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%I #37 Jul 28 2024 09:21:14
%S 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,
%T 6,105,2160,25857,37600,1,0,1,7,156,4765,121600,1311379,990784,1,0,1,
%U 8,217,8896,368145,10138880,89060065,32333824,1,0,1,9,288,14903,873936,42807605,1162426880,7778778091,1272660224,1,0
%N 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).
%C 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)))...).
%H Jianing Song, <a href="/A366834/b366834.txt">Table of n, a(n) for n = 0..5150</a> (diagonals 0..100)
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%H Jianing Song, <a href="/A366834/a366834.pdf">Formula for A366834</a>.
%F 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.
%F 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.
%F 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).
%e G.f.s of the first few rows:
%e n = 0: 1/(1-x);
%e n = 1: x/(1-x)^2;
%e n = 2: x/(1-x)^2 + 10*x^2/(1-x)^3;
%e n = 3: x/(1-x)^2 + 126*x^2/(1-x)^3 + 350*x^3/(1-x)^4:
%e n = 4: x/(1-x)^2 + 1870*x^2/(1-x)^3 + 20244*x^3/(1-x)^4 + 29400*x^4/(1-x)^5;
%e 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.
%e The explicit formulas for the first few rows:
%e T(0,k) = binomial(k,0) = 1 for k = 0, 0 for k > 0;
%e T(1,k) = binomial(k,1) = k;
%e T(2,k) = binomial(k,1) + 10*binomial(k,2) = 5*k^2 - 4*k;
%e T(3,k) = binomial(k,1) + 126*binomial(k,2) + 350*binomial(k,3) = (175/3)*k^3 - 112*k^2 + (164/3)*k;
%e 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;
%e 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.
%e Table of terms:
%e Row 0: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
%e Row 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
%e Row 2: 0, 1, 12, 33, 64, 105, 156, 217, 288, 369, 460
%e Row 3: 0, 1, 128, 731, 2160, 4765, 8896, 14903, 23136, 33945, 47680
%e Row 4: 0, 1, 1872, 25857, 121600, 368145, 873936, 1776817, 3244032, 5472225, 8687440
%e Row 5: 0, 1, 37600, 1311379, 10138880, 42807605, 130426016, 323774535, 698156544, 1358249385, 2442955360
%e Row 6: 0, 1, 990784, 89060065, 1162426880, 6937805945, 27344158016, 83303826249, 212901058560, 478937915985, 977877567040
%e Row 7: 0, 1, 32333824, 7778778091, 174394695680, 1487589904205, 7634965431296, 28668866786679, 87104014381056, 227079171721785, 527214112015360
%e Row 8: 0, 1, 1272660224, 849264442881, 33044097597440, 406373544070945, 2731282112246016, 12688038285458529, 45949019179646976, 139088689115885505, 367745105831952640
%e Row 9: 0, 1, 59527313920, 113234181108643, 7701145601933312, 137461463840219237, 1215573962763120128, 7008667055272520967, 30322784763588771840, 106757902382656031049, 321859857651846029824
%e Row 10: 0, 1, 3252626013184, 18073465545032353, 2162675327569362944, 56311245536706922889, 657730167421332884480, 4719958813316934631353, 24445625744089126797312, 100254353682662263787313, 345053755346367061654528
%e Demonstration of terms:
%e sin(sin(x)) = x - 2*x^3/3! + 12*x^5/5! - 128*x^7/7! + 1872*x^9/9! - 37600*x^11/11! + ...;
%e 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! + ...;
%e 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! + ....
%o (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)
%o 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}
%o 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
%Y Cf. A051624 (row n=2), A366827 (row n=3), A003712 (column k=2 signed), A003715 (column k=3 signed).
%Y Cf. A000447, A285018, A285019.
%K nonn,tabl
%O 0,8
%A _Jianing Song_, Oct 25 2023