[seqfan] Re: Sophie Germain pentagonal numbers
Jack Brennen
jfb at brennen.net
Sat Mar 13 00:37:31 CET 2010
Other than the trivial pair (0,1), you need to solve:
u^2 - 2*v^2 = 23
And take only those values where u & v are both congruent to 5 mod 6.
Then you have P((u+1)/6) = 2*P((v+1)/6)+1
If I'm not mistaken, the following PARI/GP will give the
non-trivial answers:
forstep(i=7,47,8,print(Vec(lift(Mod((x+1),x^2-2)^i*(4*x-3)+(x+1))/6));\
print(Vec(lift(Mod((x+1),x^2-2)^i*(4*x+3)+(x+1))/6)))
Jonathan Post wrote:
> The pentagonal number analogue of
> A124174 Sophie Germain triangular numbers tr: 2*tr+1 is also a
> triangular number.
>
> 2*(A000326(j))+1 = A000326(k)
>
> begins (denoting P(n) = n-th pentagonal number):
> 2P(0)+1 = 2*0+1 = 1 = P(1)
> 2P(75)+1 = 2*8400+1 = 16801 = P(106).
>
> What are the next such values, which seem to be obtainable through a
> Pell's equation?
>
>
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