[seqfan] Re: Sum over inverse palindromes

Robert Israel israel at math.ubc.ca
Wed Oct 20 00:06:08 CEST 2010


Yes, the sum converges. 
The number of n-digit palindromes is O(10^(n/2)) while the reciprocal
of each is less than 10^(1-n), so the sum of the reciprocals of the
n-digit palindromes is O(10^(-n/2)).

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada

On Tue, 19 Oct 2010, Richard Mathar wrote:

>
> Now something completely different:
> Does sum_(n>=1) 1/A002113(n), the sum over the inverse palindromes,
> have a (finite) series limit?
>
>> From the ailes http://list.seqfan.eu/pipermail/seqfan/2010-October/006253.html
> ap> I would not consider 3, 5, 7, 9 to be palindromes !
> By definition, a palindrome is a number which is its own base-10
> reverse, so this includes the single-digit palindromes.
>
> As an appetizer, consider the convergents for n<=Nmax:
>
> Nmax    sum
> 20000 3.34906931814132361590
> 40000 3.35600296049604565480
> 60000 3.36005794552919211990
> 80000 3.36293480655195348130
> 100000 3.36516620223804499740
> 120000 3.36534862310413434930
> 140000 3.36550281800309687960
> 160000 3.36563636311806557740
> 180000 3.36575414200193086590
> 200000 3.36585948749788792320
> 220000 3.36595482091396225520
> 240000 3.36604184380739807940
> 260000 3.36612188965792070310
> 280000 3.36619599469052528260
> 300000 3.36626498014946334860
> 320000 3.36632952856512683700
> 340000 3.36639015817873028580
> 360000 3.36644731771317088440
> 380000 3.36650138297468668600
> 400000 3.36655267184296981800
> 420000 3.36660146738233875380
> 440000 3.36664799009872165890
> 460000 3.36669244233337646060
> 480000 3.36673500055440130010
> 500000 3.36677581958818502270
> 520000 3.36681504363492976260
> 540000 3.36685278561590738180
> 560000 3.36688915346445989730
> 580000 3.36692424373637428240
> 600000 3.36695814315487578550
> 620000 3.36699093522442906980
> 640000 3.36702268501641736870
> 660000 3.36705345675153401830
> 680000 3.36708330889191076490
> 700000 3.36711229480966657250
> 720000 3.36714046728965007250
> 740000 3.36716786702692462980
> 760000 3.36719453528346254900
> 780000 3.36722051010210188990
> 800000 3.36724582663294000060
> 820000 3.36727052043760352470
>
> # my hand-crafted machine program:
> isPali := proc(n)
>        n = digrev(n) ;
> end proc:
> x := 0.0 ;
> Digits := 20 ;
> for n from 1 do
>        if isPali(n) then
>                x := x+evalf(1/n) ;
>        end if;
>        if n mod 20000 = 0 then
>                printf("%d %.20f\n",n,x) ;
>        end if;
> end do:
>
>
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>
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>




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