The Solution Treasure 179

/dev/joe's solution that he sent to me last night is found in the forwarded message below.

The nth term of your sequence is generated by (n-1)!/B(n-1), when B(n-1) is not 0, and 0 otherwise, where B(n) is the nth bernoulli number; B(0)=1, B(1)=1/2, B(n) for n odd and greater than 1 is 0, and B(n) for n even and greater than 0 is given by:

2*(n!)*(-1)^(n/2+1)   Inf
------------------- * Sum ( 1 / j^n)
(2*pi)^n              j=1

The n! cancels out with the n! in our original expression, so we can write Guy(n) (plugging in the n-1) as:

(2*pi)^(n-1) * (-1)^((n+1)/2)
 2 * Sum[j=1..Inf](1/j^(n-1))

for odd n>1; Guy(1)=1, Guy(2)=2, and Guy(even n>2)=0.


latter /dev/joe posted:

For the curious, after dividing out a factorial from each term, I found a still irregular sequence. In doing some searches for info on sequences, I eventually found an article in sci.math in which somebody was asking about a sequence somewhat similar to mine, in which the number 691 appeared very early on, and somebody identified it as Euler numbers. In doing further research I found two other sequences that were supposedly Euler numbers, so I went out to the math library on campus and looked it up in the CRC math tables, where I found that 691 is not a part of the Euler numbers; however, on the preceding page was a table of the Bernoulli numbers, which I immediately recognized as a small variation on the numbers I had -- the small variation due to dividing out the wrong factorial, and that the reciprocals of the Bernoulli numbers were actually present.

Now, then, I still needed to come up with a non-recursive formula for the numbers (CRC did not provide this). This treasure gave me a good reason to go out and find _The Book of Numbers_ by John H. Conway and Richard Guy, at the same library, and in this book the above formula is presented (without the sign term).



Maintained by K 2, Treasure Harfer of Ackanomic, Last Updated 22-Juner-98