[seqfan] Digits of "n" visible in a product
Eric Angelini
Eric.Angelini at kntv.be
Sun May 31 21:51:01 CEST 2015
a(n)=1,1,1,1,1,1,1,1,1,10,11,11,11,11,
11,11,11,11,63,10,11,11,11,11,11,11,
11,78,79,10,11,11,11,11,11,11,94,89,
82,10,11,11,11,11,11,97,91,85,102,
10,11,11,11,11,101,103,89,98,88,...
Hello SeqFans,
The last term above (88) is the 59th of the sequence.
If you compute p = [n*a(n)] you will notice
that, for any a(n), the said product shows "n" in the digits of "p" at the
odd positions of "p":
88 x 59 = 5192 showing "5" and "9"
at positions 1 and 3 in "p".
For the term before 88 we have:
98 x 58 = 5684 showing "5" and "8"
at positions 1 and 3 of "p".
For the term before 98 we have:
89 x 57 = 5073 showing 5 and 7... etc.
For 103 we have 103 x 56 = 5768
showing 5 and 6 etc.
Yes, there are a lot of "11" above --
but they will disappear very soon
from the sequence as "n" grows.
The seq was extended with the
smallest integer a(n) such that [n*a(n)]
always shows "n" at the odd positions
of "p" (the product).
I don't know if there is always an a(n)
for every "n". If not, this "hole" could
be marked with a "0" in the sequence.
Here are the products for n=1 to 59:
P=1,2,3,4,5,6,7,8,9,100,121,132,143,
154,165,176,187,198,1197,200,231,
242,253,264,275,286,297,2184,2291,
200,341,352,363,374,385,396,3478,
3382,3198,400,451,462,473,484,495,
4262,4277,4080,4998,500,561,572,
583,594,5555,5768,5073,5684,5192.
By looking at any p(n) in the sequence P
just above, you will immediately know
the index of the term p(n) in P by
looking at the digits of p(n) placed in
odd position. And you'll know that
the division of p(n) by this index
will return an integer.
The last term of the 3rd line, for instance
(2291), is the 29th term of "P" (positions
1 and 3 in 2291 are 2 and 9); and
2291 is divisible by 29 (=79).
Best,
É.
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