Erdős Problem #948 — Is there a function f(n)f(n) and a kk such that in any kk-colouring of the integers there exists a sequence a1<⋯a_1<\cdots such that an<f(n)a_n<f(n) for infinitely many nn and the set…

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Is there a function f(n)f(n) and a kk such that in any kk-colouring of the integers there exists a sequence a1<⋯a_1<\cdots such that an<f(n)a_n<f(n) for infinitely many nn and the set {∑i∈Sai:finite S}\left\{ \sum_{i\in S}a_i : \textrm{finite }S\right\} does not contain all colours?

References

Progress summary

Refreshed
Open

A claimed computer-generated construction would disprove the problem, but no independent proof has yet confirmed it.

Erdős and Galvin asked whether one can choose a function ff and a finite number of colours kk forcing a slowly growing sequence whose finite subset sums omit a colour.

Known results

  • Galvin ruled out the monochromatic version for k=2k=2 using a two-colouring based on powers of 22.
  • Erdős and Galvin showed that, for every kge2kge 2, sequences with an<22O(n)a_n<2^{2^{O(n)}} can have finite interval sums using only two colours.
  • The corresponding question for countably many colours was recorded as open.

Claimed negative answer (date not stated)

GPT Pro, prompted by Price, claimed that for every ff there is a colouring of N\mathbb{N} by N\mathbb{N} such that every sequence satisfying an<f(n)a_n<f(n) for infinitely many nn has finite subset sums of every colour. A discussion gives a rough binary-digit interval construction, but no formal verification or independent publication was found.

Current status (as of April 2026): The GPT Pro construction is claimed to settle the problem negatively, while the proposed proof remains unverified and the problem is not resolved in the published record.

Sources

Solutions 0

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