[seqfan] Re: A269254 and A034807

[Bob Selcoe]

Based on Don’s and Andrew’s observations, it appears that there is a relationship between the triangle of coefficients of Lucas polynomials A034807 and a(n) = -1, and that my observations in A269254 may be a vestige of that relationship.

Let L_m be row m in A034807 with alternating signs (Don’s output) and L_m(j) be the j-th term in the Lucas polynomials corresponding to L_m. So for example, L_3 = n^3 - 3n; L_3(1) = -2, L_3(2) = 2, L_3(3) = 18, etc.

Expect a(n) = -1 will be those terms where n = L_m(j) + 1 are composite (from Andrew’s analysis). So:

L_2 = j^2 - 2; n = a(L_2) = -1 for all j >= 3 (proved). 
L_3 = j^3 - 3j; expect n = a(L_3) = -1 when j = {5, 8, 9, 11...}, so n = {110, 488, 702, 1298...}.
L_4 = j^4 - 4j^2 + 2 repeats terms in L_2. So a(L_4) = -1 when j >= 3. 

Expect the same pattern to hold for all m, so:

L_5 = j^5 - 5j^3 + 5n; n = a(L_5) = -1 when j = {3, 4, 5...}, so expect n = {123, 724, 2510...}.

Since I can’t program and did the calculations by hand, would someone please test this for L_5(j) j > 3 and some larger m? 

Thanks,
Bob Selcoe



Sent from my iPhone

On Oct 25, 2017, at 11:38 AM, Don Reble <djr at nk.ca> wrote:

>> 123 ... remains an obstacle to a significantly larger b-file.
> 
>   The sequence for A269254(123) has a factorization pattern.
>   For 1 < k < 25000: s(k) is divisible by Lucas(2k+1), unless
>   Lucas(2k+1) is divisible by 11 (unless 5 divides 2k+1),
>   whence s(k) is divisible by Lucas(2k+1)/11.
> 
>   (Here, Lucas() is the additive sequence A000032,
>    L(0)=2,1,3,4,7,11,18,...)
> 
> 
>   It may have something to do with these rarefying polynomials:
> 
> s[k] =                 2 s[k-0] - s[k-0]
> s[k] =                 n s[k-1] - s[k-2]
> s[k] =         (n^2 - 2) s[k-2] - s[k-4]
> s[k] =        (n^3 - 3n) s[k-3] - s[k-6]
> s[k] =  (n^4 - 4n^2 + 2) s[k-4] - s[k-8]
> s[k] = (n^5 - 5n^3 + 5n) s[k-5] - s[k-10]
> s[k] =              P[j] s[k-j] - s[k-2j]
> 
>   P[j] = n P[j-1] - P[j-2]
> 
>   110, 488, 702, 1298, and 2158 are values of (n^2-2),
>   and 123 is a value of (n^5-5n^3+5n).
> 
> -- 
> Don Reble  djr at nk.ca
> 
> 
> --
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