COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

Shen Lin Sequences which form integral bases 367
method was suggested to me by Dr. R. L. Graham of the Bell Telephone
Laboratories to whom I give my sincerest thanks here.
Given a sequence S = {sl, s2, . . . , Sk, . . .} satisfying condition A.
Assume that the integer js is known such that for all k *js, we have
2sk 2 sk+l. Let P,(s) denote the set of all numbers which are representable
as a sum of distinct terms taken from the first k terms {si, ss, . . ., Sk}
of S, including zero. p,(S) may be computed recursively as follows :
P,(S) = (0, Sl},
&+1(s) = h(s) u{pk(s)+ {sk+l}}:
where, as usual, A+B = {X 1 x = a+b, a c A, be B}.
Suppose all integers from a through b (a a) belong to Pk(S) while a- 1
and bf 1 do not. Then we call [a, b] an interval in P,(S) and define its
length as bf 1 -a. For each k 3 j,, as soon as Pk(S) is computed, we
determine the interval [Q, yk] in P,(S) having the longest length 1, (if
there are two or more intervals with the same length, we pick the one with
the smallest &) and compare it with sk+l. If lk -= sk.+1, we set k to k+ 1,
and go on, repeating the above procedures. If lk * sk+l, we have proven
that S is complete and the determination of the threshold of completeness
is now a relatively simple matter. We continue to calculate P,(S) successively
until sk+l 3 xk (xk may decrease as k increases), and when this
happens, the threshold of completeness e(S) is then xk- 1.
The justification for the above procedure is easily seen. When we find
an interval [xk, &] in P,(S) such that sk+l e Ik = &-xk+ 1 with k a j,,
we are guaranteed that all integers > x, belong to P(S). For &+1(S) will
contain all integers in the interval [xk+sk+r, j&+sk+1] and hence all
integers in the interval [&, &+ ++1], since xk + sk+l < yk + 1 andlthe intervals
[xk, yk] and [xk+sk+l, yk+Sk+l] merge into one. By condition A, this
merging will continue for ever since Ik+l = ykfl-Xk+i+l > yk+Sk+ixk+
1 * 2sk+l* sk+Z. When s,&+l * xk, no further sk+i’s may be used
to represent xk- 1, and since all integers == xk belong to P(S), xk- 1 is
therefore the threshold of completeness for the sequence S.
In writing a computer program to find the thresholds of completeness
for sequences using the above procedure, the most efficient way to generate
and store the Pk(s)‘s is of major concern. Two representations for numbers
in P,(S) are used. First, in the characteristic function method, the set
P,(S) is represented as a string of binary bits, the (i+ 1)th bit being a one
if and only if the integer i belongs to Pk(S) and zero otherwise. Zero is
considered to be in P,(s) and hence the first bit of the string is always a 1.
Pk+l(S) is computed from P,(S) by shifting the entire bit string for P,(S)
an amount equal to sk+l and logically or-ing it to the original bit string
for P,(S). When the threshold of completeness is less than half a million,
this method is very fast since all computations can be done in core. When
the numbers in P,(S) get to be larger than the limit of space available, a
truncated version for Pk(S) can be used effectively as long as the threshold
is less than half the number of bits available. In the second method, Pk(S)
is stored as a sequence of intervals [ai, bi] and Pk+l(S) is obtained from
P,(S) by constructing the new sequence of intervals [ai+sk+l, bi+sk+l]
and then merging the two sequences to produce the sequence of intervals
for Pk+l(S). This method has the advantage that the number of intervals
becomes relatively constant after a while although it grows almost like a
power of two in the beginning. For large problems, the limit for storage is
exceeded very rapidly and auxiliary storages have to be used. Using the
interval method we have computed the threshold of completeness for the
sequence of fourth powers S = (1, 16, 81, 256, . . .} to be 5,134,240.
Note that if the characteristic function method were used, we would have
to carry along a bit string of about 10 million bits. Various programming
devices and techniques are employed in the program to reduce the running
time but they will not be discussed here. Also, if Z;, = si+sa+ . . . +q is
the largest number in Pk(S), the intervals are symmetric about +z,; i.e.
if [ai, bi] is an interval in Pk(S), then [zk- bi, -&- ai] is also an interval.
Observations like this help reduce the storage requirement for Pk(S) by a
substantial amount although they do make the logic for producing Pk+l(S)
from Pk(S) much harder. As is well known to computer programmers, it is
always a difficult problem to find the proper balance between storage
space, running time, and simplicity of programming logic, and this program
is no exception.
Having what we consider an efficient program to compute thresholds of
completeness for sequences satisfying condition A, we turn next to a related
problem. We say that a sequence S is essentially complete if all truncated
sequences S, = {s,, s,+~, . . .} are complete. It is not difficult to see that
all complete sequences generated by polynomials are essentially complete.
A result of Roth and Szekeres [3] also guarantees that the sequence of
primes, the sequence of squares of primes, etc., are essentially complete.
Examples of complete sequences which are not essentially complete are the
sequence of powers of 2, the Fibonacci sequence, and most Lucas sequences.
* d A study of when Lucas sequences are essentially complete is being made
by Stephen Burr, whose results will be published elsewhere. For essentially
complete sequences, e(S,) exists for every n. Using the program, we were
able to compute the thresholds of completeness for sequences such as
the sequence of primes, the sequence of squares, the sequence of pseudoprimes
(positive integers == 2 having at most 4 positive divisors), etc., for
n up to a fairly large number. Some of the results obtained are briefly
summarized in Tables 1 through 8 in Appendix A.
From the thresholds of completeness obtained, we observe that the
ratios u, E 1’3(&+J/s,, seem to settle down to a narrow region as n increases
and that for the sequence of primes, this region is very close to 3. For the