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 2016-12-07, 17:43 #1 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 BF016 Posts Primes in n-fibonacci sequence and n-step fibonacci sequence n-fibonacci sequence: n OEIS sequence 1 A000045 2 A000129 3 A006190 4 A001076 5 A052918 6 A005668 7 A054413 8 A041025 9 A099371 10 A041041 11 A049666 12 A041061 n-step fibonacci sequence: n OEIS sequence 1 A000012 2 A000045 3 A000213 4 A000288 5 A000322 6 A000383 7 A060455 8 A123526 9 A127193 10 A127194 11 A168083 12 A207539 Is there a project of searching primes in these sequences? Last fiddled with by sweety439 on 2016-12-07 at 17:44
 2016-12-07, 18:03 #2 rogue     "Mark" Apr 2003 Between here and the 144368 Posts I suggest that you take a look at MathWorld. If one exists, you would like find out about it there.
2016-12-07, 18:47   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by sweety439 Is there a project of searching primes in these sequences?
have you read up on recursive relations and parity arguments etc before posting these because with that and modular arithmetic on polynomials under the polynomial remainder theorem I bet you could do a quick scan of them first yourself.

for example we know things like:

any polynomial without a certain number of odd coefficients including the constant term have certain properties like always being even or switching back and forth etc. just based on parity arguments we can say things like:

any polynomial with an even number of odd coefficients will pair those up under half the integer x values. any with an odd number of odd coefficients including the constant term will be odd at least half the time.

we know by the pigeonhole principle that given modular remainders only can be 0 to n-1 ( n values) mod n that every n terms in a sequence has the same modular remainder mod n. etc.

for the relationship $a_n=a_{n-1}+a_{n-2}$ we have the obvious statements like unless the two values you sum are opposite parity then the nth value will be even. since the only even prime is 2 it makes it hard to be prime and have this occur.

 2017-02-01, 16:02 #4 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 24·191 Posts Primes in Lucas sequences Are there any research for primes in Lucas U(P, Q) and V(P, Q) sequences? i.e. a(0)=0, a(1)=1, a(n+2)=P*a(n+1)-Q*a(n) for all n>=0 and a(0)=2, a(1)=P, a(n+2)=P*a(n+1)-Q*a(n) for all n>=0
2017-02-01, 16:48   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by sweety439 Are there any research for primes in Lucas U(P, Q) and V(P, Q) sequences? i.e. a(0)=0, a(1)=1, a(n+2)=P*a(n+1)-Q*a(n) for all n>=0 and a(0)=2, a(1)=P, a(n+2)=P*a(n+1)-Q*a(n) for all n>=0

2017-02-01, 19:56   #6
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

27268 Posts

Quote:
 Originally Posted by sweety439 n-step fibonacci sequence: n OEIS sequence 1 A000012 [...] Is there a project of searching primes in these sequences?
Do not search primes in A000012.

2017-02-02, 13:35   #7
LaurV
Romulan Interpreter

Jun 2011
Thailand

263A16 Posts

Quote:
 Originally Posted by R. Gerbicz Do not search primes in A000012.
LOL Robert, good spot, haha.. .

 2017-02-02, 17:13 #8 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 BF016 Posts ... of course... A000012 contains no primes since it only contains 1 ... XDDD
2017-02-02, 17:19   #9
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

24×191 Posts

Quote:
 Originally Posted by sweety439 n-fibonacci sequence: n OEIS sequence 1 A000045 2 A000129 3 A006190 4 A001076 5 A052918 6 A005668 7 A054413 8 A041025 9 A099371 10 A041041 11 A049666 12 A041061 n-step fibonacci sequence: n OEIS sequence 1 A000012 2 A000045 3 A000213 4 A000288 5 A000322 6 A000383 7 A060455 8 A123526 9 A127193 10 A127194 11 A168083 12 A207539 Is there a project of searching primes in these sequences?
The 11-step fibonacci sequence should be A127624, not A168083.

2017-02-02, 17:29   #10
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by sweety439 ... of course... A000012 contains no primes since it only contains 1 ... XDDD
and if you do the math you'll see that primes could only occur in certain places in the others as well based on them being odd:

0,1,1,2,3,5,8,... notice a pattern
even+odd=odd
odd+odd=even so every third entry is even and can be eliminated from the search.( except 2)

1,1,1,3,5,9, they are all odd but technically could do other things to eliminate composites.
1,1,1,1,4,7,13,25,49,94,... every fifth number is eliminated because it's even.
1,1,1,1,1,.. all odd again.
1,1,1,1,1,1,6, every 7th number is eliminated because it's even
all odd again
every n+1th number is even and eliminated.

and that's just a start.

Last fiddled with by science_man_88 on 2017-02-02 at 17:36

2017-02-03, 06:18   #11
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

24×191 Posts

Quote:
 Originally Posted by science_man_88 and if you do the math you'll see that primes could only occur in certain places in the others as well based on them being odd: 0,1,1,2,3,5,8,... notice a pattern even+odd=odd odd+odd=even so every third entry is even and can be eliminated from the search.( except 2) 1,1,1,3,5,9, they are all odd but technically could do other things to eliminate composites. 1,1,1,1,4,7,13,25,49,94,... every fifth number is eliminated because it's even. 1,1,1,1,1,.. all odd again. 1,1,1,1,1,1,6, every 7th number is eliminated because it's even all odd again every n+1th number is even and eliminated. and that's just a start. https://en.wikipedia.org/wiki/Fibona...d_divisibility would help you with the fibonacci sequence.
For n-fibonacci sequence, if the k-th term (F(n,k)) is prime (F(n,0)=0, F(n,1)=1, F(n,2)=n), then k must be prime, the only exception is F(1,4)=3.

Last fiddled with by sweety439 on 2017-02-03 at 06:23

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