A393772 - OEIS

%I #21 Mar 17 2026 22:50:00

%S 0,3,1,4,8,0,4,2,1,3,4,0,6,9,4,5,2,3,0,9,1,6,2,9,5,8,2,2,2,4,2,2,3,4,

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

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

%N Imaginary part of Sum_{n>=0} (2*Pi*i + 1/2^n)^n / n!.

%C A related identity is Sum_{n>=0} (p + q^n)^n * r^n/n! = Sum_{n>=0} exp(p*q^n*r) * q^(n^2) * r^n/n!. Here, p = 2*Pi*i, q = 1/2, and r = 1.

%F Equals Im(Sum_{n>=0} (2*Pi*i + 1/2^n)^n / n!) (see A393771 for the real part).

%F Equals Sum_{n>=0} sin(2*Pi/2^n) / (n! * 2^(n^2)).

%F Equals Sum_{n>=0} sqrt((1 - cos(4*Pi/2^n))/2) / (n! * 2^(n^2)).

%e y = 0.03148042134069452309162958222422342516963974616558...

%e Constant y = 0 + 0/2 + 1/(2!*2^4) + sqrt(1/2)/(3!*2^9) + sqrt((1-sqrt(1/2))/2)/(4!*2^16) + sqrt((1-sqrt((1+sqrt(1/2))/2))/2)/(5!*2^25) + ... + sin(2*Pi/2^n)/(n!*2^(n^2)) + ...

%e equals the imaginary part of: Sum_{n>=0} (2*Pi*i + 1/2^n)^n/n! = (0.5002307656... + i*0.03148042134...).

%o (PARI) \\ y = imaginary( Sum_{n>=0} (2*Pi*i + 1/2^n)^n/n! ).

%o \p200 \\ set desired precision

%o {y = imag( suminf(n=0, (2*Pi*I + 1/2^n)^n/n! ) )}

%o for(n=1,120, print1(floor(y*10^n)%10,", "))

%o (PARI) \\ y = Sum_{n>=0} sin(2*Pi/2^n) / (n! * 2^(n^2)).

%o \p200 \\ set desired precision

%o {y = suminf(n=0, sin(2*Pi/2^n)/(n!*2^(n^2)) )}

%o for(n=1,120, print1(floor(y*10^n)%10,", "))

%Y Cf. A393771.

%K nonn,cons

%O 0,2

%A _Paul D. Hanna_, Mar 12 2026