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Sunday, June 12, 2016

When does n divide a_n in this sequence?

Consider the following sequence:

a(1)=0

a(2)=2

a(3)=3

for all n ≥ 4  a(n) = a(n-2)+a(n-3)

Here is a table of a(n) for 2 ≤ n ≤ 23

n       2     3     4      5      6     7       8      9    10     11     12
a(n)  2     3     2      5     5      7     10    12    17     22     29

n       13    14    15    16       17      18     19      20     21      22      23
a(n) 39    51    68    90    119   158   209   277   367   486   644

For 2≤ n ≤ 23 the n such that n divides a(n) are

n = 2,3,5,7,11,13,17,19,23.

Notice a pattern? Of course you do!

I propose the following question which I will answer in my next post (in a week or so)

PROVE or DISPROVE the following:

for all n≥ 2   n divides a(n) iff n is prime.



15 comments:

  1. I spent some time trying to prove this, but it turns out to be false! One can check that 271441 divides a(271441), but 271441 = 521^2 is not prime.

    ReplyDelete
    Replies
    1. You are correct- 561^2 is the first counterexample.
      I am surprised you found it so fast!

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    2. I wrote a program to test it out on the first half a million values and also found the counter-example at 271,441 (the program ran out of heap space before hitting 500,000 so I don't know where the next counter-example is yet). So, is there anything interesting or special about that number?

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    3. The second counterexample is 904631.

      Delete
  2. For one direction of your proof, I hope you make use of the fact that these numbers count maximal independent sets in cycles.

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  3. counter example: 134 divides a(134)
    a(134)/134 = 23149346804971524/134 = 172756319440086

    ReplyDelete
    Replies
    1. Something not quite exact here.

      23149346804971526 / 134 does yield a quotient of 172756319440086, but with a remainder of 2.

      Delete
  4. Used a python script to conclude that 271441 is indeed a counter-example.[Did not read the comment above!] But we need to see if primality is sufficient for the divisibility.

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  5. Interesting that the first counterexample is a perfect square. It suggests possibly weakening the conjecture: Maybe every n that divides a(n) is either a prime or a prime power?

    Alas no. The second counterexample is 904631 = 7 x 13 x 9941.

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  6. So, like the run of nine 8s in Pi, pointing out the difference between noticing a pattern over a finite range and proving that pattern is true on an infinite range.

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  7. I had a simpler "proof" that the conjecture is false: if it were true, it would give a very past primality testing algorithm and GASARCH would publish this algorithm somewhere other than this blog. This sequence is similar to a Fibonacci sequence and so (see Matrix Form in the wikipedia article on Fibonacci numbers) one can compute a(n) in time log(n)*(time required to multiply two numbers of size a(n)). One can compute a(n) mod n in time log(n)*(time required to do arithmetic mod(n)), so one can compute it in time O(log(n)^3).

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  8. Proof by OEIS: http://oeis.org/A013998

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  9. One observation I managed to make is this: a(n) is divisible by n if n is a prime or pseudoprime. Both 271,441 and 904,631 are Perrin's pseudoprimes.

    ReplyDelete
    Replies
    1. Aren't Perrin's pseudoprimes defined by this exact problem?

      Delete
  10. can you prove the following sentence plz?

    "If n is a prime, n divides A(n)"

    ReplyDelete