A346951 Positive integers k such that 10*k+9 is equal to the product of two integers ending with 3 (A346950).

0, 3, 6, 9, 12, 15, 16, 18, 21, 24, 27, 29, 30, 33, 36, 39, 42, 45, 48, 51, 52, 54, 55, 57, 60, 63, 66, 68, 69, 72, 75, 78, 81, 84, 87, 90, 93, 94, 96, 98, 99, 102, 105, 107, 108, 111, 114, 117, 120, 121, 123, 126, 129, 132, 133, 135, 138, 141, 144, 146, 147, 150

Offset: 1

Author

Stefano Spezia, Aug 08 2021

Keywords

• nonn
• base

Examples

			15 is a term because 3*53 = 159 = 15*10 + 9.
		

Crossrefs

Cf. A016873 (ending with 5), A017377, A324298 (ending with 6), A346950, A346952, A346953.

Programs

Mathematica

a={}; For[n=0, n<=150, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3&& 10*n+9>Max[10a+9], AppendTo[a, n]]]]; a

Python

def aupto(lim): return sorted(set(a*b//10 for a in range(3, 10*lim//3+4, 10) for b in range(a, 10*lim//a+4, 10) if a*b//10 <= lim))
print(aupto(150)) # Michael S. Branicky, Aug 11 2021

Formula

a(n) = (A346950(n) - 9)/10.

Lim_{n->infinity} a(n)/a(n-1) = 1.