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,
207] 0.000410004300047000530006100071000830009700113001310 Now observe the output!
Ignoring the "01" at the end, all of the
numbers you see hiding amongst the zeros are primes! Solution 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
decimal. If we let x = 10^5, his fraction is (41(x1)^2 + 2x)/ (x1)^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 2ndorder polynomial in n. (And by using more complex fractions we can get higherorder 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 4digit 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 fourdigit 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)





