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%I #36 Jan 14 2021 21:17:00
%S 10,22,25,63,45,50,11,9,1,18,1,2,0,0,1,18,1,0,0,9,0,0,0,1,18,1,0,0,0,
%T 1,0,9,0,0,0,0,0,0,0,9,0,0,0,0,0,0,0,1,0,9,0,0,0,0,0,0,0,0,0,1,0,0,0,
%U 9,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,9,0,0,0,0
%N The number of n-digit numbers whose last k digits are divisible by k^2 for k = 1..n.
%C The sequence is infinite.
%H Shyam Sunder Gupta, <a href="/A228010/b228010.txt">Table of n, a(n) for n = 1..1000</a>
%e There are ten one-digit numbers divisible by 1 so a(1)=10.
%e For two-digit numbers, the second digit must make it divisible by 2^2, which gives 22 numbers: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 to satisfy the requirement. So a(2)=22.
%t a = Table[j, {j, 0, 9}]; r = 2; s2 = 10; t = a; xs = {s2}; While[! a == {} , n = Length[a]; k = 1; b = {}; While[! k > n , z0 = a[[k]]; Do[z = 10^(r - 1)*j + z0; If[Mod[z, r*r] == 0 && r < 111, b = Append[b, z]; t = Append[t, z]], {j, 0, 9}]; k++]; s = Union[t]; s1 = Length[s]; If[r < 111, xs = Append[xs, s1 - s2]]; s2 = s1; a = b; r++]; xs
%Y Cf. A079238.
%K nonn,base
%O 1,1
%A _Shyam Sunder Gupta_, Aug 08 2013
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