A185896 Triangle of coefficients of (1/sec^2(x))*D^n(sec^2(x)) in powers of t = tan(x), where D = d/dx.
1, 0, 2, 2, 0, 6, 0, 16, 0, 24, 16, 0, 120, 0, 120, 0, 272, 0, 960, 0, 720, 272, 0, 3696, 0, 8400, 0, 5040, 0, 7936, 0, 48384, 0, 80640, 0, 40320, 7936, 0, 168960, 0, 645120, 0, 846720, 0, 362880, 0, 353792, 0, 3256320, 0, 8951040, 0, 9676800, 0, 3628800
Offset: 0
Examples
Table begins n\k|.....0.....1.....2.....3.....4.....5.....6 ============================================== 0..|.....1 1..|.....0.....2 2..|.....2.....0.....6 3..|.....0....16.....0....24 4..|....16.....0...120.....0...120 5..|.....0...272.....0...960.....0...720 6..|...272.....0..3696.....0..8400.....0..5040 Examples of recurrence relation T(4,2) = 3*(T(3,1) + T(3,3)) = 3*(16 + 24) = 120; T(6,4) = 5*(T(5,3) + T(5,5)) = 5*(960 + 720) = 8400. Example of integral formula (6) ... Integral_{t = -1..1} (1-t^2)*(16-120*t^2+120*t^4)*(272-3696*t^2+8400*t^4-5040*t^6) dt = 2830336/1365 = -2^13*Bernoulli(12). Examples of sign change statistic sc on snakes of type (0,0) = = = = = = = = = = = = = = = = = = = = = = .....Snakes....# sign changes sc.......t^sc = = = = = = = = = = = = = = = = = = = = = = n=1 ...0 1 -2 0...........1................t ...0 2 -1 0...........1................t yields R(1,t) = 2*t; n=2 ...0 1 -2 3 0.........2................t^2 ...0 1 -3 2 0.........2................t^2 ...0 2 1 3 0..........0................1 ...0 2 -1 3 0.........2................t^2 ...0 2 -3 1 0.........2................t^2 ...0 3 1 2 0..........0................1 ...0 3 -1 2 0.........2................t^2 ...0 3 -2 1 0.........2................t^2 yields R(2,t) = 2 + 6*t^2.
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
- K. Boyadzhiev, Derivative Polynomials for tanh, tan, sech and sec in Explicit Form, arXiv:0903.0117 [math.CA], 2009-2010.
- M-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
- A. Hodges and C. V. Sukumar, Bernoulli, Euler, permutations and quantum algebras, Proc. R. Soc. A (2007) 463, 2401-2414 doi:10.1098/rspa.2007.0001
- Michael E. Hoffman, Derivative polynomials, Euler polynomials,and associated integer sequences, Electronic Journal of Combinatorics, Volume 6 (1999), Research Paper #R21.
- Michael E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102 (1995), 23-30.
- M. Josuat-Verges, Enumeration of snakes and cycle-alternating permutations, arXiv:1011.0929 [math.CO], 2010.
- Shi-Mei Ma, Qi Fang, Toufik Mansour, Yeong-Nan Yeh, Alternating Eulerian polynomials and left peak polynomials, arXiv:2104.09374, 2021
Programs
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Maple
R = proc(n) option remember; if n=0 then RETURN(1); else RETURN(expand(diff((u^2+1)*R(n-1), u))); fi; end proc; for n from 0 to 12 do t1 := series(R(n), u, 20); lprint(seriestolist(t1)); od:
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Mathematica
Table[(-1)^(n + 1)*(-1)^((n - k)/2)*Sum[j!*StirlingS2[n + 1, j]*2^(n + 1 - j)*(-1)^(n + j - k)*Binomial[j - 1, k], {j, k + 1, n + 1}], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Jul 22 2017 *)
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PARI
{T(n, k) = if( n<0 || k<0 || k>n, 0, if(n==k, n!, (k+1)*(T(n-1, k-1) + T(n-1, k+1))))};
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PARI
{T(n, k) = my(A); if( n<0 || k>n, 0, A=1; for(i=1, n, A = ((1 + x^2) * A)'); polcoeff(A, k))}; /* Michael Somos, Jun 24 2017 */
Formula
GENERATING FUNCTION
E.g.f.:
(1)... F(t,z) = 1/(cos(z)-t*sin(z))^2 = Sum_{n>=0} R(n,t)*z^n/n! = 1 + (2*t)*z + (2+6*t^2)*z^2/2! + (16*t+24*t^3)*z^3/3! + ....
The e.g.f. equals the square of the e.g.f. of A104035.
Continued fraction representation for the o.g.f:
(2)... F(t,z) = 1/(1-2*t*z - 2*(1+t^2)*z^2/(1-4*t*z -...- n*(n+1)*(1+t^2)*z^2/(1-2*n*(n+1)*t*z -....
RECURRENCE RELATION
(3)... T(n,k) = (k+1)*(T(n-1,k-1) + T(n-1,k+1)).
ROW POLYNOMIALS
The polynomials R(n,t) satisfy the recurrence relation
(4)... R(n+1,t) = d/dt{(1+t^2)*R(n,t)} with R(0,t) = 1.
Let D be the derivative operator d/dt and U = t, the shift operator.
(5)... R(n,t) = (D + DUU)^n 1
RELATION WITH OTHER SEQUENCES
A) Derivative Polynomials A155100
The polynomials (1+t^2)*R(n,t) are the polynomials P_(n+2)(t) of A155100.
Put S(n,t) = R(n,i*t), where i = sqrt(-1). We have the definite integral evaluation
(6)... Integral_{t = -1..1} (1-t^2)*S(m,t)*S(n,t) dt = (-1)^((m-n)/2)*2^(m+n+3)*Bernoulli(m+n+2).
The case m = n is equivalent to the result of [Grosset and Veselov]. The methods used there extend to the general case.
C) Zigzag Numbers A000111
D) Eulerian Numbers A008292
The polynomials R(n,t) are related to the Eulerian polynomials A(n,t) via
(8)... R(n,t) = (t+i)^n*A(n+1,(t-i)/(t+i))
with the inverse identity
(9)... A(n+1,t) = (-i/2)^n*(1-t)^n*R(n,i*(1+t)/(1-t)),
where {A(n,t)}n>=1 = [1,1+t,1+4*t+t^2,1+11*t+11*t^2+t^3,...] is the sequence of Eulerian polynomials and i = sqrt(-1).
E) Ordered set partitions A019538
(10)... R(n,t) = (-2*i)^n*T(n+1,x)/x,
where x = i/2*t - 1/2 and T(n,x) is the n-th row po1ynomial of A019538;
F) Miscellaneous
Column 1 is the sequence of tangent numbers - see A000182.
A000670(n+1) = (-i/2)^n*R(n,3*i).
A004123(n+2) = 2*(-i/2)^n*R(n,5*i).
A080795(n+1) =(-1)^n*(sqrt(-2))^n*R(n,sqrt(-2)). - Peter Bala, Aug 26 2011
From Leonid Bedratyuk, Aug 12 2012: (Start)
T(n,k) = (-1)^(n+1)*(-1)^((n-k)/2)*Sum_{j=k+1..n+1} j! *stirling2(n+1,j) *2^(n+1-j) *(-1)^(n+j-k) *binomial(j-1,k), see A059419.
Sum_{j=i+1..n+1}((1-(-1)^(j-i))/(2*(j-i))*(-1)^((n-j)/2)*T(n,j))=(n+1)*(-1)^((n-1-i)/2)*T(n-1,i), for n>1 and 0
G.f.: 1/G(0,t,x), where G(k,t,x) = 1 - 2*t*x - 2*k*t*x - (1+t^2)*(k+2)*(k+1)*x^2/G(k+1,t,x); (continued fraction due to T. J. Stieltjes). - Sergei N. Gladkovskii, Dec 27 2013
Comments