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%I #5 Jan 09 2024 16:27:52
%S 0,121,125,625,32,64,128,256,512,0,0,219122,24344,4802,6250,31250,
%T 4374,13122,39366,0,10170397,24964,8575,2500,12500,2916,8748,26244,
%U 78732,31855013,118459,6125,2744,5000,25000,5832,17496,52488,157464,279841
%N Write n in base 10 as d1d2d3.....dk; for a list of primes P = (p1,p2,p3,....pk) with p1<p2<p3<.....<pk, let A(P,n)=(p1^d1)*(p2^d2)*(p3^d3)*.....(pk^dk) and B(P,n)= concatenation of primes and powers = p1&d1&p2&d2&p3&d3&.......&pk&dk. Then a(n) is the smallest number A(P,n) such that A(P,n)>B(p,n) if it exists, otherwise 0.
%C Computing the number of digits of A and B, it is easy to prove that for any n written only with 0 and 1, there is no solution, hence a(n) = 0.
%C The same thing is true for some other numbers such as 201,210,211,300,. . . .
%C Any number of the submitted sequence a(n) of numbers A(P,n) satisfying the condition defines univocally and the number n and the vector P.On the contrary it should be not pertinent to submit the sequence of the associated numbers B(P,n) as the value of such a number does not define always univocally n and P. For example for n=26, a(n)=2916=(2^2)*(3^6) and B({2;3},26)=2236 which should be also be associated with A=223^6, as 223 is prime.
%e For any p of q digits, p^1 contains q digits, but p&1 contains q+1 digits, hence a(1)=0
%e For p1 = 2,3,5 and 7, p1^2<p1&2 but 11^2=121<112, hence a(2)=121
%K nonn,base
%O 1,2
%A Philippe Lallouet (philip.lallouet(AT)orange.fr), Jan 16 2008
%E Edited by _N. J. A. Sloane_, Jan 15 2009