

A269299


Maximum number of times a ninedigit number with nonrepeating, nonzero digits can be divided by the nth prime until a noninteger result is returned.


0



16, 13, 7, 7, 5, 5, 5, 4, 4, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 1, 3, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2
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OFFSET

1,1


COMMENTS

Sequence was constructed with a simple game in mind, one done by hand or by calculator, where one takes a ninedigit number with nonrepeating nonzero digits and divides it by three until a noninteger result is returned. The potential best play is then of interest for this game, or the maximum number of times three divides such a number. The sequence extends this idea to all primes under the same rules and outputs the maximum number of times the nth prime divides a number in this form.
The largest such prime that would give a nonzero solution is the 6289143rd prime, namely 109739359.


LINKS



EXAMPLE

For n=2, 3 divides 618597324 13 times, and this is the highest achievable valuation. In comparison, if the rules regarding nonzero and nonrepeating were lifted, the number 3^18 = 387420489 would be allowed, and a(2) would be 18.


MAPLE

with(LinearAlgebra)*with(combinat)*with(numtheory):
A := permute(9):
B := 0:
for d to 6289143 do
a[d] := 0:
end do:
for i to factorial(9) do
B := ifactors(Determinant(Multiply(`<>`(`<, >`(100000000), `<, >`(10000000), `<, >`(1000000), `<, >`(100000), `<, >`(10000), `<, >`(1000), `<, >`(100), `<, >`(10), `<, >`(1)), Transpose(convert(A[i], Matrix))))):
C := nops([op(B[2])]):
for j to C do
if a[pi(B[2, j, 1])] < B[2, j, 2] then
a[pi(B[2, j, 1])] := B[2, j, 2]:
end if:
end do:
end do:
seq(a[i], i = 1 .. 500);


MATHEMATICA

s = FromDigits /@ Permutations@ Range@ 9; Table[Max@ Map[Length@ NestWhileList[#/Prime@ n &, #, IntegerQ] &, s], {n, 40}]  2 (* Michael De Vlieger, Feb 23 2016 *)


PROG

(PARI) a(n) = {my(p=prime(n)); my(mv = 0); for (i=0, 9!, mv = max(valuation(subst(Pol(numtoperm(9, i)), x, 10), p), mv); ); mv; } \\ Michel Marcus, Feb 26 2016


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



