View Single Post 2021-03-06, 12:49 #952 garambois   "Garambois Jean-Luc" Oct 2011 France 11·79 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 Here is the result after a few hours : 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 On the other hand, I don't know how to prove that d divides the terms of index 1 for all k. 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 Unfortunately, after several hours no c was prime even though I had tested all k from 1 to 2300. 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 2021-03-06 at 22:40 Reason: Per OP request, see post #955  