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

%I #11 Aug 21 2021 15:47:08

%S 0,3,6,9,12,15,16,18,21,24,27,29,30,33,36,39,42,45,48,51,52,54,55,57,

%T 60,63,66,68,69,72,75,78,81,84,87,90,93,94,96,98,99,102,105,107,108,

%U 111,114,117,120,121,123,126,129,132,133,135,138,141,144,146,147,150

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

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

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

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

%t 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

%o (Python)

%o 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))

%o print(aupto(150)) # _Michael S. Branicky_, Aug 11 2021

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

%K nonn,base

%O 1,2

%A _Stefano Spezia_, Aug 08 2021