Comments from ~~~~ (Start) The Comma-Successor Theorem. N. J. A. Sloane November 17 2023 The following is a proof of a slight modification of a conjecture made by Ivan N. Ianakiev in A367341. The Comma-Successor Theorem. Let D(b) denote the set of numbers k which have no comma-successor in base b ("comma-successor" is the base-b generalization of the rule that defines A121805). If a commas sequence reaches a a number in D(b) it will end there. Then D(b) consists precisely of the numbers which when written in base b have the form cc...cxy = (b^i-1)*b^2/(b-1) + 10*x + y, with i >= 0 copies of c = b-1, where x and y are in the range [1..b-2] and satisfy x+y = b-1. .... (*) For b = 10 the numbers D(10) are listed in A367341. Sketch of proof. Note that in base b = 2, no values of x satisying (*) exist, and the theorem asserts that D(2) is empty. In fact it is easy to check directly that every commas sequence in base 2 is infinite. If the initial term is 0 or 1 mod 4 then the sequence will merge with A042948, and if the initial term is 2 or 3 mod 4 then the sequence will merge with A042964. So in the proof we may assume b >= 3. In the following proof, we assume for simplicity that b = 10. The general case for any base b >= 3 is essentially the same. The proof has two parts. (i) If a number k is not of the form (*) then it has a base-b successor. (ii) Conversely, if k does have the form (*) then it does not have a base-b successor. In the proof we write often write numbers as decimal strings - that is, we write xy rather than 10*x+y, etc. It should be clear from the context when we do this! (i) We consider three subcases where k has a successor: (i)(a) k = 99...999xy has at least three digits and x+y != 9. (i)(b) k = Pxy has at least three digits, P is not all-9's, and x+y = 9. (i)(c) k = xy has two digits and x+y != 9. In each case we will show that k does have a successor k'. Case (i)(a). We will use the same notation in each case. We are given k, and are seeking a comma-successor k', The comma between k and k' will equal yz (that is, 10*y+z, which we will call the "comma-number"), for some z in the range 1 <= z <= 9. k' is then equal to k + yz, and if v denotes the leading digit of k', the comma-number must be yv. So the key question is, is there a choice for z such yz = yv? This reduces the problem to a simple question in arithmetic. We illustrate with a typical calculation in case (1)(a). Suppose k = 944. As k goes from 1 to 9 the comma-number is one of 41, 42, ..., 49. Since x+y = 4+4 = 8 < 9, there is no carry when we form the sum k + yz = 944 + yz, and the result is < 1000, so v = 9, and we can take z = 9, and we have our solution. On the other hand, if k = 946, then x+y = 10 > 9, and the leading digit of k + yz will again be constant, only now it is 1 instead of 9. But as long as the leading digit of k' = k + yz is constant, we can choose z to match it, and we have a solution. At this point we can also take care of one of the cases of the converse part of the proof. For suppose x+y was equal to 9. Now the leading digit of k' may change as z increases, and we may not be able to find a value of z that matches it. Again a sample calculation will show what happens. Suppose k = 945, with x=4, y=5, x+y = 9. Here is the calculation: xy yz k+yz yv ------------------------ k = 945 51 996 59 52 997 59 53 998 59 54 999 59 ------------------------ 55 1000 51 56 1001 51 57 1002 51 58 1003 51 59 1004 51 ------------------------ For small z, in the range 1 <= z <= 4, v is 9, while if z is in the range 5 to 9, v = 1 because of the carry. So in nether case is there a match that gives z = v. I am going to stop there. The rest of the proof (which will given in more detail when this gets written up) is no more difficult. (ii) A367346 lists the numbers with more than one successor in base 10. (End) -