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%I A346508 #17 Sep 11 2021 16:12:23
%S A346508 12,23,34,44,45,56,65,67,78,86,89,96,100,107,111,122,127,128,133,144,
%T A346508 149,155,158,166,168,170,177,188,189,191,199,209,210,212,220,221,232,
%U A346508 233,243,250,251,254,260,265,275,276,282,287,291,296,298,309,311,313,317
%N A346508 Positive integers k such that 10*k+1 is equal to the product of two integers greater than 1 and ending with 1 (A346507).
%H A346508 Stefano Spezia, <a href="/A346508/b346508.txt">Table of n, a(n) for n = 1..10000</a>
%F A346508 a(n) = (A346507(n) - 1)/10.
%F A346508 Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
%F A346508 The conjecture is true since a(n) = (A346507(n) - 1)/10 and lim_{n->infinity} A346507(n)/A346507(n-1) = 1. - _Stefano Spezia_, Aug 21 2021
%e A346508 107 is a term because 21*51 = 1071 = 107*10 + 1.
%t A346508 a={}; For[n=1, n<=350, n++, For[k=1, k<n, k++, If[Mod[10n+1, 10k+1]==0 && Mod[(10n+1)/(10k+1), 10]==1 && 10n+1>Max[10a+1], AppendTo[a, n]]]]; a
%o A346508 (Python)
%o A346508 def aupto(lim): return sorted(set(a*b//10 for a in range(11, 10*lim//11+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim))
%o A346508 print(aupto(318)) # _Michael S. Branicky_, Aug 21 2021
%Y A346508 Cf. A016873 (ending with 5), A017281, A324298 (ending with 6), A346507, A346509, A346510.
%K A346508 nonn,base
%O A346508 1,1
%A A346508 _Stefano Spezia_, Jul 21 2021

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