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, 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

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

Table of n, a(n) for n=1..40.

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

Sequence in context: A262517 A036231 A055468 * A113614 A134941 A173070

Adjacent sequences: A134325 A134326 A134327 * A134329 A134330 A134331

Keyword

nonn,base

Author

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

Extensions

Edited by N. J. A. Sloane, Jan 15 2009

Status

approved