20161207, 17:43  #1 
Nov 2016
5404_{8} Posts 
Primes in nfibonacci sequence and nstep fibonacci sequence
nfibonacci 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 nstep 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 20161207 at 17:44 
20161207, 18:03  #2 
"Mark"
Apr 2003
Between here and the
2×3^{2}×347 Posts 
I suggest that you take a look at MathWorld. If one exists, you would like find out about it there.

20161207, 18:47  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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 n1 ( n values) mod n that every n terms in a sequence has the same modular remainder mod n. etc. for the relationship 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. 
20170201, 16:02  #4 
Nov 2016
2^{2}·3·5·47 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 
20170201, 16:48  #5  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:


20170202, 13:35  #7 
Romulan Interpreter
Jun 2011
Thailand
10010000111011_{2} Posts 

20170202, 17:13  #8 
Nov 2016
2^{2}×3×5×47 Posts 
... of course... A000012 contains no primes since it only contains 1 ... XDDD

20170202, 17:19  #9  
Nov 2016
2^{2}·3·5·47 Posts 
Quote:


20170202, 17:29  #10  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
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. Last fiddled with by science_man_88 on 20170202 at 17:36 

20170203, 06:18  #11  
Nov 2016
101100000100_{2} Posts 
Quote:
Last fiddled with by sweety439 on 20170203 at 06:23 

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