A084817 Least integers that satisfy Sum_{n>=1} 1/a(n)^z = 0, where a(1)=1, a(n+1)>a(n) and z=I/log(2).

1, 5, 11, 29, 80, 222, 624, 1756, 4949, 13953, 39339, 110918, 312744, 881814, 2486368, 7010585, 19767110, 55735533, 157152449, 443108574, 1249393248, 3522801366, 9932925036, 28006972162, 78968731457, 222661004264, 627817136045, 1770199310902, 4991271216425

Offset

1,2

Comments

Sequence satisfies Sum_{n>=1} 1/a(n)^z = 0 by requiring that the modulus of the successive partial sums are monotonically decreasing in magnitude for the given z.

|Sum_{k=1..n} 1/a(k)^z| appears to converge geometrically to 0.73530592... Does this behavior continue? If this is true then the complex sum converges to a circular orbit rather than a single value. - Martin Fuller, May 24 2026

Links

Table of n, a(n) for n=1..29.

Prog

(PARI) my(z=I/log(2), S=0, w=1, a=0); for(n=1, 100, b=a+1; while(abs(S+exp(-z*log(b)))>w, b++); S=S+exp(-z*log(b)); w=abs(S); a=b; print1(b, ", "))
(PARI) \\ See Fuller link in A084813

Crossrefs

Cf. A084812, A084813, A084814, A084815, A084816, A084818.

Sequence in context: A106060 A237642 A059508 * A183382 A100965 A001632

Adjacent sequences: A084814 A084815 A084816 * A084818 A084819 A084820

Keyword

nonn

Author

Paul D. Hanna, Jun 04 2003

Extensions

a(12)-a(17) from Sean A. Irvine, May 12 2026

a(18) onward from Martin Fuller, May 24 2026

Status

approved