A001405 -id:A001405 - cp's OEIS frontend

A057977 GCD of consecutive central binomial coefficients: a(n) = gcd(A001405(n+1), A001405(n)).

1, 1, 1, 3, 2, 10, 5, 35, 14, 126, 42, 462, 132, 1716, 429, 6435, 1430, 24310, 4862, 92378, 16796, 352716, 58786, 1352078, 208012, 5200300, 742900, 20058300, 2674440, 77558760, 9694845, 300540195, 35357670, 1166803110, 129644790, 4537567650

Offset: 0

Labos Elemer, Nov 13 2000

The numbers can be seen as a generalization of the Catalan numbers, extending A000984(n)/(n+1) to A056040(n)/(floor(n/2)+1). They can also be seen as composing the aerated Catalan numbers A126120 with the aerated complementary Catalan numbers A138364. (Thus the name 'extended Catalan numbers' might be apt for this sequence.) - Peter Luschny, May 03 2011
a(n) is the number of lattice paths from (0,0) to (n,0) that do not go below the x-axis and consist of steps U=(1,1), D=(1,-1) and maximally one step H=(1,0). - Alois P. Heinz, Apr 17 2013
Equal to A063549 (see comments in that sequence). - Nathaniel Johnston, Nov 17 2014
a(n) can be computed with ballot numbers without multiplications or divisions, see Maple program. - Peter Luschny, Feb 23 2019

			This GCD equals A001405(n) for the smaller odd number gcd(C(12,6), C(11,5)) = gcd(924,462) = 462 = C(11,5).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Peter Luschny, Die schwingende Fakultät und Orbitalsysteme, August 2011.
Peter Luschny, The lost Catalan numbers.

Cf. A001405, A063549.

Bisections are A000108 and A001700.

A057977_ogf := proc(z) b := z -> (z-1)/(2*z^2);
(2 + b(z))/sqrt(1-4*z^2) - b(z) end:
seq(coeff(series(A057977_ogf(z),z,n+3),z,n), n = 0..35);
A057977_rec := n -> `if`(n=0, 1, A057977_rec(n-1)*n^modp(n,2)
*(4/(n+2))^modp(n+1,2));
A057977_int := proc(n) int((x^(2*n-1)*((4-x)^2/x)^cos(Pi*n))^(1/4),x=0..4)/(2*Pi); round(evalf(%)) end:
A057977 := n -> (n!/iquo(n,2)!^2) / (iquo(n,2)+1):
seq(A057977(n), n=0..35); # Peter Luschny, Apr 30 2011
b := proc(p, q) option remember; local S;
   if p = 0 and q = 0 then return 1 fi;
   if p < 0 or  p > q then return 0 fi;
   S := b(p-2, q) + b(p, q-2);
   if type(q, odd) then S := S + b(p-1, q-1) fi;
   S end:
seq(b(n, n), n=0..35); # Peter Luschny, Feb 23 2019

a[n_] := n! / (Quotient[n, 2]!^2 * (Quotient[n, 2]+1)); Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Feb 03 2012, after Peter Luschny *)

a(n)=if(n<0,0,(n+n%2)!/(n\2+1)!/(n\2+n%2)!/(1+n%2))
a(n)=n!/(n\2)!^2/(n\2+1) \\ Charles R Greathouse IV, May 02 2011

def A057977():
    x, n = 1, 1
    while True:
        yield x
        m = n if is_odd(n) else 4/(n+2)
        x *= m
        n += 1
a = A057977(); [next(a) for i in range(36)]   # Peter Luschny, Oct 21 2013

G.f.: (4*x^2+x-1+(1-x)*sqrt(1-4*x^2))/(2*sqrt(1-4*x^2)*x^2). E.g.f.: (1+1/x)*BesselI(1, 2*x). - Vladeta Jovovic, Jan 19 2004
From Peter Luschny, Apr 30 2011: (Start)
Recurrence: a(0) = 1 and a(n) = a(n-1)*n^[n odd]*(4/(n+2))^[n even] for n > 0.
Asymptotic formula: Let [n even] = 1 if n is even, 0 otherwise. Let N := n+1+[n even]. Then a(n) ~ 2^N /((n+1)^[n even]*sqrt(Pi*(2*N+1))).
Integral representation: a(n) = (1/(2*Pi))*Int_{x=0..4}(x^(2*n-1)* ((4-x)^2/x)^cos(Pi*n))^(1/4) (End)
E.g.f.: U(0) where U(k)= 1 + x/(1 - x/(x + (k+1)*(k+2)//U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
From R. J. Mathar, Sep 16 2016: (Start)
D-finite with recurrence: (n+2)*a(n) - n*a(n-1) + 4*(-2*n+1)*a(n-2) + 4*(n-1)*a(n-3) + 16*(n-3)*a(n-4) = 0.
D-finite with recurrence: -(n+2)*(n^2-5)*a(n) + 4*(-2*n-1)*a(n-1) + 4*(n-1)*(n^2+2*n-4)*a(n-2) = 0. (End)
Sum_{n>=0} 1/a(n) = 8/3 + 8*Pi/(9*sqrt(3)). - Amiram Eldar, Aug 20 2022

A054336 A convolution triangle of numbers based on A001405 (central binomial coefficients).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 6, 10, 9, 4, 1, 10, 22, 22, 14, 5, 1, 20, 44, 54, 40, 20, 6, 1, 35, 93, 123, 109, 65, 27, 7, 1, 70, 186, 281, 276, 195, 98, 35, 8, 1, 126, 386, 618, 682, 541, 321, 140, 44, 9, 1, 252, 772, 1362, 1624, 1440, 966, 497, 192, 54, 10, 1

Offset: 0

Wolfdieter Lang, Mar 13 2000

T(n,k) is the number of 2-Motzkin paths (i.e., Motzkin paths with blue and red level steps) with no level steps at positive height and having k blue level steps. Example: T(4,2)=9 because, denoting U=(1,1), D=(1,-1), B=blue (1,0), R=red (1,0), we have BBRR, BRBR, BRRB, RBBR, RBRB, RRBB, BBUD, BUDB, and UDBB. - Emeric Deutsch, Jun 07 2011
In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(1-(1+x)*z-z^2*c(z^2)), with c(x) the g.f. for Catalan numbers A000108.
Column sequences: A001405, A045621.
Riordan array (f(x), x*f(x)), f(x) the g.f. of A001405. - Philippe Deléham, Dec 08 2009
From Paul Barry, Oct 21 2010: (Start)
Riordan array ((sqrt(1+2x) - sqrt(1-2x))/(2x*sqrt(1-2x)), (sqrt(1+2x)-sqrt(1-2x))/(2*sqrt(1-2x))),
inverse of Riordan array ((1+x)/(1+2x+2x^2), x(1+x)/(1+2x+2x^2)) (A181472). (End)

			Fourth row polynomial (n=3): p(3,x)= 3 + 5*x + 3*x^2 + x^3.
From _Paul Barry_, Oct 21 2010: (Start)
Triangle begins
   1;
   1,  1;
   2,  2,   1;
   3,  5,   3,   1;
   6, 10,   9,   4,  1;
  10, 22,  22,  14,  5,  1;
  20, 44,  54,  40, 20,  6, 1;
  35, 93, 123, 109, 65, 27, 7, 1;
Production matrix is
   1,  1;
   1,  1,  1;
  -1,  1,  1,  1;
   1, -1,  1,  1,  1;
  -1,  1, -1,  1,  1,  1;
   1, -1,  1, -1,  1,  1,  1;
  -1,  1, -1,  1, -1,  1,  1, 1;
   1, -1,  1, -1,  1, -1,  1, 1, 1;
  -1,  1, -1,  1, -1,  1, -1, 1, 1, 1; (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A001405, A035324, A054335.

Row sums: A054341.

A053121:= function(n,k)
    if ((n-k+1) mod 2)=0 then return 0;
    else return (k+1)*Binomial(n+1, Int((n-k)/2))/(n+1);
    fi;
  end;
T:= function(n,k)
    return Sum([k..n], j-> Binomial(j,k)*A053121(n,j));
  end;
Flat(List([0..10], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Jul 21 2019

A053121:= func< n,k | ((n-k+1) mod 2) eq 0 select 0 else (k+1)*Binomial(n+1, Floor((n-k)/2))/(n+1) >;
T:= func< n,k | (&+[Binomial(j,k)*A053121(n,j): j in [k..n]]) >;
[T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Jul 21 2019

c[n_, j_] /; n < j || OddQ[n - j] = 0; c[n_, j_] = (j + 1) Binomial[n + 1, (n - j)/2]/(n + 1); t[n_, k_] := Sum[c[n, j]*Binomial[j, k], {j, 0, n}]; Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]][[;; 66]] (* Jean-François Alcover, Jul 13 2011, after Philippe Deléham *)

A053121(n,k) = if((n-k+1)%2==0, 0, (k+1)*binomial(n+1, (n-k)\2)/(n+1) );
T(n,k) = sum(j=k,n, A053121(n,j)*binomial(j,k));
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 21 2019

def A053121(n, k):
    if (n-k+1) % 2==0: return 0
    else: return (k+1)*binomial(n+1, ((n-k)//2))/(n+1)
def T(n,k): return sum(binomial(j,k)*A053121(n,j) for j in (k..n))
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 21 2019

G.f. for column m: cbi(x)*(x*cbi(x))^m, with cbi(x) := (1+x*c(x^2))/sqrt(1-4*x^2) = 1/(1-x-x^2*c(x^2)), where c(x) is the g.f. for Catalan numbers A000108.
T(n,k) = Sum_{j>=0} A053121(n,j)*binomial(j,k). - Philippe Deléham, Mar 30 2007
T(n,k) = T(n-1,k-1) + T(n-1,l) + Sum_{j>=0} T(n-1,k+1+j)*(-1)^j. - Philippe Deléham, Feb 23 2012

A018224 a(n) = binomial(n, floor(n/2))^2 = A001405(n)^2.

Original entry on oeis.org

1, 1, 4, 9, 36, 100, 400, 1225, 4900, 15876, 63504, 213444, 853776, 2944656, 11778624, 41409225, 165636900, 590976100, 2363904400, 8533694884, 34134779536, 124408576656, 497634306624, 1828114918084, 7312459672336, 27043120090000, 108172480360000, 402335398890000

Offset: 0

N. J. A. Sloane

a(n) is also the number of rooted two-vertex (or, dually, two-face) regular planar maps of valency n+1. - Valery A. Liskovets, Oct 19 2005
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic), then a(n)=(-1)^n*E[(tr(A^4))^n]. - Andrew V. Sutherland, Apr 01 2008
Number of square lattice walks with unit steps in all four directions (NSWE), starting at the origin, ending on the y-axis, and never going below the x-axis. Row sums of A378061. - Peter Luschny, Dec 08 2024

			The 9 lattice walks defined in the comments: 'NNN', 'NNS', 'NSN', 'NWE', 'NEW', 'WNE', 'WEN', 'ENW', 'EWN'.

Stephanie Anderson, Table for n, a(n) for n = 0..200
Alin Bostan, Computer Algebra for Lattice Path Combinatorics, Slides, Séminaire de Combinatoire Ph. Flajolet, March 28 2013.
Alin Bostan, Computer Algebra for Lattice Path Combinatorics, Synthesis, Séminaire de Combinatoire Ph. Flajolet, June 06 2013.
Alin Bostan, Calcul Formel pour la Combinatoire des Marches [The text is in English], Habilitation à Diriger des Recherches, Laboratoire d'Informatique de Paris Nord, Université Paris 13, December 2017.
Alin Bostan, Frédéric Chyzak, Mark van Hoeij, Manuel Kauers, and Lucien Pech, Hypergeometric expressions for generating functions of walks with small steps in the quarter plane, Eur. J. Comb. 61, 242-275 (2017).
M. Bousquet, G. Labelle and P. Leroux, Enumeration of planar two-face maps, Discrete Math., vol. 222 (2000), 1-25.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Helmut Prodinger, Two New Identities Involving the Catalan Numbers: A classical approach, arXiv:1911.07604 [math.CO], 2019.

Bisections are A002894 and A060150.

Cf. A000108, A001405, A056040, A113182, A138545, A378061.

s := x -> (1+x)*EllipticK(x)/(x*Pi/2)-1/x:
seq(4^n*coeff(series(s(x),x,n+2),x,n),n=0..23); # Peter Luschny, Oct 14 2015

(* Note that Mathematica uses a different definition of the EllipticK function. *)
CoefficientList[Series[(-Pi + (2 + 8 x) EllipticK[16 x^2])/(4 Pi x), {x,0,23}], x] (* Peter Luschny, Oct 14 2015 *)
Table[Binomial[n,Floor[n/2]]^2,{n,0,30}] (* Harvey P. Dale, Dec 02 2022 *)

vector(50, n, n--; binomial(n, n\2)^2) \\ Altug Alkan, Oct 14 2015

E.g.f.: BesselI(0, 2*x)*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Vladeta Jovovic, Jun 12 2005
G.f. (1+1/(4*x))*hypergeom([1/2, 1/2],[1],16*x^2)-1/(4*x). - Mark van Hoeij, Oct 13 2009
a(n) = (n!/(floor(n/2)!*floor((n+1)/2)!))^2. - Peter Luschny, Apr 29 2014
a(n) = A056040(n) * A056040(n+1) / (n+1). - Peter Luschny, Apr 29 2014
a(n) = 4^n*[x^n]((1+x)*EllipticK(x)/(x*Pi/2)-1/x). - Peter Luschny, Oct 14 2015
a(n) ~ 4^n*((2*n+3)/(2*n+1))^((-1)^n/2)/((n+1)*Pi/2). - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=0..n} (-1)^k*binomial(n,k)*C(k)*binomial(2*n-2*k,n-k) where C(k) are Catalan numbers (A000108), see Prodinger. - Michel Marcus, Nov 19 2019
From Peter Bala, Jul 03 2023: (Start)
Right hand side of the binomial sum identity (1/2)*Sum_{k = 0..n+1} (-1)^k*4^(n+1-k)*binomial(n+1,k)*binomial(n+k,k)*binomial(2*k,k) = a(n).
a(n) = (1/2)*4^(n+1) * hypergeom([n+1, -n-1, 1/2], [1, 1], 1).
P-recursive:
(2*n - 1)*(n + 1)^2*a(n) = 4*(2*n^2 - 1)*a(n-1) + 16*(2*n + 1)*(n - 1)^2*a(n-2) with a(0) = a(1) = 1. (End)

A034974 Number of divisors of binomial(n, floor(n/2)), the terms of A001405.

Original entry on oeis.org

1, 2, 2, 4, 4, 6, 4, 8, 12, 18, 16, 24, 24, 32, 24, 48, 32, 48, 32, 48, 96, 128, 64, 96, 144, 192, 288, 384, 384, 480, 192, 384, 512, 768, 768, 1152, 1152, 1536, 1536, 2304, 1536, 2048, 1536, 2048, 3072, 3840, 2304, 3456, 3456, 4608, 4608, 6144, 3072, 3840, 6144

Offset: 1

Labos Elemer

nonn

			a(59) = d(C(59,29)) = d(59132290782430712) = 8192 = 4*2^11.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001405, A034973.

Table[DivisorSigma[0,Binomial[n,Floor[n/2]]],{n,60}] (* Harvey P. Dale, Jun 12 2012 *)

a(n) = numdiv(binomial(n, n\2)); \\ Michel Marcus, Feb 25 2014

a(n) = A000005(A001405(n)). - Michel Marcus, Feb 25 2014

A034973 Number of distinct prime factors in central binomial coefficients C(n, floor(n/2)), the terms of A001405.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 12, 12, 13, 13, 12, 12, 12, 12, 12, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 14, 14, 15, 15, 15, 15, 16

Offset: 1

Labos Elemer

Sequence is not monotonic. E.g., a(44)=10, a(45)=9 and a(46)=10. The number of prime factors of n! is pi(n), but these numbers are lower.

Prime factors are counted without multiplicity. - Harvey P. Dale, May 20 2012

			a(25) = omega(binomial(25,12)) = omega(5200300) = 6 because the prime factors are 2, 5, 7, 17, 19, 23.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A001405, A034974, A067434.

Table[PrimeNu[Binomial[n,Floor[n/2]]],{n,90}] (* Harvey P. Dale, May 20 2012 *)

a(n)=omega(binomial(n,n\2)) \\ Charles R Greathouse IV, Apr 29 2015

A080383 Number of j (0 <= j <= n) such that the central binomial coefficient C(n,floor(n/2)) = A001405(n) is divisible by C(n,j).

Original entry on oeis.org

1, 2, 3, 4, 3, 6, 3, 6, 3, 6, 3, 6, 7, 10, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 8, 3, 6, 3, 6, 7, 10, 3, 6, 3, 6, 3, 8, 3, 6, 5, 10, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 7, 10, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 7, 10, 3, 6, 3, 6, 7, 10, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6

Offset: 0

Labos Elemer, Mar 12 2003

nonn

			For n <= 500 only a few values of a(n) arise: {1,2,3,4,5,6,7,8,10,11,14}.
From _Jon E. Schoenfield_, Sep 15 2019: (Start)
a(n)=1 occurs only at n=0.
a(n)=2 occurs only at n=1.
a(n)=3 occurs for all even n > 0 such that C(n,j) divides C(n,n/2) only at j = 0, n/2, and n. (This is the case for about 4/9 of the first 100000 terms, and there appear to be nearly as many terms for which a(n)=6.)
a(n)=4 occurs only at n=3.
For n <= 100000, the only values of a(n) that occur are 1..16, 18, 19, 22, 23, and 26.
   k | Indices n (up to 100000) at which a(n)=k
  ---+-------------------------------------------------------
   1 | 0
   2 | 1
   3 | 2, 4, 6, 8, 10, 14, 16, 18, 20, 22, 24, ...
   4 | 3
   5 | 40, 176, 208, 480, 736, 928, 1248, 1440, ... (A327430)
   6 | 5, 7, 9, 11, 15, 17, 19, 21, 23, 27, 29, ... (A080384)
   7 | 12, 30, 56, 84, 90, 132, 154, 182, 220, ...  (A080385)
   8 | 25, 37, 169, 199, 201, 241, 397, 433, ...    (A080386)
   9 | 1122, 1218, 5762, 11330, 12322, 15132, ...   (A327431)
  10 | 13, 31, 41, 57, 85, 91, 133, 155, 177, ...   (A080387)
  11 | 420, 920, 1892, 1978, 2444, 2914, 3198, ...
  12 | 1103, 1703, 2863, 7773, 10603, 15133, ...
  13 | 12324, 37444
  14 | 421, 921, 1123, 1893, 1979, 1981, 2445, ...
  15 | 4960, 6956, 13160, 16354, 18542, 24388, ...
  16 | 11289, 16483, 36657, 62653, 89183
  17 |
  18 | 4961, 6957, 12325, 13161, 16355, 18543, ...
  19 | 16356, 88510, 92004
  20 |
  21 |
  22 | 16357, 88511, 90305, 92005
  23 | 90306
  24 |
  25 |
  26 | 90307
(End)

Vaclav Kotesovec, Table of n, a(n) for n = 0..100000 (first 1000 terms from Vincenzo Librandi, terms 1001..9999 from David A. Corneth)

Cf. A001405, A000225, A057977, A022292, A020475, A067348, A042996.
Cf. A327430, A080384, A080385, A080386, A327431, A080387.
Cf. A080393.

[#[j:j in [0..n]| Binomial(n,Floor(n/2)) mod Binomial(n,j) eq 0]:n in [0..100]]; // Marius A. Burtea, Sep 15 2019

Table[Count[Table[IntegerQ[Binomial[n, Floor[n/2]]/Binomial[n, j]], {j, 0, n}], True], {n, 0, 500}] (* adapted by Vincenzo Librandi, Jul 29 2017 *)

a(n) = my(b=binomial(n, n\2)); sum(i=0, n, (b % binomial(n, i)) == 0); \\ Michel Marcus, Jul 29 2017

a(n) = {if(n==0, return(1)); my(bb = binomial(n, n\2), b = n); res = 2 + !(n%2) + 2 * (n>2 && n%2 == 1); for(i = 2, (n-1)\2, res += 2*(bb%b==0); b *= (n + 1 - i) / i); res} \\ David A. Corneth, Jul 29 2017

Edited by Dean Hickerson, Mar 14 2003

Offset corrected by David A. Corneth, Jul 29 2017

A056175 Number of nonunitary prime divisors of the central binomial coefficient C(n, floor(n/2)) (A001405).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 2, 3, 3, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 3, 3, 2, 3, 3, 3, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 0, 1, 1, 1, 2, 2, 3, 3, 1, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 3, 3, 3, 3, 4, 3, 3, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Labos Elemer, Jul 27 2000

nonn

Number of prime divisors of the largest square dividing A001405(n). (A prime divisor is nonunitary iff its exponent exceeds 1.)

			For n=10, binomial(10, 5) = 252 = 2*2*3*3*7 has 3 prime divisors of which only one, p=7, is unitary, while 2 and 3 are not. So a(10)=2.
For n=256, binomial(256, 128) also has only 2 prime divisors (3 and 13) whose exponents exceed 1 (4 and 2, respectively), thus a(256)=2.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A001221, A001405, A034444, A034973, A039593, A056057, A056173.

Table[Count[FactorInteger[Binomial[n, Floor[n/2]]][[All, -1]], e_ /; e > 1], {n, 105}] (* Michael De Vlieger, Mar 05 2017 *)

a(n)=omega(core(binomial(n, n\2), 1)[2]) \\ Charles R Greathouse IV, Mar 09 2017

a(n) = A001221(A000188(A001405(n))).

a(n) = A001221(A056057(n)).

Edited by Jon E. Schoenfield, Mar 05 2017

A056173 Number of unitary prime divisors of central binomial coefficient C(n, floor(n/2)) (A001405).

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 3, 2, 1, 4, 3, 3, 3, 3, 4, 5, 4, 5, 4, 5, 5, 6, 5, 4, 4, 3, 3, 5, 5, 6, 7, 7, 6, 8, 7, 7, 7, 9, 8, 9, 9, 9, 9, 6, 6, 8, 7, 7, 7, 7, 7, 8, 8, 11, 11, 12, 12, 11, 11, 11, 11, 10, 11, 13, 12, 13, 12, 12, 12, 14, 13, 13, 13, 13, 13, 11, 11, 14, 13, 12, 12, 14, 14, 13, 13, 13

Offset: 1

Labos Elemer, Jul 27 2000

nonn

A prime divisor is unitary iff its exponent equals 1.

			For n = 10: binomial(10,5) = 252 = 2*2*3*3*7 has 3 prime factors of which only one, p = 7, is unitary. So a(10) = 1.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A001221, A001405, A034973, A056169, A056175.

Array[Function[k, Count[FactorInteger[k][[All, 1]], ?(CoprimeQ[#, k/#] &)]]@ Binomial[#, Floor[#/2]] &, 87]  (* _Michael De Vlieger, Oct 26 2017 *)

a(n) = vecsum(apply(x -> if(x > 1, 0, 1), factor(binomial(n, n\2))[,2])); \\ Amiram Eldar, Jul 22 2024

a(n) = A056169(A001405(n)). - Michel Marcus, Oct 27 2017 [corrected by Amiram Eldar, Jul 22 2024]

A205573 Array M read by antidiagonals in which successive rows evidently converge to A001405 (central binomial coefficients).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 3, 5, 1, 1, 1, 2, 3, 6, 8, 1, 1, 1, 2, 3, 6, 10, 13, 1, 1, 1, 2, 3, 6, 10, 19, 21, 1, 1, 1, 2, 3, 6, 10, 20, 33, 34, 1, 1, 1, 2, 3, 6, 10, 20, 35, 61, 55, 1, 1, 1, 2, 3, 6, 10, 20, 35, 69, 108, 89, 1

Offset: 0

L. Edson Jeffery, Jan 29 2012

CONJECTURE 1. Let M(n,k) (n,k >= 0) denote the entry in row n and column k of the array. For all n, M(n,j) = A001405(j), j=0,...,2*n+1; hence row n of M -> A001405 as n -> infinity.

Taking finite differences of even numbered columns from the top -> down yields triangle A205946 with row sums A000984, central binomial coefficients; while odd numbered columns yield triangle A205945 with row sums A001700. A205946 and A205945 represent the bisection of A191314. - Gary W. Adamson, Feb 01 2012

			Array begins:
  1, 1, 1, 1, 1,  1,  1,  1,  1,   1,   1,...
  1, 1, 2, 3, 5,  8, 13, 21, 34,  55,  89,...
  1, 1, 2, 3, 6, 10, 19, 33, 61, 108, 197,...
  1, 1, 2, 3, 6, 10, 20, 35, 69, 124, 241,...
  1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 251,...
  1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252,...
  ...
According to Conjecture 2, row n = 3 has g.f. F_3(x) = (1-2*x^2)/(1-x-3*x^2+2*x^3+x^4).

L. E. Jeffery, Unit-primitive matrices

Cf. A001405, A108299, A115139, A191314.

Cf. also A205945, A205946, A001700, A000984.

Let N=2*n+3. For each n>0, define the (n+1) X (n+1) tridiagonal unit-primitive matrix (see [Jeffery]) B_n = A_{N,1} = [0,1,0,...,0; 1,0,1,0,...,0; 0,1,0,1,0,...,0; ...; 0,...,0,1,0,1; 0,...,0,1,1], and put B_0 = [1]. Then, for all n, M(n,k)=[(B_n)^k]{n+1,n+1}, k=0,1,..., where X{n+1,n+1} denotes the lower right corner entry of X.
CONJECTURE 2 (Rows of M). Let S(n,i) denote term i in row n of A115139, i=0,...,floor(n/2), and let T(n,j) denote term j in row n of A108299, j=0,...,n.  The generating function for row n of M is of the form F_n(x) =sum[i=0,...,floor(n/2) S(n,i)*x^(2*i)]/sum[j=0,...,n T(n,j)*x^j].
CONJECTURE 3 (Columns of M). Let D(m,k) denote term m in column k of A191314, m=0,...,floor(k/2). The generating function for column k of M is of the form G_k(x)=sum[m=0,...,floor(k/2) D(m,k)*x^m]/(1-x).

A048621 a(n) = A001222(A001405(n)).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 2, 3, 4, 5, 4, 5, 5, 6, 5, 6, 5, 6, 5, 6, 7, 8, 6, 7, 8, 9, 10, 11, 10, 11, 8, 9, 10, 11, 10, 11, 11, 12, 11, 12, 11, 12, 11, 12, 14, 15, 12, 13, 13, 14, 14, 15, 13, 14, 13, 14, 15, 16, 14, 15, 15, 16, 14, 15, 15, 16, 15, 16, 16, 17, 14, 15, 15, 16, 17, 18, 18, 19

Offset: 1

Labos Elemer

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A001222, A001221, A046660, A034973, A034974.

Table[PrimeOmega[Binomial[n, Floor[n/2]]], {n, 1, 50}] (* G. C. Greubel, May 12 2017 *)

for(n=1,100, print1(bigomega(binomial(n,floor(n/2))), ", ")) \\ G. C. Greubel, May 12 2017

a(n) = A001222(A001405(n)).

Simpler name using formula by Joerg Arndt, May 14 2017

cp's OEIS Frontend

A057977 GCD of consecutive central binomial coefficients: a(n) = gcd(A001405(n+1), A001405(n)).

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A054336 A convolution triangle of numbers based on A001405 (central binomial coefficients).

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A018224 a(n) = binomial(n, floor(n/2))^2 = A001405(n)^2.

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A034974 Number of divisors of binomial(n, floor(n/2)), the terms of A001405.

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A034973 Number of distinct prime factors in central binomial coefficients C(n, floor(n/2)), the terms of A001405.

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A080383 Number of j (0 <= j <= n) such that the central binomial coefficient C(n,floor(n/2)) = A001405(n) is divisible by C(n,j).

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A056175 Number of nonunitary prime divisors of the central binomial coefficient C(n, floor(n/2)) (A001405).

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