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Triangle read by rows: T(n, k) = binomial(2n, 2k).
+10
39
1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760
COMMENTS
Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005 Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005 Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005 Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008 Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes. The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term). (End)
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010 Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)* A086646(n,k). - R. J. Mathar, Mar 12 2013 Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013 Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.
FORMULA
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008 Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009 Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013 n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459. R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)
EXAMPLE
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)
MAPLE
A086645:=(n, k)->binomial(2*n, 2*k): seq(seq( A086645(n, k), k=0..n), n=0..12);
MATHEMATICA
Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)
PROG
(PARI) {T(n, k) = binomial(2*n, 2*k)};
(PARI) {T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */ (Maxima) create_list(binomial(2*n, 2*k), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */ (Magma) /* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016
Triangle of even numbered entries of odd numbered rows of Pascal's triangle A007318.
+10
24
1, 1, 3, 1, 10, 5, 1, 21, 35, 7, 1, 36, 126, 84, 9, 1, 55, 330, 462, 165, 11, 1, 78, 715, 1716, 1287, 286, 13, 1, 105, 1365, 5005, 6435, 3003, 455, 15, 1, 136, 2380, 12376, 24310, 19448, 6188, 680, 17, 1, 171, 3876, 27132, 75582, 92378, 50388, 11628, 969, 19, 1, 210, 5985, 54264, 203490, 352716, 293930, 116280, 20349, 1330, 21
COMMENTS
The row polynomials Pe(n, x) := Sum_{m=0..n} a(n, m)*x^m appear as numerators of the generating functions for the even numbered column sequences of array A034870. Elements have the same parity as those of Pascal's triangle.
All zeros of polynomial Pe(n, x) are negative. They are -tan^2(Pi/2*n+1), -tan^2(2*Pi/2*n+1), ..., -tan^2(n*Pi/2*n+1). Moreover, for m >= 1, Pe(m, -x^2) is the characteristic polynomial of the linear difference equation with constant coefficients for differences between multiples of 2*m+1 with even and odd digit sum in base 2*m in the interval [0,(2*m)^n). - Vladimir Shevelev and Peter J. C. Moses, May 22 2012 The row polynomial Pe(d, x), multiplied by (2*d)!/d! = A001813(d), gives the numerator polynomial of the o.g.f. of the sequence of the diagonal d, for d >= 0, of the Sheffer triangle Lah[4,1] given in A048854. - Wolfdieter Lang, Oct 12 2017
REFERENCES
A. M. Yaglom and I. M. Yaglom, An elementary proof of the Wallis, Leibniz and Euler formulas for pi. Uspekhi Matem. Nauk, VIII (1953), 181-187(in Russian).
FORMULA
T(n, m) = binomial(2*n+1, 2*m) = A007318(2*n+1, 2*m), n >= m >= 0, otherwise 0. E.g.f.: sinh(t)*cosh(sqrt(x)*t) = t + (1 + 3*x)*t^3/3! + (1 + 10*x + 5*x^2)*t^5/5! + (1 + 21*x + 35*x^2 + 7*x^3)*t^7/7! + ....
O.g.f.: A(x,t) = (1 + (x - 1)*t)/( (1 + (x - 1)*t)^2 - 4*t*x ) = 1 + (1 + 3*x)*t + (1 + 10*x + 5*x^2)*t^2 + ...
The function A( x/(x + 4), t*(x + 4)/4 ) = 1 + (1 + x)*t + (1 + 3*x + x^2)*t^2 + ... is the o.g.f. for A085478. O.g.f. for n-th diagonal: ( Sum_{k = 0..n} binomial(2*n,2*k)*x^k )/(1 - x)^(2*n).
n-th row polynomial R(n,x) = (1/2)*( (1 + sqrt(x))^(2*n+1) - (sqrt(x) - 1)^(2*n+1) ).
T(n, k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2) with T(0,0)=T(1,0)=1, T(1,1)=3, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 26 2013 Define S(r,N) = Sum_{j = 1..N} j^r. Then the following identity holds for n >= 0: (1/2)*(N^2 + N)^(2*n+1) = T(n,0)*S(2*n+1,N) + T(n,1)*S(2*n+3,N) + ... + T(n,n)* S(4*n+1,N). Some examples are given below. (End)
EXAMPLE
Triangle a(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 1 3
2: 1 10 5
3: 1 21 35 7
4: 1 36 126 84 9
5: 1 55 330 462 165 11
6: 1 78 715 1716 1287 286 13
7: 1 105 1365 5005 6435 3003 455 15
8: 1 136 2380 12376 24310 19448 6188 680 17
9: 1 171 3876 27132 75582 92378 50388 11628 969 19
10: 1 210 5985 54264 203490 352716 293930 116280 20349 1330 21
(1/2)*(N^2 + N) = Sum_{j = 1..N} j.
(1/2)*(N^2 + N)^3 = Sum_{j = 1..N} j^3 + 3*Sum_{j = 1..N} j^5.
(1/2)*(N^2 + N)^5 = Sum_{j = 1..N} j^5 + 10*Sum_{j = 1..N} j^7 + 5*Sum_{j = 1..N} j^9.
(1/2)*(N^2 + N)^7 = Sum_{j = 1..N} j^7 + 21*Sum_{j = 1..N} j^9 + 35*Sum_{j = 1..N} j^11 + 7*Sum_{j = 1..N} j^13. (End)
MAPLE
f := (x, t) -> cosh(sqrt(x)*t)*sinh(t); seq(seq(coeff(((2*n+1)!*coeff(series(f(x, t), t, 2*n+2), t, 2*n+1)), x, k), k=0..n), n=0..9); # Peter Luschny, Jul 29 2013
MATHEMATICA
T[n_, k_] /; 0 <= k <= n := T[n, k] = 2T[n-1, k] + 2T[n-1, k-1] + 2T[n-2, k-1] - T[n-2, k] - T[n-2, k-2]; T[0, 0] = T[1, 0] = 1; T[1, 1] = 3; T[_, _] = 0;
Table[Binomial[2*n+1, 2*k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Aug 01 2019 *)
PROG
(PARI) T(n, k) = binomial(2*n+1, 2*k); \\ G. C. Greubel, Aug 01 2019 (Magma) [[Binomial(2*n+1, 2*k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Aug 01 2019 (Sage) [[binomial(2*n+1, 2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019 (GAP) Flat(List([0..12], n-> List([0..n], k-> Binomial(2*n+1, 2*k) ))); # G. C. Greubel, Aug 01 2019
Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1).
+10
13
1, 3, 1, 5, 10, 1, 7, 35, 21, 1, 9, 84, 126, 36, 1, 11, 165, 462, 330, 55, 1, 13, 286, 1287, 1716, 715, 78, 1, 15, 455, 3003, 6435, 5005, 1365, 105, 1, 17, 680, 6188, 19448, 24310, 12376, 2380, 136, 1, 19, 969, 11628, 50388, 92378, 75582, 27132, 3876, 171, 1
COMMENTS
A subset of Pascal's triangle A007318. Elements have the same parity as those of Pascal's triangle.
Let E(y) = cosh(sqrt(y)) = 1 + 3*y/3! + 5*y^2/5! + 7*y^3/7! + .... Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n+1)! as defined in Wang and Wang. Cf. A086645. - Peter Bala, Aug 06 2013 The row polynomial P(d, x) = Sum_{k=0..d} T(d, k)*x^k, multiplied by (2*d)!/d! = A001813(d), gives the numerator polynomial of the o.g.f. of the sequence of the diagonal d, for d >= 0, of the Sheffer triangle Lah[4,3] given in A292219. - Wolfdieter Lang, Oct 12 2017
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.
FORMULA
G.f. for column k: Sum_{j=0..k+1} C(2*(k+1), 2*j)x^j)/(1-x)^(2*(k+1)). - Paul Barry, Feb 24 2005 G.f.: A(x, y) = (1 + x*(1-y))/( (1 + x*(1-y))^2 - 4*x ). - Paul D. Hanna, Feb 28 2005 E.g.f.: 1/sqrt(x)*sinh(sqrt(x)*t)*cosh(t) = t + (3 + x)*t^3/3! + (5 + 10*x + x^2)*t^5/5! + .... - Peter Bala, Jul 29 2013 T(n+2,k+2) = 2*T(n+1,k+2) + 2*T(n+1,k+1) - T(n,k+2) + 2*T(n,k+1) - T(n,k). - Emanuele Munarini, Jul 05 2017
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 3 1
2: 5 10 1
3: 7 35 21 1
4: 9 84 126 36 1
5: 11 165 462 330 55 1
6: 13 286 1287 1716 715 78 1
7: 15 455 3003 6435 5005 1365 105 1
8: 17 680 6188 19448 24310 12376 2380 136 1
9: 19 969 11628 50388 92378 75582 27132 3876 171 1
10: 21 1330 20349 116280 293930 352716 203490 54264 5985 210 1
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n+1)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + 3*y/3! + 5*y^2/5! + 7*y^3/7! + 9*y^4/9! + ....
2nd col: 1/3!*y*cosh(sqrt(y)) = y/3! + 10*y^2/5! + 35*y^3/7! + 84*y^4/9! + ....
3rd col: 1/5!*y^2*cosh(sqrt(y)) = y^2/5! + 21*y^3/7!! + 126*y^4/9! + 462*y^5/11! + .... (End)
MATHEMATICA
Flatten[Table[Binomial[2n+1, 2k+1], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Jun 19 2014 *)
PROG
(PARI) {T(n, k)=local(X=x+x*O(x^n), Y=y+y*O(y^k)); polcoeff(polcoeff((1+X*(1-Y))/((1+X*(1-Y))^2-4*X), n, x), k, y)} \\ Paul D. Hanna, Feb 28 2005 (PARI) T(n, k) = binomial(2*n+1, 2*k+1);
for(n=0, 12, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Aug 01 2019 (Maxima) create_list(binomial(2*n+1, 2*k+1), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */ (Magma) [Binomial(2*n+1, 2*k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019 (Sage) [[binomial(2*n+1, 2*k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019 (GAP) Flat(List([0..12], n-> List([0..n], k-> Binomial(2*n+1, 2*k+1) ))); # G. C. Greubel, Aug 01 2019
Numbers j such that 8*(j^2) + 9 = k^2 for some positive number k.
+10
13
0, 3, 18, 105, 612, 3567, 20790, 121173, 706248, 4116315, 23991642, 139833537, 815009580, 4750223943, 27686334078, 161367780525, 940520349072, 5481754313907, 31950005534370, 186218278892313, 1085359667819508, 6325939728024735, 36870278700328902, 214895732473948677
COMMENTS
The ratio k(n) /(2*j(n)) tends to sqrt(2) as n increases.
The squares of the numbers in this sequence are one less than a triangular number: a(n)^2 = A164080(n). For example, 18^2 is 324, and 325 is a triangular number. a(n)^2 + 1 = A164055(n). a(n)^2 = A072221(n)( A072221(n)+1)/2 - 1. - Tanya Khovanova & Alexey Radul, Aug 09 2009 For n > 0, a(n+1) is the n-th almost balancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022
FORMULA
a(1)=0, a(2)=3 then a(n) = 6*a(n-1) - a(n-2).
a(n) = ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1))*3/4/sqrt(2). - Max Alekseyev, Jan 11 2007 E.g.f.: 3 - 3*exp(3*x)*(4*cosh(2*sqrt(2)*x) - 3*sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Nov 25 2022
MATHEMATICA
Rest@ CoefficientList[Series[3 x^2/(1 - 6 x + x^2), {x, 0, 24}], x] (* Michael De Vlieger, Nov 02 2020 *)
PROG
(Haskell)
a106328 n = a106328_list !! (n-1)
a106328_list = 0 : 3 : zipWith (-) (map (* 6) (tail a106328_list)) a106328_list
(PARI) concat(0, Vec(3*x^2/(1-6*x+x^2) + O(x^40))) \\ Michel Marcus, Sep 07 2016
Numbers k such that k^2 = 8*j^2 + 9.
+10
4
3, 9, 51, 297, 1731, 10089, 58803, 342729, 1997571, 11642697, 67858611, 395508969, 2305195203, 13435662249, 78308778291, 456417007497, 2660193266691, 15504742592649, 90368262289203, 526704831142569, 3069860724566211, 17892459516254697, 104284896372961971
COMMENTS
The ratio a(n)/(2*j(n)) tends to sqrt(2) as n increases.
For n > 0, a(n+1) is the n-th almost Lucas-balancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022
FORMULA
a(1)=3, a(2)=9 then a(n) = 6*a(n-1)-a(n-2).
a(n) = 3*((3-2*sqrt(2))^(n-1) + (3+2*sqrt(2))^(n-1))/2. - Colin Barker, Oct 13 2015 E.g.f.: 3*exp(3*x)*(3*cosh(2*sqrt(2)*x) - 2*sqrt(2)*sinh(2*sqrt(2)*x)) - 9. - Stefano Spezia, Nov 25 2022
MATHEMATICA
CoefficientList[Series[3 x (1 - 3 x)/(1 - 6 x + x^2), {x, 0, 23}], x] (* Michael De Vlieger, Nov 02 2020 *)
Triangle T read by rows: n-th row (n>=0) gives the non-vanishing coefficients of the polynomial q(n,x) = 2^(-n)*((x+1)^(2^n) - (x-1)^(2^n))/2.
+10
4
1, 1, 1, 1, 1, 7, 7, 1, 1, 35, 273, 715, 715, 273, 35, 1, 1, 155, 6293, 105183, 876525, 4032015, 10855425, 17678835, 17678835, 10855425, 4032015, 876525, 105183, 6293, 155, 1, 1, 651, 119133, 9706503, 430321633, 11618684091, 205263418941, 2492484372855, 21552658988805, 136248095712855
COMMENTS
Row n has length 1 for n = 1 and 2^(n-1) = A000079(n-1) for n >= 1. The triangle T gives the non-vanishing coefficients of the polynomial q(0,x) = 1 and q(n,x) = 2^(-n)*Sum_{k=0..2^(n-1)-1} A281122(n,k) * x^(2^n-1-2*k), n >= 1. The polynomial q(n,x) = product_{k=0..n-1} p(k,x) with polynomial p(n,x) = ((x+1)^(2^n) + (x-1)^(2^n))/2, whose coefficients are tabulated in A201461. The algorithm r(n) = 1/2*(r(n-1) + A/r(n-1)), starting with r(0) = A, used for approximating sqrt(A), which is known as the Babylonian method or Hero's method after the first-century Greek mathematician Hero of Alexandria and which can be derived from Newton's method, generates fractions beginning with (A+1)/2, (A^2 + 6*A + 1)/(4*(A + 1)), (A^4 + 28*A^3 + 70*A^2 + 28*A + 1)/(8*(A^3 + 7*A^2 + 7*A + 1), ... This is sqrt(A)*p(n,sqrt(A))/(2^n*q(n,sqrt(A))) with the given polynomials p(n,x) and q(n,x).
FORMULA
T(n, k) = 1 for n=0, k=0, and T(n, k) = 2^(-n) * binomial(2^n,2*k+1) = A103328(2^(n-1),k) for k = 0..2^(n-1)-1 and n >= 1. - Wolfdieter Lang, Jan 20 2017
EXAMPLE
The triangle T(n, k) starts with
1
1
1, 1
1, 7, 7, 1
1, 35, 273, 715, 715, 273, 35, 1
1, 155, 6293, 105183, 876525, 4032015, 10855425, 17678835, 17678835, 10855425, 4032015, 876525, 105183, 6293, 155, 1
etc., since the first few polynomials are
q(0,x) = 1,
q(1,x) = x,
q(2,x) = x^3 + x = x*(x^2 + 1),
q(3,x) = x^7 + 7*x^5 + 7*x^3 + x = x*(x^2 + 1)*(x^4 + 6*x^2 + 1),
q(4,x) = x^15 + 35*x^13 + 273*x^11 + 715*x^9 + 715*x^7 + 273*x^5 + 35*x^3 + x = x*(x^2 + 1)*(x^4 + 6*x^2 + 1)*(x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1),
etc.
MATHEMATICA
t={1}; T[n_, k_]:=Table[2^(-n)Binomial[2^n, 2k+1], {n, 1, 6}, {k, 0, 2^(n-1)-1}]; Do[AppendTo[t, T[n, k]]]; Flatten[t] (* Indranil Ghosh, Feb 22 2017 *)
Triangle T read by rows: n-th row (n>=0) gives the non-vanishing coefficients of the polynomial q(n,x) = ((x+1)^(2^n) - (x-1)^(2^n))/2.
+10
3
1, 2, 4, 4, 8, 56, 56, 8, 16, 560, 4368, 11440, 11440, 4368, 560, 16, 32, 4960, 201376, 3365856, 28048800, 129024480, 347373600, 565722720, 565722720, 347373600, 129024480, 28048800, 3365856, 201376, 4960, 32
COMMENTS
The length of row n is 1 for n = 0 and 2^(n-1) = A000079(n-1) for n >= 1. The polynomial q(n,x) = 2^n*Product_{k=0..n-1} p(k,x) with polynomial p(n,x) = ((x+1)^(2^n) + (x-1)^(2^n))/2, whose coefficients are tabulated in A201461. The row polynomials are q(n, x) = Sum_{k=0..2^(n-1)-1} T(n, k)*x^(2^n-1-2*k) for n >= 1, and q(0,x) = 1. - Wolfdieter Lang, Jan 19 2017 A201461 and T(n,k) are a bisection of row 2^n of Pascal's triangle A007318.
The algorithm r(n) = 1/2*(r(n-1) + A/r(n-1)), starting with r(0) = A, used for approximating sqrt(A), which is known as the Babylonian method or Hero's method after the first-century Greek mathematician Hero of Alexandria and which can be derived from Newton's method, generates fractions beginning with (A+1)/2, (A^2 + 6*A + 1)/(4*A + 4), (A^4 + 28*A^3 + 70*A^2 + 28*A + 1)/(8*A^3 + 56*A^2 + 56*A + 8), ... This is sqrt(A)*p(n,sqrt(A))/q(n,sqrt(A)) with the given polynomials p(n,x) and q(n,x).
FORMULA
T(n, k) = 1 if n = 0 and k = 0, and T(n, k) = binomial(2^n,2*k+1) = A103328(2^(n-1),k) for k = 0..2^(n-1)-1 and n >= 1. - Wolfdieter Lang, Jan 19 2017
EXAMPLE
The triangle of non-vanishing coefficients starts with
1
2
4, 4
8, 56, 56, 8
16, 560, 4368, 11440, 11440, 4368, 560, 16
etc., since the first few polynomials are
q(0,x) = 1,
q(1,x) = 2*x,
q(2,x) = 4*x^3 + 4*x = 4*x*(x^2 + 1),
q(3,x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x = 8*x*(x^2 + 1)*(x^4 + 6*x^2 + 1),
q(4,x) = 16*x^15 + 560*x^13 + 4368*x^11 + 11440*x^9 + 11440*x^7 + 4368*x^5 + 560*x^3 + 16*x = 16*x*(x^2 + 1)*(x^4 + 6*x^2 + 1)*(x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1),
etc.
MAPLE
CoeffList := p -> remove(n->n=0, [op(PolynomialTools:-CoefficientList(p, x))]):
Tpoly := n -> (1/2)*((x+1)^(2^n) - (x-1)^(2^n));
seq(print(CoeffList(Tpoly(n))), n=0..5); # Peter Luschny, Feb 04 2021
MATHEMATICA
q[n_] := DeleteCases[ CoefficientList[ Expand[((x +1)^(2^n) - (x -1)^(2^n))/2], x], 0]; Array[q, 7, 0] // Flatten (* Robert G. Wilson v, Jan 16 2017 *) t={1}; T[n_, k_]:=Table[Binomial[2^n, 2k+1], {n, 1, 6}, {k, 0, 2^(n-1)-1}]; Do[AppendTo[t, T[n, k]]]; Flatten[t] (* Indranil Ghosh, Feb 22 2017 *)
Degree of SO(n,C), the special orthogonal group, as an algebraic variety.
+10
2
2, 8, 40, 384, 4768, 111616, 3433600, 196968448, 14994641408, 2112561610752, 397713919469568, 137785594909556736, 64120367727755108352, 54666180849611078369280, 62864933930402036994048000, 131959858152100309567348408320, 374913851106401853810511580364800, 1938349609799484523235647407112847360, 13603397258157549964912652571654029312000
LINKS
M. Brandt, D. Bruce, T. Brysiewicz, R. Krone, E. Robeva, The degree of SO(n), arXiv:1701.03200 [math.AG], 2017
FORMULA
a(n) = 2^(n-1)*det(binomial(2n-2i-2j, n-2i))_{i,j=1..floor(n/2)}.
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i+2*j-2, 2*i-1) = A103328(i+j-1, i-1); then a(2*n+1) = 2^(2*n)*det(M_n). Let M_n be the n X n matrix M_n(i,j) = binomial(2*i+2*j-4, 2*i-2) = A086645(i+j-2, i-1); then a(2*n) = 2^(2*n-1)*det(M_n).
EXAMPLE
For n = 4 we have a(4) = 2^3*det({6,1},{1,1}) = 2^3*(6-1) = 40.
MATHEMATICA
a[n_] := 2^(n-1) Det[Table[Binomial[2n-2i-2j, n-2i], {i, n/2}, {j, n/2}]];
PROG
(PARI) a(n) = 2^(n-1)*matdet(matrix(n\2, n\2, i, j, binomial(2*n-2*i-2*j, n-2*i))); \\ Michel Marcus, Jan 14 2017
Degree of Sp(n,C), the symplectic group, as an algebraic variety.
+10
2
2, 24, 1744, 769408, 2063048448, 33639061257216, 3336558889746769920, 2013547640260319665029120, 7394216956327379315321530941440, 165246096715086213509958939917335920640, 22475501333841331145301219459764999435840913408
LINKS
M. Brandt, D. Bruce, T. Brysiewicz, R. Krone, E. Robeva, The degree of SO(n), arXiv:1701.03200 [math.AG], 2017.
FORMULA
a(n) = det(binomial(2*i+2*j-2,2*i-1))_{i,j=1}^n.
Let M_n be the n X n matrix M_n(i,j) = binomial(2*i+2*j-2,2*i-1) = A103328(i+j-1,i-1); then a(n) = det(M_n).
EXAMPLE
For n=2 we have a(2)=det({2,4},{4,20})=24.
MATHEMATICA
a[n_] := Det[Table[Binomial[2i+2j-2, 2i-1], {i, 1, n}, {j, 1, n}]]
PROG
(PARI) a(n) = matdet(matrix(n, n, i, j, binomial(2*i+2*j-2, 2*i-1))); \\ Michel Marcus, Jan 14 2017
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