[seqfan] Re: Two Sequences

Ali Sada pemd70 at yahoo.com
Thu Jul 4 18:10:08 CEST 2019


 Thank you Maximilian. I really appreciate your thorough feedback. And you are right. I checked manually and found that 83711*27720 is 2320468920, and 1145993*360360 is 412970037480. Thank you! 

Could you please check if the connection to primary Stirling numbers will persist 
 "this sequence is related to A120299 (Largest prime factor of Stirling numbers of first kind s(n,2) A000254.)  A120299 also represents the largest prime factor of M, except for M(1.)"

 Best,
Ali

    On Wednesday, July 3, 2019, 2:36:23 PM EDT, M. F. Hasler <seqfan at hasler.fr> wrote:  
 
 On Wed, Jul 3, 2019 at 4:19 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu> wrote:

The first sequence is 1, 66, 6, 12, 8220, 20, 420,213080, 17965080, 153720, 210951720, 14109480, 31766925960,....
A(n) is the least integer that when multiplied by the harmonic number sum of n generates a square number (M*M.)
H(1)=1, A(1)=1, M(1)=1
H(2)=3/2, A(2)=6, M(2)=3 ((6*3/2)^0.5)
H(3)=11/6, A(3)=66, M(3)=11 ((66*11/6)^0.5)
H(4)=25/12, A(4)=12, M(4)=5 ((12*25/12)^0.5)
And so on.
The second sequence is M itself. 
1, 3, 11, 5, 137, 7, 33, 761, 7129, 671, 83771, 6617,....
I searched OEIS and found that this sequence is related to A120299 (Largest prime factor of Stirling numbers of first kind s(n,2) A000254.)  A120299 also represents the largest prime factor of M, except for M(1.)


in order to get an integer square from a fraction p/q (in lowest terms)one must "complete" by the prime factors of p which occur to an odd power(the product of these primes (without multiplicity) is A007913, called squarefree part of p, a.k.a. "core",not to be confused with the squarefree kernel A007947 a.k.a. radical of p) and multiply by q (which cannot have any prime factor in common with p).In the PARI/gp language,A(n,H=sum(k=1,n,1/k))=core(numerator(H))*denominator(H)} /*=A007913(A001008(n))*A002805(n)*/orA_vec(Nmax, H=0)= vector(Nmax, k, H+=1/k; core(numerator(H))*denominator(H))A_vec(13)
%21 = [1, 6, 66, 12, 8220, 20, 420, 213080, 17965080, 153720, 2320468920, 14109480, 412970037480]

You may notice that [in addition to swapped 6,66]you have wrong values instead of 2320468920 and 412970037480,I assume hat this is due to lack of precision in the BASIC dialect you used.

Then, M = sqrt(H*A) = sqrt(N*core(N)) = A019554(N) (= "outer square root") of N = numerator(H),this does not depend at all on the denominator,and it could be simpler to use  A(n) = M(n)^2/H(n) .)
M_vec(Nmax, H=0)=vector(Nmax, k, H+=1/k; sqrtint(core(N=numerator(H))*N))
 = [1, 3, 11, 5, 137, 7, 33, 761, 7129, 671, 83711, 6617, 1145993]

(So far this yields the same result as the radical of N, A007947, but this will no more be the case as soon as N has a cube as factor. However, this isn't the case soon - you'll have to go beyond 150 terms or more!)
- Maximilian
 

I would really appreciate any feedback and support.
Best,
Ali Sada 

  



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