A266344 a(n) = number of ways n can be divided into two factors that have the same number of digits in factorial base representation (the two different orders for unequal factors are counted only once).

1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 2, 0, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 2, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 2, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 1, 3

Offset: 1

Author

Antti Karttunen, Dec 28 2015

Keywords

• nonn
• base

Examples

			1 can be factored just one way, as 1*1, and thus a(1) = 1.
4 can be factored as 2*2, and thus also a(4) = 1, and generally for all perfect squares k, a(k) >= 1.
14 can be factored as 2*7, but as A007623(2) = 2 and A007623(7) = 101, with different number of digits in factorial base (and 1*14 fares even less well), a(14) = 0.
72 can be factored to two divisors so that the factorial base representations are of equal length as 6*12 or 8*9 (where the corresponding factorial base representations are "100" * "200" and "110" * "111"), thus a(72) = 2.
120 can be similarly factored as 6*20 ("100" * "310"), 8*15 ("110" * "211") and 10*12 ("120" * "200"), thus a(120) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..14161
• Index entries for sequences related to factorial base representation

Crossrefs

Cf. A084558.

Cf. A266345 (positions of records).

Cf. also A078781, A266342.

Programs

Mathematica

r = Most@ Reverse@ Range@ 10; Map[Length, Table[Flatten@ Map[Differences@ IntegerLength[#, MixedRadix@ r] &, Transpose@ {#, n/#}] &@ TakeWhile[Divisors@ n, # <= Sqrt@ n &], {n, 120}] /. k_ /; k > 0 -> Nothing] (* Michael De Vlieger, Dec 30 2015, Version 10.2 *)

PARI

A084558(n) = { my(m=1); if(0==n,n,while(m!<=n,m++);return(m-1)); }
A266344(n) = sumdiv(n, d, ((d <= (n/d)) && (A084558(d)==A084558(n/d))));
for(n=1, 14161, write("b266344.txt", n, " ", A266344(n)));

Formula

a(n) = Sum_{d|n} [(d <= (n/d)) and (A084558(d) = A084558(n/d))].

(In the above formula [ ] stands for Iverson bracket, giving as its result 1 only if d is less than or equal to n/d and in factorial base representation d and n/d require equal number of digits, and 0 otherwise.)