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Decimal expansion of 1 - DawsonF(1/2).
0

%I #10 Nov 23 2023 09:50:21

%S 5,7,5,5,6,3,6,1,6,4,9,7,9,7,7,7,0,4,0,6,5,9,5,7,6,4,7,5,1,0,3,3,0,4,

%T 2,8,9,0,3,5,7,0,5,2,2,6,4,0,3,0,7,9,6,1,8,4,8,6,6,0,3,0,3,3,6,6,7,5,

%U 4,8,4,5,2,4,0,4,0,8,0,5,2,3,8,3,2,2,8,7,9,8,7,1,5,2,1,3,8,7,7,7,8,5,7,4,0,3,8,3,0,2

%N Decimal expansion of 1 - DawsonF(1/2).

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DawsonsIntegral.html">Dawson's Integral</a>.

%F Equals 1 - sqrt(Pi/4) * erfi(1/2) / exp(1/4) = 1 - A019704 * A367563 / A092042.

%F Let C denote the constant. Then:

%F 2*C - 1 = Sum_{n>=0} (-1)^n / Pochhammer(n, n).

%F 2*(C - 1) = Sum_{n>=1} (-1)^n*Gamma(n) / Gamma(2*n).

%e 0.57556361649797770406595764751033042890357052264030796184866030336675484524040...

%p 1 - sqrt(Pi/4)*erfi(1/2)/exp(1/4): evalf(%, 109);

%t N[1 - DawsonF[1/2], 110] // RealDigits // First

%Y Cf. A019704, A092042, A367563.

%K nonn,cons

%O 0,1

%A _Peter Luschny_, Nov 23 2023