Displaying 1-10 of 11 results found.
a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.
+20
3
1, 12, 660, 48720, 4005540, 349260912, 31626298704, 2940502593600, 278788387440420, 26831860080682800, 2613367831568654160, 257012469788428710720, 25479526081439438845200, 2543092744417831625342400, 255292245777771431285140800, 25755871314484468746363582720
COMMENTS
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?
REFERENCES
D. Husemöller, Elliptic Curves, 2nd ed., New York: Springer, 2004.
J. H. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., New York: Springer, 2009.
LINKS
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.
FORMULA
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).
MATHEMATICA
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]]
Central sextinomial coefficients.
+10
6
1, 6, 146, 4332, 135954, 4395456, 144840476, 4836766584, 163112472594, 5542414273884, 189456975899496, 6507792553644256, 224442843729333276, 7766945604528200460, 269557528994032024080, 9378595792117360310832
COMMENTS
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
REFERENCES
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.
FORMULA
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
MATHEMATICA
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 *)
PROG
(PARI) 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 (GAP) 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
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.
+10
6
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
COMMENTS
Equivalently, T(k,n,h) is the number of ordered sets of n nonnegative integers < k with the sum equal to h.
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
LINKS
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).
FORMULA
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] (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)
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)
EXAMPLE
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.
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)
MATHEMATICA
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 *)
CROSSREFS
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.
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.
+10
3
1, 492, 707220, 1298204880, 2654173160100, 5765723073622512, 13021894087331233104, 30217387890886676251200, 71532102917478013611243300, 171944976047709681477985038000, 418347201888204996027087975427920
COMMENTS
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<x<1/108". 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.
LINKS
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.
FORMULA
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
MAPLE
a := n -> 36^n*(3*n)!/n!^3*hypergeom([-2*n, n+1/2], [n+1], -2/3):
MATHEMATICA
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 *)
Array read by ascending antidiagonals: the s-th column gives the central s-binomial coefficients.
+10
3
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
FORMULA
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.
EXAMPLE
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 ...
...
MATHEMATICA
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}]]
G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3*n} 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.
+10
2
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
COMMENTS
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)).
EXAMPLE
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.
PROG
(PARI) /* 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), ", "))
G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3*n} binomial(3*n,k)^2 * x^k] / A(x)^n * x^n/n ).
+10
2
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
COMMENTS
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)).
EXAMPLE
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).
PROG
(PARI) /* 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), ", "))
Coefficient of z^(3*n) in the expansion of (1 + 9*z + 9*z^2 + z^3)^(2*n).
+10
2
1, 164, 47148, 15454820, 5361965980, 1919987703504, 701459496193236, 259867456921970040, 97260263038893462300, 36686877800581349096240, 13924013746979490475444528, 5311128944356277793155688612, 2034235241375650519750351973188
FORMULA
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).
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
MATHEMATICA
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 *)
PROG
(PARI) a(n) = polcoeff((1 + 9*z + 9*z^2 + z^3)^(2*n), 3*n); \\ Michel Marcus, Mar 06 2018 (GAP) 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
Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of ( Sum_{j=1..k} x_j^(2*j-1) + x_j^(-(2*j-1)) )^(2*n).
+10
2
1, 1, 0, 1, 2, 0, 1, 4, 6, 0, 1, 6, 44, 20, 0, 1, 8, 146, 580, 70, 0, 1, 10, 344, 4332, 8092, 252, 0, 1, 12, 670, 18152, 135954, 116304, 924, 0, 1, 14, 1156, 55252, 1012664, 4395456, 1703636, 3432, 0, 1, 16, 1834, 137292, 4816030, 58199208, 144840476, 25288120, 12870, 0
FORMULA
T(n,k) = Sum_{j=0..floor((2*k-1)*n/(2*k))} (-1)^j * binomial(2*n,j) * binomial((2*k+1)*n-2*k*j-1,(2*k-1)*n-2*k*j) for k > 0.
EXAMPLE
(x^3 + x + 1/x + 1/x^3)^2 = x^6 + 2*x^4 + 3*x^2 + 4 + 3/x^2 + 2/x^4 + 1/x^6. So T(1,2) = 4.
Square array begins:
1, 1, 1, 1, 1, 1, ...
0, 2, 4, 6, 8, 10, ...
0, 6, 44, 146, 344, 670, ...
0, 20, 580, 4332, 18152, 55252, ...
0, 70, 8092, 135954, 1012664, 4816030, ...
0, 252, 116304, 4395456, 58199208, 432457640, ...
MATHEMATICA
T[n_, 0] = Boole[n == 0]; T[n_, k_] := Sum[(-1)^j * Binomial[2*n, j] * Binomial[(2*k + 1)*n - 2*k*j - 1, (2*k - 1)*n - 2*k*j], {j, 0, Floor[(2*k - 1)*n/(2*k)]}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 06 2021 *)
Central pentanomial coefficients.
+10
2
1, 5, 85, 1751, 38165, 856945, 19611175, 454805755, 10651488789, 251345549849, 5966636799745, 142330448514875, 3408895901222375, 81922110160246231, 1974442362935339179, 47705925773278538281, 1155170746105476171285, 28025439409568101909625, 681077893998769910221225
COMMENTS
Largest coefficient of (Sum_{j=0..4} x^j)^(2*n).
FORMULA
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). 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)
MATHEMATICA
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}]
CROSSREFS
Central coefficients in triangle A035343.
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