login
A228010
The number of n-digit numbers whose last k digits are divisible by k^2 for k = 1..n.
1
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, 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, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 9, 0, 0, 0, 0
OFFSET
1,1
COMMENTS
The sequence is infinite.
LINKS
EXAMPLE
There are ten one-digit numbers divisible by 1 so a(1)=10.
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.
MATHEMATICA
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
CROSSREFS
Cf. A079238.
Sequence in context: A110367 A104865 A063555 * A303745 A303746 A303747
KEYWORD
nonn,base
AUTHOR
Shyam Sunder Gupta, Aug 08 2013
STATUS
approved