[seqfan] Proof For A269254
Brad Klee
bradklee at gmail.com
Sun Oct 22 01:20:29 CEST 2017
After: http://list.seqfan.eu/pipermail/seqfan/2017-October/018015.html
For all integer m>2, excluding initial condition a_{0}=1, we prove
composite every term of the sequence
a_{n} = (m^2-2)*a_{n-1} - a_{n-2} ; a_{0} = 1, a_{1}=m^2-1 .
a_{n}: 1, -1 + m^2, 1 - 3 m^2 + m^4, -1 + 6 m^2 - 5 m^4 + m^6 . . .
Certainly a_{n} will never reach a prime if there exists a factorization by
sequences L and R:
a_{n} = L_{n}*R_{n},
With L_{n}!=1, R_{n}!=1 when n>0. To suggest a form for L & R, we observe
that a_{n} = a'_{2n}, where,
a'_{n} = m*a_{n-1}-a_{n-2} ; a_{0} = 1, a_{1}=m.
a'_{n}: 1, m, -1 + m^2, -2 m + m^3, 1 - 3 m^2 + m^4, 3 m - 4 m^3 + m^5 . . .
The proposed factors L & R have the same signature = (m,-1), with slightly
different initial conditions,
L_{n} = m*L_{n-1}-L_{n-2}; L(0)=1, L(1)=m-1.
R_{n} = m*R_{n-1}-R_{n-2}; R(0)=1, R(1)=m+1.
L_{n}: 1, -1 + m, -1 - m + m^2, 1 - 2 m - m^2 + m^3 . . .
R_{n}: 1, 1 + m, -1 + m + m^2, -1 - 2 m + m^2 + m^3 . . .
For m>2, these sequences strictly increase. They will factor a_{n} if they
satisfy a zero-sum,
L_{n}*R_{n} - (m^2-2) * L_{n-1}*R_{n-1} + L_{n-2}*R_{n-2} = 0.
With the recursion relations, the zero-sum reduces to,
L_{n} R_{n-2} - 2 L_{n-1} R_{n-1} + L_{n-2} R_{n} = 0.
Satisfaction of this condition depends on the particular initial
conditions. We write an induction on index n:
L_{n} R_{n-2} - 2 L_{n-1} R_{n-1} + L_{n-2} R_{n} = 0
===> L_{n+1} R_{n-1} - 2 L_{n} R_{n} + L_{n-1} R_{n+1} = 0 .
By another application of the recursion relations, the second condition
returns to the first. The case n=2 holds,
(-1 - m + m^2) + (-1 + m + m^2) - 2*(m-1)*(m+1) = 0.
===> 0 = 0, True.
Then the induction completes, thus proving that L_{n} and R_{n} factor
a_{n}. The strictly increasing property of L & R ensures that neither L_{n}
or R_{n} ever equal to one after n=0. Ultimately this proves that
A269254(m^2-2)=-1 for all m>2.
Regards,
Brad
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