A005721 -id:A005721 - cp's OEIS frontend

A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

1, 12, 660, 48720, 4005540, 349260912, 31626298704, 2940502593600, 278788387440420, 26831860080682800, 2613367831568654160, 257012469788428710720, 25479526081439438845200, 2543092744417831625342400, 255292245777771431285140800, 25755871314484468746363582720

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Compare with EllipticK A002894 and the Ramanujan period-energy functions A113424, A006480, A000897. The series expansion "T(x) = 2*Pi*Sum_{n>=0} a_n*x^n" determines the real period T of elliptic curves in the family "x=p^2+q^2-4*(q^2-p^2)*q, 0 < x < 1/108". This sequence serves as a counterexample to the naive idea that elliptic integrals will always evaluate to a hypergeometric function such as 2F1(a,b;c;x).
A300058 is the complex period-energy function, after scaling energy and time dimensions such that all a(n) are integers and a(0)=1. The Picard-Fuchs equation is "(12-288*x+9216*x^2)*T(x) + (-1+232*x-8160*x^2+82944*x^3)*T'(x) + (-x+164*x^2-6432*x^3+41472*x^4)*T''(x)".
Although the sequence is not generated by a hypergeometric function, it can be formulated in terms of Hypergeometric numbers, specifically the binomial coefficients. Then Zeilberger's algorithm outputs a second order recurrence with polynomial coefficients.
The contour plot is nice to look at, with reflection symmetry, three critical points, and two separatrices dividing the phase plane into eight distinct regions.
Hyperbolic Critical points are located at (q,p) locations (1/6,0) and (-1/4,sqrt(5)/4) and (-1/4,-sqrt(5)/4). Is it possible to use chord-and-tangent addition rules to produce an exponentially-convergent Diophantine approximation to sqrt(5) that moves along the upper separatrix x=1/8?
Does there exist a period-preserving transformation that takes any one of the curves with 0 < x < 1/108 into a particular Weierstrass curve from the L-function and Modular Forms Database?

D. Husemöller, Elliptic Curves, 2nd ed., New York: Springer, 2004.
J. H. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., New York: Springer, 2009.

J. Cremona, Elliptic Curves over Q, LMFDB 2017.
B. Klee, The Virtues of X_{n+1} = (4+3*X_{n})/(3+2*X_{n}), seqfans mailing list, 2017.
B. Klee, Geometric G.F. for Ramanujan Periods, seqfans mailing list, 2017.
Brad Klee, Deriving Hypergeometric Picard-Fuchs Equations, Wolfram Demonstrations Project (2018).
Bradley Klee, Phase Plane Geometry.
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.
P. Paule and M. Schorn, FastZeil: the Paule/Schorn implementation of Gosper's and Zeilberger's algorithm, RISC 2017; Local copy, pdf file only, no active links
D. Zeilberger, The Method of Creative Telescoping, Journal of Symbolic Computation, 11.3 (1991), 195-204.

Cf. A000897, A002894, A006480, A113424.
Factors: A005190, A005809, A005721.
Complex Period: A300058.

b[NN_]:=Total/@Table[((-1)^k)*Binomial[3*n,n]*Binomial[2*n,k]*Binomial[5*n-4*k-1,3*n-4*k],{n,0,NN},{k,0,Floor[3*n/4]}];
c1=8*(-30+201*n-319*n^2+145*n^3);c2=-8640*(n-5/3)*(n-4/3)*(n-1/5);c3=10*(n-6/5)*n^2;a[0]=1;a[1]=12;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],{n->n0}];
({#,SameQ[a/@Range[0, 15],#]}&@b[15])[[1]]

a(n) = A005809(n)*A005721(n).
a(n) = Sum_{k=0..floor(3n/4)} ((-1)^k)*binomial(3*n,n)*binomial(2 *n, k)*binomial(5*n - 4*k - 1, 3*n - 4*k).
c1 = 8 *(-30 + 201*n - 319*n^2 + 145*n^3); c2 = -8640*(n - 5/3)*(n - 4/3)*(n - 1/5); c3 = 10*(n - 6/5)*n^2; a(0)=1; a(1)=12; a(n) = (c1/c3)*a(n-1) + (c2/c3)*a(n-2).

A005190 Central quadrinomial coefficients: largest coefficient of (1 + x + x^2 + x^3)^n.

Original entry on oeis.org

1, 1, 4, 12, 44, 155, 580, 2128, 8092, 30276, 116304, 440484, 1703636, 6506786, 25288120, 97181760, 379061020, 1463609356, 5724954544, 22187304112, 86981744944, 338118529539, 1327977811076, 5175023913008, 20356299454276

Offset: 0

N. J. A. Sloane

The maximal coefficient is that of x^[3n/2]. - M. F. Hasler, Jul 23 2007

Let f(m) = ceiling((q+log(q))/log(16)), where q = -log(log(256)/(5*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..200
V. E. Hoggatt, Jr. and M. Bicknell, Diagonal sums of generalized Pascal triangles, Fib. Quart., 7 (1969), 341-358, 393.
T. Neuschel, A Note on Extended Binomial Coefficients, J. Int. Seq. 17 (2014) # 14.10.4.
Claudia Smith and Verner E. Hoggatt, Jr., A Study of the Maximal Values in Pascal's Quadrinomial Triangle, Fibonacci Quart. 17 (1979), no. 3, 264-269.
Index entries for sequences of k-nomial coefficients

Cf. A005721 (bisection), A005723 (bisection).

Row 4 of A077042.

P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2+x^3)^n)): n in [0..26]]; // Vincenzo Librandi, Aug 09 2014

With[{exp=Total[x^Range[0,3]]},Table[Max[CoefficientList[Expand[exp^n], x]],{n,0,30}]] (* Harvey P. Dale, Nov 24 2011 *)

a(n)=vecmax(vector(3*n,i,polcoeff((1+x+x^2+x^3)^n,i,x)))

A005190(n)=polcoeff((1+x+x^2+x^3)^n,(3*n)>>1) \\ M. F. Hasler, Jul 23 2007

Limit_{n -> infinity} a(n+1)/a(n) = 4; for n>2, a(n+1) < 4*a(n). - Benoit Cloitre, Sep 28 2002
a(n) ~ 4^n * sqrt(2/(5*Pi*n)). - Vaclav Kotesovec, Aug 09 2013
Recurrence: 3*n*(3*n-1)*(3*n+1)*(75*n^3 - 390*n^2 + 635*n - 348)*a(n) = 12*(675*n^5 - 4095*n^4 + 8405*n^3 - 7925*n^2 + 3548*n - 664)*a(n-1) + 16*(n-1)*(2175*n^5 - 13335*n^4 + 29275*n^3 - 27707*n^2 + 11334*n - 2814)*a(n-2) - 640*(n-2)*(n-1)*(15*n^3 - 66*n^2 + 52*n - 15)*a(n-3) - 512*(n-3)*(n-2)*(n-1)*(75*n^3 - 165*n^2 + 80*n - 28)*a(n-4). - Vaclav Kotesovec, Aug 09 2013

A063419 Central sextinomial coefficients.

Original entry on oeis.org

1, 6, 146, 4332, 135954, 4395456, 144840476, 4836766584, 163112472594, 5542414273884, 189456975899496, 6507792553644256, 224442843729333276, 7766945604528200460, 269557528994032024080, 9378595792117360310832

Offset: 0

Wolfdieter Lang, Jul 24 2001

Largest coefficient of (Sum_{j=0..5} x^j)^(2*n). a(n)= A018901(2*n).
The exponent of 2 in prime factorization of a(n) for n=1,2,...,7 is in A023520. - Roman Witula, Oct 07 2012
Central coefficients in triangle A063260. - Zagros Lalo, Sep 25 2018

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 605, 606.
R. Witula and D. Slota, Central trinomial coefficients and convolution type identities, Congressus Numerantium 201 (2010), 109-126.

G. C. Greubel, Table of n, a(n) for n = 0..500

Central q-nomial coefficients (appearing once) for q=2..5: A000984, A002426, A005721, A005191. For q=7: A025012.

Cf. A063260.

Concatenation([1],List([1..15],n->Sum([0..Int(5*n/6)],k->(-1)^k*Binomial(2*n,k)*Binomial(7*n-6*k-1,2*n-1)))); # Muniru A Asiru, Sep 26 2018

Table[Sum[(-1)^k*Binomial[2*n, k]*Binomial[7*n - 6*k - 1, 2*n - 1], {k, 0, Floor[5*n/6]}], {n, 0, 50}] (* G. C. Greubel, Mar 02 2017 *)

concat([1], for(n=1,25, print1(sum(k=0,floor(5*n/6),(-1)^(k)*binomial(2*n,k)* binomial(7*n-6*k-1, 2*n-1)), ", "))) \\ G. C. Greubel, Mar 02 2017

a(n) =  A063260(2*n, 5*n)= [x^(5*n)](Sum_{j=0..5} x^j)^(2*n).
a(n) = Sum_{k=0..floor(5*n/6)} ((-1)^(k)*binomial(2*n,k)*binomial(7*n-6*k-1, 2*n-1)). - Warut Roonguthai, May 22 2006
2*Pi*a(n) = Integral[-Pi;Pi] (sin(6*x)/sin(x))^(2*n) dx. The proof of this fact is in the Witula/Slota paper. - Roman Witula, Oct 07 2012
The Almkvist-Zeilberger algorithm in EKHAD establishes the following recurrence: -31104*(2*n+5)*(2*n+3)*(2*n+1)*(7*n+19)*(5*n+11)*(7*n+20)*(7*n+13)*(n+2)*(n+1)*a(n)+ 864*(7*n+20)*(2*n+5)*(2*n+3)*(n+2)*(25480*n^5+ 223496*n^4+755066*n^3+1223233*n^2+946889*n+279936)*a(n+1)- 6*(5*n+6)*(2*n+5)*(7*n+6)*(499359*n^6+ 6777015*n^5+38079431*n^4+113390385*n^3+18872398*n^2+ 166469280*n+60800544)*a(n+2)+ 5*(5*n+14)*(5*n+13)*(5*n+12)*(7*n+12)*(5*n+11)*(5*n+6)*(7*n+13)*(7*n+6)*(n+3)*a(n+3) = 0. - Doron Zeilberger, Apr 02 2013
a(n) ~ sqrt(3) * 36^n / sqrt(35*Pi*n). - Vaclav Kotesovec, Dec 09 2021

A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 4, 10

Offset: 1

Andrey Zabolotskiy, Nov 10 2016

Equivalently, T(k,n,h) is the number of ordered sets of n nonnegative integers < k with the sum equal to h.
From Juan Pablo Herrera P., Nov 21 2016: (Start)
T(k,n,h) is the number of possible ways of randomly selecting h cards from k-1 sets, each with n different playing cards. It is also the number of lattice paths from (0,0) to (n,h) using steps (1,0), (1,1), (1,2), ..., (1,k-1).
Shallow diagonal sums of each triangle with fixed k give the k-bonacci numbers. (End)
T(k,n,h) is the number of n-dimensional grid points of a k X k X ... X k grid, which are lying in the (n-1)-dimensional hyperplane which is at an L1 distance of h from one of the grid's corners, and normal to the corresponding main diagonal of the grid. - Eitan Y. Levine, Apr 23 2023

			For first few k and for first few n, the rows with h = 0..n*(k-1) are given:
k=1:  1;  1;  1;  1;  1; ...
k=2:  1;  1, 1;  1, 2, 1;  1, 3, 3, 1;  1, 4, 6, 4, 1; ...
k=3:  1;  1, 1, 1;  1, 2, 3, 2, 1;  1, 3, 6, 7, 6, 3, 1; ...
k=4:  1;  1, 1, 1, 1;  1, 2, 3, 4, 3, 2, 1; ...
For example, (1 + x + x^2)^3 = 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6, hence T(3,3,2) = T(3,3,4) = 6.
From _Eitan Y. Levine_, Apr 23 2023: (Start)
Example for the repeated cumulative sum formula, for (k,n)=(3,3) (each line is the cumulative sum of the previous line, and the first line is the padded, alternating 3rd row from Pascal's triangle):
  1  0  0 -3  0  0  3  0  0 -1
  1  1  1 -2 -2 -2  1  1  1
  1  2  3  1 -1 -3 -2 -1
  1  3  6  7  6  3  1
which is T(3,3,h). (End)

Andrey Zabolotskiy, slices with k+n = 1..20, flattened
Florentin Smarandache, K-Nomial Coefficients, arXiv:math/0612062 [math.GM], 2006 (originally published in French in: F. Smarandache, Généralisations et Généralités, Ed. Nouvelle, 1984, pp. 24-26).

k-nomial arrays for fixed k=1..10: A000012, A007318, A027907, A008287, A035343, A063260, A063265, A171890, A213652, A213651.
Arrays for fixed n=0..6: A000012, A000012, A004737, A109439, A277949, A277950, A277951.
Central n-nomial coefficients for n=1..9, i.e., sequences with h=floor(n*(k-1)/2) and fixed n: A000012, A000984 (A001405), A002426, A005721 (A005190), A005191, A063419 (A018901), A025012, (A025013), A025014, A174061 (A025015), A201549, (A225779), A201550. Arrays: A201552, A077042, see also cfs. therein.
Triangle n=k-1: A181567. Triangle n=k: A163181.

a = Table[CoefficientList[Sum[x^(h-1),{h,k}]^n,x],{k,10},{n,0,9}];
Flatten@Table[a[[s-n,n+1]],{s,10},{n,0,s-1}]
(* alternate program *)
row[k_, n_] := Nest[Accumulate,Upsample[Table[((-1)^j)*Binomial[n,j],{j,0,n}],k],n][[;;n*(k-1)+1]] (* Eitan Y. Levine, Apr 23 2023 *)

T(k,n,h) = Sum_{i = 0..floor(h/k)} (-1)^i*binomial(n,i)*binomial(n+h-1-k*i,n-1). [Corrected by Eitan Y. Levine, Apr 23 2023]
From Eitan Y. Levine, Apr 23 2023: (Start)
(T(k,n,h))_{h=0..n*(k-1)} = f(f(...f(g(P))...)), where:
(x_i)_{i=0..m} denotes a tuple (in particular, the LHS contains the values for 0 <= h <= n*(k-1)),
f repeats n times,
f((x_i){i=0..m}) = (Sum{j=0..i} x_j)_{i=0..m} is the cumulative sum function,
g((x_i){i=0..m}) = (x(i/k) if k|i, otherwise 0)_{i=0..m*k} is adding k-1 zeros between adjacent elements,
and P=((-1)^i*binomial(n,i))_{i=0..n} is the n-th row of Pascal's triangle, with alternating signs. (End)
From Eitan Y. Levine, Jul 27 2023: (Start)
Recurrence relations, the first follows from the sequence's defining polynomial as mentioned in the Smarandache link:
T(k,n+1,h) = Sum_{i = 0..s-1} T(k,n,h-i)
T(k+1,n,h) = Sum_{i = 0..n} binomial(n,i)*T(k,n-i,h-i*k) (End)

A300058 a(n) = binomial(3n,n)/(2Pi)Integral_{x=0..2Pi} (12cos^2(x)sin(x) + 20sin^3(x))^(2n) dx.

Original entry on oeis.org

1, 492, 707220, 1298204880, 2654173160100, 5765723073622512, 13021894087331233104, 30217387890886676251200, 71532102917478013611243300, 171944976047709681477985038000, 418347201888204996027087975427920

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Compare with A295870. The series expansion "T(x)=2*Pi*sqrt(3/5)*Sum_{n>=0} a_n*(x/25)^n" determines the period T of anharmonic oscillation along a contour of the Hamiltonian energy surface "x=2H=(5/3)*p^2+q^2+4*(p^2+q^2)*q,0

The period-energy function T(x) satisfies the Picard-Fuchs equation "(2460+28512*x+2239488*x^2)*T(x)-(125-24840*x-1423008*x^2-20155392*x^3)*T'(x)+(-125*x+1620*x^2+1189728*x^3+10077696 x^4)*T''(x)", also the P.F.Eq. of A295870 under transformation x->x'=1/108-x.

A300057 has a similar definition to A005721, with a couple of extra integers appearing in the integrand. This makes a nice analogy between real and complex periods A295870, A300058. Second-order recurrences with polynomial coefficients define both sequences.

Bradley Klee, Phase Plane Geometry.
Brad Klee, Deriving Hypergeometric Picard-Fuchs Equations, Wolfram Demonstrations Project (2018).
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.

Cf. A002894, A113424, A006480, A000897. Factors: A005809, A300057. Real Period: A295870.

a := n -> 36^n*(3*n)!/n!^3*hypergeom([-2*n, n+1/2], [n+1], -2/3):
seq(simplify(a(n)), n=0..10); # Peter Luschny, Apr 19 2018

c1=12*(-230+2259*n-3933*n^2+1863*n^3);c2=5248800*(n-5/3)*(n-4/3)*(n-1/9);c3=9*n^2*(n-10/9);a[0]=1;a[1]=492;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0];
b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[3*n,n]*Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n,3*n-k1-k2]*((4+Sqrt[15])^(2*n-k1))*((4-Sqrt[15])^(2*n-k2)),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]; ({#,SameQ[#,a/@Range[0,10]]}&@b[10])[[1]]
Table[Binomial[3*n, n] * SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *)

a(n) = A005809(n)*A300057(n).
a(n) = Sum_{k1=0..2n} Sum_{k2=0..2n} binomial(3*n,n)*binomial(2*n,k2)*binomial(2*n,k1)*binomial(2*n,3*n-k1-k2)*((4+sqrt(15))^(2*n-k1))*((4-sqrt(15))^(2*n-k2))
a(0) = 1; a(1) = 492; a(n):=(c1/c3)*a(n-1)+(c2/c3)*a(n-2); with
  c1 = 12*(-230+2259*n-3933*n^2+1863*n^3);
  c2 = 5248800*(n-5/3)*(n-4/3)*(n-1/9);
  c3 = 9*n^2*(n-10/9);
a(n) ~ 2^(2*n - 1) * 3^(3*n - 1/2) * 5^(2*n + 1/2) / (Pi*n). - Vaclav Kotesovec, Apr 18 2018

A349933 Array read by ascending antidiagonals: the s-th column gives the central s-binomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 20, 19, 4, 1, 1, 70, 141, 44, 5, 1, 1, 252, 1107, 580, 85, 6, 1, 1, 924, 8953, 8092, 1751, 146, 7, 1, 1, 3432, 73789, 116304, 38165, 4332, 231, 8, 1, 1, 12870, 616227, 1703636, 856945, 135954, 9331, 344, 9, 1, 1, 48620, 5196627, 25288120, 19611175, 4395456, 398567, 18152, 489, 10, 1

Offset: 0

Stefano Spezia, Dec 06 2021

			The array begins:
n\s |   0     1     2     3     4
----+----------------------------
  0 |   1     1     1     1     1 ...
  1 |   1     2     3     4     5 ...
  2 |   1     6    19    44    85 ...
  3 |   1    20   141   580  1751 ...
  4 |   1    70  1107  8092 38165 ...
  ...

William Linz, s-Catalan numbers and Littlewood-Richardson polynomials, arXiv:2110.12095 [math.CO], 2021. See p. 2.

Cf. A000984 (s=1), A082758 (s=2), A005721 (s=3), A349936 (s=4), A063419 (s=5), A270918 (n=s), A163269 (s>0).
Cf. A007318, A008287, A025012, A027907, A035343, A063260.
Cf. A349934, A349935.

T[n_,k_,s_]:=If[k==0,1,Coefficient[(Sum[x^i,{i,0,s}])^n,x^k]]; A[n_,s_]:=T[2n,s n,s]; Flatten[Table[A[n-s,s],{n,0,9},{s,0,n}]]

A(n, s) = T(2*n, s*n, s), where T(n, k, s) is the s-binomial coefficient defined as the coefficient of x^k in (Sum_{i=0..s} x^i)^n.

A349936 Central pentanomial coefficients.

Original entry on oeis.org

1, 5, 85, 1751, 38165, 856945, 19611175, 454805755, 10651488789, 251345549849, 5966636799745, 142330448514875, 3408895901222375, 81922110160246231, 1974442362935339179, 47705925773278538281, 1155170746105476171285, 28025439409568101909625, 681077893998769910221225

Offset: 0

Stefano Spezia, Dec 06 2021

Largest coefficient of (Sum_{j=0..4} x^j)^(2*n).

Vaclav Kotesovec, Table of n, a(n) for n = 0..200

Central coefficients in triangle A035343.
Column s = 4 in A349933.
Cf. A005721, A063419, A082758.

T[n_,k_,s_]:=If[k==0,1,Coefficient[(Sum[x^i,{i,0,s}])^n,x^k]]; Table[T[2n,4n,4],{n,0,18}]

a(n) = T(2*n, 4*n, 4), where T(n, k, s) is the s-binomial coefficient defined as the coefficient of x^k in (Sum_{i=0..s} x^i)^n.
a(n) = A035343(2*n, 4*n) = [x^(4*n)] (Sum_{j=0..4} x^j)^(2*n).
From Vaclav Kotesovec, Dec 09 2021: (Start)
Recurrence: 2*n*(2*n - 1)*(3*n - 4)*(4*n - 7)*(4*n - 3)*(4*n - 1)*(6*n - 13)*(6*n - 7)*a(n) = 3*(4*n - 7)*(6*n - 13)*(10584*n^6 - 47628*n^5 + 84190*n^4 - 73965*n^3 + 33531*n^2 - 7272*n + 570)*a(n-1) - 75*(n-1)*(2*n - 3)*(4*n - 5)*(6*n - 1)*(504*n^4 - 2520*n^3 + 4160*n^2 - 2525*n + 476)*a(n-2) + 625*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(3*n - 1)*(4*n - 3)*(6*n - 7)*(6*n - 1)*a(n-3).
a(n) ~ 25^n / sqrt(8*Pi*n). (End)

A248876 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3n} T(n,k)^2 x^k] / A(x)^n * x^n/n ) where T(n,k) equals the coefficient of x^k in (1+x+x^2+x^3)^n.

Original entry on oeis.org

1, 1, 1, 2, 4, 8, 13, 24, 45, 85, 161, 305, 582, 1116, 2149, 4152, 8049, 15653, 30528, 59695, 117012, 229880, 452565, 892703, 1764099, 3492029, 6923494, 13747483, 27335873, 54427621, 108505081, 216568556, 432740907, 865610375, 1733227339, 3473805680, 6968708734, 13991916510, 28116598325

Offset: 0

Paul D. Hanna, Mar 04 2015

nonn

Compare the definition of this sequence to G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / G(x)^n * x^n/n ) where T(n,k) = [x^k] (1+x+x^2)^n, which is satisfied by the rational function: G(x) = (1+x^3)*(1+x^6)/((1-x)*(1-x^4)).

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 8*x^5 + 13*x^6 + 24*x^7 +...
where
log(A(x)) = (1 + x + x^2 + x^3)/A(x) * x +
(1 + 2^2*x + 3^2*x^2 + 4^2*x^3 + 3^2*x^4 + 2^2*x^5 + x^6)/A(x)^2 * x^2/2 +
(1 + 3^2*x + 6^2*x^2 + 10^2*x^3 + 12^2*x^4 + 12^2*x^5 + 10^2*x^6 + 6^2*x^7 + 3^2*x^8 + x^9)/A(x)^3 * x^3/3 +
(1 + 4^2*x + 10^2*x^2 + 20^2*x^3 + 31^2*x^4 + 40^2*x^5 + 44^2*x^6 + 40^2*x^7 + 31^2*x^8 + 20^2*x^9 + 10^2*x^10 + 4^2*x^11 + x^12)/A(x)^4 * x^4/4 +
which involves the squares of the coefficients in (1 + x + x^2 + x^3)^n.

Paul D. Hanna, Table of n, a(n) for n = 0..100

Cf. A255839, A200537, A251687, A251688, A005721.

/* By Definition: */
{T(n,k)=polcoeff((1 + x + x^2 + x^3 + x*O(x^k))^n, k)}
{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, min(3*m,n-m), T(m,k)^2 * x^k) / (A +x*O(x^n))^m * x^m/m)+x*O(x^n))); polcoeff(A, n)}
for(n=0, 40, print1(a(n), ", "))

A255839 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3n} binomial(3n,k)^2 * x^k] / A(x)^n * x^n/n ).

Original entry on oeis.org

1, 1, 9, 18, 64, 172, 477, 1368, 3681, 10485, 28701, 80829, 225090, 632160, 1778553, 5010948, 14181849, 40161357, 114151716, 324873027, 926918784, 2649218580, 7585705665, 21758756931, 62508649059, 179859399129, 518234494662, 1495275239115, 4319808231645, 12495043092609, 36183457564425

Offset: 0

Paul D. Hanna, Mar 07 2015

nonn

Compare the definition of this sequence to G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} binomial(2*n,k)^2 * x^k] / G(x)^n * x^n/n ), which is satisfied by the rational function: G(x) = (1+x^2)^2*(1+x^3)/((1-x)*(1-x^2)).

			G.f.: A(x) = 1 + x + 9*x^2 + 18*x^3 + 64*x^4 + 172*x^5 + 477*x^6 +...
where
log(A(x)) = (1 + 3^2*x + 3^2*x^2 + x^3)/A(x) * x +
(1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)/A(x)^2 * x^/2 +
(1 + 9^2*x + 36^2*x^2 + 84^2*x^3 + 126^2*x^4 + 126^2*x^5 + 84^2*x^6 + 36^2*x^7 + 9^2*x^8 + x^9)/A(x)^3 * x^3/3 +
(1 + 12^2*x + 66^2*x^2 + 220^2*x^3 + 495^2*x^4 + 792^2*x^5 + 924^2*x^6 + 792^2*x^7 + 495^2*x^8 + 220^2*x^9 + 66^2*x^10 + 12^2*x^11 + x^12)/A(x)^4 * x^4/4 +...
which involves the squares of the coefficients in (1 + x)^(3*n).

Paul D. Hanna, Table of n, a(n) for n = 0..100

Cf. A248876, A200537, A251687, A251688, A005721.

/* By Definition: */
{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, min(3*m,n-m), binomial(3*m,k)^2 * x^k) / (A +x*O(x^n))^m * x^m/m)+x*O(x^n))); polcoeff(A, n)}
for(n=0, 40, print1(a(n), ", "))

A300057 Coefficient of z^(3n) in the expansion of (1 + 9z + 9z^2 + z^3)^(2n).

Original entry on oeis.org

1, 164, 47148, 15454820, 5361965980, 1919987703504, 701459496193236, 259867456921970040, 97260263038893462300, 36686877800581349096240, 13924013746979490475444528, 5311128944356277793155688612, 2034235241375650519750351973188

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Muniru A Asiru, Table of n, a(n) for n = 0..100

Cf. A300058, A295870, A005721.

List([0..15],n->6^(2*n)Sum([0..2*n],k->Binomial(2*n,k)*Binomial(2*(n+k),n+k)*(1/6)^k)); # Muniru A Asiru, Apr 07 2018

c1=16*(n-1/2)*(-230+2259*n-3933*n^2+1863*n^3);c2=1036800*(n-1)*(n-3/2)*(n-1/2)*(n-1/9);c3=81*n*(n-2/3)*(n-1/3)*(n-10/9);a[0]=1;a[1]=164;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0]
b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n, 3*n-k1-k2]*(4 + Sqrt[15])^(2*n-k1)*(4-Sqrt[15])^(2*n-k2),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]
({#,SameQ[Coefficient[(1+9*z+9*z^2+z^3)^(2*#),z,3*#]&/@Range[0,10],#],SameQ[a/@Range[0,10],#]}&@b[10])[[1]]
Table[SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *)

a(n) = polcoeff((1 + 9*z + 9*z^2 + z^3)^(2*n), 3*n); \\ Michel Marcus, Mar 06 2018

a(n) = 1/(2*Pi)*Integral_{0..2*Pi}(12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
a(n) = Sum_{k1=0..2*n} Sum_{k2=0..2*n} binomial(2*n,k1)*binomial(2*n,k2)*binomial(2*n,3*n-k1-k2)*((4-sqrt(15))^(2*n-k1))*((4+sqrt(15))^(2*n-k2)).
a(n) = (c1/c3)*a(n-1)+(c2/c3)*a(n-2); with a(0)=1; a(1)=164; and
  c1=16*(n-1/2)*(-230+2259*n-3933*n^2+1863*n^3);
  c2=1036800*(n-1)*(n-3/2)*(n-1/2)*(n-1/9);
  c3=81*n*(n-2/3)*(n-1/3)*(n-10/9).
From Wolfdieter Lang, Apr 06 2018: (Start)
a(n) = 4^(2*n)*(2/Pi)*Integral_{0..Pi/2} sin(x)^(2*n)*(3 + 2*sin(x)^2)^(2*n) dx. With the binomial formula and integrals over even powers of sin(x) this becomes
a(n) = 6^(2*n)*Sum_{k=0..2*n} binomial(2*n, k)*binomial(2*(n+k), n+k)*(1/6)^k = 6^(2*n)*binomial(2*n, n)*hypergeometric([-2*n, n+1/2], [n+1], -2/3). (End)
a(n) ~ 2^(4*n) * 5^(2*n + 1/2) / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 18 2018

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cp's OEIS Frontend

A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

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A005190 Central quadrinomial coefficients: largest coefficient of (1 + x + x^2 + x^3)^n.

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Magma

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A063419 Central sextinomial coefficients.

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GAP

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A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

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A300058 a(n) = binomial(3*n,n)/(2*Pi)*Integral_{x=0..2*Pi} (12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.

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A349933 Array read by ascending antidiagonals: the s-th column gives the central s-binomial coefficients.

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A349936 Central pentanomial coefficients.

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A300058 a(n) = binomial(3n,n)/(2Pi)Integral_{x=0..2Pi} (12cos^2(x)sin(x) + 20sin^3(x))^(2n) dx.

A248876 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3n} T(n,k)^2 x^k] / A(x)^n * x^n/n ) where T(n,k) equals the coefficient of x^k in (1+x+x^2+x^3)^n.

A255839 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3n} binomial(3n,k)^2 * x^k] / A(x)^n * x^n/n ).

A300057 Coefficient of z^(3n) in the expansion of (1 + 9z + 9z^2 + z^3)^(2n).