A000984 -id:A000984 - cp's OEIS frontend

A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1

Offset: 0

Wolfdieter Lang, Mar 13 2000

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/(sqrt(1-4*z)-x*z).
The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.
Riordan array (1/sqrt(1-4*x),x/sqrt(1-4*x)). - Paul Barry, May 06 2009
The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013

			Triangle begins:
    1;
    2,    1;
    6,    4,   1;
   20,   16,   6,   1;
   70,   64,  30,   8,  1;
  252,  256, 140,  48, 10,  1;
  924, 1024, 630, 256, 70, 12, 1; ...
Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
From _Paul Barry_, May 06 2009: (Start)
Production matrix begins
    2,   1;
    2,   2,  1;
    0,   2,  2,  1;
   -2,   0,  2,  2,  1;
    0,  -2,  0,  2,  2,  1;
    4,   0, -2,  0,  2,  2, 1;
    0,   4,  0, -2,  0,  2, 2, 1;
  -10,   0,  4,  0, -2,  0, 2, 2, 1;
    0, -10,  0,  4,  0, -2, 0, 2, 2, 1; (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.

Cf. A000984, A035324, A054336.

Row sums: A026671.

T:= function(n, k)
    if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2),Int(k/2))/Binomial(k,Int(k/2));
    else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
    fi;
  end;
Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # G. C. Greubel, Jul 20 2019

T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2),Floor(k/2))/Binomial(k,Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
[[T(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019

A054335 := proc(n,k)
    if k <0 or k > n then
        0 ;
    elif type(k,odd) then
        kprime := floor(k/2) ;
        binomial(n-kprime-1,kprime)*4^(n-k) ;
    else
        kprime := k/2 ;
        binomial(2*n-k,n-kprime)*binomial(n-kprime,kprime)/binomial(k,kprime) ;
    end if;
end proc: # R. J. Mathar, Mar 12 2013
# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022

Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)

T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 20 2019

def T(n, k):
    if (mod(k,2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2)
    else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019

a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n

Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.

G.f. for column m: cbie(x)*(x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).

G.f.: 1/(1-x*y-2*x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009

Sum_{k>=0} T(n,2*k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012

Sum_{k=0..floor(n/2)} T(n-k,n-2*k) = A098615(n). - Philippe Deléham, Feb 01 2012

T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012

Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015

A232606 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (2*n)!^2/n!^4, the square of the central binomial coefficients (A000984), for n>=0.

Original entry on oeis.org

1, 3, 10, 42, 221, 1379, 9678, 73666, 594326, 5007958, 43641702, 390632678, 3573598539, 33289289533, 314871186248, 3017358158132, 29242725947318, 286209134234602, 2825613061237808, 28111283170770480, 281598654896870051, 2838309465080014489, 28767973963085929656, 293059625830028920012

Offset: 0

Paul D. Hanna, Nov 26 2013

nonn

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = (2*n)!/n!^2 = A000984(n).

a(n+1)/a(n) tends to 11.3035... - Vaclav Kotesovec, Jan 23 2014

			G.f.: A(x) = 1 + 3*x + 10*x^2 + 42*x^3 + 221*x^4 + 1379*x^5 + 9678*x^6 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1], 0,   0,    0,     0,      0,       0,        0,         0, ...;
A^1: [1,  3], 10,   42,   221,   1379,    9678,    73666,    594326, ...;
A^2: [1,  6,  29], 144,   794,   4924,   33814,   251544,   1988885, ...;
A^3: [1,  9,  57,  333], 1989,  12669,   86935,   639123,   4979499, ...;
A^4: [1, 12,  94,  636,  4157], 27728,  193504,  1423120,  11006058, ...;
A^5: [1, 15, 140, 1080,  7730,  54538], 391970,  2915490,  22558825, ...;
A^6: [1, 18, 195, 1692, 13221,  99102,  739547], 5612016,  43767477, ...;
A^7: [1, 21, 259, 2499, 21224, 169232, 1317722, 10267666], 81223912, ...;
A^8: [1, 24, 332, 3528, 32414, 274792, 2238492, 17990904, 145096413], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals the square of the central binomial coefficients:
1^1 = 1;
2^2 = 1 + 3;
6^2 = 1 +  6 +  29;
20^2 = 1 +  9 +  57 +  333;
70^2 = 1 + 12 +  94 +  636 +  4157;
252^2 = 1 + 15 + 140 + 1080 +  7730 +  54538;
924^2 = 1 + 18 + 195 + 1692 + 13221 +  99102 +  739547;
3432^2 = 1 + 21 + 259 + 2499 + 21224 + 169232 + 1317722 + 10267666; ...
RELATED SERIES.
From a main diagonal in the above array we can derive sequence A232607:
[1/1, 6/2, 57/3, 636/4, 7730/5, 99102/6, 1317722/7, 17990904/8, ...] =
[1, 3, 19, 159, 1546, 16517, 188246, 2248863, 27844369, 354576634, ...];
from which we can form the series G(x) = A(x*G(x)):
G(x) = 1 + 3*x + 19*x^2 + 159*x^3 + 1546*x^4 + 16517*x^5 + 188246*x^6 +...
such that
(G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = 1 + 2^2*x + 6^2*x^2 + 20^2*x^3 + 70^2*x^4 + 252^2*x^5 +...+ A000984(n)^2*x^n +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..350

Cf. A232683, A232607, A232687, A000984, A002894.

terms = 24; a[0] = 1; A[x_] = Sum[a[n]*x^n, {n, 0, terms - 1}];
c[n_] := Sum[Coefficient[B[x], x, k], {k, 0, n}] == (2*n)!^2/n!^4 // Solve // First;
Do[B[x_] = A[x]^n + O[x]^(n+1) // Normal; A[x_] = (A[x] /. c[n]) + O[x]^terms, {n, 0, terms-1}];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)

/* By Definition: */
{a(n)=if(n==0, 1, ((2*n)!^2/n!^4 - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0,20,print1(a(n)*1!,", "))

/* Faster, using series reversion: */
{a(n)=local(CB2=sum(k=0,n,binomial(2*k,k)^2*x^k)+x*O(x^n), G=1+x*O(x^n));
for(i=1,n,G = 1 + intformal( (CB2-1)*G/x - CB2*G^2));polcoeff(x/serreverse(x*G),n)}
for(n=0,30,print1(a(n),", "))

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (2*n)!^2/n!^4 = A000984(n)^2.

Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (2*n)!^2/n!^4 * x^n.

A187364 Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0.

Original entry on oeis.org

1, 35, 1716, 92378, 5200300, 300540195, 17672631900, 1052049481860, 63205303218876, 3824345300380220, 232714176627630544, 14226520737620288370, 873065282167813104916, 53753604366668088230810, 3318776542511877736535400, 205397724721029574666088520

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187363 concerning trisection.

This appears also in the trisection of A001700 (central binomials in the odd numbered Pascal rows): binomial(2*(3*n)+1,3*n+1).

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 (binomial(6n,3n)), A187365 (binomial(2(3n+2),3n+2)/3!).

Cf. A002458, A100033.

Table[c=3n+1;Binomial[2c,c]/2,{n,0,20}] (* Harvey P. Dale, May 10 2012 *)

a(n) = binomial(2*(3*n+1),3*n+1)/2, n>=0.
a(n) = binomial(2*(3*n)+1,3*n+1), n>=0.
O.g.f.: (cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1+8*x^(1/3))/2))/(6*x^(1/3)), with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/2)*Sum_{k = 0..3*n+1} binomial(3*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 3*n, -1 - 3*n], [1], 1).
a(n) = 8*(2*n - 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n - 1)) * a(n-1). (End)
Right-hand side of the binomial sum identity (1/18) * Sum_{k = 0..6*n+3} (-1)^(n+k) * (k/(2*n + 1))^2 * binomial(6*n+3, k)^2 = a(n). - Peter Bala, Nov 05 2024

A081387 Number of non-unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of prime factors in C(2n,n) whose exponent is greater than one.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 2, 2, 2, 2, 1, 1, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 3, 3, 3, 2, 4, 2, 2, 3, 3, 3, 3, 4, 4, 5, 4, 4, 2, 3, 3, 2, 2, 2, 4, 3, 3, 4, 3, 4, 5, 4, 2, 2, 2, 3, 5, 5, 5, 5, 3, 2, 3, 2, 3, 3, 3

Offset: 1

Labos Elemer, Mar 27 2003

nonn

			For n = 14: binomial(28,14) = 40116600 = 2*2*2*3*3*3*5*5*17*19*23; unitary prime divisors: {17,19,23}; non-unitary prime divisors: {2,3,5}, so a(14) = 3.

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

Cf. A000984, A034444, A048105, A056169, A056170, A067434, A081387, A081388, A081389.

Table[Count[Transpose[FactorInteger[Binomial[2n,n]]][[2]],?(#>1&)],{n,110}] (* _Harvey P. Dale, Oct 08 2012 *)

a(n) = {my(e = factor(binomial(2*n, n))[, 2]); sum(k = 1, #e, e[k] > 1);} \\ Amiram Eldar, Oct 05 2024

a(n) = A056170(A000984(n)) = A001221(A000984(n)) - A081386(n) = A067434(n) - A081386(n).

A201555 a(n) = C(2*n^2,n^2) = A000984(n^2), where A000984 is the central binomial coefficients.

Original entry on oeis.org

1, 2, 70, 48620, 601080390, 126410606437752, 442512540276836779204, 25477612258980856902730428600, 23951146041928082866135587776380551750, 365907784099042279561985786395502921046971688680, 90548514656103281165404177077484163874504589675413336841320

Offset: 0

Paul D. Hanna, Dec 02 2011

nonn

Central coefficients of triangle A228832.

			L.g.f.: L(x) = 2*x + 70*x^2/2 + 48620*x^3/3 + 601080390*x^4/4 + ...
where exponentiation equals the g.f. of A201556:
exp(L(x)) = 1 + 2*x + 37*x^2 + 16278*x^3 + 150303194*x^4 + ... + A201556(n)*x^n + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..40
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
R. Oblath, Congruences with binomial coefficients, Proceedings of the Indian Academy of Science, Section A, Vol. 1 No. 6, 383-386

Cf. A000984, A007814, A159918, A201556, A214441, A228832, A285388, A285717.

Table[Binomial[2n^2,n^2],{n,0,10}] (* Harvey P. Dale, Dec 10 2011 *)

PARI
```
a(n) = binomial(2*n^2,n^2)
```

from math import comb
def A201555(n): return comb((m:=n**2)<<1,m) # Chai Wah Wu, Jul 08 2022

L.g.f.: ignoring initial term, equals the logarithmic derivative of A201556.
a(n) = (2*n^2)! / (n^2)!^2.
a(n) = Sum_{k=0..n^2} binomial(n^2,k)^2.
For primes p >= 5: a(p) == 2 (mod p^3), Oblath, Corollary II; a(p) == binomial(2*p,p) (mod p^6) - see Mestrovic, Section 5, equation 31. - Peter Bala, Dec 28 2014
A007814(a(n)) = A159918(n). - Antti Karttunen, Apr 27 2017, based on Vladimir Shevelev's Jul 20 2009 formula in A000984.

A226078 Table read by rows: prime power factors of central binomial coefficients, cf. A000984.

Original entry on oeis.org

1, 2, 2, 3, 4, 5, 2, 5, 7, 4, 9, 7, 4, 3, 7, 11, 8, 3, 11, 13, 2, 9, 5, 11, 13, 4, 5, 11, 13, 17, 4, 11, 13, 17, 19, 8, 3, 7, 13, 17, 19, 4, 7, 13, 17, 19, 23, 8, 25, 7, 17, 19, 23, 8, 27, 25, 17, 19, 23, 16, 9, 5, 17, 19, 23, 29, 2, 9, 5, 17, 19, 23, 29, 31

Offset: 0

Reinhard Zumkeller, May 25 2013

			.   n        initial rows               A000984(n)   A226047(n)
.  ---+------------------------------+-------------+------------
.   0   [1]                                      1
.   1   [2]                                      2            2
.   2   [2,3]                                    6            3
.   3   [4,5]                                   20            5
.   4   [2,5,7]                                 70            7
.   5   [4,9,7]                                252            9
.   6   [4,3,7,11]                             924           11
.   7   [8,3,11,13]                           3432           13
.   8   [2,9,5,11,13]                        12870           13
.   9   [4,5,11,13,17]                       48620           17
.  10   [4,11,13,17,19]                     184756           19
.  11   [8,3,7,13,17,19]                    705432           19
.  12   [4,7,13,17,19,23]                  2704156           23
.  13   [8,25,7,17,19,23]                 10400600           25
.  14   [8,27,25,17,19,23]                40116600           27
.  15   [16,9,5,17,19,23,29]             155117520           29
.  16   [2,9,5,17,19,23,29,31]           601080390           31
.  17   [4,27,5,11,19,23,29,31]         2333606220           31
.  18   [4,3,25,7,11,19,23,29,31]       9075135300           31
.  19   [8,3,25,7,11,23,29,31,37]      35345263800           37
.  20   [4,9,5,7,11,13,23,29,31,37]   137846528820           37 .

Reinhard Zumkeller, Rows n = 0..250 of triangle, flattened

Cf. A067434 (row lengths), A001316 (left edge), A060308 (right edge), A226047 (row maxima), A226083 (row minima), A000984 (row products).

Cf. A267823.

a226078 n k = a226078_tabf !! n !! k
a226078_row n = a226078_tabf !! n
a226078_tabf = map a141809_row a000984_list

f:= n-> add(i[2]*x^i[1], i=ifactors(n)[2]):
b:= proc(n) local p;
      p:= add(f(n+i) -f(i), i=1..n);
      seq(`if`(coeff(p, x, i)>0,
             i^coeff(p, x, i), NULL), i=1..degree(p))
    end:
T:= n-> `if`(n=0, 1, b(n)):
seq(T(n), n=0..30);  # Alois P. Heinz, May 25 2013

Table[Power @@@ FactorInteger[(2n)!/n!^2] , {n, 0, 30}] // Flatten (* Jean-François Alcover, Jul 29 2015 *)

row(n)= if(n<1, [1], [ e[1]^e[2] |e<-Col(factor(binomial(2*n, n)))]); \\ Ruud H.G. van Tol, Nov 18 2024

T(n,k) = A141809(A000984(n),k) for k = 0..A067434(n)-1.

A081386 Number of unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of those prime factors in C(2n,n), whose exponent equals one.

Original entry on oeis.org

1, 2, 1, 3, 1, 3, 3, 4, 4, 4, 5, 5, 4, 3, 5, 7, 6, 7, 7, 8, 9, 9, 6, 7, 7, 7, 8, 11, 12, 11, 11, 11, 12, 12, 12, 13, 13, 13, 11, 13, 12, 14, 13, 13, 15, 14, 14, 14, 15, 16, 16, 16, 17, 19, 18, 17, 18, 19, 18, 19, 18, 18, 18, 20, 18, 21, 22, 20, 20, 20, 20, 20, 20, 19, 21, 21, 24, 23

Offset: 1

Labos Elemer, Mar 27 2003

nonn

			n=10: C(20,10) = 184756 = 2*2*11*13*17*19; unitary-p-divisors = {11,13,17,19}, so a(10)=4.

T. D. Noe, Table of n, a(n) for n=1..2000

Cf. A000984, A067434, A081387-A081389, A034444, A048105, A056170, A056169.

Table[Function[m, Count[Divisors@ m, k_ /; And[PrimeQ@ k, GCD[k, m/k] == 1]]]@ Binomial[2 n, n], {n, 50}] (* Michael De Vlieger, Dec 17 2016 *)

a(n) = my(f=factor(binomial(2*n, n))); sum(k=1, #f~, f[k,2] == 1); \\ Michel Marcus, Dec 18 2016

a(n) = A056169(A000984(n)).

A134760 a(n) = 2*A000984(n) - 1.

Original entry on oeis.org

1, 3, 11, 39, 139, 503, 1847, 6863, 25739, 97239, 369511, 1410863, 5408311, 20801199, 80233199, 310235039, 1202160779, 4667212439, 18150270599, 70690527599, 275693057639, 1076515748879, 4208197927439, 16466861455199, 64495207366199, 252821212875503

Offset: 0

Gary W. Adamson, Nov 09 2007

Inverse binomial transform of this is A134761: (the sequence interpolated with ones): (1, 1, 3, 1, 11, 1, 39, 1, 139, ...).

Alois P. Heinz, Table of n, a(n) for n = 0..500
C. J. Fewster and D. Siemssen, Enumerating Permutations by their Run Structure, arXiv preprint arXiv:1403.1723 [math.CO], 2014.

Cf. A000984, A000108, A134761.

[2*(n+1)*Catalan(n)-1: n in [0..40]]; // G. C. Greubel, Apr 06 2024

a:= proc(n) option remember; `if`(n<2, 2*n+1,
       ((12-31*n+15*n^2) *a(n-1)
        -2*(3*n-2)*(2*n-3)*a(n-2)) / (n*(3*n-5)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 16 2013

a[n_] := 2 Binomial[2n, n] - 1; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jul 21 2017 *)

[2*binomial(2*n,n)-1 for n in range(41)] # G. C. Greubel, Apr 06 2024

From R. J. Mathar, Mar 23 2015: (Start)
n*a(n) = 2*(3*n-2)*a(n-1) - (9*n-14)*a(n-2) + 2*(2*n-5)*a(n-3).
n*(3*n-5)*a(n) = (15*n^2-31*n+12)*a(n-1) - 2*(3*n-2)*(2*n-3)*a(n-2). (End)
From G. C. Greubel, Apr 06 2024: (Start)
a(n) = 2*(n+1)*A000108(n) - 1.
G.f.: 2/sqrt(1 - 4*x) - 1/(1 - x).
E.g.f.: 2*exp(2*x)*BesselI(0, 2*x) - exp(x). (End)

A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.

Original entry on oeis.org

1, 42, 2145, 117572, 6686100, 388934370, 22974421470, 1372238454600, 82653088824684, 5011211083256840, 305437356823765089, 18697712969443807572, 1148770108115543559100, 70797430141465286938140, 4374750896947475198160300, 270950190057528375091435920

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187357 concerning trisection.

This appears also in the trisection of A001700: binomial(2*(3*n+1)+1,(3*n+1)+1)/3.

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 binomial(6n,3n), A187364 binomial(2*(3n+1),3n+1)/2, A002458, A100033.

a(n)=binomial(2*(3*n+2),3*n+2)/3!, n>=0.
a(n)=binomial(3*(2*n+1),3*n+2)/3, n>=0.
O.g.f.:(cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1-4*x^(1/3))/2))/(18*x^(2/3)),
with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z)the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/6)*Sum_{k = 0..3*n+2} binomial(3*n+2,k)^2.
a(n) = (1/6)*hypergeom([-2 - 3*n, -2 - 3*n], [1], 1).
a(n) = 8*(2*n + 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n + 2)) * a(n-1). (End)

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cp's OEIS Frontend

A060165 Number of orbits of length n under the map whose periodic points are counted by A000984.

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A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

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A232606 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (2*n)!^2/n!^4, the square of the central binomial coefficients (A000984), for n>=0.

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A187364 Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0.

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A081387 Number of non-unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of prime factors in C(2n,n) whose exponent is greater than one.

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A201555 a(n) = C(2*n^2,n^2) = A000984(n^2), where A000984 is the central binomial coefficients.

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A226078 Table read by rows: prime power factors of central binomial coefficients, cf. A000984.

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