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The Masked Avenger's user avatar
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The Masked Avenger
  • Member for 11 years, 5 months
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How large must $A$ be if $\{1, \ldots, N\} \subseteq A-A$?
In fact, the comment was also intended as a response to Seva, who remarked about an unsuccessful search.
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How large must $A$ be if $\{1, \ldots, N\} \subseteq A-A$?
@QinJianbin, yes, but the problems are related and a literature search on one should help with the other.
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revised
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A prime sequence can be partitioned into two sets of equal or consecutive sum
One wonders how thin a subset of Pn can be to produce a thick subset of SSn.
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A prime sequence can be partitioned into two sets of equal or consecutive sum
This will also follow from the assertion that for n large enough, all but 6 positive integers between 0 and S_n, the sum of the first n primes, are realized as a subset sum, with the exceptions smaller than 7 or bigger than S_n - 7.
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Conjecture on the square root of the sum of the squares of the prime factors of a number
I think adding the triplet near the link will encourage other readers to follow the link, especially if one knows that Greg Martin rendered an opinion on the problem.
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Conjecture on the square root of the sum of the squares of the prime factors of a number
Indeed. For clarity, and to maintain cleanliness and simplicity, you could start the post with "For $n \gt 0$," which would make your considered domain more explicit. In any case, this would be of greater interest if you listed three largish consecutive integers with integral values of A.
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Conjecture on the square root of the sum of the squares of the prime factors of a number
I don't see it excluded, and I see two interpretations of A_0 and A_1 that are integral. Since A_p=p for positive primes p, n=0 gives you the sequence of four consecutive integers with integral values of A_n.
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Looking for interesting, natural models of this algebraic theory in which $x^\dagger$ is not always the multiplicative inverse of $x$
There are a lot of changes one can ring on such algebras. I misread the problem, not noticing the join meet interchange. I thought dagger might serve as reverse (transpose?) and then one could pick a commutative and distributive subalgebra. But for the interchange, I don't know now. I still think something can be done with relational algebras, because Tarski et al did it with relational and cylindric algebras.
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A generalized diagonal?
In algebra, if f is a homomorphism, then E is a congruence. If f is not a homomorphism, E can still be an equivalence relation of interest.
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Is the set of numbers $\{ [n^{3/2}] \mid n\text{ an integer}\}$ a basis of order 3?
Altwrnatively, consider $k(\alpha,\beta)$, which represents the order of building $S(\beta)$ from $S(\alpha)$. That might lead to weak but easy upper bounds.
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Families of Sets with Two Intersection Numbers
You are right. I missed the cardinality symbol. Maybe constructing a partial design of a few sets, and then showing that only so many copies of that part can be present will help.
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Mod sequences that seem to become constant; and the number 316
@Istvan, that doesn't work when a_1 is a reasonably sized factorial. There are other exceptions when a_n is larger than n^2.
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Families of Sets with Two Intersection Numbers
The approach above seems so natural to me that I must be missing something. Perhaps you can tell me what.
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Families of Sets with Two Intersection Numbers
Given F, look at the multiset G given by f-I for all f in F. For distinct g, h in G, either g=h as sets or they are disjoint. Your parameters suggest much smaller bounds like 4 times 5 or 4 times 2.
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