Pre-Q&A

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
agentredlum said:
Please allow me to post a question.

Q. Find a relation between primes that generates the first 31 numbers in the following integer sequence

1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ...

Bonus Question: Why is this sequence not in OEIS?

If no one gets it in 24 hours i will post the answer unless someone asks for more time or asks for a hint.

:D
Solution: For positive integer n let the sum of the first 2n primes be defined by

\(\displaystyle f(2n) \ = \ 1 \ \ \text{iff} \ \ f(2n) \text{ is \ prime}\)

\(\displaystyle f(2n) \ = \ 0 \ \ \text{iff} \ \ f(2n) \text{ is \ composite}\)

Examples:

n = 1

\(\displaystyle f(2) \ = \ 2 \ + \ 3 \ = \ 5 \ = \ \text{prime} \ = \ 1\)

n = 2

\(\displaystyle f(4) \ = \ 2 \ + \ 3 \ + \ 5 \ + \ 7 \ = \ 17 \ = \ \text{prime} \ = \ 1\)

n = 3

\(\displaystyle f(6) \ = \ 17 \ + \ 11 \ + \ 13 \ = \ 41 \ = \ \text{prime} \ = \ 1\)

n = 4

\(\displaystyle f(8) \ = \ 41 \ + \ 17 \ + \ 19 \ = \ 77 \ = \ \text{composite} \ = \ 0\)

For integer n from 1 to 4 , we have thus generated the first 4 terms of the sequence ,

1 1 1 0

continue in like fashion to generate as many terms as you wish.

Bonus Question: There are related sequences at OEIS , for example ,

http://oeis.org/A013918

But none that use my scheme for 2n that i could find.

Justification for the use of 2n: It is obvious that if we sum an odd number of consecutive primes , beginning with the first prime 2 , the sum cannot be prime so it is natural to ignore those sums , hence we sum primes with an even number of terms 2n.

:D
 

johnr

Math Team
Apr 2012
1,579
22
CRGreathouse said:
johnr said:
Messing with the Fibonacci question empirically cost me hours of sleep! The more I sought, the more I found. Anyone get anywhere interesting with it?
Well, two key components are that the Fibonacci sequence is a gcd sequence: gcd(F(m), F(n)) = F(gcd(m, n)) and that Fibonacci numbers greater than, uh, 144 have a primitive prime divisor: p | F(n) but p does not divide F(m) for 0 < m < n. So sometimes you find that two factors are 'stuck together' for Fibonacci sequences and you can't get them apart, which means you can't make those numbers separately.

As usual there is an OEIS sequence spoiling the answer further if you care to look.
Since no one has posted an answer, I'd love to see the OEIS spoiler. I tried to find it myself, but failed.
 

johnr

Math Team
Apr 2012
1,579
22
Let me post a question. The letters below correspond to musical notes in a little tune or "call" I sort of composed. I say "sort of" because this is actually generated by a number sequence. What number sequence? (It helps to no just a smidge about elementary music theory: this piece uses a pentatonic scale in the key of G.)

A A C D G

D D A E G

E E D C G

C C A E G
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
johnr said:
Since no one has posted an answer, I'd love to see the OEIS spoiler. I tried to find it myself, but failed.
https://oeis.org/A178772

The Luca-Pomerance-Wagner paper is excellent. I seem to remember that there are relevant conference notes online somewhere.

Edit: They're actually in the entry, marked (strangely enough) as the preprint to the paper I mentioned.
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
I ran a hideous command
Code:
egrep ',(-?[0-9]+),\1,(-?[0-9]+),(-?[0-9]+),(-?[0-9]+),\3,\3,\1,(-?[0-9]+),\4,\5,\5,\3,\2,\4,\2,\2,\1,\5,\4,' stripped | perl -e 'while(<>){my %c; my $s=$_;if(/,(-?[0-9]+),\1,(-?[0-9]+),(-?[0-9]+),(-?[0-9]+),\3,\3,\1,(-?[0-9]+),\4,\5,\5,\3,\2,\4,\2,\2,\1,\5,\4,/){$c{$1}++;$c{$2}++;$c{$3}++;$c{$4}++;$c{$5}++;if(scalar(keys(%c))>4){print "$s\n"}}}'
on a local copy of the terms of the OEIS, but it didn't find any matches.
 

johnr

Math Team
Apr 2012
1,579
22
agentredlum said:
Greeting's johnr , glad to see you back to MMF. Will this link below help to answer your question?

http://www.bgfl.org/bgfl/custom/resourc ... sic/piano/

I played the notes but didn't recognize the tune.

:D
It'll help you hear the tune or, better, call. But it won't help you solve the math. While it would have been interesting if the tune was a recognizable one and perhaps even worth a short paper in a certain journal (hint, hint), it's not surprising that it is not recognizable.

Further hint. G is the lowest note in any G scale. Pentatonic scales use 5 of the 7 notes in the normal heptatonic scale. Commonly, the third and seventh note (B and F# in G) are left out. So the remaining scale is G, A, C, D, E. Further hint: After the fourth line here, the call repeats.
 

johnr

Math Team
Apr 2012
1,579
22
Ok, so the scale is G, A, C, D, E.

Set G=0, A=1, C=2, D=3, E=4

Perhaps this "tune" is more familiar: 1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0 - which then repeats.