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%I #31 Jan 04 2018 01:32:21
%S 1,1,1,0,1,1,0,0,1,1,0,0,1,0,0,1,0,1,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,1,
%T 0,0,0,1,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,0,0,0,
%U 0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,1,0
%N Irregular triangle read by rows: T(n,k) = 1 if k is the largest divisor of n <= sqrt(n), otherwise T(n,k) = 0. With n >= 1, and 1 <= k <= A000196(n).
%C The first element of column k is in the row k^2.
%F T(n, A033676(n)) = 1.
%F T(n,k) = 0 if k is not equal to A033676(n), n >= 1, and 1 <= k <= A000196(n).
%F a(n) = A057427(A294721(n)).
%e Triangle begins:
%e 1;
%e 1;
%e 1;
%e 0, 1;
%e 1, 0;
%e 0, 1;
%e 1, 0;
%e 0, 1;
%e 0, 0, 1;
%e 0, 1, 0;
%e 1, 0, 0;
%e 0, 0, 1;
%e 1, 0, 0;
%e 0, 1, 0;
%e 0, 0, 1;
%e 0, 0, 0, 1;
%e 1, 0, 0, 0;
%e 0, 0, 1, 0;
%e 1, 0, 0, 0;
%e 0, 0, 0, 1;
%e 0, 0, 1, 0;
%e 0, 1, 0, 0;
%e 1, 0, 0, 0;
%e 0, 0, 0, 1;
%e 0, 0, 0, 0, 1;
%e ...
%t Table[ReplacePart[ConstantArray[0, IntegerPart@ Sqrt@ n], SelectFirst[Reverse@ Divisors@ n, # <= Sqrt@ n &] -> 1], {n, 32}] // Flatten (* _Michael De Vlieger_, Nov 13 2017 *)
%o (PARI) row(n) = {d = divisors(n); kmax = vecmax(select(x->(x^2 <= n), d)); vector(sqrtint(n), k, k==kmax);}
%o tabf(nn) = for (n=1, nn, print(row(n))); \\ _Michel Marcus_, Dec 12 2017
%Y Row n has length A000196(n).
%Y Row sums give A000012.
%Y Cf. A033676, A057427, A163280, A237273, A294721.
%Y Sequences related to columns 1..12: A008578, A161344, A161345, A161424, A161835, A162527, A162527, A162528, A162529, A162530, A162531, A162532.
%K nonn,tabf
%O 1
%A _Omar E. Pol_, Nov 09 2017
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