A367338 Comma-successor to n: second term of commas sequence if initial term is n, or -1 if there is no second term.

12, 24, 36, 48, 61, 73, 85, 97, 100, 11, 23, 35, 47, 59, 72, 84, 96, -1, 110, 22, 34, 46, 58, 71, 83, 95, -1, 109, 120, 33, 45, 57, 69, 82, 94, -1, 108, 119, 130, 44, 56, 68, 81, 93, -1, 107, 118, 129, 140, 55, 67, 79, 92, -1, 106, 117, 128, 139, 150, 66, 78, 91, -1, 105, 116

Offset: 1

Author

N. J. A. Sloane, Nov 15 2023

Keywords

• sign
• base

Comments

Construct the commas sequence as in A121805, but take the first term to be n. Then a(n), the comma-successor to n, is the second term, or -1 if no second term exists.

More generally, we define a comma-child of n to be any number m with the property that m-n = 10*x+y, where x is the least significant digit of n and y is the most significant digit of m.

A positive number can have 0, 1, or 2 comma-children. In accordance with the Law of Primogeniture, the first-born child (i.e. the smallest), if there is one, is the comma-successor.

Comment from N. J. A. Sloane, Nov 19 2023: (Start)

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 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) + b*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.

For an outline of the proof, see the attached text-file.

Note that in base b = 2, no values of x satisfying (*) 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.

(End)

Examples

			a(1) = A121803(2) = 12,
a(2) = A139284(2) = 24,
a(3) = 36, since the full commas sequence starting with 3 is [3, 36] (which also implies a(36) = -1),
a(4) = A366492(2) = 48, and so on.
60 is the first number that is a comma-child (a member of A367312) but is missing from the present sequence (it is a comma-child but not a comma-successor, since it loses out to 59).
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Eric Angelini, Michael S. Branicky, Giovanni Resta, N. J. A. Sloane, and David W. Wilson, The Comma Sequence: A Simple Sequence With Bizarre Properties, arXiv:2401.14346, Fibonacci Quarterly 62:3 (2024), 215-232.
• Eric Angelini, Michael S. Branicky, Giovanni Resta, N. J. A. Sloane, and David W. Wilson, The Comma Sequence: A Simple Sequence With Bizarre Properties, Local copy.
• N. J. A. Sloane, Eric Angelini's Comma Sequence, Experimental Math Seminar, Rutgers Univ., January 18, 2024, Youtube video; Slides

Crossrefs

A367346 lists those n for which there is more than one choice for the second term.

A367612 lists the numbers that are comma-children of some number k.

Cf. A121805, A139284, A366492, A367337.

See also A042948, A042964, A367339, A367340, A367341.

Programs

Maple

Ldigit:=proc(n) local v; v:=convert(n, base, 10); v[-1]; end;
A367338 := proc(n) local f,i,d;
f := (n mod 10);
d:=10*f;
for i from 1 to 9 do
d := d+1;
if Ldigit(n+d) = i then return(n+d); fi;
od:
return(-1);
end;
for n from 1 to 50 do lprint(n, A367338(n)); od: # N. J. A. Sloane, Dec 06 2023

Mathematica

a[n_] := a[n] = Module[{l = n, y = 1, d}, While[y < 10, l = l + 10*(Mod[l, 10]); y = 1; While[y < 10, d = IntegerDigits[l + y][[1]]; If[d == y, l = l + y; Break[];]; y++;]; If[y < 10, Return[l]];]; Return[-1];];
Table[a[n], {n, 1, 65}] (* Robert P. P. McKone, Dec 18 2023 *)

Python

from itertools import islice
def a(n):
    an, y = n, 1
    while y < 10:
        an, y = an + 10*(an%10), 1
        while y < 10:
            if str(an+y)[0] == str(y):
                an += y
                break
            y += 1
        if y < 10:
            return an
    return -1
print([a(n) for n in range(1, 66)]) # Michael S. Branicky, Nov 15 2023