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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.
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%I #34 Jun 25 2022 10:01:29

%S 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,

%T 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,

%U 0,1,0,7,6,0,0,3,0,0,0,8,0,0,1,0,1,6,0

%N 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.

%C First row is all zeros since n=1 has all zeros after the decimal point.

%F 1/n = Sum_{k>=1} T(n, k)*10^-k, for n > 1.

%e Array begins:

%e k=1 2 3 4 5 6 7 8

%e n=1: 0, 0, 0, 0, 0, 0, 0, 0,

%e n=2: 5, 0, 0, 0, 0, 0, 0, 0,

%e n=3: 3, 3, 3, 3, 3, 3, 3, 3,

%e n=4: 2, 5, 0, 0, 0, 0, 0, 0,

%e n=5: 2, 0, 0, 0, 0, 0, 0, 0,

%e n=6: 1, 6, 6, 6, 6, 6, 6, 6,

%e n=7: 1, 4, 2, 8, 5, 7, 1, 4,

%e n=8: 1, 2, 5, 0, 0, 0, 0, 0,

%e 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, ...

%o (PARI) T(n,k) = my(r=lift(Mod(10,n)^(k-1))); floor(10*r/n)%10;

%o (Python) def T(n,k): return (10*pow(10,k-1,n)//n)%10

%Y Cf. A020806, A021017, A021018.

%Y Cf. A048962, A257927.

%Y Cf. A061480 (diagonal).

%Y Cf. A355202 (binary).

%K base,easy,nonn,tabl

%O 1,2

%A _Chittaranjan Pardeshi_, Jun 17 2022