A152561 -id:A152561 - cp's OEIS frontend

A374923 a(n) is the least k such that 2^k begins with n!.

0, 0, 1, 6, 81, 80, 56, 7284, 33889, 2044921, 8151937, 127668791, 258943304, 19207561921, 189815680859, 2687562198191, 75909586168557, 512148453482307, 5376323935222903, 502774568129731130, 1053338431686717460, 122114339415457901831, 2120280158164651048122

Offset: 0

Zhining Yang, Jul 23 2024

			a(4) = 81 because 2^81 = 2417851639229258349412352 is the smallest power of 2 beginning with 4! = 24.

Zhao Hui Du, Table of n, a(n) for n = 0..100

Cf. A018856, A000142, A152561, A374922.

a[n_] := Module[{target = IntegerDigits[n!], k = 0},
   While[UnsameQ[Take[IntegerDigits[2^k], Length@target], target],
    k++]; k];
Table[a[n], {n, 0, 8}]

a(n) = A018856(n!).

a(13) onwards from Zhao Hui Du, Oct 02 2024

A385971 Smallest m such that 5^m begins with n 9's after the first digit.

Original entry on oeis.org

0, 8, 195, 799, 28737, 167821, 325146, 6432162, 543157237, 1807789217, 3731189547, 3731189547

Offset: 0

Giulio Bonfissuto, Jul 13 2025

a(n) is also the smallest m such that 1/2^m begins with n 9's after the first nonzero digit.
a(n) is equal to A152561(n)-1 for n=2, 6, 7 and possibly for many other terms.
When summing a series with dominant term 1/2^m (such as the Riemann zeta function), the n 9's here show how small further terms must be to avoid changing the initial decimal digit from 1/2^m.

			5^a(0) = 5^0      = 1
5^a(1) = 5^8      = 390625
5^a(2) = 5^195    = 1991364888915565346...
5^a(3) = 5^799    = 2999393627791261909...
5^a(4) = 5^28737  = 1999929120817815105...
5^a(5) = 5^167821 = 6999994116858573262...

Marco Ripà, Is the leading digit of the decimal expansion of the prime zeta function at n equal to the first digit of 5^n, for all integers n≥10?, Stack Exchange (2025)

Cf. A152561, A000351, A111395, A011557.

a(n) = Min_{d=1..9} S(d*10^(n+1)-1) where 5^S(k) is the smallest power of 5 beginning with k.

cp's OEIS Frontend

A374923 a(n) is the least k such that 2^k begins with n!.

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