A364881 First significant digit of the decimal expansion of n/(2^n).

5, 5, 3, 2, 1, 9, 5, 3, 1, 9, 5, 2, 1, 8, 4, 2, 1, 6, 3, 1, 1, 5, 2, 1, 7, 3, 2, 1, 5, 2, 1, 7, 3, 1, 1, 5, 2, 1, 7, 3, 1, 9, 4, 2, 1, 6, 3, 1, 8, 4, 2, 1, 5, 2, 1, 7, 3, 2, 1, 5, 2, 1, 6, 3, 1, 8, 4, 2, 1, 5, 3, 1, 7, 3, 1, 1, 5, 2, 1, 6, 3, 1, 8, 4, 2, 1, 5

Offset: 1

Author

Ejder Aysun, Aug 10 2023

Keywords

• easy
• nonn
• base

Comments

a(n) is also the first digit of n*5^n = A036291(n).

Examples

			n     n/(2^n)
1     0.5                            a(1) = 5
2     0.5                            a(2) = 5
3     0.375                          a(3) = 3
4     0.25                           a(4) = 2
5     0.15625                        a(5) = 1
6     0.9375                         a(6) = 9
7     0.0546875                      a(7) = 5
8     0.03125                        a(8) = 3
9     0.017578125                    a(9) = 1
10    0.009765625                    a(10) = 9
...
		

Crossrefs

Cf. A000030, A000079, A000799, A036291, A111395.

Programs

Maple

a:= n-> parse((""||(n*5^n))[1]):
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 18 2023

Mathematica

Table[Floor[n/(2^n)/10^Floor[Log10[n/(2^n)]]], {n, 100000}]

Python

def A364881(n): return (n*5**(m:=len(str((1<>n-m) % 10 # Chai Wah Wu, Aug 24 2023

Formula

a(n) = floor(n/(2^n)/10^floor(log_10(n/(2^n)))), for n > 0.

a(n) = floor(n/A000079(n)/10^floor(log_10(n/A000079(n)))).

a(n) = floor(A036291(n)/10^floor(log_10(A036291(n)))).

a(n) = A000030(A036291(n)).