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Least product of any two numbers whose concatenation is n, excluding 0*n for n > 9.
2

%I #10 Oct 22 2021 23:45:23

%S 0,0,0,0,0,0,0,0,0,0,0,1,2,3,4,5,6,7,8,9,0,2,4,6,8,10,12,14,16,18,0,3,

%T 6,9,12,15,18,21,24,27,0,4,8,12,16,20,24,28,32,36,0,5,10,15,20,25,30,

%U 35,40,45,0,6,12,18,24,30,36,42,48,54,0,7,14,21,28,35,42,49,56,63,0,8,16,24,32,40,48,56,64,72,0,9,18,27,36,45,54,63,72,81,0,10,20,30,40,50,60,70,80,90,0,11,12

%N Least product of any two numbers whose concatenation is n, excluding 0*n for n > 9.

%C Leading zeros are not allowed: e.g., 101 = concat(10,1) but not concat(1,01). Although 0 is a valid number, we don't allow the trivial decomposition n = concat(0, n) except for the single-digit n < 10, otherwise the minimal product would always be 0.

%C For n < 111, this sequence coincides with A035930 (same with "largest"), because there is only one possible concatenation, but it differs for n > 111, cf. examples.

%H Eric Angelini, <a href="https://cinquantesignes.blogspot.com/2021/08/surface-of-number.html">Surface of a number</a>, Sep 01 2021.

%e The number n = 112 is the concatenation of 1 and 12, or of 11 and 2, with respective products 1*12 = 12 and 11*2 = 22. Hence, a(112) = 12, while A035930(112) = 22.

%o (PARI) apply( {A347470(x,t(b,c)=if(c\10<=b%c,b\c*(b%c),c>10,oo))= if(x>9,vecmin(vector(logint(x,10),j,t(x,10^j))))}, [0..112])

%Y Cf. A035930, A007954, A088117, A171765 and A257297.

%K nonn,base

%O 0,13

%A _M. F. Hasler_, Sep 03 2021