A326722 - OEIS

%I #28 Aug 12 2019 05:33:19

%S 1,0,1,-1,0,1,0,-1,0,1,2,0,-4,0,2,0,5,0,-10,0,5,-16,0,48,0,-48,0,16,0,

%T -61,0,183,0,-183,0,61,272,0,-1088,0,1632,0,-1088,0,272,0,1385,0,

%U -5540,0,8310,0,-5540,0,1385,-7936,0,39680,0,-79360,0,79360,0,-39680,0,7936

%N T(n, k) = n! * [x^k] [y^n] sec(z)(x + z*sin(z)/y) where z = y*sqrt(x^2 - 1) for 0 <= k <= n+1 and T(-1, 0) = 1, triangle read by rows.

%F T(n, k) = A178111(n, k)*A000111(n-1) if n > 1 else k^n, assuming 0 based triangle.

%F Exponential generating functions for the columns (n >= 0) are:

%F egf_col0(x) = -tanh(x).

%F egf_col1(x) = sech(x).

%F egf_col2(x) = (tanh(x) + x*sech(x)^2)/2.

%F egf_col3(x) = x*tanh(x)*sech(x)/2.

%F egf_col4(x) = (tanh(x) + x*(2*x*tanh(x) - 1)*sech(x)^2)/8.

%F egf_col5(x) = x*sech(x)*(x + tanh(x) - 2*x*sech(x)^2)/8.

%F egf_col6(x) = (3*tanh(x) + x*sech(x)^2*(4*x^2 - 6*x^2*sech(x)^2 - 3))/48.

%F A recurrence of the row polynomials based on offset 0 is given by the recurrence of the Euler-Bernoulli-Entringer numbers A008281 combined with Paul Barry's A178111. See the Maple script.

%e [-1]     1;

%e [ 0]     0,    1;

%e [ 1]    -1,    0,     1;

%e [ 2]     0,   -1,     0,     1;

%e [ 3]     2,    0,    -4,     0,      2;

%e [ 4]     0,    5,     0,   -10,      0,    5;

%e [ 5]   -16,    0,    48,     0,    -48,    0,    16;

%e [ 6]     0,  -61,     0,   183,      0, -183,     0,    61;

%e [ 7]   272,    0, -1088,     0,   1632,    0, -1088,     0,    272;

%e [ 8]     0, 1385,     0, -5540,      0, 8310,     0, -5540,      0, 1385;

%e [ 9] -7936,    0, 39680,     0, -79360,    0, 79360,     0, -39680,    0, 7936;

%p z := y*sqrt(x^2 - 1): gf := sec(z)*(x + z*sin(z)/y):

%p ser := series(gf, y, 16): cy :=  n -> convert(n!*coeff(ser, y, n), polynom):

%p Trow := n -> `if`(n=-1, [1], PolynomialTools:-CoefficientList(cy(n), x)):

%p ListTools:-Flatten([seq(Trow(n), n=-1..9)]);

%p # Alternatively, compute the row polynomials based on offset 0 by recurrence.

%p RowPoly := proc(n) local E, P, L;

%p     E := proc(n, k) option remember; if k = 0 then return(`if`(n = 0, 1, 0)) fi;

%p         E(n, k-1) + E(n-1, n-k) end:

%p     P := proc(n) option remember; `if`(n < 2, x^n,

%p         x*P(n-1) - ((1 + (-1)^n)/2)*P(n-2)) end:

%p     # `if`(n = 0, 1, sort(expand(P(n)*E(n-1,n-1)), x, ascending)):

%p     L := n -> PolynomialTools:-CoefficientList(P(n), x):

%p     `if`(n = 0, [1], L(n)*E(n-1, n-1)):

%p end: for n from 0 to 9 do RowPoly(n) end;

%p # Alternative:

%p T := (n, k) -> if n <= 1 then k^n else A178111(n, k)*A000111(n-1) fi:

%p seq(seq(T(n,k), k=0..n), n=0..10);

%t z := y Sqrt[x^2 - 1]; gf := Sec[z](x + z Sin[z]/y); ser := Series[gf, {y, 0, 16}];

%t cy[-1] := {1}; cy[n_] := n! Coefficient[ser, y, n];

%t row[n_] := CoefficientList[cy[ n], x]; Table[row[n], {n, -1, 9}] // Flatten

%o (SageMath)

%o def A326722(n,k):

%o     if n == 0: return 1

%o     if is_odd(n-k): return 0

%o     b = I^(n-k)*binomial(floor(n/2),floor(k/2))

%o     if is_odd(n): return b*I^(n-1)*euler_number(n-1)

%o     return 2*b*psi(n-1, 1/2)/pi^n

%o for n in range(11): print([A326722(n,k) for k in range(n+1)])

%Y Cf. A000364, A028296, A000182, A000111, A008281, A178111, A012816, A261042, A326721, A326723, A326724.

%Y T(n, 0) = -A155585(n) for n >= 1.

%Y T(n, 1) =  A122045(n) for n >= 0.

%Y |T(2*n-1, 2)| = A024255(n) for n >= 0.

%Y T(n, 3) = A326719(n) for n >= 0.

%Y T(n, 4) = A326718(n) for n >= 0.

%K sign,tabl

%O -1,11

%A _Peter Luschny_, Aug 08 2019