k, k) = A106269(n) (diagonal sums).
Bivariate g.f.: Sum_{n, k >= 0} T(n,k)*x^n*y^k = (1/2) * (1/(1 - x*y)) * (1 + 1/sqrt(1 - 4*x)). - Petros Hadjicostas, Jul 15 2019

A343233 Triangle read by rows: Riordan triangle T = (1 - x*c(x), x), with the generating function c of A000108 (Catalan).

Original entry on oeis.org

1, -1, 1, -1, -1, 1, -2, -1, -1, 1, -5, -2, -1, -1, 1, -14, -5, -2, -1, -1, 1, -42, -14, -5, -2, -1, -1, 1, -132, -42, -14, -5, -2, -1, -1, 1, -429, -132, -42, -14, -5, -2, -1, -1, 1, -1430, -429, -132, -42, -14, -5, -2, -1, -1, 1

Offset: 0

Gary W. Adamson and Wolfdieter Lang, Apr 12 2021

As an unsigned sequence a(n) this is identical with the one of A155586(n+1), for n >= 0, but the triangle is not a simple signed version of A155586. See the formula.

This lower triangular Riordan matrix T of Toeplitz type is the inverse of the Riordan matrix (c(x), x) = |A106270|, also of Toeplitz type.

			The triangle matrix T begins:
  n/m     0    1    2   3   4   5   6   7   8   9 ...
  --------------------------------------------------
  0:      1
  1:     -1    1
  2:     -1   -1    1
  3:     -2   -1   -1   1
  4:     -5   -2   -1  -1   1
  5:    -14   -5   -2  -1  -1   1
  6:    -42  -14   -5  -2  -1  -1   1
  7:   -132  -42  -14  -5  -2  -1  -1   1
  8:   -429 -132  -42 -14  -5  -2  -1  -1   1
  9:  -1430 -429 -132 -42 -14  -5  -2  -1  -1   1
  ...

Cf. A106270 (unsigned), A155586.

The lower triangular matrix T satisfies: T = I - L^{tr}*|A106270|, also for the finite N X N version, with the unit matrix I and the lower triangular matrix L^{tr}(i, j) = delta_{i, j-1} (Kronecker symbol delta) with first lower diagonal of 1s and 0 otherwise.
T(n, n) = 1, and for T(n, m) = -C_{n - 1 - m } = - |A106270(n-1, m)|, for 0 <= m <= n-1, with the Catalan numbers C(n) = A000108, and T(n, m) = 0 for n < m.
O.g.f. of column m: (1/c(x))*x^m = (1 - x*c(x))*x^m (Riordan matrix of Toeplitz type), with the o.g.f. c of A000108.
O.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m, that is the o.g.f. of the triangle. G(z, x) = c(z)/(1 - x*z).

cp's OEIS Frontend

A106271 Row sums of number triangle A106270.

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A106272 Antidiagonal sums of number triangle A106270.

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A014318 Convolution of Catalan numbers and powers of 2.

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A014140 Apply partial sum operator twice to Catalan numbers.

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A106268 Number triangle T(n,k) = (-1)^(n-k)*binomial(k-n, n-k) = (0^(n-k) + binomial(2*(n-k), n-k))/2 if k <= n, 0 otherwise; Riordan array (1/(2-C(x)), x) where C(x) is g.f. for Catalan numbers A000108.

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A343233 Triangle read by rows: Riordan triangle T = (1 - x*c(x), x), with the generating function c of A000108 (Catalan).

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A106268 Number triangle T(n,k) = (-1)^(n-k)binomial(k-n, n-k) = (0^(n-k) + binomial(2(n-k), n-k))/2 if k <= n, 0 otherwise; Riordan array (1/(2-C(x)), x) where C(x) is g.f. for Catalan numbers A000108.