A366834 Square array read by descending antidiagonals: (-1)^n*T(n,k)/n! is the coefficient of x^(2*n+1) in the Taylor expansion of the k-th iteration of sin(x).

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 12, 1, 0, 1, 4, 33, 128, 1, 0, 1, 5, 64, 731, 1872, 1, 0, 1, 6, 105, 2160, 25857, 37600, 1, 0, 1, 7, 156, 4765, 121600, 1311379, 990784, 1, 0, 1, 8, 217, 8896, 368145, 10138880, 89060065, 32333824, 1, 0, 1, 9, 288, 14903, 873936, 42807605, 1162426880, 7778778091, 1272660224, 1, 0

Offset: 0

Author

Jianing Song, Oct 25 2023

Keywords

• nonn
• tabl

Comments

T(n,k)/n! is the coefficient of x^(2*n+1) in the Taylor expansion of the k-th iteration of sinh(x). This is most easily seen from the relation -i*sin(...sin(sin(sin(i*x)))...) = -i*sin(...sin(sin(i*sinh(x)))...) = -i*sin(...sin(i*sinh(sinh(x)))...) = ... = sinh(...sinh(sinh(sinh(x)))...).

Examples

			G.f.s of the first few rows:
n = 0: 1/(1-x);
n = 1: x/(1-x)^2;
n = 2: x/(1-x)^2 + 10*x^2/(1-x)^3;
n = 3: x/(1-x)^2 + 126*x^2/(1-x)^3 + 350*x^3/(1-x)^4:
n = 4: x/(1-x)^2 + 1870*x^2/(1-x)^3 + 20244*x^3/(1-x)^4 + 29400*x^4/(1-x)^5;
n = 5: x/(1-x)^2 + 37598*x^2/(1-x)^3 + 1198582*x^3/(1-x)^4 + 5118960*x^4/(1-x)^5 + 4851000*x^5/(1-x)^6.
The explicit formulas for the first few rows:
T(0,k) = binomial(k,0) = 1 for k = 0, 0 for k > 0;
T(1,k) = binomial(k,1) = k;
T(2,k) = binomial(k,1) + 10*binomial(k,2) = 5*k^2 - 4*k;
T(3,k) = binomial(k,1) + 126*binomial(k,2) + 350*binomial(k,3) = (175/3)*k^3 - 112*k^2 + (164/3)*k;
T(4,k) = binomial(k,1) + 1870*binomial(k,2) + 20244*binomial(k,3) + 29400*binomial(k,4) = 1225*k^4 - 3976*k^3 + 4288*k^2 - 1536*k;
T(5,k) = binomial(k,1) + 37598*binomial(k,2) + 1198582*binomial(k,3) + 5118960*binomial(k,4) + 4851000*binomial(k,5) = 40425*k^5 - 190960*k^4 + (1004696/3)*k^3 - 255552*k^2 + (213568/3)*k.
Table of terms:
Row 0: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Row 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Row 2: 0, 1, 12, 33, 64, 105, 156, 217, 288, 369, 460
Row 3: 0, 1, 128, 731, 2160, 4765, 8896, 14903, 23136, 33945, 47680
Row 4: 0, 1, 1872, 25857, 121600, 368145, 873936, 1776817, 3244032, 5472225, 8687440
Row 5: 0, 1, 37600, 1311379, 10138880, 42807605, 130426016, 323774535, 698156544, 1358249385, 2442955360
Row 6: 0, 1, 990784, 89060065, 1162426880, 6937805945, 27344158016, 83303826249, 212901058560, 478937915985, 977877567040
Row 7: 0, 1, 32333824, 7778778091, 174394695680, 1487589904205, 7634965431296, 28668866786679, 87104014381056, 227079171721785, 527214112015360
Row 8: 0, 1, 1272660224, 849264442881, 33044097597440, 406373544070945, 2731282112246016, 12688038285458529, 45949019179646976, 139088689115885505, 367745105831952640
Row 9: 0, 1, 59527313920, 113234181108643, 7701145601933312, 137461463840219237, 1215573962763120128, 7008667055272520967, 30322784763588771840, 106757902382656031049, 321859857651846029824
Row 10: 0, 1, 3252626013184, 18073465545032353, 2162675327569362944, 56311245536706922889, 657730167421332884480, 4719958813316934631353, 24445625744089126797312, 100254353682662263787313, 345053755346367061654528
Demonstration of terms:
sin(sin(x)) = x - 2*x^3/3! + 12*x^5/5! - 128*x^7/7! + 1872*x^9/9! - 37600*x^11/11! + ...;
sin(sin(sin(x))) = x - 3*x^3/3! + 33*x^5/5! - 731*x^7/7! + 25857*x^9/9! - 1311379*x^11/11! + ...;
sin(sin(sin(sin(x)))) = x - 4*x^3/3! + 64*x^5/5! - 2160*x^7/7! + 121600*x^9/9! - 10138880*x^11/11! + ....
		

Links

• Jianing Song, Table of n, a(n) for n = 0..5150 (diagonals 0..100)
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
• Jianing Song, Formula for A366834.

Crossrefs

Cf. A051624 (row n=2), A366827 (row n=3), A003712 (column k=2 signed), A003715 (column k=3 signed).

Cf. A000447, A285018, A285019.

Programs

PARI

A160562(n,k) = (-1)^k / (2*k+1)! * sum(j=0, k, (-1)^j * binomial(2*k+1, k-j) * (2*j+1)^(2*n+1)) / 2^(2*k)
coeff_of_n_gfs(n) = my(M=matid(1)); for(k=1, n, M = matconcat([concat(M, matrix(k, 1)); concat(0, matrix(1, k, i, j, A160562(k, j-1))*M)])); M \\ The lower triangle matrix (C(i,j))_{0<=j<=i<=n}
T_mat(n,k) = coeff_of_n_gfs(n)*matrix(n+1, k+1, i, j, binomial(j-1,i-1)) \\ gives T(i,j) for i=0..n and j=0..k

Formula

T(0,0) = 1, T(n,0) = 0 for n >= 1; T(n,k) = Sum_{i=0..n} A160562(n,i)*T(i,k-1) for k >= 1, where A160562(n,k) = ((-1)^(n-k)*(2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sin(x)^(2*k+1). Note that this is not very efficient to calculate the terms.

A more efficient way would be to calculate the g.f. for each row: the g.f. of the n-th row is C(n,0)/(1-x) + C(n,1)*x/(1-x)^2 + ... + C(n,n)*x^n/(1-x)^(n+1), where C(0,0) = 1, C(n,0) = 0 for n >= 1; C(n,k) = A160562(n,k-1)*C(k-1,k-1) + ... + A160562(n,n-1)*C(n-1,k-1) for n >= k >= 1, so we have T(n,k) = C(n,0)*binomial(k,0) + C(n,1)*binomial(k,1) + ... + C(n,n)*binomial(k,n). See my pdf in the link section for the proof.

From the formula above we see that the n-th row is a degree-n polynomial of k with leading coefficient C(n,n)/n!. We have C(n,n) = A160562(n,n-1)*C(n-1,n-1) = A000447(n)*C(n-1,n-1) for n >= 1, so it can be shown that C(n,n)/n! = n! * A285018(n)/A285019(n).