A175386 a(n) = denominator of Sum_{i=1..n} (1/i)*C(2n-i-1,i-1).

1, 2, 6, 4, 5, 4, 7, 8, 18, 10, 11, 24, 13, 14, 30, 16, 17, 12, 19, 20, 42, 22, 23, 48, 25, 26, 54, 28, 29, 20, 31, 32, 66, 34, 35, 72, 37, 38, 78, 40, 41, 28, 43, 44, 90, 46, 47, 96, 49, 50, 6, 52, 53, 36, 55, 56, 114, 58, 59, 120, 61, 62, 126, 64, 65, 44, 67, 68, 138, 70, 71

Offset

1,2

Comments

We conjecture that sum((1/i)*C(2n-i-1,i-1),i=1..n) is not an integer for n>1.

This was proved by an autonomous AI agent, see the Tsoukalas paper and the Lean file. The proof uses the closed-form identity 2n * S(n) = L(2n) - 1, where L is the Lucas sequence and S the above sum, reducing non-integrality of the sum to showing 2n does not divide L(2n) - 1. This is done via Fibonacci identities for L(2n), and L(2n) = L(n)^2 + 2 and a Cassini/Fibonacci-divisibility argument on the smallest prime factor of n, ruling out n | L(n)^2 + 1 and forcing the denominator to exceed 1 (Summary by Opus 4.7). - Ralf Stephan, May 25 2026

Links

Table of n, a(n) for n=1..71.

Google Deepmind, AlphaProof Nexus: A175386 Lean file

George Tsoukalas et al., Advancing Mathematics Research with AI-Driven Formal Proof Search, arXiv:2605.22763 [cs.AI], 2026.

Formula

According to Mathematica, Sum_{i=1..n} (1/i)*C(2n-i-1,i-1) = (Hypergeometric2F1[1/2-n,-n,1-2 n,-4]-1)/(2 n).

Mathematica

Table[Denominator[Sum[(1/i)*Binomial[2n-i-1, i-1], {i, 1, n}]], {n, 1, 150}]

Crossrefs

Cf. A175385.

Sequence in context: A222236 A351001 A334171 * A021382 A356309 A182505

Adjacent sequences: A175383 A175384 A175385 * A175387 A175388 A175389

Keyword

frac,nonn,changed

Author

Zak Seidov, Vladimir Shevelev, Apr 24 2010

Status

approved