A393772 Imaginary part of Sum_{n>=0} (2*Pi*i + 1/2^n)^n / n!.

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, 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, 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

Offset

0,2

Comments

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.

Links

Table of n, a(n) for n=0..119.

Formula

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

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

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

Example

y = 0.03148042134069452309162958222422342516963974616558...
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)) + ...
equals the imaginary part of: Sum_{n>=0} (2*Pi*i + 1/2^n)^n/n! = (0.5002307656... + i*0.03148042134...).

Prog

(PARI) \\ y = imaginary( Sum_{n>=0} (2*Pi*i + 1/2^n)^n/n! ).
\p200 \\ set desired precision
{y = imag( suminf(n=0, (2*Pi*I + 1/2^n)^n/n! ) )}
for(n=1, 120, print1(floor(y*10^n)%10, ", "))
(PARI) \\ y = Sum_{n>=0} sin(2*Pi/2^n) / (n! * 2^(n^2)).
\p200 \\ set desired precision
{y = suminf(n=0, sin(2*Pi/2^n)/(n!*2^(n^2)) )}
for(n=1, 120, print1(floor(y*10^n)%10, ", "))

Crossrefs

Cf. A393771.

Sequence in context: A217151 A081255 A005371 * A210739 A193605 A193667

Adjacent sequences: A393769 A393770 A393771 * A393773 A393774 A393775

Keyword

nonn,cons

Author

Paul D. Hanna, Mar 12 2026

Status

approved