20210305, 03:26  #947 
"Ed Hall"
Dec 2009
Adirondack Mtns
1254_{16} Posts 
I was updating my set of sequences and I found the cycle for 38^9 (1184/1210). I also see yoyo has finished 439^46 into the same cycle (1184/1210). In fact, they merge at the previous term (2362) to 1184.

20210305, 04:08  #948  
"Alexander"
Nov 2008
The Alamo City
1100111001_{2} Posts 
Quote:
Last fiddled with by Happy5214 on 20210305 at 04:08 

20210305, 07:44  #949  
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
2^{3}·7·107 Posts 
Quote:


20210305, 14:35  #950  
"Ed Hall"
Dec 2009
Adirondack Mtns
11124_{8} Posts 
Quote:
Code:
46^9:i231 merges with 11408:i15 46^21:i452 merges with 9852:i11 46^29:i913 merges with 67260:i2 

20210305, 17:49  #951  
"Garambois JeanLuc"
Oct 2011
France
2×463 Posts 
Quote:
Quote:
OK, many thanks ! We will visualize all of this in the next update ... Last fiddled with by garambois on 20210305 at 17:53 Reason: Addition of the second quote 

20210306, 12:49  #952 
"Garambois JeanLuc"
Oct 2011
France
2×463 Posts 
Two new conjectures
and general method to find increasing sequences from index 1 for all bases (with bases and exponents of the same parity) After reading all your exchanges on the abundances about base 38, I had some new ideas. It helped me find several thousand sequences for base 6 with even exponents with abundant index 1 term. I remind you that until today, we had nothing for base 6. I therefore set out below two new conjectures : (138) et (139). I explain my approach for each of the two conjectures. This lead me to propose a general method which makes it possible to find for each base sequences with exponents of the same parity as the base with terms of index 1 which are abundant. Conjecture (138) : Base 6 sequences starting with 6^((2^3 * 3^2 * 5 * 7) * k) are increasing at least from index 1 to 2. This is proved for all k integers such that 0 < k < 5001. It remains to prove it for all integer values of k, but this is too difficult for me. Indeed, s(6^(2^3 * 3^2 * 5 * 7)) = d * c, with d = 5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 41 * 43 * 61 * 71 * 73 * 127 * 181 * 211 * 281 * 337 * 421 * 631. But, d is abundant, because s(d) / d > 1.2. And using a very small program, it is very easy to verify for all k from 1 to 5000 that d is a divisor of s(6^((2^3 * 3^2 * 5 * 7) * k)). Here is the little program in python : Code:
k=1 d=5^2*7^2*11*13*19*29*31*37*41*43*61*71*73*127*181*211*281*337*421*631 print('d = 5^2*7^2*11*13*19*29*31*37*41*43*61*71*73*127*181*211*281*337*421*631') print('s(d)/d =',(sigma(d)d)*1.0/d) while True: n=6^((2^3*3^2*5*7)*k) m=sigma(n)n if m%d!=0: print ('***** ',k) if k%1000==0: print (k) k+=1 Code:
sage: load("conj_6.sage") d=5^2*7^2*11*13*19*29*31*37*41*43*61*71*73*127*181*211*281*337*421*631 s(d)/d = 1.20464241970184 1000 2000 3000 4000 5000 Now let me introduce you a general method to find increasing sequences from index 1 for all bases with bases and exponents of the same parity. You can proceed with any base as with base 6, as in the example above. The problem is to find a good "starting exponent" i = 2^3 * 3^2 * 5 * 7 in our example. This is done by trial and error, but i think the search for a good starting exponent i can be automated. It must be chosen so that we can find a compound factor d which is abundant and which then divides the terms of index 1 for all the sequences for k from 1 to 10. We can finally verify using of a program similar to the one shown above that d divides well all the multiple exponents of i for k from 1 to 1000 for example. Then, we can formulate a solid conjecture, like the conjecture (138). For conjecture 137, for the sequences 2^(12 * k), we have d = 3^2 * 5 * 7 * 13 and s(d) / d > 1.13, so d is abundant. For the sequences 2^(40 * k), we have d = 3 * 5^2 * 11 * 17 * 31 * 41 and s(d) / d > 1.01. I did not check for the other starting exponents of conjecture 137 (90, 140, 210, 220 and 330) if the corresponding d factors are indeed abundant. Strictly speaking, if the factor d corresponding to one of these exponent was not abundant, it would be necessary to remove this exponent with its multiples from the conjecture. To find this starting exponent i, we noticed that for almost all bases, the more the exponent breaks down into small prime numbers, the more chance we have of finding an abundant value of d. But, we don't know how to prove it for even bases. It's just an observation that seems to be true. Thus, even for base 3 and in the general case, for base b, we could take a starting exponent of this type : i = 3 * 5 * 7 * 11 * 13 * 17 * 19 ... and we conjecture that after several tries, we would find a abundant d that would divide the b^(i * k) for the first values of k. But the most difficult will certainly then be to prove that d divides s (b^(i * k)) for any integer k ! And also to find the i as small as possible. This general method should work in theory. But in practice, I have not been able to find an exponent i and an abundant compound factor d for the base 3. The numbers grow to astronomical size very quickly ! But we feel that we have enough good reasons to state the following more general conjecture. Conjecture (139) : There exists for each base b a starting exponent i, such that for any integer k, the sequences b^(i * k) are increasing from index 1 during at least one iteration. One last little idea To finish, I would like to present to you an idea which should allow us to demonstrate that the conjecture for the base 38 was false, with the d = 3 * 5 * 7 * 13 which is deficient. I wrote a little program. I calculate for as many k as possible the numbers c = s(38^(12 * k)) / d = s(38^(12 * k)) / (3 * 5 * 7 * 13). And then I do a primality test on c. If c is a prime number for a single value of k, the conjecture is invalidated. Here is the little program : Code:
k=1 while True: n=38^(12*k) m=sigma(n)n c=m/(3*5*7*13) if is_pseudoprime(c): print ('***** ',k) if k%100==0: print (k) k+=1 But in principle, this test should work, to my mind, if we let the program run longer ... Here is the result of the program : Code:
sage: load("conj_38_12k.sage") 100 200 300 ... ... 2000 2100 2200 2300 Last fiddled with by Uncwilly on 20210306 at 22:40 Reason: Per OP request, see post #955 
20210306, 16:18  #953  
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
2^{3}·7·107 Posts 
Quote:
2^(2^3 * 3^2 * 5 * 7 * k + 1)  1 == 1 mod d 3^(2^3 * 3^2 * 5 * 7 * k + 1)  1) / 2 == 1 mod d 6^(2^3 * 3^2 * 5 * 7 * k) == 1 mod d If this is true then s(6^(2^3 * 3^2 * 5 * 7 * k)) = (2^(2^3 * 3^2 * 5 * 7 * k + 1)  1)*3^(2^3 * 3^2 * 5 * 7 * k + 1)  1) / 2  6^(2^3 * 3^2 * 5 * 7 * k) = 1*11 mod d Quote:
I suspect that the way to attack this for individual b at least would be to find an i that has a d that is abundant. As far as I can tell it is usually quite easy. Finding that for 22^(180 * k), d= 3^3*7*13*19*31*5*37*61 which is abundant was quite easy. d=3^2*5*7*13 is abundant. 2^(180 * k + 1)  1 == 1 mod d p^(180 * k) == 1 mod d for a lot of p. I haven't worked out the logic behind this yet. http://factordb.com/index.php?query=...at=1&sent=Show (2*p)^(180 * k) == 1 mod d if gcd(d,p)==1 as indicated by http://factordb.com/index.php?query=...at=1&sent=Show I would be very interested if someone could come up with a case where (2*p)^(180*k) does not have an abundant divisor for some p. 

20210306, 18:19  #954  
Aug 2020
19 Posts 
Quote:
So if p1 divides 2^3 * 3^2 * 5 * 7, it means 2^(2^3 * 3^2 * 5 * 7 * k) == 1^k == 1 mod p, hence 2^(2^3 * 3^2 * 5 * 7 * k+1)1 == 1^k*21 == 1 mod p. Same thing for 3 and 6. That take care of p=5+,7+,11,13,19,29,31,37,41,43,61,71,73,127,181,211,281,421,631 (+ This only prove that 5,7 divide s(6^(2^3 * 3^2 * 5 * 7 * k)), not 5^2,7^2) For p=337, we have (p1)/2 divides 2^3 * 3^2 * 5 * 7. Possible value for a^((p1)/2) is +1 mod p. I guess that mean 2^((3371)/2),3^((3371)/2),6^((3371)/2) are happened to be 1 mod 337. To prove that 5^2,7^2 divide s(6^(2^3 * 3^2 * 5 * 7 * k)), we use the fact that if a == 1 mod p, then a^p == 1 mod p^2. (51) divides 2^3 * 3^2 * 7, so a^(2^3 * 3^2 * 5 * 7) == 1 mod 5^2 for (a,p)=1 (71) divides 2^3 * 3^2 * 5, so a^(2^3 * 3^2 * 5 * 7) == 1 mod 7^2 for (a,p)=1 

20210306, 18:23  #955  
"Garambois JeanLuc"
Oct 2011
France
926_{10} Posts 
Many thanks henryzz ! Too much fatigue after all this work ! Please, would it be possible for someone (Edwin maybe?) to make the following correction in my post # 952 : Quote:
Quote:


20210306, 18:51  #956  
"Garambois JeanLuc"
Oct 2011
France
1636_{8} Posts 
Quote:
Yes, that is exactly what I am saying above the statement of the conjecture (139). The small paragraph below titled "One last little idea" was just intended to attempt to provide a method to invalidate a conjecture that would use a deficient d in its statement. I did not have fun looking for a conjecture true with a larger i and an abundant d for base 38. But it would indeed not be difficult ... And in my opinion, this is easy for any base in the form 2*p. The difficulty is much greater for an odd base like 3, I have tried and we have to take values of i which give starting numbers of sequences of several million digits. Quote:
It seems difficult to me ! I'll think about it... Last fiddled with by garambois on 20210306 at 18:56 Reason: Addition of the second quote and my answer 

20210306, 19:21  #957 
"Garambois JeanLuc"
Oct 2011
France
39E_{16} Posts 
@henryzz and warachwe :
It is difficult for me to understand your reasoning, I am not used to this theoretical reasoning ! But does this mean that conjecture (138) is fully proven ? I will try to understand all this, because these mechanisms interest me a lot, like everything you taught us in the summer of 2020. I often use your explanations to understand phenomena on aliquot sequences. 
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