A355068 Square array read by upwards antidiagonals: T(n,k) = k-th digit after the decimal point in decimal expansion of 1/n, for n >= 1 and k >= 1.
0, 5, 0, 3, 0, 0, 2, 3, 0, 0, 2, 5, 3, 0, 0, 1, 0, 0, 3, 0, 0, 1, 6, 0, 0, 3, 0, 0, 1, 4, 6, 0, 0, 3, 0, 0, 1, 2, 2, 6, 0, 0, 3, 0, 0, 1, 1, 5, 8, 6, 0, 0, 3, 0, 0, 0, 0, 1, 0, 5, 6, 0, 0, 3, 0, 0, 0, 9, 0, 1, 0, 7, 6, 0, 0, 3, 0, 0, 0, 8, 0, 0, 1, 0, 1, 6, 0
Offset: 1
Examples
Array begins:
k=1 2 3 4 5 6 7 8
n=1: 0, 0, 0, 0, 0, 0, 0, 0,
n=2: 5, 0, 0, 0, 0, 0, 0, 0,
n=3: 3, 3, 3, 3, 3, 3, 3, 3,
n=4: 2, 5, 0, 0, 0, 0, 0, 0,
n=5: 2, 0, 0, 0, 0, 0, 0, 0,
n=6: 1, 6, 6, 6, 6, 6, 6, 6,
n=7: 1, 4, 2, 8, 5, 7, 1, 4,
n=8: 1, 2, 5, 0, 0, 0, 0, 0,
Row n=7 is 1/7 = .142857142857..., whose digits after the decimal point are 1,4,2,8,5,7,1,4,2,8,5,7, ...
Crossrefs
Programs
-
PARI
T(n,k) = my(r=lift(Mod(10,n)^(k-1))); floor(10*r/n)%10;
-
Python
def T(n,k): return (10*pow(10,k-1,n)//n)%10
Formula
1/n = Sum_{k>=1} T(n, k)*10^-k, for n > 1.
Comments