THEORY AND PRACTICE OF SYLVER COINAGE

George Sicherman 912 Deal Road, Wayside, N.J. 07712

                           colonel@mail.monmouth.com

Received: 6/5/02, Revised: 8/27/02, Accepted: 9/10/02, Published: 9/11/02

Abstract

Sylver Coinage is J. H. Conway’s game of denominating coins. Two players take turns naming numbers that cannot be expressed as sums with non-negative multiples of numbers previously named. The player who chooses 1 loses. This paper presents a variety of general and specific results about Sylver Coinage.

“Look well to your change! Ring the silver on the four-pound weight!” —Hawthorne, The House of the Seven Gables

1. Basic Knowledge

Sylver Coinage is J. H. Conway’s game of denominating coins, familiar to readers of Winning Ways [1]. Two players take turns choosing numbers. Each number must not be a sum with multiples of numbers previously chosen. The player who chooses 1 loses.

Notation in Sylver Coinage varies from number-theoretic convention. The greatest common divisor of x_{1}, ..., x_{j} is written g(x_{1}, ..., x_{j}), or g. If g = 1, the greatest number that cannot be expressed as a sum with multiples of members of {x_{1}, ..., x_{j}} is denoted t(x_{1}, ..., x_{j}), or t. If S = {x_{1}, ..., x_{j}}, g(S) denotes g(x_{1}, ..., x_{j}) and t(S) denotes t(x_{1}, ..., x_{j}).

Sylver Coinage is named after J. J. Sylvester, who showed [2] that if g(m, n) = 1, then t(m, n) = (m − 1)(n − 1) − 1. For example, the greatest number that cannot be expressed as a sum of 5’s and 6’s is (5 − 1)(6 − 1) − 1, or 19.

Every game of Sylver Coinage must end. The value of g cannot increase as moves are added, and at most finitely many legal moves preserve the value of g. Thus every position has a well-defined Sprague-Grundy ordinal, which may be transfinite.

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While Sprague-Grundy ordinals have been studied for some Sylver Coinage positions, most people care only who wins. Following Grundy and Smith [3], let a position be an 𝒩-position if the player next to play can win; that is, if its Sprague-Grundy ordinal is nonzero. Otherwise let it be a 𝒫-position (for previous).

A position S in Sylver Coinage is determined by a set of previous moves {x_{1}, ..., x_{j}}. The expression Sn denotes the Sylver Coinage position obtained by adjoining n to the set S. This notation does not imply that n is a legal move in the position S, only that n is not an element of the set S.

It is often useful to multiply or divide an entire set of moves by some number. If S = {x_{1}, ..., x_{n}}, then S × k = {kx_{1}, ..., kx_{n}}. Similarly, if S = {kx_{1}, ..., kx_{n}}, then S ÷ k = {x_{1}, ..., x_{n}}. Regardless of the value of g(S), let t̄(S) = g(S)t(S ÷ g(S)). For example, t̄(6, 8) = 10, because t(3, 4) = 5. This extends the concept of t to positions with g > 1.

Different sets of moves may be equal as Sylver Coinage positions; for example, the position {4, 5, 10, 13} equals the position {4, 5}. The moves 10 and 13 can be expressed as sums of 4’s and 5’s, so they do not affect play in the position. They are said to be eliminated by 4 and 5. A set of moves is a canonical representation of a position if it contains no superfluous moves. Any position can be made canonical by deleting the moves that are eliminated by the other moves. Mutual elimination is impossible: no number can help eliminate a smaller number!

Sylver Coinage positions have a natural partial order. If S and T are positions, write T > S (T is later than S) if S and T are unequal and T = Sx_{1}, ..., x_{j} for some numbers x_{1}, ..., x_{j}.

Every position partitions the set of numbers into legal and illegal moves. For example, in the position {4, 5} the moves 1, 2, 3, 6, 7, and 11 are legal; all other numbers are illegal. In any position, all moves that lower the value of g are legal, so the set of legal moves is infinite if and only if g > 1.

In a position with g = 1, a legal move is called an end if it does not eliminate any other legal move. In particular, t is always an end. A position S where t(S) is the only end is called an ender. By this definition, in an ender every legal move other than t eliminates t. Enders are important because they allow strategy stealing. Provided that t > 1, the player on the move in an ender must be able to win. Suppose he plays t. If the second player has a winning reply u, the first player can play u instead of t and reach the same winning position as the second player, because Stu = Su. Thus every ender with t > 1 is 𝒩.

A quiet ender is an ender in which each legal move n less than t can eliminate t without multiples of n. For a counterexample let S = {4, 5, 7}. The legal moves in S are 1, 2, 3, and 6, so t = 6. Every legal move less than 6 eliminates 6, so S is an ender. But 3 does not quietly eliminate 6, because 6 cannot be expressed as a sum of a single 3 and multiple 4’s, 5’s, and 7’s. Thus S is not a quiet ender. An ender that is not a quiet ender is an unquiet ender. Any position with g = 1 is thus a quiet ender, an unquiet ender, or a non-ender.

R. L. Hutchings has shown [1] that every position {m, n} with g = 1 is an ender.^{1} The only such position with t = 1 is {2, 3}, so every other position of the form {m, n} is 𝒩. It follows at once that if p is a prime greater than 3, {p} is 𝒫. This is Hutchings’s p-theorem [1].

A powerful theorem in the analysis of Sylver Coinage is the Quiet End Theorem [1]: if z is coprime with each of x and y, then (S × x)z is a quiet ender if and only if (S × y)z is. In particular this theorem applies when S is itself a quiet ender and x = 1. Then when z > t(S), z is illegal; that is, Sz = S, so Sz is a quiet ender. By the Quiet End Theorem, (S × y)z is also a quiet ender for values of z that are coprime with y and greater than t.

Now, quiet enders are enders, so those other than {2, 3} are 𝒩-positions. This can let us eliminate all but finitely many possible winning moves in a position with prime g. For example, let S = {2, 3} and y = 2, so S × 2 = {4, 6}. Sufficiently large moves in {2, 3} produce quiet enders, so sufficiently large odd moves in {4, 6} produce quiet enders. Thus none of 5, 7, 9, ... can win in the position {4, 6}. Since the moves 2 and 3 lose to each other, it follows that {4, 6} is 𝒫.

A prime multiple of a quiet ender is said to be short, because in looking for a winning move in such a position, we need consider only finitely many moves. A prime multiple of an unquiet ender or non-ender is said to be long. By the Quiet End Theorem, sufficiently large moves in long positions produce positions that are not quiet enders. In fact, it is easy to show that sufficiently large moves produce positions that are not enders at all. Let T = (S × p)z. It is routine to verify that for sufficiently large z, t(T) = z(p − 1) + pt(S). There is some s that does not quietly eliminate t(S) in S, so t(S) − s cannot be expressed as a sum with multiples of members of S. But then ps cannot be expressed as a sum with multiples of members of S × p. If z is large enough that t(T) − ps > t(T)/2, then t(T) − ps cannot eliminate t(T) in T.

2. Computing Sylver Coinage

From here on Guy’s notation [4] for winning moves is used. The statement {x_{1}, ..., x_{j}} [w_{1}, ..., w_{g}] means that w_{1}, ..., w_{g} are all the winning moves in the position {x_{1}, ..., x_{j}}.

Any position with g = 1 has finitely many legal move sequences and can be analyzed exhaustively. While these computations are not particularly interesting, they raise some general questions, such as the asymptotic behavior of the unique winning move in {4, x} for odd x.

1The only known positions {m, n} in which t wins are: {2, 5}; {4, 5}; {5, 6}; {5, 9}; {8, 15}; {13, 14}; {13, 21}.

Are there infinitely many others? (And do they all involve Fibonacci numbers?)

Guy’s “Twenty Questions” paper [4] is a list of such questions and others. Some early research in Sylver Coinage entailed finding all the winning moves in positions {m, n} for small m and n; see for example Nowakowski’s paper [5]. As m and n increase, the computation grows rapidly. From the starting position {12, 17} a total of 158793 positions can be reached, counting {12, 17} itself.

Positions with g > 1 are more interesting because they have infinitely many legal moves. The case g = 2 is special because of the Periodicity Theorem [1]: if g(S) = 2, then the computation of the winning moves in Sx for odd x must eventually become periodic. The proof shows that the analysis can be computed by a finite-state automaton because the number of potential winning moves in Sx is bounded over x. For example, let S = {8, 10, 22}; then g(S) = 2 and t(S) = 14. In analyzing an odd move n we need consider no odd replies greater than n + 14 (because n eliminates them) and no odd replies less than n − 14 (because they eliminate n). It suffices to analyze the odd replies n − 14, n − 12, ..., n − 2, n + 2, ..., n + 12, n + 14, along with the finitely many even replies.

The Periodicity Theorem implies that a finite computation suffices to determine whether a position with g = 2 has any winning moves.

A short position S with g = 2 can have few possible winning moves for the next player, for the Quiet End Theorem assures us that S has no odd winning move greater than t(S ÷2). Thus it is not unusual to find short positions that are 𝒫; for example {4, 6}, {8, 14}, and {10, 16, 24}. On the other hand, in long positions with g = 2 the only known limit on possible winning moves is imposed by the Periodicity Theorem—usually a very high limit! For this reason, most long positions with g = 2 are 𝒩, and their winning moves may be quite large. A few long positions with g = 2 are known to be 𝒫, notably {8, 10, 22} and the members of the series {8, 12, 8n + 2, 8n + 6} for n = 1, 2, 3, .... (From this it can be shown that {8, 12} is also 𝒫.) Computationally speaking, a position with g = 2 may be called Probably 𝒫 if it has no odd winning moves and every even move either is known to lose or produces a long position.

While a finite analysis suffices to find any winning moves in a long position with g = 2, the computation may be very long. The unique winning move in {8, 30, 34} is 49337. In {6, 44, 82} it is obvious that 4 wins, reducing the position to {4, 6}; it is less obvious that 5993171 wins. The position {6, 40, 74}, though long, has no odd winning move; a period of length 403200 sets in at x = 381091. The long positions {16, 22, 24, 26, 28, 30, 34} and {20, 22, 24, 26, 28, 36} also have no odd winning moves, and are in fact 𝒫. The long position {6, 50, 94} has neither period nor odd winning move below 10^{8}.

When g = 3, short positions are still often 𝒫, but long ones are generally intractable. The number of legal moves in Sx is o(2x), so, intuitively, the chance of finding no winning reply to x decreases as x increases. For example, in {12, 15, 21}, the move 100 has 63 possible winning replies that are not multiples of 3, ranging from 41 to 218; the move 200 has 113 possible winning replies that are not multiples of 3, ranging from 91 to 418. The short positions {6, 9}, {12, 15, 18}, and {12, 18, 21} are 𝒫, but nobody knows the status of the long position {12, 15, 21}, nor whether the long position {9, 15, 21} has a winning move other than 6.

For nontrivial positions with prime g ≥ 5, g itself wins by Hutchings’s Theorem. In long positions the chance of finding a winning move not divisible by g is even poorer than with g = 3.

When S is a multiple of a short position with g = 2, it has infinitely many legal moves that produce short positions with g = 2. For example, in {8, 14} × 3 = {24, 42}, all the moves 38, 40, 44, 46, 50, 52, ... produce short positions with g = 2. At least one of these short positions will probably be 𝒫, in which case S will be 𝒩. Other cases with composite g are generally intractable.

Which moves win in the empty position {}? Any prime p ≥ 5 wins by Hutchings’s Theorem. Any multiple of such a prime loses to the prime. That leaves only numbers of the form 2^{a}3^{b}, where a and b are nonnegative. We know that 1 loses at once, 2 loses to 3 and vice versa, 4 loses to 6, 6 loses to 4 and 9, 8 loses to 12 and 14, and 12 loses to 8 at least. Opening moves 2^{a}3^{b} greater than 12 are undecided, and Conway has offered a prize for the first to be decided.

Practically, it is not likely that anybody will find a winning response to an opening move of 4n for n > 3. The only practically verifiable responses are of the form 4n + 2, which produce short positions. These short positions allow many moves producing new short positions, and at least one of them generally turns out to be 𝒫. The search for a winning move in {12} with g = 2 has eliminated 10, 14, 22, 26, 34, 38, 46, 50, and 58. The position {12, 32, 62} is Probably 𝒫; if 𝒫, it eliminates 62 as a winning move in {12}.

3. Enders and Antisymmetry

The definition of a quiet ender entails an important structural property, the Strong Antisymmetry Principle:

Proposition. Let S be a quiet ender. Then for every n such that 0 < n < t(S), exactly one of n, t − n is a legal move in S.

An example will help make this clear. Consider the quiet ender {7, 8, 10, 13}, for which t = 19. The legal and illegal moves from 1 to 18 form an antisymmetric pattern:

legal 1 2 3 4 5 6  9  11 12
illegal  7 8  10  13 14 15 16 17 18

Proof. Both of n, t−n cannot be illegal, for then their sum would be illegal; but t is by definition the greatest legal move in S. Now suppose that n is a legal move in S and is less than t. By the definition of a quiet ender, t is the sum of n once and elements of S with multiples allowed. Thus t − n can be expressed as a sum with multiples of elements of S. In other words, t − n is an illegal move. □

Since the proof works in reverse, the Strong Antisymmetry Principle exactly characterizes quiet enders.

Proposition. If S is a quiet ender, then t(S) is odd.

Proof. If t is even, the Antisymmetry Principle implies that t/2 is both legal and illegal! □

It turns out that unquiet enders are much like quiet ones.

Proposition. Let S be an ender. Then for every n such that 0 < n < t and n ≠ t/2, exactly one of n, t − n is a legal move in S.

Proof. We have seen that n and t−n cannot both be illegal. Suppose without loss of generality that n > t/2. By the definition of an ender, t is the sum with multiples of n and elements of S. Since n > t/2, n must occur only once in this sum. Thus t − n can be expressed as a sum with multiples of elements of S. In other words, t − n is an illegal move. □

The proof of this Weak Antisymmetry Principle looks very like that of the Strong Antisymmetry Principle. The only difference is that antisymmetry may fail at t/2 for unquiet enders.

Proposition. Let S be an ender and t(S) be odd. Then S is a quiet ender.

Proof. When t/2 is not a number, the Weak Antisymmetry Principle reduces to the Strong Antisymmetry Principle. A position that satisfies the Strong Antisymmetry Principle is a quiet ender. □

Combining these results gives a neat characterization of quietness in enders, the Odds and Enders Theorem:

Theorem. Let S be an ender. Then S is a quiet ender if and only if t(S) is odd.

We can now prove a stronger form of Hutchings’s basic theorem.

Proposition. If m and n are coprime, {m, n} is a quiet ender.

Proof. Hutchings has shown that the position is an ender, and Sylvester has shown that t(m, n) = (m − 1)(n − 1) − 1. Since m and n are coprime, at least one of them must be odd. It follows that t is odd, so the position is a quiet ender. □

The Antisymmetry Principles suggest an infallible method of generating enders:

Definition. Let S be a position with g = 1. The enclosure of S, written E(S), is the position obtained by adjoining to S every legal move n such that t/2 < n < t and t − n is a legal move.

For example, let S = {7, 11, 13, 15}. Then t(S) = 23 and the legal moves in S are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 16, 17, 19 and 23. The pairs of legal moves that sum to 23 are 4+19 and 6+17. Adjoining 19 and 17 to S gives E(S) = {7, 11, 13, 15, 17, 19}, which is an ender.

Thus the enclosure of a position is obtained by attempting to impose antisymmetry on it. But does this always produce an ender?

Proposition. Let S be a position with g = 1. Then t(E(S)) = t(S).

Proof. Suppose that t(E(S)) < t(S). Then t(S) can be expressed as a sum with multiples of elements of E(S). Such a sum must include one or more elements of E(S)−S, since t(S) cannot be expressed as a sum with multiples of elements of S. The numbers adjoined to S to form E(S) are all greater than t(S)/2, so the sum can include at most one of them. Call it n, and let m = t(S) − n. Then m is also a sum with multiples of elements of E(S). By the definition of E, both n and m are legal in S, and m < t(S)/2. Being legal in S, m cannot be expressed as a sum with multiples of members of S. But all the adjoined members in E(S) are greater than t(S)/2 and thus greater than m, so they cannot be included in a sum to m. This shows that m cannot be expressed as a sum with multiples of members of E(S), and we have a contradiction. □

Theorem. Let S be a position with g = 1. Then E(S) is an ender.

Proof. This now follows from the definition of E. For t(E(S)) = t(S), and every possible violation of weak antisymmetry with respect to t is removed by the process of enclosure. □

Obviously when S is an ender, E(S) = S. For arbitrary S enclosure satisfies a minimality condition:

Proposition. Let S be a position with g = 1. If S ≤ T ≤ E(S) and T is an ender, then T = E(S).

Proof. Since the t operator is monotone, t(S) ≥ t(T) ≥ t(E(S)). We have shown that t(E(S)) = t(S), so t(T) = t(S). Suppose that T is a proper subposition of E(S), so that for some n, n is legal in T and illegal in E(S). Since S ≤ T, n is legal in S. Thus n is one of the numbers adjoined to S to produce E(S). By definition of E(S), m = t − n is also a legal move in S, and m < t/2. Since m is also legal in E(S), it must be legal in T. Thus T has two distinct legal moves m and n that sum to t, violating the Weak Antisymmetry Principle. Such a position cannot be an ender. □

Like t, the function E can be extended to a function over all positions: Ē(S) = E(S ÷g(S))× g(S). For example, Ē(14, 22, 26, 30) = {14, 22, 26, 30, 34, 38}.

4. Arithmetic Progressions

This remarkable theorem is due to J. B. Roberts [6]. Let a_{0}, ..., a_{s} be an increasing arithmetic progression with a_{0} ≥ 2 and successive difference d, and suppose that g(a_{0}, ..., a_{s}) = 1. Then

t(a_{0}, ..., a_{s}) = (⌊\frac{a_{0} − 2}{s}⌋ + 1) · a_{0} + (d − 1)(a_{0} − 1) − 1.

When s = 1 this reduces to Sylvester’s Formula. Like Sylvester’s Formula, Roberts’s Formula has an analogue in Sylver Coinage:

Theorem. Let S = {a_{0}, ..., a_{s}} satisfy the conditions of Roberts’s Theorem. Assume also that s ≤ a_{0} − 1. Then

A) if s = 1, S is a quiet ender;

B) if a_{0} = 3 and s = 2, S is an unquiet ender;

C) if a_{0} > 3, S is a quiet ender if s | a_{0} − 2 and not an ender otherwise.

Proof. I only sketch the proof, omitting demonstrations of the values of t and the moves that eliminate them.

In case A the set has only two members and is covered by Hutchings’s Theorem.

In case B, S = {3, 3 + d, 3 + 2d}, and the value of t is 2d, which is even, so S is not a quiet ender. The legal moves are just the positive values of d − 3k and 2d − 3k for k ≥ 0. By congruence modulo 3, the only two such moves that can sum to 2d are d and d, so weak antisymmetry holds.

Case C has two subcases.

I. If s = a_{0} − 1, then clearly s does not divide a_{0} − 2. In this subcase, t = sd, and the moves that fail to eliminate t are the values of kd where 1 ≤ k ≤ s and k does not divide s. In particular s − 1 does not divide s, so S is a non-ender.

II. If s < a_{0} − 1, then the moves that fail to eliminate t are t − kd for 1 ≤ k ≤ r, where r is the remainder on dividing a_{0} − 2 by s. If r = 0 then S is an ender; otherwise it is not. Now if r = 0, then

i) if a_{0} is odd, a_{0} − 2 is also odd; since this is divisible by s, (a_{0} − 2)/s is odd, and the first term in Roberts’s Formula is even;

ii) if a_{0} is even, the first term in Roberts’s Formula is even because it contains a_{0} as a factor.

In both cases the second term (d − 1)(a_{0} − 1) is even because d and a_{0} are coprime and cannot both be even. Thus t is odd and S is a quiet ender. □

5. Synthetic Methods

A natural problem in Sylver Coinage is: given a set of numbers, is there a position in which those numbers are winning moves? If so, is there a position in which those numbers are the only winning moves?

For a single number the answer to both questions is yes. This is the Separation Lemma:

Lemma. Suppose that x and y both win in position S. Then there is a position T such that T > S, x wins in T, and y does not.

Proof. We are given that Sx and Sy are 𝒫-positions. First, Sxy ≠ Sx; that is, x cannot eliminate y. If it did, then x would be a winning move in Sy. Therefore Sxy is an 𝒩-position; let z be a winning move in it, and let T = Syz. Since y is not a legal move in T, it is not a winning move. But x is a legal move in T, for otherwise Sxyz = Syz, whence Syz is 𝒫. But z is a legal move in the 𝒫-position Sy, so Syz must be 𝒩. □

This suffices to prove the Single Win Theorem:

Theorem. Let n be a number greater than 1. Then there is a position in which n alone wins.

Proof. We know that 2 alone wins in {3}, and 3 alone wins in {2}. For any n greater than 3, let p be a prime greater than 3 such that p ≠ n. By Hutchings’s Theorem, {n, p} is 𝒩, so it has a winning move r. Then {n, p, r} is 𝒫. Let S_{0} = {p, r}. Then S_{0}n ≠ S_{0}, for S_{0}n is 𝒫 while S_{0} = {p, r} is 𝒩 by Hutchings’s Theorem. Thus n is a legal and winning move in S_{0}.

If n is the only winning move in S_{0}, we are done. Suppose that S_{0} has some other winning move n′. By the Separation Lemma there exists S_{1} > S_{0} in which n wins and n′ does not.

If n is the only winning move in S_{1}, we are done. If not, repeat the process. Eventually it must terminate, since there cannot be an infinite series of positions

S_{0} < S_{1} < S_{2} < ...

When it terminates in some S_{j}, n alone wins in S_{j}. □

To illustrate, let n = 14. Using p = 5 gives {5, 14}[18]. Solving S_{0} = {5, 18} gives [14, 16, 17]. Let us eliminate 16 by solving {5, 14, 16, 18}; the unique winning move is 17. Then S_{1} = {5, 16, 17, 18}, and this has the desired unique winning move 14.

Conjecture. Any set of mutually indivisible numbers greater than 3 is the set of winning moves in some position.

Here is an example of an attempt at constructing such a position. Let [4, 5, 6, 7] be the desired set of winning moves. Start with any number bigger than all of them and indivisible by them; 9 will do. Solve {4, 9} to get 19, so 4 wins in {9, 19}. Does 5 win? No: {5, 9, 19} [31]. However, {4, 9, 19, 31} = {4, 9, 19}, so we can add 31 to the target position. Now we have {9, 19, 31} [4, 5, ...].

Does 6 win in {9, 19, 31}? No: {6, 9, 19, 31} [17, 20, 22]. But adjoining 20 to the position has no effect on 4 or 5. We now have {9, 19, 20, 31} [4, 5, 6, ...].

Does 7 win in {9, 19, 20, 31}? No: {9, 19, 20, 31} [15]. Adjoining 15 to the position has no effect on 5 (15 = 5 + 5 + 5) or 6 (15 = 6 + 9), but 15 is not eliminated by 4.

Instead of adjoining 15, can we adjoin two moves that do not affect 4, 5, and 6? No; the only such move that does affect 7 is 24.

When we reach a dead end, the thing to do is backtrack. At any stage of the search we can try adjoining several moves instead of one. As a last resort we can choose a new starting move; there are infinitely many to choose from!

Probabilistically, the search is certain to succeed—unless it can somehow be proven to be certain to fail. As for [4, 5, 6, 7], one position that works is {17, 18, 27, 33, 43}. By good fortune, adjoining 56 to this position lets us add 9 to the set of winning moves: {17, 18, 27, 33, 43, 56} [4, 5, 6, 7, 9].

A more speculative conjecture is:

Conjecture. For any n ≥ 3 there is a position with g = 2 in which n wins.

Other values of g produce other conjectures.

References

1. Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for Your Mathematical Plays, Academic Press, London, 1985, pp. 575–606.

2. Joseph J. Sylvester, Mathematical Questions from the Educational Times 41 (1884), 21 (q. 7382).

3. P. M. Grundy and C. A. B. Smith, Disjunctive Games with the Last Player Losing, Proceedings of the Cambridge Philosophical Society 52 (1956), 527–533.

4. R. K. Guy, Twenty Questions Concerning Conway’s Sylver Coinage, American Mathematical Monthly 83 (1976-10), 634–637.

5. Richard J. Nowakowski, …, Welter’s Game, Sylver Coinage, Dots-and-Boxes, …, Proceedings of Symposia on Applied Mathematics (Columbus, Ohio, 1990) (R. K. Guy, ed.), vol. 43, 1991, pp. 155–182.

6. J. B. Roberts, Note on Linear Forms, Proceedings of the American Mathematical Society 7 (1956-06), 465–469.

Acknowledgments

I would like to thank the referee for many valuable suggestions for improving this article.