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 2022-03-25, 00:58 #1 pvaldivia   Mar 2022 17 Posts Question on a sequence Hello, hope this is the right place to post this.... I noticed that in the first sequence that the entries in the sequence A077262(A144769(n)) appear to be divisible by the nth prime for n>=2. I.e. 5 is divisible by 5, 14 divisible by 7, 99 divisible by 11, etc; and wanted to ask if there is a good reason for this? As already noted in OEIS, there is a simple recursion relation for the first sequence, and the second sequence listed below is fairly simple https://oeis.org/A077262 https://oeis.org/A144769
2022-03-25, 12:47   #2
Dr Sardonicus

Feb 2017
Nowhere

135238 Posts

Quote:
 Originally Posted by pvaldivia Hello, hope this is the right place to post this.... I noticed that in the first sequence that the entries in the sequence A077262(A144769(n)) appear to be divisible by the nth prime for n>=2. I.e. 5 is divisible by 5, 14 divisible by 7, 99 divisible by 11, etc; and wanted to ask if there is a good reason for this?
Yes. The "law of small numbers."

Keep going further into the sequence. The "pattern" breaks down fairly soon.

Last fiddled with by Dr Sardonicus on 2022-03-26 at 12:28 Reason: I misread the question!

2022-03-25, 14:04   #3
charybdis

Apr 2020

32×5×19 Posts

Quote:
 Originally Posted by Dr Sardonicus Keep going further into the sequence. The "pattern" breaks down fairly soon.
Does it? Where?

This looks real to me. The sequence is really two sequences woven together, and each one on its own looks a bit like a Lucas sequence. Lucas sequences are known to have congruence properties like this, e.g. the Lucas numbers themselves have L(p) == 1 mod p for all primes p, and these properties are used in Lucas pseudoprimality tests. What OP has spotted is likely a similar phenomenon.

 2022-03-26, 13:13 #4 Dr Sardonicus     Feb 2017 Nowhere 7×853 Posts Oops, I misread the question. Your statement is essentially correct. (There may be an offset of 1 in the index, but that's not important.) There is a simple formula for A077262 [Second member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = k] in terms of Lucas numbers. But the index of the Lucas number giving the n-th term of the sequence isn't n. It's the n-th term of the sequence 1, 5, 7, 11,13,... of numbers congruent to 1 or 5 (mod 6). And when the index of the Lucas number is a prime p > 3, the corresponding term of A077262 is indeed divisible by p. (The value of Lp (mod p) for p prime is well known; I leave it to the reader to look it up.) In the following table, the index of the sequence A077262 is in the first column. The index of the corresponding Lucas number is in the second column. The Lucas number is in the third column. The term in the sequence is in the last column. For any number N in the second column (index of Lucas number), compare floor(N/3) to the number in the first column (index of sequence A077262) just to its left. You should be able to prove that this relationship always holds. Code: 1 1 1 0 2 5 11 5 3 7 29 14 4 11 199 99 5 13 521 260 6 17 3571 1785 7 19 9349 4674 8 23 64079 32039 9 25 167761 83880 10 29 1149851 574925 There is an obvious question: Why does the index of the Lucas number have to be relatively prime to 6? Well, there's one reason the index can't be divisible by 2, and a different reason it can't be divisible by 3. I will leave these as exercises for the reader.
 2022-03-27, 17:11 #5 pvaldivia   Mar 2022 100012 Posts Thanks for this excellent answer! I appreciate the insight. There are so many fascinating properties of the Fibonacci and Lucas sequences that it’s no wonder there is an entire journal devoted to their study.

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