A341514 Number of trailing zeros in A097801-base.

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

Offset: 1

Author

Antti Karttunen, Feb 25 2021

Keywords

• nonn
• base

Comments

A097801-base uses values 1, 2, 2*3, 2*3*5, 2*3*5*7, 2*3*5*7*9, 2*3*5*7*9*11, 2*3*5*7*9*11*13, 2*3*5*7*9*11*13*15, ..., for its digit-positions, instead of primorials (A002110), thus up to 1889 = 2*3*5*7*9 - 1 = 9*A002110(4) - 1 its representation is identical with the primorial base A049345.

From Amiram Eldar, Mar 10 2021: (Start)

The asymptotic density of the occurrences of k is 1/2 if k=0, and 2*k/(A097801(k+1)) otherwise.

The asymptotic mean of this sequence is sqrt(e*Pi/2)*erf(1/sqrt(2))/2 = 0.7053430673..., where erf(x) is the error function. (End)

Examples

			In A097801-base number 1890 = 2*3*5*7*9 is expressed as "100000", thus a(1890) = 5.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A097801, A341356, A341513.

Differs from A276084 for the first time at n=1890, as a(1890) = 5, while A276084(1890) = 4.

Programs

Mathematica

Block[{nn = 105, b}, b = MixedRadix@ NestWhile[Prepend[#1, 2 #2 - 1] & @@ {#, Length[#] + 1} &, {2}, Times @@ # < nn &]; Array[LengthWhile[Reverse@ IntegerDigits[#, b], # == 0 &] &, nn]] (* Michael De Vlieger, Feb 25 2021 *)

PARI

A341514(n) = { my(m=2,k=3,i=0); while(!(n%m), n /= m; m = k; k += 2; i++); (i); };

Formula

For odd n, a(n) = 0; for even n, a(n) = the largest k such that A097801(k) divides n.