A367346 Numbers k such that there is more than one possible solution for A367338(k).

14, 33, 52, 71, 118, 227, 336, 445, 554, 663, 772, 881, 1918, 2927, 3936, 4945, 5954, 6963, 7972, 8981, 19918, 29927, 39936, 49945, 59954, 69963, 79972, 89981, 199918, 299927, 399936, 499945, 599954, 699963, 799972, 899981, 1999918, 2999927, 3999936, 4999945, 5999954, 6999963, 7999972, 8999981

Offset: 1

Author

N. J. A. Sloane, Nov 15 2023

Keywords

• nonn
• base

Comments

The number of solutions is either 0, 1, or 2.

The definition of A121805 instructs us to pick the smallest solution, so there is no ambiguity in the definition of A121805. The present sequence shows that there are very few cases where there is any possible ambiguity.

The sequence begins with the four exceptional terms 14, 33, 52, 71. It also includes all numbers with decimal expansions of the form d 9^i d (9-d), where juxtaposition is concatenation, ^ denotes repeated concatenation of digits, 1 <= d <= 8, and i >= 0, with associated next terms in the commas sequence being either d 9^(i+2) or (d+1) 0^(i+2). It is conjectured that there are no other terms. - Michael S. Branicky, Nov 16 2023

The conjecture is true; see link. - Michael S. Branicky, Nov 21 2023

Examples

			In the commas sequence starting at 14, the next term could be either 59 or 60, because both 14,59 and 14,60 satisfy the "commas" rule (since both 14 + 45 = 59 and 14 + 46 = 60).
		

Links

• 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

Cf. A121805, A367338, A367341.

Programs

Mathematica

fQ[n_]:=Module[{k=n+10*Last[IntegerDigits[n]]+Range[9]},Length[Select[k,#-n==FromDigits[{Last[IntegerDigits[n]],First[IntegerDigits[#]]}]&]]]>1;
Select[Range[10^6],fQ[#]&] (* Ivan N. Ianakiev, Dec 16 2023 *)

Extensions

a(30) and beyond from Michael S. Branicky, Nov 16 2023

Second comment edited by N. J. A. Sloane, Nov 20 2023