[math-fun] Sum of squared factorials (question by Asimov/Wolfram)
From: Dan Asimov <asimov@msri.org> Also, Wolfram ( http://mathworld.wolfram.com/FactorialSums.html ) states that the sequence
fsq_n := f_n := Sum_{1<=k<=n} (k!)^2
contains only a finite set of primes among its terms. Amazing!
Is there some easy way to see this?
--1, 1+4=5, 1+4+36=41, 617, 15017, 533417, ... At first I thought this was going to be trivial, but the initial values of fsq_n often contain giant prime factors, sometimes also with no small primes involved. Such as fsq_31 whose smallest prime factor is 1751935243. I would presume that some n exists such that fsq_n is divisible by a prime P<=n+1, and if so, then fsq_m will be divisible by P for all m>n. So I would suggest, for each prime P>31 in succession, computing fsq_(P-1) mod P; until you get zero; then declare victory. Heuristically (since the sum over primes p of 1/p diverges) this attack ought to succeed, but it might take a while. I haven't done the computation; the odds seem about even that the first such P will be below 10000, but it seems to me entirely possible that the first such P is unreachably large, such as 10^30. The underlying problem here is the divergence I mentioned is very slow, doubly-logarithmic. To make it even worse: I hereby proclaim that I am "heuristically certain" that fsq_n is, for all except a finite number of n, divisible both by an upper, and by a lower, TWIN prime. (If p and p+2 both are prime, than p is the lower and p+2 is the upper twin.) The same sort of proof must exist, but now finding that proof is likely to be so hard that nobody will ever do it!
sq_n := f_n := Sum_{1<=k<=n} (k!)^2 is A104344 (or A061062 if k starts at 0). Comments could be added there. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Fri, Apr 10, 2015 at 10:02 AM, Warren D Smith <warren.wds@gmail.com> wrote:
From: Dan Asimov <asimov@msri.org> Also, Wolfram ( http://mathworld.wolfram.com/FactorialSums.html ) states that the sequence
fsq_n := f_n := Sum_{1<=k<=n} (k!)^2
contains only a finite set of primes among its terms. Amazing!
Is there some easy way to see this?
--1, 1+4=5, 1+4+36=41, 617, 15017, 533417, ... At first I thought this was going to be trivial, but the initial values of fsq_n often contain giant prime factors, sometimes also with no small primes involved. Such as fsq_31 whose smallest prime factor is 1751935243.
I would presume that some n exists such that fsq_n is divisible by a prime P<=n+1, and if so, then fsq_m will be divisible by P for all m>n.
So I would suggest, for each prime P>31 in succession, computing fsq_(P-1) mod P; until you get zero; then declare victory. Heuristically (since the sum over primes p of 1/p diverges) this attack ought to succeed, but it might take a while. I haven't done the computation; the odds seem about even that the first such P will be below 10000, but it seems to me entirely possible that the first such P is unreachably large, such as 10^30. The underlying problem here is the divergence I mentioned is very slow, doubly-logarithmic.
To make it even worse: I hereby proclaim that I am "heuristically certain" that fsq_n is, for all except a finite number of n, divisible both by an upper, and by a lower, TWIN prime. (If p and p+2 both are prime, than p is the lower and p+2 is the upper twin.) The same sort of proof must exist, but now finding that proof is likely to be so hard that nobody will ever do it!
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To make it even worse: I hereby proclaim that I am "heuristically certain" that fsq_n is, for all except a finite number of n, divisible both by an upper, and by a lower, TWIN prime. (If p and p+2 both are prime, than p is the lower and p+2 is the upper twin.) The same sort of proof must exist, but now finding that proof is likely to be so hard that nobody will ever do it!
--CORRECTION: sorry, my brain switched to wrong. I was basing the above on Brun's theorem that the sum of 1/t, where t are twin primes, diverges. But actually Brun says the opposite: this sum is finite. OK, so to fix this: I am heuristically certain that for all but a finite set of n, fsq_n is divisible by a prime P<n such that P is of the form P=1000000000*k+1. [Pick your favorite number of zeros, I'm still certain.] And a proof of this should be available findable by the technique I mentioned... but that proof will be extremely hard to find. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
I found that 1248829 | fsq_n for n > 1248827. Searching the OEIS I found that this was already known in 2004, see https://oeis.org/A100289. I agree with your heuristic -- there should be infinitely many fixed prime divisors, and so any appropriately natural class of primes with positive relative density should have infinitely many instances. Charles Greathouse Analyst/Programmer Case Western Reserve University On Fri, Apr 10, 2015 at 10:14 AM, Warren D Smith <warren.wds@gmail.com> wrote:
To make it even worse: I hereby proclaim that I am "heuristically certain" that fsq_n is, for all except a finite number of n, divisible both by an upper, and by a lower, TWIN prime. (If p and p+2 both are prime, than p is the lower and p+2 is the upper twin.) The same sort of proof must exist, but now finding that proof is likely to be so hard that nobody will ever do it!
--CORRECTION: sorry, my brain switched to wrong. I was basing the above on Brun's theorem that the sum of 1/t, where t are twin primes, diverges. But actually Brun says the opposite: this sum is finite.
OK, so to fix this: I am heuristically certain that for all but a finite set of n, fsq_n is divisible by a prime P<n such that P is of the form P=1000000000*k+1. [Pick your favorite number of zeros, I'm still certain.] And a proof of this should be available findable by the technique I mentioned... but that proof will be extremely hard to find.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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participants (3)
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Charles Greathouse -
Neil Sloane -
Warren D Smith