[seqfan] Re: Fractions, digits, decimal expansion

Bob Selcoe rselcoe at entouchonline.net
Tue Dec 15 18:36:16 CET 2015


Hi Lars, Hans, Neil, et al.

Lars, I was the one that missed something.  89/9 works, of course.  I 
over-thought
the problem (wasn't the first time, won't be the last LOL) and made an
obvious logical error initially.  I don't program and didn't check your
output, but I have no doubt a(63)=9.

I'm inclined to agree that this probably is a permutation.  A proof would be
interesting.

I would like to see this sequence and Hans' variant when they are posted.

Cheers,
Bob


--------------------------------------------------
From: "Lars Blomberg" <larsl.blomberg at comhem.se>
Sent: Tuesday, December 15, 2015 9:39 AM
To: "'Sequence Fanatics Discussion list'" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Fractions, digits, decimal expansion

> Hello,
>
> Maybe I am misunderstanding something, but I find a(63)=9.
>
> 1
> 6 0.1(6)
> 13 0.(461538)
> 10 1.3(0)
> 14 0.(714285)
> 17 0.(8235294117647058)
> 7 2.(428571)
> 8 0.875(0)
> 19 0.(421052631578947368)
> 23 0.(8260869565217391304347)
> 21 1.0(952380)
> 29 0.(7241379310344827586206896551)
> 34 0.8(5294117647058823)
> 31 1.0(967741935483870)
> 3 10.(3)
> 38 0.0(789473684210526315)
> 28 1.3(571428)
> 46 0.(6086956521739130434782)
> 47 0.(9787234042553191489361702127659574468085106382)
> 35 1.3(428571)
> 39 0.(897435)
> 49 0.(795918367346938775510204081632653061224489)
> 43 1.(139534883720930232558)
> 51 0.(8431372549019607)
> 42 1.2(142857)
> 41 1.0(24390)
> 48 0.8541(6)
> 53 0.(9056603773584)
> 26 2.0(384615)
> 12 2.1(6)
> 57 0.(210526315789473684)
> 58 0.9(8275862068965517241379310344)
> 59 0.(9830508474576271186440677966101694915254237288135593220338)
> 2 29.5(0)
> 61 0.0(327868852459016393442622950819672131147540983606557377049180)
> 24 2.541(6)
> 68 0.(3529411764705882)
> 11 6.(18)
> 52 0.21(153846)
> 63 0.(825396)
> 22 2.8(63)
> 69 0.(3188405797101449275362)
> 62 1.1(129032258064516)
> 71 0.(87323943661971830985915492957746478)
> 56 1.267(857142)
> 65 0.8(615384)
> 76 0.85(526315789473684210)
> 81 0.(938271604)
> 44 1.84(09)
> 67 0.(656716417910447761194029850746268)
> 64 1.046875(0)
> 83 0.(77108433734939759036144578313253012048192)
> 85 0.9(7647058823529411)
> 78 1.0(897435)
> 77 1.0(129870)
> 79 0.(9746835443037)
> 72 1.097(2)
> 84 0.(857142)
> 86 0.(976744186046511627906)
> 87 0.(9885057471264367816091954022)
> 80 1.0875(0)
> 89 0.(89887640449438202247191011235955056179775280)
> 9 9.(8)
>
> As to the question of the sequence being a permutation,
> the following table shows:
> C=Number of terms calculated
> F=First term that is missing
> C F F/C
> 1000 5 0.005
> 2000 50 0.025
> 5000 1480 0.296
> 10000 1650 0.165
> 20000 2475 0.124
> 50000 24750 0.495
> 100000 83250 0.833
> 200000 198000 0.990
> 500000 495000 0.990
> which seems to support the conjecture.
>
> /Lars
>
> -----Ursprungligt meddelande-----
> Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Hans Havermann
> Skickat: den 15 december 2015 14:19
> Till: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Ämne: [seqfan] Re: Fractions, digits, decimal expansion
>
> Bob Selcoe:
>
>> I may have spoken too soon:  80190/9 = 8910.  It may be impossible for
>> S to equal 80190 or another "precursor" of 9 but I'm not so sure.
>>
>> Perhaps someone who knows how to program could expand the sequence and
>> see if 9 appears?
>
> A no-leading-or-trailing-zeros version of the sequence has 9 appearing 
> after 89, earlier than the appearances of either 4 or 5. Allowing the 
> leading and trailing zeros makes the sequence *less* constrained.
>
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