

A134328


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.


0



0, 121, 125, 625, 32, 64, 128, 256, 512, 0, 0, 219122, 24344, 4802, 6250, 31250, 4374, 13122, 39366, 0, 10170397, 24964, 8575, 2500, 12500, 2916, 8748, 26244, 78732, 31855013, 118459, 6125, 2744, 5000, 25000, 5832, 17496, 52488, 157464, 279841
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OFFSET

1,2


COMMENTS

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.
The same thing is true for some other numbers such as 201,210,211,300,. . . .
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.


LINKS



EXAMPLE

For any p of q digits, p^1 contains q digits, but p&1 contains q+1 digits, hence a(1)=0
For p1 = 2,3,5 and 7, p1^2<p1&2 but 11^2=121<112, hence a(2)=121


CROSSREFS



KEYWORD

nonn,base


AUTHOR

Philippe Lallouet (philip.lallouet(AT)orange.fr), Jan 16 2008


EXTENSIONS



STATUS

approved



