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Old 2021-03-09, 03:05   #969
EdH
 
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Quote:
Originally Posted by RichD View Post
Might have a merge with 58^7.
For base 58, I found these:
Code:
58^3:i63 merges with 65436:i2
58^7:i1035 merges with 2352868:i1
58^39:i878 merges with 2484:i9
I ran through i=100.
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Old 2021-03-09, 17:00   #970
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OK, mergers verified.
I will add the base 58 during the next update, the next weekend.
Many thanks !

Last fiddled with by garambois on 2021-03-09 at 17:01
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Old 2021-03-12, 19:42   #971
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Another sequence bites the dust thanks to a collaboration of yafu@home and my Kubuntu Focus laptop. 21^98 terminates at 43.
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Old 2021-03-12, 20:20   #972
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I will not post every terminated sequence here ;-)
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Old 2021-03-12, 21:10   #973
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Quote:
Originally Posted by yoyo View Post
I will not post every terminated sequence here ;-)
I'm sure you would need your own thread for the number of sequences you terminate/merge, assuming opportunistic people like myself don't poach them first. ;) In all seriousness, base 21 was one I had originally reserved (it's my favorite number), so I'm happy to see another termination there.
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Old 2021-03-12, 21:15   #974
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By the way, this thread is now almost a thousand posts long. Can someone stick this? It's now a long-running subproject and being a sticky thread is long overdue IMO.
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Old 2021-03-13, 14:08   #975
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Quote:
Originally Posted by Happy5214 View Post
Another sequence bites the dust thanks to a collaboration of yafu@home and my Kubuntu Focus laptop. 21^98 terminates at 43.
Another non trivial sequence bites the dust.
Thanks !


Quote:
Originally Posted by Happy5214 View Post
By the way, this thread is now almost a thousand posts long. Can someone stick this? It's now a long-running subproject and being a sticky thread is long overdue IMO.
Thanks also for sticking this thread !
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Old 2021-03-13, 18:24   #976
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Quote:
Originally Posted by henryzz View Post
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.

Henryzz, I don't know exactly what you are looking for, but I have done some work on the subject.
I have not found a prime number p for which n = (2 * p)^(180 * k) does not have an abundant dividor for any integer k.
But I have found good candidates for some p such that the term of index 1 of sequences that start on n is most likely deficient for k = 1 and even some for k = 1, 2, 3 .
Why do I say that my finds are only candidates ?
Because I only factored n by considering all its prime factors less than 10^4.
But in my opinion, it is unlikely that n becomes abundant if we consider the larger prime numbers of its factorization.

Here is the result after about 72 hours of running my program :
Code:
b = 2 * 3   k 2
b = 2 * 3   k 3
b = 2 * 3   k 4
b = 2 * 300301   k 2
b = 2 * 390391   k 2
b = 2 * 660661   k 2
b = 2 * 798799   k 2
b = 2 * 930931   k 2
b = 2 * 1179179   k 2
b = 2 * 1360591   k 2
b = 2 * 1623931   k 2
b = 2 * 1939939   k 2
b = 2 * 2372371   k 2
b = 2 * 2402401   k 2
b = 2 * 2482481   k 2
b = 2 * 2548547   k 2
b = 2 * 2662661   k 2
b = 2 * 3233231   k 2
b = 2 * 3333331   k 2
b = 2 * 3663661   k 2
b = 2 * 3723721   k 2
b = 2 * 3873871   k 2
b = 2 * 3993991   k 2
b = 2 * 3993991   k 3
b = 2 * 3993991   k 4
b = 2 * 4792789   k 2
b = 2 * 6066061   k 2
b = 2 * 6276271   k 2
b = 2 * 6516511   k 2
b = 2 * 6846841   k 2
b = 2 * 6846841   k 3
b = 2 * 7087081   k 2
b = 2 * 7417411   k 2
b = 2 * 8368361   k 2
b = 2 * 8558551   k 2
b = 2 * 8558551   k 3
b = 2 * 8558551   k 4
b = 2 * 8708701   k 2
b = 2 * 8786779   k 2
b = 2 * 8978971   k 2
b = 2 * 8998991   k 2
b = 2 * 10420411   k 2
b = 2 * 10498489   k 2
b = 2 * 10840831   k 2
b = 2 * 11111101   k 2
b = 2 * 11221211   k 2
b = 2 * 11297287   k 2
b = 2 * 11791781   k 2
b = 2 * 11791781   k 3
b = 2 * 11791781   k 4
b = 2 * 12192181   k 2
b = 2 * 12222211   k 2
b = 2 * 12438427   k 2
b = 2 * 12552541   k 2
b = 2 * 12642631   k 2
How to read this data ?

For example, for the first three lines in bold :
Code:
b = 2 * 3     k 2
b = 2 * 3     k 3
b = 2 * 3     k 4
This means that if p = 3 and therefore the base b = 2 * 3, then we have s(n) = s(b^(180 * k)) = s(6^(180 * k)) which is deficient for k = 1, 2, 3.
And s(n) = s(b^(180 * k)) = s(6^(180 * k)) is abundant only from k = 4.
I did not then test for k> 4.

The same goes for all the other bases noted in bold :
b = 2 * 3993991 (abundant only from k = 4 in this case)
b = 2 * 6846841 (abundant only from k = 3 in this case)
b = 2 * 8558551 (abundant only from k = 4 in this case)
b = 2 * 11791781 (abundant only from k = 4 in this case)
And for all the other bases which are not noted in bold, n is deficient only for k = 1 and becomes abundant for k = 2.

I carried out this test for the first 853683 prime numbers, that is to say for all the prime numbers p < 13073363.
And for all the prime numbers p < 13073363 which do not appear in the table above, we therefore have s(n) = s((2 * p)^(180 * k)) which is abundant (proved by calculation) already from k = 1.
So I did not test the k > 1 for all these prime numbers which are not in the table above.

An extremely curious finding :
It would seem that some prime numbers in the table above are arithmetic progressions.
For example, if we look at the prime numbers that appear at the beginning of the table, we can see that :
390391 + 270270 = 660661
And after that :
660661 + 270270 = 930931
Indeed, the prime numbers 390391, 660661 and 930931 appear in the table.
But unfortunately, the next one which is also prime does not appear in the table (930931 + 270270 = 1201201 is prime).
I do not have time to occupy myself with this question of the presence of arithmetic progressions in this table of prime numbers, but perhaps there are completely new conjectures to formulate ?

Henryzz, I don't know if this is what you were looking for ?
Do you need me to let the program run any further ?


The program I wrote is powerful at finding sequences starting with numbers whose base and exponent have the same parity and which have an abundant index 1 term.
We do not need to decompose the term.
A week ago, in post #952, I presented the first even exponent i = 2^3 * 3^2 * 5 * 7 such that sequences starting with the number n = 6^(i * k) have a index term 1 abundant for all k and with the following abundant factor d which was suitable for all these sequences :
d = 5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 41 * 43 * 61 * 71 * 73 * 127 * 181 * 211 * 281 * 337 * 421 * 631
This gave the conjecture (138) that several of you have subsequently proved.
With this new program, in just 10 seconds, I have already found a dozen of exponents that were suitable for base 6.
And after 1 hour and 15 minutes of operation, I found all of the following exponents that were suitable :
Code:
[720, 1080, 1260, 1440, 1680, 1800, 1980, 2160, 2340, 2520, 2640, 2700, 2772, 2880, 3120, 3240, 3360, 3600, 3780, 3960, 4032, 4200, 4320, 4680, 5040, 5280, 5400, 5760, 5940, 6240, 6480, 6720, 7020, 7200, 7560, 7920, 8100, 8400, 8640, 9360, 9720, 10080, 10800, 11340, 11880, 12600]
And simultaneously during these 1 hour and 15 minutes of calculation, the program generated the second following table :
Code:
b = 6   k=1   exposant e = 720 = 2^4 * 3^2 * 5   i = 720 = 2^4 * 3^2 * 5
b = 6   k=2   exposant e = 1440 = 2^5 * 3^2 * 5   i = 720 = 2^4 * 3^2 * 5
b = 6   k=3   exposant e = 2160 = 2^4 * 3^3 * 5   i = 720 = 2^4 * 3^2 * 5
Sequence multiples pour base b = 6  i = 720 = 2^4 * 3^2 * 5  d = 4132565474853920632953425 = 5^2 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 * 61 * 73 * 97 * 181 * 241 * 577
b = 6   k=1   exposant e = 1080 = 2^3 * 3^3 * 5   i = 1080 = 2^3 * 3^3 * 5
b = 6   k=3   exposant e = 3240 = 2^3 * 3^4 * 5   i = 1080 = 2^3 * 3^3 * 5
Sequence multiples pour base b = 6  i = 1080 = 2^3 * 3^3 * 5  d = 64946984492713940892505843187575 = 5^2 * 7 * 11 * 13 * 19 * 31 * 37 * 41 * 61 * 73 * 109 * 181 * 241 * 271 * 433 * 541 * 2161
b = 6   k=1   exposant e = 1260 = 2^2 * 3^2 * 5 * 7   i = 1260 = 2^2 * 3^2 * 5 * 7
b = 6   k=2   exposant e = 2520 = 2^3 * 3^2 * 5 * 7   i = 1260 = 2^2 * 3^2 * 5 * 7
b = 6   k=3   exposant e = 3780 = 2^2 * 3^3 * 5 * 7   i = 1260 = 2^2 * 3^2 * 5 * 7
Sequence multiples pour base b = 6  i = 1260 = 2^2 * 3^2 * 5 * 7  d = 4888951109215725280985556125250425 = 5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 43 * 61 * 71 * 73 * 127 * 181 * 211 * 421 * 631 * 2521
b = 6   k=2   exposant e = 2880 = 2^6 * 3^2 * 5   i = 1440 = 2^5 * 3^2 * 5
b = 6   k=3   exposant e = 4320 = 2^5 * 3^3 * 5   i = 1440 = 2^5 * 3^2 * 5
Sequence multiples pour base b = 6  i = 1440 = 2^5 * 3^2 * 5  d = 797585136646806682160011025 = 5^2 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 * 61 * 73 * 97 * 181 * 193 * 241 * 577
b = 6   k=1   exposant e = 1680 = 2^4 * 3 * 5 * 7   i = 1680 = 2^4 * 3 * 5 * 7
b = 6   k=2   exposant e = 3360 = 2^5 * 3 * 5 * 7   i = 1680 = 2^4 * 3 * 5 * 7
b = 6   k=3   exposant e = 5040 = 2^4 * 3^2 * 5 * 7   i = 1680 = 2^4 * 3 * 5 * 7
Sequence multiples pour base b = 6  i = 1680 = 2^4 * 3 * 5 * 7  d = 1027478644809677744369628450729130985075 = 5^2 * 7^2 * 11 * 13 * 17 * 29 * 31 * 41 * 43 * 61 * 71 * 97 * 113 * 211 * 241 * 281 * 337 * 421 * 673 * 3361
b = 6   k=1   exposant e = 1800 = 2^3 * 3^2 * 5^2   i = 1800 = 2^3 * 3^2 * 5^2
b = 6   k=2   exposant e = 3600 = 2^4 * 3^2 * 5^2   i = 1800 = 2^3 * 3^2 * 5^2
b = 6   k=3   exposant e = 5400 = 2^3 * 3^3 * 5^2   i = 1800 = 2^3 * 3^2 * 5^2
Sequence multiples pour base b = 6  i = 1800 = 2^3 * 3^2 * 5^2  d = 430549205282764555621103122027375 = 5^3 * 7 * 11 * 13 * 19 * 31 * 37 * 41 * 61 * 73 * 101 * 151 * 181 * 241 * 601 * 1201 * 1801
b = 6   k=1   exposant e = 1980 = 2^2 * 3^2 * 5 * 11   i = 1980 = 2^2 * 3^2 * 5 * 11
b = 6   k=2   exposant e = 3960 = 2^3 * 3^2 * 5 * 11   i = 1980 = 2^2 * 3^2 * 5 * 11
b = 6   k=3   exposant e = 5940 = 2^2 * 3^3 * 5 * 11   i = 1980 = 2^2 * 3^2 * 5 * 11
Sequence multiples pour base b = 6  i = 1980 = 2^2 * 3^2 * 5 * 11  d = 500921715483185619108910812206761117175 = 5^2 * 7 * 11^2 * 13 * 19 * 23 * 31 * 37 * 61 * 67 * 73 * 181 * 199 * 331 * 397 * 661 * 991 * 1321 * 2971
b = 6   k=3   exposant e = 6480 = 2^4 * 3^4 * 5   i = 2160 = 2^4 * 3^3 * 5
Sequence multiples pour base b = 6  i = 2160 = 2^4 * 3^3 * 5  d = 1122456169041960873091755904130496194140035570711503206523075 = 5^2 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 * 61 * 73 * 97 * 109 * 163 * 181 * 241 * 271 * 433 * 541 * 577 * 811 * 1297 * 1621 * 2161 * 2593 * 3889 * 6481
b = 6   k=1   exposant e = 2340 = 2^2 * 3^2 * 5 * 13   i = 2340 = 2^2 * 3^2 * 5 * 13
b = 6   k=2   exposant e = 4680 = 2^3 * 3^2 * 5 * 13   i = 2340 = 2^2 * 3^2 * 5 * 13
b = 6   k=3   exposant e = 7020 = 2^2 * 3^3 * 5 * 13   i = 2340 = 2^2 * 3^2 * 5 * 13
Sequence multiples pour base b = 6  i = 2340 = 2^2 * 3^2 * 5 * 13  d = 2592519558190028920204970920911756521975 = 5^2 * 7 * 11 * 13^2 * 19 * 31 * 37 * 53 * 61 * 73 * 79 * 131 * 157 * 181 * 313 * 937 * 1171 * 2341 * 6553
b = 6   k=3   exposant e = 7560 = 2^3 * 3^3 * 5 * 7   i = 2520 = 2^3 * 3^2 * 5 * 7
Sequence multiples pour base b = 6  i = 2520 = 2^3 * 3^2 * 5 * 7  d = 149724897769239702501470160794374161071134869124345288345668279421025 = 5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 41 * 43 * 61 * 71 * 73 * 109 * 127 * 181 * 211 * 241 * 271 * 281 * 337 * 379 * 421 * 433 * 541 * 631 * 757 * 1009 * 2161 * 2521 * 7561
b = 6   k=1   exposant e = 2640 = 2^4 * 3 * 5 * 11   i = 2640 = 2^4 * 3 * 5 * 11
b = 6   k=2   exposant e = 5280 = 2^5 * 3 * 5 * 11   i = 2640 = 2^4 * 3 * 5 * 11
b = 6   k=3   exposant e = 7920 = 2^4 * 3^2 * 5 * 11   i = 2640 = 2^4 * 3 * 5 * 11
Sequence multiples pour base b = 6  i = 2640 = 2^4 * 3 * 5 * 11  d = 1564216444318060267573094415936051275 = 5^2 * 7 * 11^2 * 13 * 17 * 23 * 31 * 41 * 61 * 67 * 89 * 97 * 241 * 331 * 661 * 881 * 1321 * 5281
b = 6   k=1   exposant e = 2700 = 2^2 * 3^3 * 5^2   i = 2700 = 2^2 * 3^3 * 5^2
b = 6   k=3   exposant e = 8100 = 2^2 * 3^4 * 5^2   i = 2700 = 2^2 * 3^3 * 5^2
Sequence multiples pour base b = 6  i = 2700 = 2^2 * 3^3 * 5^2  d = 696329743891563806780087534740162625 = 5^3 * 7 * 11 * 13 * 19 * 31 * 37 * 61 * 73 * 101 * 109 * 151 * 181 * 271 * 541 * 601 * 1201 * 1801
b = 6   k=1   exposant e = 2772 = 2^2 * 3^2 * 7 * 11   i = 2772 = 2^2 * 3^2 * 7 * 11
b = 6   k=2   exposant e = 5760 = 2^7 * 3^2 * 5   i = 2880 = 2^6 * 3^2 * 5
b = 6   k=3   exposant e = 8640 = 2^6 * 3^3 * 5   i = 2880 = 2^6 * 3^2 * 5
Sequence multiples pour base b = 6  i = 2880 = 2^6 * 3^2 * 5  d = 919615662553768104530492711825 = 5^2 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 * 61 * 73 * 97 * 181 * 193 * 241 * 577 * 1153
b = 6   k=1   exposant e = 3120 = 2^4 * 3 * 5 * 13   i = 3120 = 2^4 * 3 * 5 * 13
b = 6   k=2   exposant e = 6240 = 2^5 * 3 * 5 * 13   i = 3120 = 2^4 * 3 * 5 * 13
b = 6   k=3   exposant e = 9360 = 2^4 * 3^2 * 5 * 13   i = 3120 = 2^4 * 3 * 5 * 13
Sequence multiples pour base b = 6  i = 3120 = 2^4 * 3 * 5 * 13  d = 1284529589404309867171183810386986553775 = 5^2 * 7 * 11 * 13^2 * 17 * 31 * 41 * 53 * 61 * 79 * 97 * 131 * 157 * 241 * 313 * 521 * 1249 * 2341 * 3121
b = 6   k=3   exposant e = 9720 = 2^3 * 3^5 * 5   i = 3240 = 2^3 * 3^4 * 5
Sequence multiples pour base b = 6  i = 3240 = 2^3 * 3^4 * 5  d = 27148037961226222991113680636686083867943088958727935867644872425 = 5^2 * 7 * 11 * 13 * 19 * 31 * 37 * 41 * 61 * 73 * 109 * 163 * 181 * 241 * 271 * 433 * 487 * 541 * 811 * 1297 * 1621 * 2161 * 2593 * 3889 * 4861 * 6481 * 9721
b = 6   k=2   exposant e = 6720 = 2^6 * 3 * 5 * 7   i = 3360 = 2^5 * 3 * 5 * 7
b = 6   k=3   exposant e = 10080 = 2^5 * 3^2 * 5 * 7   i = 3360 = 2^5 * 3 * 5 * 7
Sequence multiples pour base b = 6  i = 3360 = 2^5 * 3 * 5 * 7  d = 198303378448267804663338290990722280119475 = 5^2 * 7^2 * 11 * 13 * 17 * 29 * 31 * 41 * 43 * 61 * 71 * 97 * 113 * 193 * 211 * 241 * 281 * 337 * 421 * 673 * 3361
b = 6   k=2   exposant e = 7200 = 2^5 * 3^2 * 5^2   i = 3600 = 2^4 * 3^2 * 5^2
b = 6   k=3   exposant e = 10800 = 2^4 * 3^3 * 5^2   i = 3600 = 2^4 * 3^2 * 5^2
Sequence multiples pour base b = 6  i = 3600 = 2^4 * 3^2 * 5^2  d = 788670033040908691634515988885664479014123375 = 5^3 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 * 61 * 73 * 97 * 101 * 151 * 181 * 241 * 401 * 577 * 601 * 1201 * 1801 * 4801
b = 6   k=3   exposant e = 11340 = 2^2 * 3^4 * 5 * 7   i = 3780 = 2^2 * 3^3 * 5 * 7
Sequence multiples pour base b = 6  i = 3780 = 2^2 * 3^3 * 5 * 7  d = 82404142607783838376929183648309830176928742845471755163471825 = 5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 43 * 61 * 71 * 73 * 109 * 127 * 163 * 181 * 211 * 271 * 379 * 421 * 541 * 631 * 757 * 811 * 1621 * 2269 * 2521 * 7561
b = 6   k=3   exposant e = 11880 = 2^3 * 3^3 * 5 * 11   i = 3960 = 2^3 * 3^2 * 5 * 11
Sequence multiples pour base b = 6  i = 3960 = 2^3 * 3^2 * 5 * 11  d = 15657616943680432898365837997947429516858002526388553596160425 = 5^2 * 7 * 11^2 * 13 * 19 * 23 * 31 * 37 * 41 * 61 * 67 * 73 * 89 * 109 * 181 * 199 * 241 * 271 * 331 * 397 * 433 * 541 * 661 * 991 * 1321 * 2161 * 2377 * 2971
b = 6   k=1   exposant e = 4032 = 2^6 * 3^2 * 7   i = 4032 = 2^6 * 3^2 * 7
b = 6   k=1   exposant e = 4200 = 2^3 * 3 * 5^2 * 7   i = 4200 = 2^3 * 3 * 5^2 * 7
b = 6   k=2   exposant e = 8400 = 2^4 * 3 * 5^2 * 7   i = 4200 = 2^3 * 3 * 5^2 * 7
b = 6   k=3   exposant e = 12600 = 2^3 * 3^2 * 5^2 * 7   i = 4200 = 2^3 * 3 * 5^2 * 7
Sequence multiples pour base b = 6  i = 4200 = 2^3 * 3 * 5^2 * 7  d = 2616729487213159500641282093659007108501832335890375 = 5^3 * 7^2 * 11 * 13 * 29 * 31 * 41 * 43 * 61 * 71 * 101 * 151 * 211 * 241 * 281 * 337 * 421 * 601 * 701 * 1051 * 1201 * 4201 * 6301
For example, the line :
Code:
Sequence multiples pour base b = 6  i = 720 = 2^4 * 3^2 * 5  d =  4132565474853920632953425 = 5^2 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 *  61 * 73 * 97 * 181 * 241 * 577
means that we can formulate the following conjecture which is very similar to the conjecture (138) :
Base 6 sequences starting with 6^(720 * k) are increasing at least from index 1 to 2.
And we can see that the GCD (with only primes < 10^4) of all the terms of index 1 of sequences 6^(720 * k) for k from 1 to 3 is d = 4132565474853920632953425 = 5^2 * 7 * 11 * 13 * 17 * 19 * 31 * 37 * 41 * 61 * 73 * 97 * 181 * 241 * 577.
Thus, the same program found conjecture (138) by displaying the following line :
Code:
Sequence multiples pour base b = 6  i = 2520 = 2^3 * 3^2 * 5 * 7  d =  149724897769239702501470160794374161071134869124345288345668279421025 =  5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 41 * 43 * 61 * 71 * 73 * 109 *  127 * 181 * 211 * 241 * 271 * 281 * 337 * 379 * 421 * 433 * 541 * 631 *  757 * 1009 * 2161 * 2521 * 7561
And yet, one can still see a multitude of other conjectures of the same type with all the lines beginning with "Sequence multiples pour base b = ..."

The same phenomenon occurs if I run the program with bases 12 and 24 for which we did not have a suitable exponent in post #921.
After a few minutes, we can formulate a multitude of conjectures similar to conjecture (138) also for these bases.


My new holy grail :

On the other hand : nothing for odd bases which are prime numbers, like the base 3 for example.
Right now, I'm letting the program run.
This is my new holy grail : a base 3 sequence with an odd exponent whose term at index 1 would be abundant !
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Old 2021-03-13, 21:12   #977
henryzz
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I suspect if i is a suitable multiplier for a base to make it abundant then any multiple of i will be.

The thing I was wanting was an example such as 300301 that is deficient for (2 * p)^(180 * k). You have found many of them although they aren't that common. I suspect that for any multiplier there will be counter examples although I would imagine there will be a multiplier suitable for any p.
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Old 2021-03-14, 16:23   #978
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Quote:
Originally Posted by garambois View Post
The same phenomenon occurs if I run the program with bases 12 and 24 for which we did not have a suitable exponent in post #921.
After a few minutes, we can formulate a multitude of conjectures similar to conjecture (138) also for these bases.

My new holy grail :

On the other hand : nothing for odd bases which are prime numbers, like the base 3 for example.
Right now, I'm letting the program run.
This is my new holy grail : a base 3 sequence with an odd exponent whose term at index 1 would be abundant !

I tried my program with the base 30.
Nothing at all after 24 hours of calculation !!!
While it only took a few seconds for bases 6, 12 and 24 ...

I will be doing some other tests in the next few days and try to figure out which are the even and odd bases for which it is very difficult to find sequences (with base and exponent of the same parity) that have abundant index 1 terms !
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Old 2021-03-14, 16:26   #979
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Quote:
Originally Posted by henryzz View Post
You have found many of them although they aren't that common.
Have you had enough or do you still need more ?
Should I run the program even further ?
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