| Displaying 1-8 of 8 results found.
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1
a(n) = 2^n * (2*n)! / (n!)^2.
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29
1, 4, 24, 160, 1120, 8064, 59136, 439296, 3294720, 24893440, 189190144, 1444724736, 11076222976, 85201715200, 657270374400, 5082890895360, 39392404439040, 305870434467840, 2378992268083200, 18531097667174400
COMMENTS
Number of lattice paths from (0,0) to (n,n) using steps (0,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011 The convolution square root of this sequence is A004981. - T. D. Noe, Jun 11 2002 Also main diagonal of array: T(i,1)=2^(i-1), T(1,j)=1, T(i,j) = T(i,j-1) + 2*T(i-1,j). - Benoit Cloitre, Feb 26 2003 The Hankel transform (see A001906 for definition) of this sequence with interpolated zeros(1, 0, 4, 0, 24, 0, 160, 0, 1120, ...) = is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005 Equals number of permutations whose reverse shares the same recording tableau in the Robinson-Schensted correspondence with n=(k-1)/2 for k odd. - Dang-Son Nguyen, Jul 02 2024
FORMULA
a(n) = C(2*n,n) * 2^n.
D-finite with recurrence a(n) = a(n-1)*(8-4/n).
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+1)/(4*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013 E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 4*x/(4*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013 G.f.: Q(0)/(1+2*sqrt(x)), where Q(k) = 1 + 2*sqrt(x)/(1 - 2*sqrt(x)*(2*k+1)/(2*sqrt(x)*(2*k+1) + (k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 09 2013 a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(3*n-2*k,n)* binomial(n+k,n). - Peter Bala, Aug 04 2016 Sum_{n>=0} 1/a(n) = 8/7 + 8*sqrt(7)*arcsin(1/sqrt(8))/49.
Sum_{n>=0} (-1)^n/a(n) = (8/27)*(3 - arcsinh(1/sqrt(8))). (End)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)*binomial(k,n). In general, for m >= 1, Sum_{k = n..m*n} binomial(m*n,k)*binomial(k,n) = 2^((m-1)*n)*binomial(m*n,n). - Peter Bala, Mar 25 2023 Conjecture: a(n) = Sum_{0 <= j, k <= n} binomial(n, j)*binomial(n, k)* binomial(k+j, n). - Peter Bala, Jul 16 2024
MATHEMATICA
Table[2^n Binomial[2n, n], {n, 0, 30}] (* Harvey P. Dale, Dec 16 2014 *)
PROG
(PARI) {a(n)=if(n<0, 0, 2^n*(2*n)!/n!^2)} /* Michael Somos, Jan 31 2007 */ (PARI) { for (n = 0, 200, write("b059304.txt", n, " ", 2^n * (2*n)! / n!^2); ) } \\ Harry J. Smith, Jun 25 2009 (PARI) /* as lattice paths: same as in A092566 but use */ steps=[[1, 0], [1, 0], [0, 1]]; /* note the double [1, 0] */
(Magma) [2^n*Factorial(2*n)/Factorial(n)^2: n in [0..25]]; // Vincenzo Librandi, Oct 08 2015
a(n) = binomial(4*n+1, 2*n).
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20
1, 10, 126, 1716, 24310, 352716, 5200300, 77558760, 1166803110, 17672631900, 269128937220, 4116715363800, 63205303218876, 973469712824056, 15033633249770520, 232714176627630544, 3609714217008132870, 56093138908331422716, 873065282167813104916
REFERENCES
The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1982, (3.109), page 35.
FORMULA
a(n) = Sum_{k=0..n} 4^k * binomial( n + k, n) * binomial( 2*n - 2*k, n - k). - Michael Somos, Feb 25 2012 G.f.: (4 - (1+4*y)*c(y) - (1-4*y)*c(-y))/(2*(1 - (4*y)^2)) with y^2 = x, c(y) = g.f. for A000108 (Catalan). - Wolfdieter Lang, Dec 13 2001 a(n) ~ 2^(1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - 5/16*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
D-finite with recurrence n*(2*n + 1)*a(n) - 2*(4*n - 1)*(4*n + 1)*a(n-1) = 0. - R. J. Mathar, Aug 10 2015 a(n) = 4^n*binomial(2*n + 1/2, n).
O.g.f.: sqrt(c(4*x)/(1 - 16*x)) = sqrt(2/(1 - 16*x)/(1 + sqrt(1 - 16*x))), where
c(y) = g.f. for A000108 (Catalan). In general, c(x)^k/sqrt(1 - 4*x) is the o.g.f. for the sequence binomial(2*n + k, n). (End) [Edited by Petros Hadjicostas, May 25 2020] E.g.f.: 2F2(3/4,5/4; 1,3/2; 16*x).
Sum_{n>=0} 1/a(n) = 3F2(1,1,3/2; 3/4,5/4; 1/16) = 1.108563435104316693... (End)
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} 4^(n-k) * C(2*n+1, 2*k) * C(2*k, k) = a(n).
a(n) = 4^n*hypergeom([-n, -n-1/2], [1], 1). (End)
a(n) = Sum_{k = 0..n} binomial(2*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 2*n, -1 - 2*n], [1], 1). (End)
EXAMPLE
1 + 10*x + 126*x^2 + 1716*x^3 + 24310*x^4 + 352716*x^5 + 5200300*x^6 + ...
MATHEMATICA
Table[Binomial[4n+1, 2n], {n, 0, 30}] (* Harvey P. Dale, Apr 04 2011 *) 4^Range[0, 22] Simplify[ CoefficientList[ Series[ Sqrt[2]/(((Sqrt[1 - 4 x] + 1)^(1/2))*Sqrt[1 - 4 x]), {x, 0, 22}], x]] (* Robert G. Wilson v, Aug 08 2011 *)
PROG
(PARI) a(n) = binomial( 4*n + 1, 2*n)
a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 3).
+10
9
1, 6, 42, 308, 2310, 17556, 134596, 1038312, 8046918, 62587140, 488179692, 3816677592, 29897307804, 234578876616, 1843119744840, 14499208659408, 114181268192838, 900017055167076, 7100134546318044, 56053693786721400, 442824180915099060, 3500419715805068760, 27685137752276452920
COMMENTS
Conjecture: a(p*n) = a(n) (mod p^2) for prime p == 1 (mod 4) and all positive integers n. Cf. A004981. - Peter Bala, Dec 22 2019
FORMULA
G.f.: (1 - 8*x)^(-3/4).
a(n) ~ Gamma(3/4)^-1*n^(-1/4)*2^(3*n)*{1 - 3/32*n^-1 + ...}
a(n) = (-8)^n*binomial(-3/4,n).
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([3/4], [1], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n) = Integral_{x=0..8} ( x^n*(2^(1/4)/(8*Pi*x^(1/4)*(1-x/8)^(3/4)) ) dx, n >= 0. This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)
D-finite with recurrence: n*a(n) +2*(-4*n+1)*a(n-1)=0. - R. J. Mathar, Jan 16 2020
MAPLE
A004982 := n -> (-8)^n*binomial(-3/4, n):
MATHEMATICA
Table[2^n/n! Product[4k+3, {k, 0, n-1}], {n, 0, 30}] (* Harvey P. Dale, Oct 03 2011 *) Table[Sum[2^k*Binomial[2*n-2*k, n-k]*Binomial[n+k, n], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Sep 15 2013 *) FullSimplify[Table[8^n*Gamma[n+3/4]/(n!*Gamma[3/4]), {n, 0, 25}]] (* Vaclav Kotesovec, Sep 15 2013 *) max = 30; s = Hypergeometric1F1[3/4, 1, 8x] + O[x]^(max+1);
PROG
(PARI) a(n)=2^n/n!*prod(k=0, n-1, 4*k+3)
for(n=0, 25, print(a(n)))
(PARI) x='x+O('x^66); Vec((1-8*x)^(-3/4)) \\ Joerg Arndt, Apr 20 2013 (Magma) [1] cat [2^n*&*[4*k+3: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019 (Sage) [8^n*rising_factorial(3/4, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019 (GAP) List([0..25], n-> 2^n*Product([0..n-1], k-> 4*k+3)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
AUTHOR
Joe Keane (jgk(AT)jgk.org)
Triangle T(n,k), read by rows, giving the numerator of the coefficient of x^k in the Boros-Moll polynomial P_n(x) for n >= 0 and 0 <= k <=n.
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6
1, 3, 1, 21, 15, 3, 77, 43, 35, 5, 1155, 885, 1095, 315, 35, 4389, 8589, 7161, 777, 693, 63, 33649, 80353, 42245, 12285, 16485, 3003, 231, 129789, 91635, 233001, 170145, 152625, 20889, 6435, 429, 4023459, 3283533, 9804465, 8625375, 9695565, 1772199, 819819, 109395, 6435, 15646785, 58019335, 49782755, 25638305, 69324255, 31726695, 9794785, 245245, 230945, 12155
COMMENTS
Let P_n(x) = Sum_{k=0..n} (T(n,k)/ A223550(n,k))*x^k be the Boros-Moll polynomial. It follows from the theory in Comtet (1967, pp. 81-83 and 85-86) that the polynomial Q_n(x) = 2^n*n!*P_n(x) has integer coefficients and satisfies the recurrence (x-1)*Q_n(x) = 2*(2*n - 1)*(x^2 - 2)*Q_{n-1}(x) + (16*(n-1)^2 - 1)*(x + 1)*Q_{n-2}(x). We have integral_{y = 0..infinity} dy/(y^4 + 2*x*y + 1)^(n + 1) = Pi * P_n(x)/(2^(n + (3/2)) * (x + 1)^(n + (1/2))) = Pi * Q_n(x)/(2^(2*n + (3/2)) * n! * (x + 1)^(n + (1/2))) for x > -1 and n integer >= 0.
It also follows from the theory in Comtet (1967, pp. 81-83) that g(t) = (sqrt(x + sqrt(x^2 - 1 + t)) - sqrt(x - sqrt(x^2 - 1 + t))) / sqrt((1 - t) * (x^2 - 1 + t)) = Sum_{n >= 0} t^n * P_n(x)/(2^(n - (1/2)) * (x + 1)^(n + (1/2))) for x >= 1 and 0 <= t < 1.
From Comtet's result, we get g(t)^2 = 2*(x - sqrt(1-t))/((1-t) * (x^2 - 1 + t)) = 2/((1-t) * (x + sqrt(1-t))) = Sum_{n >= 0} (Sum_{k=0..n} P_k(x) * P_{n-k}(x)) / (2^(n-1) * (x+1)^(n+1)) * t^n for 0 <= t < 1 and x > 1. (End)
FORMULA
T(n,k)/ A223550(n,k) = 2^(-2*n)*Sum_{j=k..n} 2^j*binomial(2*n - 2*j, n - j)*binomial(n + j, j)*binomial(j, k) = 2^(-2*n)* A067001(n,n-k) for n >= 0 and k = 0..n. P_n(x) = Sum_{k=0..n} (T(n, k)/ A223550(n,k))*x^k = ((2*n)!/4^n/(n!)^2)*2F1([-n, n + 1], [1/2 - n], (x + 1)/2). Recurrence for the polynomial: 4*n*(n - 1)*(x - 1)*P_n(x) = 4*(2*n - 1)*(n - 1)*(x^2 - 2)*P_{n-1}(x) + (16*(n - 1)^2 - 1)*(x + 1)*P_{n-2}(x).
O.g.f. for P_n(x): sqrt((x + 1)/(1 - 2*(x + 1)*w)/(x + sqrt(1 - 2*(x + 1)*w))). [It follows from Comtet's theory and my comments.]
P_n(1) = Sum_{k=0..n} T(n,k)/ A223550(n,k) = A334907(n)/(2^n*n!). (End)
EXAMPLE
P_3(x) = 77/16 + 43*x/4 + 35*x^2/4 + 5*x^3/2.
As a result, integral_{y = 0..infinity} dy/(y^4 + 2*x*y + 1)^4 = Pi * P_3(x)/(2^(3 + (3/2)) * (x + 1)^(3 + (1/2))) = Pi * (40*x^3 + 140*x^2 + 172*x + 77)/(32 * sqrt(2*(x + 1)^7)) for x > -1. - Petros Hadjicostas, May 22 2020 Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins as follows:
1;
3, 1;
21, 15, 3;
77, 43, 35, 5;
1155, 885, 1095, 315, 35;
4389, 8589, 7161, 777, 693, 63;
33649, 80353, 42245, 12285, 16485, 3003, 231;
129789, 91635, 233001, 170145, 152625, 20889, 6435, 429;
... (End)
MATHEMATICA
t[n_, k_] := 2^(-2*n)*Sum[ 2^j*Binomial[2*n - 2*j, n-j]*Binomial[n+j, j]*Binomial[j, k], {j, k, n}]; Table[t[n, k] // Numerator, {n, 0, 9}, {k, 0, n}] // Flatten
PROG
(Magma) /* As triangle: */ [[Numerator(2^(-2*n)*&+[2^j*Binomial(2*n-2*j, n-j)*Binomial(n+j, j)*Binomial(j, k): j in [k..n]]): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Mar 22 2013
Triangle T(n,k), read by rows, giving the denominator of the coefficient of x^k in the Boros-Moll polynomial P_n(x) for n >= 0 and 0 <= k <= n.
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6
1, 2, 1, 8, 4, 2, 16, 4, 4, 2, 128, 32, 32, 16, 8, 256, 128, 64, 8, 16, 8, 1024, 512, 128, 32, 64, 32, 16, 2048, 256, 256, 128, 128, 32, 32, 16, 32768, 4096, 4096, 2048, 2048, 512, 512, 256, 128, 65536, 32768, 8192, 2048, 4096, 2048, 1024, 64, 256, 128
COMMENTS
As Chen and Xia (2009) state, the Boros-Moll polynomial P_n(x) can be viewed as a Jacobi polynomial P_n^{a,b}(x) with a = n + (1/2) and b = -(n + (1/2)). For more information about the relation of this polynomial P_n(x) to the theory in Comtet (1967, pp. 81-83 and 85-86), see my comments for A223549. - Petros Hadjicostas, May 22 2020
FORMULA
A223549(n,k)/T(n,k) = 2^(-2*n)*Sum_{j=k..n} 2^j*binomial(2*n - 2*j, n - j)*binomial(n + j, j)*binomial(j, k) = 2^(-2*n)* A067001(n,n-k) for n >= 0 and k = 0..n.
P_n(x) = Sum_{k=0..n} ( A223549(n,k)/T(n,k))*x^k = ((2*n)!/4^n/(n!)^2)*2F1([-n, n + 1], [1/2 - n], (x + 1)/2). Recurrence for the polynomial: 4*n*(n - 1)*(x - 1)*P_n(x) = 4*(2*n - 1)*(n - 1)*(x^2 - 2)*P_{n-1}(x) + (16*(n - 1)^2 - 1)*(x + 1)*P_{n-2}(x).
P_n(1) = Sum_{k=0..n} A223549(n,k)/T(n,k) = A334907(n)/(2^n*n!). (End)
EXAMPLE
P_3(x) = 77/16 + 43*x/4 + 35*x^2/4 + 5*x^3/2.
Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins as follows:
1;
2, 1;
8, 4, 2;
16, 4, 4, 2;
128, 32, 32, 16, 8;
256, 128, 64, 8, 16, 8;
1024, 512, 128, 32, 64, 32, 16;
2048, 256, 256, 128, 128, 32, 32, 16;
32768, 4096, 4096, 2048, 2048, 512, 512, 256, 128;
65536, 32768, 8192, 2048, 4096, 2048, 1024, 64, 256, 128;
... (End)
MATHEMATICA
t[n_, k_] := 2^(-2*n)*Sum[ 2^j*Binomial[2*n - 2*j, n-j]*Binomial[n+j, j]*Binomial[j, k], {j, k, n}]; Table[t[n, k] // Denominator, {n, 0, 9}, {k, 0, n}] // Flatten
PROG
(Magma) /* As triangle: */ [[Denominator(2^(-2*n)*&+[2^j*Binomial(2*n-2*j, n-j)*Binomial(n+j, j)*Binomial(j, k): j in [k..n]]): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Mar 22 2013
Numerator of Sum_{k=0..n} 2^(k-2*n) * binomial(2*n-2*k,n-k) * binomial(n+k,n).
+10
4
1, 3, 21, 77, 1155, 4389, 33649, 129789, 4023459, 15646785, 122044923, 477084699, 7474326951, 29322359577, 230389968105, 906200541213, 57090634096419, 225004263791769, 1775033636579511, 7006711723340175, 110706045228774765
COMMENTS
Numerator of e(0,n) (see the Maple line).
The generating function of the full fraction is (1-2*x)^(-3/4). - R. J. Mathar, Nov 06 2011
FORMULA
Numerator of 2^n*Gamma(n + 3/4)/(Gamma(3/4)*n!). - R. J. Mathar, Nov 06 2011 If fr(n) = A067002(n)/ A046161(n), then fr(n) = P_n(0), where P_n(x) is the Boros-Moll polynomial mentioned in A223549 and A223550 (and whose coefficients are the numbers e(l,n) = A067001(n,n-l)/2^(2*n) that are mentioned in the Maple line below with l = 0..n). Recurrence for fr(n): 4*n*(n - 1)*fr(n) = 8*(2*n - 1)*(n - 1)*fr(n-1) - (16*(n-1)^2 - 1)*fr(n-2) for n >= 2 with fr(0) = 1 and fr(1) = 3/2. (End)
EXAMPLE
1, 3/2, 21/8, 77/16, 1155/128, 4389/256, 33649/1024, 129789/2048, 4023459/32768, ... = A067002/ A046161.
MAPLE
e := proc(l, m) local k; add(2^(k-2*m)*binomial(2*m-2*k, m-k)*binomial(m+k, m)*binomial(k, l), k=l..m); end;
MATHEMATICA
Numerator[Table[Sum[2^(k-2n) Binomial[2n-2k, n-k]Binomial[n+k, n], {k, 0, n}], {n, 0, 30}]] (* Harvey P. Dale, Oct 19 2012 *)
Coefficients of a polynomial representation of the integral of 1/(x^4 + 2*a*x^2 + 1)^(n+1) from x = 0 to infinity.
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2
1, 6, 4, 42, 60, 24, 308, 688, 560, 160, 2310, 7080, 8760, 5040, 1120, 17556, 68712, 114576, 99456, 44352, 8064, 134596, 642824, 1351840, 1572480, 1055040, 384384, 59136, 1038312, 5864640, 14912064, 21778560, 19536000, 10695168, 3294720
COMMENTS
The integral N(a;n) = Integral_{x=0..infinity} 1/(x^4 + 2*a*x^2 + 1)^(n+1) has a polynomial representation P_n(a) = 2^(n + 3/2) * (a+1)^(n + 1/2) * N(a;n) / Pi (known as the Boros-Moll polynomial). The table contains the coefficients T(n,l) of P_n(a) = 2^(-2*n)*Sum_{l=0..n} T(n,l)*a^l in row n and column l (with n >= 0 and 0 <= l <= n).
FORMULA
Bivariate o.g.f.: Sum_{n,l >= 0} T(n,l)*x^n*y^l = sqrt((1 + y)/(1 - 8*x*(1 + y))/(y + sqrt(1 - 8*x*(1 + y)))). (End)
EXAMPLE
The table T(n,l) (with rows n >= 0 and columns l = 0..n) starts:
1;
6, 4;
42, 60, 24;
308, 688, 560, 160;
2310, 7080, 8760, 5040, 1120;
17556, 68712, 114576, 99456, 44352, 8064;
...
For n = 2, N(a;2) = Integral_{x=0..oo} dx/(x^4 + 2*a*x + 1)^3 = 2^(-2*2)*(Sum_{l=0..2} T(2,l)*a^l) * Pi/(2^(2 + 3/2) * (a + 1)^(2 + 1/2) = (42 + 60*a + 24*a^2) * Pi/(32 * (2*(a+1))^(5/2)) for a > -1. - Petros Hadjicostas, May 25 2020
MAPLE
add(2^k*binomial(2*m-2*k, m-k)*binomial(m+k, m)*binomial(k, l), k=l..m):
end:
MATHEMATICA
t[m_, l_] := Sum[2^k*Binomial[2*m-2*k, m-k]*Binomial[m+k, m]*Binomial[k, l], {k, l, m}]; Table[t[m, l], {m, 0, 11}, {l, 0, m}] // Flatten (* Jean-François Alcover, Jan 09 2014, after Maple, adapted May 2020 *)
T(n,k) = Sum_{j=1..n} 2^j*binomial(2*n-2*j, n-j)*binomial(n+j, n)*binomial(j, k), triangle read by rows (n >= 0 and 0 <= k <= n).
+10
0
0, 4, 4, 36, 60, 24, 288, 688, 560, 160, 2240, 7080, 8760, 5040, 1120, 17304, 68712, 114576, 99456, 44352, 8064, 133672, 642824, 1351840, 1572480, 1055040, 384384, 59136, 1034880, 5864640, 14912064, 21778560, 19536000, 10695168, 3294720, 439296
COMMENTS
This was the original version of A126936.
FORMULA
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = -1/sqrt(1 - 4*x) + sqrt((1 + y)/(1 - 8*x*(1 + y))/(y + sqrt(1 - 8*x*(1 + y)))).
EXAMPLE
Table T(n,k) (with rows n >= 0 and columns k = 0..n) begins as follows:
0;
4, 4;
36, 60, 24;
288, 688, 560, 160;
2240, 7080, 8760, 5040, 1120;
17304, 68712, 114576, 99456, 44352, 8064;
133672, 642824, 1351840, 1572480, 1055040, 384384, 59136;
...
MATHEMATICA
t[l_, m_] := Sum[2^k*Binomial[2*m-2*k, m-k]*Binomial[m+k, m]*Binomial[k, l], {k, 1, m}]; Table[t[l, m], {m, 0, 11}, {l, 0, m}] // Flatten (* Jean-François Alcover, Jan 09 2014_ from the original version of A126936 *)
PROG
(PARI) T(n, k) = sum(j=1, n, 2^j*binomial(2*n-2*j, n-j)*binomial(n+j, n)*binomial(j, k));
tabl(nn) = {for (n=0, nn, for (k=0, n, print1(T(n, k), ", "); ); print(); ); }
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