A003715 -id:A003715 - cp's OEIS frontend

A366834 Square array read by descending antidiagonals: (-1)^nT(n,k)/n! is the coefficient of x^(2n+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

Jianing Song, Oct 25 2023

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)))...).

			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! + ....

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.

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

Cf. A000447, A285018, A285019.

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

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).

A212261 Array A(i,j) read by antidiagonals: A(i,j) is the (2i-1)-th derivative of sin(sin(sin(...sin(x)))) nested j times evaluated at 0.

Original entry on oeis.org

1, 1, -1, 1, -2, 1, 1, -3, 12, -1, 1, -4, 33, -128, 1, 1, -5, 64, -731, 1872, -1, 1, -6, 105, -2160, 25857, -37600, 1, 1, -7, 156, -4765, 121600, -1311379, 990784, -1, 1, -8, 217, -8896, 368145, -10138880, 89060065, -32333824, 1

Offset: 1

John M. Campbell, May 12 2012

The determinant of the n X n such matrix has a closed form given in the formula section (and the Mathematica code below).

Rows appear to be given by polynomials (see formula section).

			Evaluate the fifth derivative of sin(sin(sin(x))) at 0, which is 33. So the (3,3) entry of the array is 33. The array begins as:
|  1      1        1         1         1          1 |
| -1     -2       -3        -4        -5         -6 |
|  1     12       33        64       105        156 |
| -1   -128     -731     -2160     -4765      -8896 |
|  1   1872    25857    121600    368145     873936 |
| -1 -37600 -1311379 -10138880 -42807605 -130426016 |

Cf. A003712, A003715.

A:= (i, j)-> (D@@(2*i-1))(sin@@j)(0):
seq(seq(A(i, 1+d-i), i=1..d), d=1..9); # Alois P. Heinz, May 14 2012

A[a_, b_] :=
  A[a, b] =
   Array[D[Nest[Sin, x, #2], {x, 2*#1 - 1}] /. x -> 0 &, {a, b}];
Print[A[7, 7] // MatrixForm];
Table2 = {};
k = 1;
While[k < 8, Table1 = {};
  i = 1;
  j = k;
  While[0 < j,
   AppendTo[Table1, First[Take[First[Take[A[7, 7], {i, i}]], {j, j}]]];
   j = j - 1;
   i = i + 1];
  AppendTo[Table2, Table1];
  k++];
Print[Flatten[Table2]]
Print[Table[Det[A[n, n]], {n, 1, 7}]];
Print[Table[(
  I^(n + n^2) 2^(-(1/12) + n^2) 3^(n/2 - n^2/2)
    Glaisher^3 (-(1/\[Pi]))^
   n BarnesG[1/2 + n] BarnesG[1 + n] BarnesG[3/2 + n])/E^(1/4), {n, 1, 7}]]

A(i,j) = ((d/dx)^(2i-1) sin^j(x))_{x=0}.
Let A_n denote the n X n such matrix. Then:
det(A_n)=(i^(n + n^2) 2^(-(1/12) + n^2) 3^(n/2 - n^2/2) G^3 (-(1/pi))^n B(1/2 + n) B(1 + n) B(3/2 + n))/e^(1/4), where B is the Barnes G-function and G is the Glaisher-Kinkelin constant (and i is the imaginary unit). (This can be shown by evaluating recurrence relations for det(A_n)). See Mathematica code below.
First row: 1.
Second row: -x.
Third row: x (5 x - 4).
Fourth row: -(1/3) x (164 + 7 x (-48 + 25 x)).
Fifth row: (8 - 7 x)^2 x (-24 + 25 x).
Sixth row: -(1/3) x (213568 - 766656 x + 1004696 x^2 - 572880 x^3 + 121275 x^4).
Seventh row: 1/3 x (-14371328 + 65012064 x - 111160192 x^2 + 91291200 x^3 - 36552516 x^4 + 5780775 x^5).
Second column: A003712.
Third column: A003715.

A302452 a(n) = coefficient of x^(2*n-1) in the n-th iteration (n-fold self-composition) of e.g.f. sinh(x).

Original entry on oeis.org

1, 2, 33, 2160, 368145, 130426016, 83303826249, 87104014381056, 139088689115885505, 321859857651846029824, 1036109938469605247521009, 4490275483028481600517832704, 25503692273369769781221175069521, 185636732310716855091866841134243840, 1699077450890747555020338066545506541145

Offset: 1

Ilya Gutkovskiy, Apr 08 2018

nonn

a(n) = coefficient of x^(2*n-1) in the n-th iteration (n-fold self-composition) of e.g.f. sin(x) (absolute values).

			The initial coefficients of successive iterations of e.g.f. A(x) = sinh(x) (odd powers only) are as follows:
n = 1: (1), 1,    1,     1,       1,  ... e.g.f. A(x)
n = 2:  1, (2),  12,   128,    1872,  ... e.g.f. A(A(x))
n = 3:  1,  3,  (33),  731,   25857,  ... e.g.f. A(A(A(x)))
n = 4:  1,  4,   64, (2160)  121600,  ... e.g.f. A(A(A(A(x))))
n = 5:  1,  5,  105,  4765, (368145), ... e.g.f. A(A(A(A(A(x)))))
...
More explicitly, the successive iterations of e.g.f. A(x) = sinh(x) begin:
sinh(x) = x/1! + x^3/3! + x^5/5! + x^7/7! + x^9/9! + ...
sinh(sinh(x)) = x/1! + 2*x^3/3! + 12*x^5/5! + 128*x^7/7! + 1872*x^9/9! + ...
sinh(sinh(sinh(x))) = x/1! + 3*x^3/3! + 33*x^5/5! + 731*x^7/7! + 25857*x^9/9! + ...
sinh(sinh(sinh(sinh(x)))) = x/1! + 4*x^3/3! + 64*x^5/5! + 2160*x^7/7! + 121600*x^9/9! + ...
sinh(sinh(sinh(sinh(sinh(x))))) = x/1! + 5*x^3/3! + 105*x^5/5! + 4765*x^7/7! + 368145*x^9/9! + ...

N. J. A. Sloane, Transforms

Cf. A003712, A003715.

Table[(2 n - 1)! SeriesCoefficient[Nest[Function[x, Sinh[x]], x, n], {x, 0, 2 n - 1}], {n, 15}]

cp's OEIS Frontend

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

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A212261 Array A(i,j) read by antidiagonals: A(i,j) is the (2i-1)-th derivative of sin(sin(sin(...sin(x)))) nested j times evaluated at 0.

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A302452 a(n) = coefficient of x^(2*n-1) in the n-th iteration (n-fold self-composition) of e.g.f. sinh(x).

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cp's OEIS Frontend

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).

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PARI

Formula

A212261 Array A(i,j) read by antidiagonals: A(i,j) is the (2i-1)-th derivative of sin(sin(sin(...sin(x)))) nested j times evaluated at 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A302452 a(n) = coefficient of x^(2*n-1) in the n-th iteration (n-fold self-composition) of e.g.f. sinh(x).

Views

Author

Keywords

Comments

Examples

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Crossrefs

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Mathematica

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