A370411 Square array T(n, k) = denominator( zeta_r(2*n) * sqrt(A003658(k + 2)) / Pi^(4*n) ), read by antidiagonals, where zeta_r is the Dedekind zeta-function over r and r is the real quadratic field with discriminant A003658(k + 2).
1, 75, 1, 16875, 24, 1, 221484375, 34560, 18, 1, 116279296875, 116121600, 58320, 39, 1, 12950606689453125, 780337152000, 440899200, 296595, 51, 1, 4861333986053466796875, 8899589151129600, 6666395904000, 68420017575, 663255, 63, 1, 677114376628875732421875
Offset: 0
Examples
The array begins:
1, 1, 1, 1, 1
75, 24, 18, 39, 51
16875, 34560, 58320, 296595, 663255
221484375, 116121600, 440899200, 68420017575, 20126472975
116279296875, 780337152000, 6666395904000, 93393323989875, 10382542981248375
Crossrefs
Programs
-
PARI
\p 700 row(n) = {v=[]; for(k=2, 30, if(isfundamental(k), v=concat(v, denominator(bestappr(sqrt(k)*lfun(x^2-(k%2)*x-floor(k/4), 2*n)/Pi^(4*n)))))); v} z(n,d) = if(n == 0, 0,(1/(-2*n))*bernfrac(2*n)*d^(2*n-1)*sum(k=1,d-1, kronecker(d, k)*subst(bernpol(2*n),x,k/d)*(1/(-2*n)))) row(n) = {v=[]; for(k=2, 100, if(isfundamental(k), v=concat(v, denominator((2^(n*4)*n^2*z(n,k))/((2*n)!^2 * (k^(2*n-1))))))); v} \\ more accuracy here -
Sage
# Only suitable for small n and k def T(n, k): discs = [fundamental_discriminant(i) for i in range(1, 4*k+10)] D = sorted(list(set(discs)))[k+1] zetaK = QuadraticField(D).zeta_function(1000) val = (zetaK(2*n)*sqrt(D)/(pi^(4*n))).n(1000).nearby_rational(2^-900) return val.denominator() # Robin Visser, Mar 19 2024
Formula
T(n, k) = denominator( 2^(n*4) * n^2 * zeta_r(1 - 2*n) /((2*n)!^2 * A003658(k + 2)^(2*n - 1)) ), where zeta_r is the Dedekind zeta-function over r and r is the real quadratic field with discriminant A003658(k + 2).
T(n, 0) = denominator((5^(-2*n)*(zeta(2*n, 1/5) - zeta(2*n, 2/5) - zeta(2*n, 3/5) + zeta(2*n, 4/5) ))*zeta(2*n)*sqrt(5)*Pi^(-4*n)). A sum of Hurwitz zeta functions with signs according A080891.
T(n, 1) = denominator( 2^(n*4) * n^2 * zeta(1 - 2*n) * (-1)^n * A000464(n+1) /((2*n)!^2 * 8^(2*n - 1)) ).
T(n, 2) = denominator( 2^(n*4) * n^2 * zeta(1 - 2*n) * (-1)^n * A000191(n+1) /((2*n)!^2 * 12^(2*n - 1)) ).
T(n, 3) = denominator((13^(-2*n)*(zeta(2*n, 1/13) - zeta(2*n, 2/13) + zeta(2*n, 3/13) + zeta(2*n, 4/13) - zeta(2*n, 5/13) - zeta(2*n, 6/13) - zeta(2*n, 7/13) - zeta(2*n, 8/13) + zeta(2*n, 9/13) + zeta(2*n, 10/13) - zeta(2*n, 11/13) + zeta(2*n, 12/13) ))*zeta(2*n)*sqrt(13)*Pi^(-4*n)). A sum of Hurwitz zeta functions with signs according the Dirichlet character X13(12,.).
T(n, 6) = denominator( 2^(n*4) * n^2 * zeta(1 - 2*n) * (-1)^n * A000411(n+1) /((2*n)!^2 * 24^(2*n - 1)) ).
T(n, 7) = denominator( 2^(n*4) * n^2 * zeta(1 - 2*n) * (-1)^n * A064072(n+1) /((2*n)!^2 * 28^(2*n - 1)) ).
T(n, 11) = denominator( 2^(n*4) * n^2 * zeta(1 - 2*n) * (-1)^n * A064075(n+1) /((2*n)!^2 * 40^(2*n - 1)) ).
T(n, k) = denominator( 2^(n*4) * n^2 * zeta(1 - 2*n) * (-1)^n * d(A003658(k+2)/4, n+1) /((2*n)!^2 * 40^(2*n - 1)) ), for all k where A003658(k+2) is a multiple of four (The discriminant of the quadratic field is from 4*A230375). d() are the generalized tangent numbers.
T(0, k) = 1, because for a real quadratic number field the discriminant D is positive, hence the Kronecker symbol (D/-1) = 1. This means the associated Dirichlet L-function will be zero at s = 0 inside the expression zeta_r(s) = zeta(s)*L(s, x).