Problems & Puzzles: Problems
Problem 16.- A little puzzle from Mike Keith (12/12/98)
In reply to the flurry of posts I've received this morning (Saturday), I was inspired to construct the following little puzzle. I'm sure someone else has thought of this, but I haven't seen it before.
In UBASIC, enter
point 43 [which sets how many decimals to print - namely,
Now observe the output!
Ignoring the "01" at the end, all of the
numbers you see hiding amongst the zeros are primes!
At 8/1/99 Bob Mead
from Plymouth State College, NH, USA wrote: ", I
noticed the pattern in the "prime" digits is
the same as the diagonal pattern of primes if you make a
square spiral wrapping around the number 41, counting by
wholes. This is modeled by the function i^2 + i +
41 for i = 0, 1, 2, ... if we divide these primes
by an incremented power of 10^5 and sum, we get Mike's
If we let x = 10^5, his fraction is (41(x-1)^2 + 2x)/ (x-1)^3. The infinite expansion of this rational is the infinite sum (41 + i(i+1)) / (10^(5(i+1))) for i = 0, 1, 2, ... and that's why his fraction equals his decimal. The answer did not come quickly, but the exercise was fun."
This is the Mike Keith's solution (20/12/98):
The explanation is that it is possible to construct a fraction whose decimal expansion spells out the successive values of ANY integer sequence S(n) that's a polynomial in n, using the theory of generating functions. In particular, the fraction (x+2)/x^3 gives the squares (S(n)=n^2), where x is any integer of the form (10^d)-1 (i.e., one less than a power of 10, i.e.\line 999....9), 1/x^2 gives the sequence S(n)=n, and 1/x gives the constant sequence S(n)=1. By taking a linear combination of these three fractions we can get any 2nd-order polynomial in n. (And by using more complex fractions we can get higher-order polynomials).
Now...we just figure out what fraction gives the famous series n^2 + n + 41, which as you no doubt know is prime for the values n=0 to 40. That fraction turns out to be the one above. Note that the denominator is just 99999^3, but I chose not to write it that way to make the answer harder to guess. Also, the numerator is 200000 + 41*(99999^2).
Note that the value of 'd' is a free choice, which determines how many \line decimals apart each term of the sequence appears in the fraction. I chose d=5 since that makes the period equal to 5 so that even when we get to the 4-digit primes there is still one separator zero between them. d=4 would also work (but be less aesthetically pleasing, in my opinion); d=3 would not because then the four-digit primes would "collide" with each other.
p.s. I've submitted an article on this kind of stuff - though I didn't include\line this fraction, not having thought of it yet! - to REC.
In spanish we say that "Hay muchos modos de matar pulgas"...(very free translation to English: "each one kills fleas its own way", CBRF)