

A330083


a(n) is the smallest number k > 0 such that for each b = 2..n the baseb expansion of k has exactly n  b zeros.


0




OFFSET

2,2


COMMENTS

This list is complete. Proof: When converting base 2 to base 4, we can group the digits in base 2 into pairs from the least significant bit. We then convert pairs into single digits in base 4 as 00 > 0, 01 > 1, 10 > 2, 11 > 3. This always causes the number of zeros to go to half or less than half. For all n >= 7, n4 is greater than (n2)/2, so the condition is impossible.  Christopher Cormier, Dec 08 2019
Does k exist for every n >= 2?
a(7) > 10^7, if it exists.


LINKS



EXAMPLE

For n = 6: The baseb expansions of 271 for b = 2..6 are shown in the following table:
b  baseb expansion  number of zeros

2  100001111  4
3  101001  3
4  10033  2
5  2041  1
6  1131  0


PROG

(PARI) count_zeros(vec) = #setintersect(vecsort(vec), vector(#vec))
a(n) = for(k=1, oo, for(b=2, n, if(count_zeros(digits(k, b))!=nb, break, if(b==n, return(k)))))


CROSSREFS



KEYWORD

nonn,base,fini,full


AUTHOR



EXTENSIONS



STATUS

approved



