[seqfan] Re: Seq. A118064.
Robert G. Wilson, v
rgwv at rgwv.com
Mon Jun 1 20:25:19 CEST 2009
Dear Robert,
Thank you for your calculations. They match mine exactly. I took the digits
out the 64 digits and found:
= 1.323982145891606234777299440047139038371441916546100653011463101...
I will putting these results togther and sent it to Eric first and then
after a reply the OEIS.
Sincerely yours, Bob.
Robert Gerbicz wrote:
> 2009/5/31 Robert G. Wilson, v <rgwv at rgwv.com>
>
>>Sequence Fans,
>>
>> Would someone please send me the recent discussion on
>>sequence A118064: Decimal expansion of the sum of the reciprocals
>> of the palindromic primes (Honaker's constant)?
>>
>>Bob.
>>
>>
>>_______________________________________________
>>
>>Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> Hi!
>
> I don't remember for that discussion, but there should be problem with
> A118064. Let f(L) to be the sum of reciprocal of L digits palindromic
> primes, then my computation gives:
>
> f(1)=1.17619047619047619047619047619
> f(2)=0.0909090909090909090909090909091
> f(3)=0.0536236774907238646456937673383
> f(5)=0.00302515766035758995163241184216
> f(7)=0.000215703420553903851823404296181
> f(9)=0.0000166123943226029201053839857831
> f(11)=0.0000013084139501429137664563612017
> f(13)=0.000000109957987520944535638817533488
> f(15)=0.0000000094541435099826428103062275838
>
> Using this and bounds for the error term gives that the constant begins
> with: 1.323982146
>
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>
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>
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