[seqfan] Re: Composite such that the number of non-divisors of n divides their sum
Jack Brennen
jfb at brennen.net
Thu Feb 13 02:15:01 CET 2014
Just my analysis, which I must leave unfinished; although I spent
about an hour so far, I'm not able to reach a nice conclusion,
but perhaps one of you can take it further than I did.
If you search among numbers of the form Xp, where X is a constant
and p is a prime which does not divide X, it seems that the
fractional part of the ratio:
antisigma(Xp)/antitau(Xp)
converges on a constant fraction as p -> infinity.
For instance, given the largest known member of the sequence,
692945344, you can write this as Xp where X is 250432 and where
p is 2767.
Note that if you look at values of antisigma(Xp)/antitau(Xp)
for this value of X, you get this behavior:
? antisigma(x)=return(x*(x+1)/2-sigma(x));
? antitau(x)=return(x-numdiv(x));
? X=250432;
? forprime(p=2700,2800,print([p,antisigma(X*p)/antitau(X*p)*1.0]));
[2707, 338959765.99998017463175617572611650279]
[2711, 339460629.99998152362460294351755544356]
[2713, 339711061.99998219662932739360136967265]
[2719, 340462357.99998420970303879330550622458]
[2729, 341714517.99998754515495875725556178480]
[2731, 341964949.99998820931415054249436871931]
[2741, 343217109.99999151557178111806816095939]
[2749, 344218837.99999414325881768989629175551]
[2753, 344719701.99999545137544720695770607665]
[2767, 346472726.00000000000000000000000000000]
[2777, 347724886.00000322093815943676098593561]
[2789, 349227478.00000705557532363843832537203]
[2791, 349477910.00000769147568171712194594335]
[2797, 350229206.00000959372033441790341713928]
? for(z=12,30,p=nextprime(2^z);\
print([p,antisigma(X*p)/antitau(X*p)*1.0]));
[4099, 513260438.00029065951888648600170009633]
[8209, 1027898198.0005929615076781814190902552]
[16411, 2054919830.0007436436494949788111430455]
[32771, 4103453590.0008189316672843090206057362]
[65537, 8206281046.0008566901443940524631339528]
[131101, 16415942870.000875576152410999269346868]
[262147, 32824998806.000885013287252317641341118]
[524309, 65651875798.000889733970093969571704524]
[1048583, 131299368982.00089209409768136848017309]
[2097169, 262599113558.00089327424194236764086031]
[4194319, 525195847958.00089386430971150285079205]
[8388617, 1050389066326.0008941593455304128055018]
[16777259, 2100781262998.0008943068643982676119349]
[33554467, 4201556139926.0008943806235002648410939]
[67108879, 8403105392918.0008944175031331453586830]
[134217757, 16806210660566.000894435942979673278117]
[268435459, 33612414434198.000894445162901185500290]
[536870923, 67224829494422.000894449772863873612473]
[1073741827, 134449656609686.00089445207784513394512]
The fractional portion of the ratio seems to be converging and would
indicate that there are no more members of the sequence of the
form 250432*p.
The behavior seems to be convergence for every value of X (but to
different fractional parts of course); the convergence also seems
to happen when we replace p with p^2 or p^3, etc., showing why this is
true, and how fast the fractional part converges, would seem to lead
to much more efficient searches than brute force while still allowing
the search to be exhaustive.
On 2/12/2014 11:59 AM, Don Reble wrote:
>> A230605
>> Is there a better way to search for members of this sequence...?
>
> Exercise: prove that each member has a squared factor.
> (BTW, it's easy to prove that for the even members.)
>
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