UncleEbenezer wrote:Replying to my own observation, googling "Fibonacci 89" tells us it's a well-known result.
Yes, indeed - I try not to post research problems to this board, or at least to warn people upfront that the problem doesn't have a known answer if I do post one!
Some of the answers I've seen from that bit of googling are a bit cumbersome, with lots of matrix equations, etc, that may make people's eyes glaze over - and it is in fact pretty straightforward. Basically, if one starts the Fibonacci sequence with F(1) = 1, F(2) = 1 and then continues it with F(n) = F(n-1) + F(n-2) for all n >= 3, the addition of the Fibonacci numbers that I gave is calculating the following sum:
X = F(1)/10 + F(2)/100 + F(3)/1000 + F(4)/10000 + F(5)/100000 + ...
Now use the defining equations of the Fibonacci sequence on this sum:
F(1)/10 = 1/10
F(2)/100 = 1/100
F(3)/1000 = F(1)/1000 + F(2)/1000
F(4)/10000 = F(2)/10000 + F(3)/10000
F(5)/100000 = F(3)/100000 + F(4)/100000
...
Add up all the first parts of those sums and then all the second parts, and we see that:
X = 1/10 + 1/100
+ F(1)/1000 + F(2)/10000 + F(3)/100000 + ...
+ F(2)/1000 + F(3)/10000 + F(4)/100000 + ...
But look at the second line of that addition as I've now written it out, and it's clearly just X divided by 100. And similarly, the third line is X minus its first term F(1)/10 = 1/10, all divided by 10, i.e. X/10 - 1/100. So:
X = 1/10 + 1/100 + X/100 + X/10 - 1/100
= 1/10 + X/100 + X/10
Multiplying through by 100 gives 100X = 10 + X + 10X, then collecting together the X's gives (100-10-1)X = 10, so 89X = 10, i.e. X = 10/89. So I get the decimal places of 10/89, which are of course the decimal places of 1/89 excluding the initial 0.
It's quite easy to adapt that argument for any base B other than 10 - basically just put Bs in place of the 10s, B^2s in place of the 100s, B^3s in place of the 1000s, B^4s in place of the 10000s, etc. So the answer to UncleEbenezer's "
Further wild thought: do any similar properties arise if we pose the problem in numbers to a base other than 10?" is "yes - in base B, the sum is related to the decimal places of B/(B^2-B-1)". This arguably works out most spectacularly in binary, where B/(B^2-B-1) = 2. The finite sums do converge to 2 from below, so they're of the form 1.111... with more and more consecutive 1s, but the number of consecutive 1s rises quite slowly compared with the number of digits in the addition result. E.g. you'd have to go quite a bit further than the analogue of the addition of the first 15 terms that I did earlier:
.
1
1
10
11
101
1000
1101
10101
100010
110111
1011001
10010000
11101001
101111001
1001100010
---------------- +
1111010111101000
before the fact that the sums were starting with more and more consecutive 1s really became all that noticeable...
Gengulphus