A055392 -id:A055392 - cp's OEIS frontend

A055113 Number of bracketings of 0^0^0^...^0, with n 0's, giving the result 0, with conventions that 0^0 = 1^0 = 1^1 = 1, 0^1 = 0.

0, 1, 0, 1, 1, 5, 11, 41, 120, 421, 1381, 4840, 16721, 59357, 210861, 759071, 2744393, 10000437, 36609977, 134750450, 498016753, 1848174708, 6882643032, 25715836734, 96365606679, 362102430069, 1364028272451, 5150156201026, 19486989838057, 73880877535315

Offset: 0

Jeppe Stig Nielsen, Jun 15 2000

Total number of bracketings of 0^0^...^0 is A000108(n-1) (this is Catalan's problem). So the number of bracketings giving 1 is A000108(n-1) - a(n).
Also bracketings of f => f => ... => f where f is "false" and "=>" is implication.
Self-convolution yields A187430. - Paul D. Hanna, May 31 2015
Also, number of nonnegative walks of n steps with step sizes 1 and 2, starting at 0 and ending at 1. - Andrew Howroyd, Dec 23 2017
Series reversion is related to A001006. - F. Chapoton, Jul 14 2021

			Number of bracketings of 0^0^0^0^0^0 giving 0 is 11, so a(6) = 11.
From _Jon E. Schoenfield_, Dec 24 2017: (Start)
The 11 ways of parenthesizing 0^0^0^0^0^0 to obtain 0 are
0^(0^(0^((0^0)^0))) = 0^(0^(0^(1^0))) = 0^(0^(0^1)) = 0^(0^0) = 0^1 = 0;
0^((0^0)^(0^(0^0))) = 0^(1^(0^1)) = 0^(1^0) = 0^1 = 0;
0^((0^0)^((0^0)^0)) = 0^(1^(1^0)) = 0^(1^1) = 0^1 = 0;
0^(((0^0)^0)^(0^0)) = 0^((1^0)^1) = 0^(1^1) = 0^1 = 0;
0^((0^(0^(0^0)))^0) = 0^((0^(0^1))^0) = 0^((0^0)^0) = 0^(1^0) = 0^1 = 0;
0^((0^((0^0)^0))^0) = 0^((0^(1^0))^0) = 0^((0^1)^0) = 0^(0^0) = 0^1 = 0;
0^(((0^0)^(0^0))^0) = 0^((1^1)^0) = 0^(1^0) = 0^1 = 0;
0^(((0^(0^0))^0)^0) = 0^(((0^1)^0)^0) = 0^((0^0)^0) = 0^(1^0) = 0^1 = 0;
0^((((0^0)^0)^0)^0) = 0^(((1^0)^0)^0) = 0^((1^0)^0) = 0^(1^0) = 0^1 = 0;
(0^(0^0))^((0^0)^0) = (0^1)^(1^0) = 0^1 = 0;
(0^((0^0)^0))^(0^0) = (0^(1^0))^1. (End)

Thanks to Soren Galatius Smith, Jesper Torp Kristensen et al.

Alois P. Heinz, Table of n, a(n) for n = 0..1671 (terms n = 1..200 from T. D. Noe)
E. A. Bender and S. G. Williamson, Foundations of Combinatorics with Applications (see Chap. 11, Example 11.3, pp. 312-313 and Example 11.31, pp. 351-352).
V. Čačić and V. Kovač, On the fraction of IL formulas that have normal forms, arXiv preprint arXiv:1309.3408 [math.LO], 2013-2015.
D. G. Rogers, Comments on A111160, A055113 and A006013
Wikipedia, Zero to the power of zero

Column k=2 of A185286.

Cf. A000108, A001006, A055392, A055395, A006632, A111160, A187430, A296619, A306668.

a:= proc(n) option remember; `if`(n<3, n*(2-n),
      ((n-1)*(115*n^3-689*n^2+1332*n-840) *a(n-1)
       +(8*n-20)*(5*n^3+12*n^2-113*n+126) *a(n-2)
       -36*(n-3)*(5*n-8)*(2*n-5)*(2*n-7)  *a(n-3))
      /((2*(2*n-1))*(5*n-13)*n*(n-1)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Mar 04 2019

Rest[ CoefficientList[ Series[(-1 - Sqrt[1 - 4x] + Sqrt[2]Sqrt[1 + Sqrt[1 - 4x] + 6x])/4, {x, 0, 28}], x]] (* Robert G. Wilson v, Oct 28 2005 *)
a[n_] := (-1)^(n+1)*Binomial[2n-1, n]*HypergeometricPFQ[{1-n, (n+1)/2, n/2}, {n, n+1}, 4]/(2n-1);
Array[a, 27] (* Jean-François Alcover, Dec 26 2017, after Vladimir Kruchinin *)

a(n):= sum(binomial(2*j+n-1,j+n-1)*(-1)^(n-j-1)*binomial(2*n-1,j+n), j,0,n-1)/(2*n-1); /* Vladimir Kruchinin, May 10 2011 */

a(n)={sum(j=0, n-1, binomial(2*j+n-1, j+n-1)*(-1)^(n-j-1)*binomial(2*n-1, j+n))/(2*n-1)} \\ Andrew Howroyd, Dec 23 2017

first(n) = x='x+O('x^(n+1)); Vec(-((1 - 4*x)^(1/2) + 1)/4 + (2 + 2*(1 - 4*x)^(1/2) + 12*x)^(1/2)/4) \\ Iain Fox, Dec 23 2017

G.f.: - 1/4 - (1/4)*(1 - 4*x)^(1/2) + (1/4)*(2 + 2*(1 - 4*x)^(1/2) + 12*x)^(1/2).
The ratio a(n)/A000108(n-1) converges to (5-sqrt(5))/10 as n->oo.
a(n) = (Sum_{j=0..n-1} binomial(2*j+n-1, j+n-1)*(-1)^(n-j-1)*binomial(2*n-1, j+n))/(2*n-1). - Vladimir Kruchinin, May 10 2011
a(n) ~ (1-1/sqrt(5))*2^(2*n-3)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 09 2013
D-finite with recurrence: 2*n*(2*n-1)*(n-1)*a(n) -(n-1)*(19*n^2-60*n+48)*a(n-1) +(-31*n^3+173*n^2-346*n+264)*a(n-2) +4*(2*n-7)*(17*n^2-83*n+102)*a(n-3) +36*(2*n-7)*(2*n-9)*(n-4)*a(n-4)=0. - R. J. Mathar, Feb 20 2020
a(n) + A111160(n) = A000108(n). - F. Chapoton, Jul 14 2021

a(0)=0 prepended by Alois P. Heinz, Mar 04 2019

A055395 Number of bracketings of 0#0#0#...#0 giving result 0, where 0#0 = 0#1 = 1#0 = 1, 1#1 = 0.

Original entry on oeis.org

1, 0, 0, 1, 4, 12, 36, 116, 392, 1350, 4696, 16500, 58572, 209824, 757440, 2752185, 10057636, 36943044, 136319052, 505086728, 1878395920, 7009239644, 26235435248, 98475145476, 370584275964, 1397918543552, 5284861554816

Offset: 1

Jeppe Stig Nielsen, Jun 24 2000

nonn

Operation # can be interpreted as NOT AND. The ratio a(n)/A000108(n-1) converges to (2-sqrt(2))/2. Thanks to Soren Galatius Smith

G. C. Greubel, Table of n, a(n) for n = 1..500

Cf. A055113, A055392, A273958.

f[x_] := (1 - Sqrt[1 - 4*x])/2; CoefficientList[Series[(1 + 2*f[x] - Sqrt[1 + 4*(f[x])^2])/(2*x), {x, 0, 50}], x] (* G. C. Greubel, Jun 10 2016 *)

G.f.: 1 - (1/2)*(1 - 4*x)^(1/2) - (1/2)*(3 - 2*(1 - 4*x)^(1/2) - 4*x)^(1/2).
G.f.: (1 + 2*C(x) - sqrt(1 + 4*C(x)^2))/2, where C(x) = (1 - sqrt(1 - 4*x))/2 is the g.f. of the Catalan numbers (A000108). - Paul D. Hanna, Jun 10 2016
G.f. A(x) satisfies: A(x) = x + (A(x) - C(x))^2, where C(x) = x + C(x)^2 is a g.f. of the Catalan numbers (A000108). - Paul D. Hanna, Jun 11 2016

A112519 Riordan array (1, xc(x)c(-x*c(x))), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 4, 0, 1, 0, 12, 2, 6, 0, 1, 0, 14, 28, 3, 8, 0, 1, 0, 100, 32, 48, 4, 10, 0, 1, 0, 180, 249, 54, 72, 5, 12, 0, 1, 0, 990, 440, 455, 80, 100, 6, 14, 0, 1, 0, 2310, 2552, 792, 726, 110, 132, 7, 16, 0, 1, 0, 10920, 5876, 4836, 1248, 1070, 144, 168, 8, 18, 0, 1

Offset: 0

Paul Barry, Sep 09 2005

Row sums are A112520. Second column is essentially A055392. Inverse is A112517. Riordan array product (1, x*c(x))*(1, x*c(-x)).

			Triangle begins
  1;
  0,   1;
  0,   0,   1;
  0,   2,   0,   1;
  0,   1,   4,   0,  1;
  0,  12,   2,   6,  0,   1;
  0,  14,  28,   3,  8,   0,  1;
  0, 100,  32,  48,  4,  10,  0,  1;
  0, 180, 249,  54, 72,   5, 12,  0, 1;
  0, 990, 440, 455, 80, 100,  6, 14, 0, 1;

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

Cf. A000108, A055392, A112517, A112520.

A112519:= func< n,k | n eq 0 and k eq 0 select 1 else (k/n)*(&+[(-1)^j*Binomial(2*n-k-j-1, n-k-j)*Binomial(2*j+k-1, j): j in [0..n-k]]) >;
[A112519(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Jan 12 2022

(* First program *)
c[x_]:= (1 - Sqrt[1-4x])/(2x);
(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, # c[#] c[-# c[#]]&, 12] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
(* Second program *)
T[n_, k_]:= If[k==n, 1, (k/n)*Binomial[2*n-k-1, n-1]*HypergeometricPFQ[{k-n, k/2, (1+k)/2}, {k-2*n+1, k}, -4]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Jan 12 2022 *)

@CachedFunction
def A112519(n,k):
    if (k==n): return 1
    else: return (k/n)*sum( (-1)^j*binomial(2*n-k-j-1, n-k-j)*binomial(2*j+k-1, j) for j in (0..n-k) )
flatten([[A112519(n,k) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Jan 12 2022

Riordan array (1, (sqrt(3-2*sqrt(1-4*x)) - 1)/2).
T(n, k) = (k/n)*Sum_{j=0..n} (-1)^(j-k)*C(2*n-j-1, n-j)*C(2*j-k-1, j-k), with T(0, 0) = 1.
T(n, k) = (k/n)*binomial(2*n-k-1, n-1)*Hypergoemetric3F2([k-n, k/2, (1+k)/2], [k-2*n+1, k], -4), with T(0, 0) = 1. - G. C. Greubel, Jan 12 2022

A112521 Sequence related to NOR bracketings.

Original entry on oeis.org

0, 1, 0, 6, 4, 60, 84, 700, 1440, 8910, 23100, 120120, 360360, 1684956, 5552064, 24302520, 85101456, 357502860, 1302562404, 5333981796, 19947127200, 80408748420, 305922388200, 1221485157360, 4701015343440, 18664243014300

Offset: 0

Paul Barry, Sep 09 2005

Conjecture: Starting with n=1, a(n) is the main diagonal of the array defined as T(1,1) = 1, T(i,j) = 0 if i<1 or j<1, T(n,k) = T(n,k-2) + T(n,k-1) -2*T(n-1,k-1) + T(n-1,k) + T(n-2,k). - Gerald McGarvey, Oct 07 2008

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A055392.

a[n_]:= Sum[(-1)^j*Binomial[2*j, j]*Binomial[2*n-j-2, n-j-1], {j,0,n-1}];
Table[a[n], {n,0,30}] (* G. C. Greubel, Jan 11 2022 *)

a(n) = sum(j=0,n, (-1)^(j-1)*binomial(2*n-j-1, n-j)*binomial(2*(j-1), j-1)); \\ Michel Marcus, Aug 19 2014

def a(n): return n if (n<2) else binomial(2*n-2, n-1)*simplify( hypergeometric([-(n-1), 1/2], [2-2*n], -4) )
[a(n) for n in (0..30)] # G. C. Greubel, Jan 11 2022

a(n) = Sum_{j=0..n} (-1)^(j-1)*C(2*n-j-1, n-j)*C(2*(j-1), j-1). - corrected by Peter Bala, Aug 19 2014
a(n) = n*A055392(n), n>1.
a(n) = binomial(2*n-2, n-1)*Hypergeometric([-(n-1), 1/2], [2-2*n], -4) with a(0) = 0, a(1) = 1. - G. C. Greubel, Jan 11 2022

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A055113 Number of bracketings of 0^0^0^...^0, with n 0's, giving the result 0, with conventions that 0^0 = 1^0 = 1^1 = 1, 0^1 = 0.

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A055395 Number of bracketings of 0#0#0#...#0 giving result 0, where 0#0 = 0#1 = 1#0 = 1, 1#1 = 0.

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A112519 Riordan array (1, x*c(x)*c(-x*c(x))), c(x) the g.f. of A000108.

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A112521 Sequence related to NOR bracketings.

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A112519 Riordan array (1, xc(x)c(-x*c(x))), c(x) the g.f. of A000108.