[seqfan] a(n) + (next digit) = a(n+1)
Eric.Angelini at kntv.be
Sun Nov 16 12:19:05 CET 2014
To build a sequence S including the term
<1000000> (one million) that satisfies:
a(n) + (next digit of S) = a(n+1)
is easy. Start with 1000000 (one million)
and work backwards, subtracting at each
step the first digit of a(n) to get the next integer, and so on.
You will reach 9 at the end. So the seq S above
will look like:
The funny thing here is that the lexico-first
infinite such sequence is very difficult
to compute "upwards"! What is the
next term after 18, in the example
above? Is it 19 ou 20?
Building S "upwards" starting with 9
leads to a lot of such crossroads!
The contrast between "backwards from any a(n), no matter how huge
a(n) is" and the "upwards" method
to reach said a(n) is stunning!
Backtracking mon amour!
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