[seqfan] Re: The last digit "d" of a(n) is present in a(n+1+d)
Eric Angelini
Eric.Angelini at kntv.be
Wed Jan 28 14:38:56 CET 2009
Hello Paolo,
> 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)?
... you are right, Paolo -- this was not clear enough; my
"natural" way to see this would be:
- the order in which the digits must be present in an integer
has no importance: a(5) = 102 is ok (the first writing cons-
traint on a(5) is given by the last digit of a(2) which is
"2"; a(5) doesn't have to start with this "2"; a(5) has to
contain "2" [in the minimum number a(n+1)>a(n), yes]
(BTW, I made a mistake in the rule (a) -- which should read now:
>Two more easy rules:
>
a) When there is no writing constraint on a(n),
take a(n) = a(n-1) + 1
>b) When there is a writing constraint on a(n),
> take the smallest a(n) > a(n-1) not leading
> to a contradiction
... I'll check yr seq tonight -- many, many thanks for yr post!
Best,
É.
-----Message d'origine-----
De : seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] De la part de Paolo Lava
Envoyé : mercredi 28 janvier 2009 9:28
À : Sequence Fanatics Discussion list
Objet : [seqfan] Re: The last digit "d" of a(n) is present in a(n+1+d)
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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