[seqfan] Re: The last digit "d" of a(n) is present in a(n+1+d)

[Paolo Lava]

Dear Eric,

I tried to follow your indications and I have computed the sequence up to 50 terms (free from mistakes?):

1,2,10,20,102,103,104,112,113,123,124,134,143,153,154,164,234,243,244,254,264,334,335,340,404,414,424,425,435,445,454,464,465,475,485,545,554,555,565,575,585,645,646,650,705,715,725,735,736,746

I think you need to better specify the rules. For instance if more than one digit have  to be present in a number (case of a(5)) should we keep the precedence and put the first digit to insert as less significant than the second an so on? Or may we adjust the digits in any order, trying to insert the minimum number a(n+1)>a(n)?

Ciao

Paolo

--- Eric.Angelini at kntv.be wrote:

From: "Eric Angelini" <Eric.Angelini at kntv.be>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan]  The last digit "d" of a(n) is present in a(n+1+d)
Date: Mon, 26 Jan 2009 18:57:51 +0100


Hello SeqFans,

... as said above, the last digit "d" of a(n)
must be present in a(n+1+d).

Two more easy rules:

a) When there is no writing constraint on a(n),
   take a(n) = a(n) + 1
b) When there is a writing constraint on a(n),
   take the smallest a(n) > a(n-1) not leading
   to a contradiction

---

We then have (if I don't miss smtg):

     n = 1  2   3   4   5    6    7    8    9    10   11   12   13   14   
     S = 1, 2, 10, 20, 102, 103, 104, 112, 113, 123, 124, ...
ObDigit:    -   1   0   02   -    -     2   -     3    2    4    3    3    

Could please someone compute S (as this is the 
third time I make a mistake in the first terms :-(

Best,
É.



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