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- From: tadpole@drysdale.ucr.edu (tad white)
- Newsgroups: sci.math
- Subject: digit puzzles (was Re: Math puzzles - answer)
- Message-ID: <23725@galaxy.ucr.edu>
- Date: 6 Nov 92 18:43:36 GMT
- References: <19561@galois.mit.edu>
- Sender: news@galaxy.ucr.edu
- Organization: University of California, Riverside
- Lines: 31
- Nntp-Posting-Host: drysdale.ucr.edu
-
- John Baez (jbaez@riesz.mit.edu) posted two math puzzles
- <1992Oct28.170601.19561@galois.mit.edu>, to which he later gave
- partial answers. He writes:
-
- > The second question concerned the sum from 1 to infinity of o(2^n)/2^n.
- > Here o(2^n) is the number of odd digits in the decimal representation of
- > 2^n. The sum is surprisingly simple - it's 1/9.
-
- Along these lines, I can't resist posting the following (due to
- Doug Bowman and myself):
-
- For each nonnegative $n$, let $f(n)$ denote the sum of the Fibonacci
- numbers $F_i$, over the integers $i$ such that the $i$th digit of
- $2^n$ (counting from the right) is at least 5. e.g., 2^26=67108864,
- so f(26) = 21+13+0+0+3+2+1+0 = 40.
-
- (For you Mathematica types, f[n_] := Block[ {zz},
- (zz=Reverse[Floor[IntegerDigits[2^n]/5]]).Fib[Range[Length[zz]]]]
- where Fib is a {Listable} Fibonacci-number function.)
-
- Then the sum of f(n)/2^n, over all nonnegative n, is 20/89.
-
- /********************************************************\
- * Tad White UCR Dept. of Mathematics *
- * tadpole@ucrmath.ucr.edu Riverside, CA 92521 *
- \********************************************************/
- --
- /********************************************************\
- * Tad White UCR Dept. of Mathematics *
- * tadpole@ucrmath.ucr.edu Riverside, CA 92521 *
- \********************************************************/
-