For large n, the sum of the base-5 digits of n hovers around 2*log_5(n) ~= 1.242 log(n), but this is not an asymptotic relationship. Similarly, the sum of the base-2 digits of n hovers around 0.5*log_2(n) = 0.721 log(n). This means that as n grows larger, the sum of the base-5 digits will tend grow faster than the sum of the base-2 digits. This means that the proportion of n satisfying [1] 0 <= sum of base-5 digits of n - sum of base-2 digits of n <= 3 should tend to decrease n grows larger. The numbers satisfying [1] are precisely the n such that n! has the same number of trailing 0's in bases 10 and 16. Hence we should expect these numbers to thin out as n increases, I strongly suspect the limit density is 0 over Z. On the other hand, there are arbitrarily large numbers n (specifically, the powers of 5), whose base-5 digit sum is smaller than the base-2 digit sum. This would indicate that there are arbitrarily large numbers n satisfying [1], i.e., an infinitude of such n. Let S(a, b) = {n | n! has the same number of trailing zeroes in bases a and b}. I suspect that S(a, b) is infinite iff A090624(a) = A090624(b). In some cases, S(a, b) = N, e.g. a = 5, b = 10. -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of David Wilson Sent: Saturday, September 03, 2016 6:21 PM To: 'math-fun' Subject: Re: [math-fun] peculiar sequence The number of zeroes at the end of n! is equal in bases 10 and 16 iff 0 <= sum of base-5 digits of n - sum of base-2 digits of n <= 3. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun