# [seqfan] Re: Composite such that the number of non-divisors of n divides their sum

C Boyd cb1 at gmx.co.uk
Fri Feb 14 09:42:11 CET 2014

```Dear Jack and other Seqfans,

It seems possible that the fractional part is converging to 1/1118 (= 0.00089445438282647584973...) or its neighbourhood. Note that 1118 = 2*13*43 which divides 250432 = 2^6*7*13*43.

Furthermore, if R is all but the first of the antisigma/antitau ratios presented below, 1118|(round(R*1118)-1). Pari snippet to go with Jack's code:

for(z=12,30,p=nextprime(2^z);R=antisigma(X*p)/antitau(X*p)*1.0;q=round(R*1118)-1;print("R="R" : round(R*1118-1) mod(1118) = "q%1118));

R=513260438.00029065951888648600170009633 : round(R*1118-1) mod(1118) = 1117
R=1027898198.0005929615076781814190902552 : round(R*1118-1) mod(1118) = 0
R=2054919830.0007436436494949788111430455 : round(R*1118-1) mod(1118) = 0
R=4103453590.0008189316672843090206057362 : round(R*1118-1) mod(1118) = 0
R=8206281046.0008566901443940524631339528 : round(R*1118-1) mod(1118) = 0
R=16415942870.000875576152410999269346868 : round(R*1118-1) mod(1118) = 0
R=32824998806.000885013287252317641341118 : round(R*1118-1) mod(1118) = 0
R=65651875798.000889733970093969571704524 : round(R*1118-1) mod(1118) = 0
R=131299368982.00089209409768136848017309 : round(R*1118-1) mod(1118) = 0
R=262599113558.00089327424194236764086031 : round(R*1118-1) mod(1118) = 0
R=525195847958.00089386430971150285079205 : round(R*1118-1) mod(1118) = 0
R=1050389066326.0008941593455304128055018 : round(R*1118-1) mod(1118) = 0
R=2100781262998.0008943068643982676119349 : round(R*1118-1) mod(1118) = 0
R=4201556139926.0008943806235002648410939 : round(R*1118-1) mod(1118) = 0
R=8403105392918.0008944175031331453586830 : round(R*1118-1) mod(1118) = 0
R=16806210660566.000894435942979673278117 : round(R*1118-1) mod(1118) = 0
R=33612414434198.000894445162901185500290 : round(R*1118-1) mod(1118) = 0
R=67224829494422.000894449772863873612473 : round(R*1118-1) mod(1118) = 0
R=134449656609686.00089445207784513394512 : round(R*1118-1) mod(1118) = 0

For the smallest R above, 1118|round(R*1118).

(Please forgive purely empirical observations.)

Chris

> ----- Original Message -----
> From: Jack Brennen
> Sent: 13/02/14 01:15 AM
> To: seqfan at list.seqfan.eu
> Subject: [seqfan] Re: Composite such that the number of non-divisors of n divides their sum
>
> 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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>
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