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-1 votes

Is there a more efficient way to find the least prime factor?

For a general number $n$, the fastest known way to find the smallest prime number of $n$ is to use ECM factoring. This is fairly efficient if there exists a small factor, but can take an extremely ...
D.W.'s user avatar
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-1 votes

Is there a more efficient way to find the least prime factor?

You can reverse the task and get some information. For any prime $p$ you can compute $n\# + 1 \pmod p$ without tears. Let's see, a big loop on primes $p$ up to some bound. A list of the ...
Will Jagy's user avatar
  • 141k
2 votes

Find sum of factorials divisible by the largest possible prime squared

I used such approach: for given $n$ (currently, $n=32$), loop through prime numbers $p$ starting from certain value $p_0$ to, theoretically, $\sqrt{\sum_{k=1}^n k!}$; and for these $(n,p)$ construct 2 ...
Oleg567's user avatar
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3 votes

Smallest "diamond-number" above some power of ten?

The smallest "diamond-number" (if my calculations are correct), related to $10^{37}$, is $$ 10^{37} + 1483238923930317 = 1102689419521^2 \times 8224198521037. $$ The smallest "diamond-...
Oleg567's user avatar
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0 votes

Computing the radical of an integer's equality

Generalising on Peter's comment, I don't think there is an 'efficient' (polynomial time) way to confirm that an arbitrary candidate integer m is the radical of an arbitrary integer n, but if you know ...
Simon Goater's user avatar
2 votes
Accepted

Does the monoid of non-zero representations with the tensor product admit unique factorization?

Counterexamples are much easier to produce than this and exist already when $G = C_2$ and with two $2$-dimensional representations, see here. Abstractly the problem is that the representation ring is ...
Qiaochu Yuan's user avatar
2 votes

Does the monoid of non-zero representations with the tensor product admit unique factorization?

A counterexample is given by Nate at https://math.stackexchange.com/a/4436073/491450 : Taking $G = A_5$, we have $$ V_4 \otimes V_5 \otimes V_3 \cong V_4 \otimes V_5 \otimes {V_3}' $$ where the ...
Smiley1000's user avatar
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