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A347471 a(n) is the least number k such that n = A347470(k) := least a*b with concat(a,b) = k.

%I A347471 #14 Oct 22 2021 23:45:39
%S A347471 0,11,12,13,14,15,16,17,18,19,25,111,26,113,27,35,28,117,29,119,45,37,
%T A347471 122,123,38,55,126,39,47,129,56,311,48,133,134,57,49,137,138,139,58,
%U A347471 411,67,431,144,59,146,147,68,77,105,173,226,531,69,155,78,157,158,159,106,611,621
%N A347471 a(n) is the least number k such that n = A347470(k) := least a*b with concat(a,b) = k.
%F A347471 a(n) = 10 + n for 0 < n < 10.
%F A347471 a(n) <= concat(1,n) with equality when n is prime.
%e A347471 a(3) = a(1*3) = 13 and similar for 1 <= n <= 9, cf. first formula.
%e A347471 a(11) = a(1*11) = 111, a(13) = a(1*13) = 113, a(17) = a(1*17) = 117 etc. according to the second formula with prime n.
%e A347471 a(10) = a(2*5) = 25, a(14) = a(2*7) = 27, a(15) = a(3*5) = 35 etc. for semiprime indices; in these cases a(p*q) = concat(p,q) where p is the lexicographic smaller factor, but this is not always the case.
%e A347471 a(22) = 122, not concat(11,2), although 11*2 = 22, but the smallest product that can be formed by slicing 112 in two parts is A347470(112) = 1*12 = 12, less than 22.
%e A347471 a(93) = a(3*31) = concat(93,1) because concat(1,93) gives 19*3, concat(3,31) gives 33*1 and concat(31,3) gives 3*13 as smaller products.
%o A347471 (PARI) apply( {A347471(s,m=oo)=if(s, fordiv(s,d, my(t=eval(Str(d,s/d))); s==A347470(t) && m>t && m=eval(Str(d,s/d)));m)}, [0..111])
%Y A347471 Cf. A347470.
%K A347471 nonn,base
%O A347471 0,2
%A A347471 _M. F. Hasler_, Sep 03 2021