Sequence

Citrix · 2005-09-13, 07:00

Generate a sequence where the N+1 term is the product of all previous terms +1 and the t(N+1) is divisible byp(N+1) ie the Nth prime.

SO t(1) divisible by 1
t(2)=t(1)+1 and divisble by 2
t(3)=t(2)*t(1)+1 and divisble by 3
t(4)=t(3)*t(2)*t(1)+1 and divisble by 5.

and so on...

Good luck this is a hard problem, lets see who can generate the longest sequence.

Can you find a algorithm to generate this sequence to some prime p?
Citrix

Citrix · 2005-09-13, 07:24

There are multiple solution till each p but the smallest t(1) will be the correct one.

Citrix

Jushi · 2005-09-14, 13:07

Let T = t(1). Then:
t(2) = T + 1
t(3) = t(2)*t(1) + 1 = (T + 1)*T + 1 = T^2 + T + 1
t(4) = t(3)*t(2)*t(1) + 1 = (T^2 + T + 1)*(T + 1)*T + 1 = T^4 + 2T^3 + 2T^2 + T + 1

By looking modulo 5, the condition that t(4) is divisible by 5 is impossible.
The best sequence is thus:
t(1) = 1
t(2) = 2
t(3) = 3

Citrix · 2005-09-14, 23:00

cool solution, but the formula you generated for t4 is prime for 2,4,16,256.

All 2^2^n. Pretty cool. Are there any other primes of the form 2^2^n?

Citrix

Citrix · 2005-09-14, 23:23

Also prime for 2^(2^6)

Citrix · 2005-09-14, 23:33

no primes upto n=18

2005-09-13, 07:00	#1
Citrix Jun 2003 1605₁₀ Posts	Sequence Generate a sequence where the N+1 term is the product of all previous terms +1 and the t(N+1) is divisible byp(N+1) ie the Nth prime. SO t(1) divisible by 1 t(2)=t(1)+1 and divisble by 2 t(3)=t(2)t(1)+1 and divisble by 3 t(4)=t(3)t(2)*t(1)+1 and divisble by 5. and so on... Good luck this is a hard problem, lets see who can generate the longest sequence. Can you find a algorithm to generate this sequence to some prime p? Citrix

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2005-09-13, 07:24	#2
Citrix Jun 2003 1605₁₀ Posts	There are multiple solution till each p but the smallest t(1) will be the correct one. Citrix

2005-09-14, 13:07	#3
Jushi Sep 2005 UGent 74₈ Posts	Let T = t(1). Then: t(2) = T + 1 t(3) = t(2)t(1) + 1 = (T + 1)T + 1 = T^2 + T + 1 t(4) = t(3)t(2)t(1) + 1 = (T^2 + T + 1)(T + 1)T + 1 = T^4 + 2T^3 + 2T^2 + T + 1 By looking modulo 5, the condition that t(4) is divisible by 5 is impossible. The best sequence is thus: t(1) = 1 t(2) = 2 t(3) = 3

2005-09-14, 23:00	#4
Citrix Jun 2003 645₁₆ Posts	cool solution, but the formula you generated for t4 is prime for 2,4,16,256. All 2^2^n. Pretty cool. Are there any other primes of the form 2^2^n? Citrix

2005-09-14, 23:23	#5
Citrix Jun 2003 3105₈ Posts	Also prime for 2^(2^6)

2005-09-14, 23:33	#6
Citrix Jun 2003 3105₈ Posts	no primes upto n=18