A093726 Given the infinite continued fraction (1+i)+1/((1+i)+1/((1+i)+...)), where i is the square root of (-1), this is the numerator of the imaginary part of the convergents.

1, 1, 4, 3, 11, 97, 280, 101, 2337, 2251, 19516, 14101, 163009, 16245, 30256, 245929, 11371969, 32865601, 94983348, 22875581, 19349753, 2292794785, 6626299912, 2393795271, 2635503517, 159951677089, 462268926316, 11517086141, 3861059617665, 247970431013

Offset

1,3

Links

Table of n, a(n) for n=1..30.

Formula

Conjecture: a(n)/A093727(n) = A317974(n+1)/A138573(n). - Zhuorui He, Jun 16 2026

Example

From Zhuorui He, Jun 16 2026: (Start)
(1+i) = 1+i, so A093725(1)=1, a(1)=1 and A093727(1)=1.
(1+i)+1/(1+i) = (3+i)/2, so A093725(2)=3, a(2)=1 and A093727(2)=2.
(1+i)+1/((1+i)+1/(1+i)) = (8+4*i)/5, so A093725(3)=8, a(3)=4 and A093727(3)=5.
(1+i)+1/((1+i)+1/((1+i)+1/(1+i))) = (6+3*i)/4, so A093725(4)=6, a(4)=3 and A093727(4)=4.
(1+i)+1/((1+i)+1/((1+i)+1/((1+i)+1/(1+i)))) = (23+11*i)/15, so A093725(5)=23, a(5)=11 and A093727(5)=15. (End)

Mathematica

Table[ Im[ Numerator[ FromContinuedFraction[ Table[1 + I, {n}]]]], {n, 30}]

Crossrefs

Cf. A091806, A091807, A091808, A091809, A093725, A093727, A317974, A138573.

Sequence in context: A065337 A257078 A072802 * A126604 A191617 A169704

Adjacent sequences: A093723 A093724 A093725 * A093727 A093728 A093729

Keyword

frac,nonn

Author

Robert G. Wilson v, Mar 11 2004

Extensions

Definition corrected by and a(29)-a(30) from Zhuorui He, Jun 16 2026

Status

approved