# [seqfan] Re: Sophie Germain pentagonal numbers

Jack Brennen jfb at brennen.net
Sat Mar 13 00:55:32 CET 2010

```Realized that I should probably post some output to be polite.  :)
This is for a loop upper bound of 87:

[75, 106]
[244, 345]
[86359, 122130]
[281384, 397937]
[99658019, 140937722]
[324716700, 459218761]
[115005267375, 162642008866]
[374722790224, 529938052065]
[132715978892539, 187688737293450]
[432429775201604, 611548052864057]
[153154124636722439, 216592640194632242]
[499023585859860600, 705725923067069521]
[176739727114798801875, 249947719095868313626]
[575872785652503930604, 814407103671345362985]
[203957491936353180641119, 288439451243991839291970]
[664556695619403676056224, 939825091910809481814977]
[235366768954824455661049259, 332858876787847486674619562]
[766897850872006189664951700, 1084557341657970470669120281]
[271613047416375485479670203575, 384118855373724755630671682386]
[884999455349599523469678205384, 1251578232448206012342682989105]
[313441221351728355419083753876099, 443272826242401580150308446853690]
[1021288604575586978077818984061244, 1444320195687888080272985500306697]

Jack Brennen wrote:
> 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
>
> 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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>>
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>>
>>
>
>
>
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>

```