[seqfan] Sum over inverse palindromes

Richard Mathar mathar at strw.leidenuniv.nl
Tue Oct 19 23:04:08 CEST 2010


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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