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%I #23 Feb 07 2023 05:57:37
%S 1,2,0,3,1,0,4,0,0,0,5,3,2,0,0,6,2,0,0,0,0,7,3,1,0,0,0,0,8,0,0,0,0,0,
%T 0,0,9,7,6,4,0,0,0,0,0,10,6,4,0,0,0,0,0,0,0,11,7,3,2,0,0,0,0,0,0,0,12,
%U 4,0,0,0,0,0,0,0,0,0,0,13,9,7,2,0,0,0,0,0,0,0,0,0
%N T(n,k): number of binomial coefficients C(n,r), for 0 <= r <= n, divisible by 2^k.
%C Since the sum of binomial coefficients of the n-th row of Pascal's triangle is 2^n, T(n, k)=0 for k >= n. So only n elements, from k=0 to n-1, will be displayed at row n, giving a triangle instead of a table.
%C The number A119387(n) gives the position of the last positive number in each row. - _T. D. Noe_, Dec 18 2012
%H T. D. Noe, <a href="/A220645/b220645.txt">Rows n = 0..100 of triangle, flattened</a>
%H F. T. Howard, <a href="http://dx.doi.org/10.1090/S0002-9939-1971-0302459-9">The number of binomial coefficients divisible by a fixed power of 2</a>, Proc. Amer. Math. Soc. 29 (1971), 236-242
%e Triangle starts:
%e 0: 1
%e 1: 2 0
%e 2: 3 1 0
%e 3: 4 0 0 0
%e 4: 5 3 2 0 0
%e 5: 6 2 0 0 0 0
%e For n=4, the corresponding Pascal's triangle row is:
%e 1 4 6 4 1,
%e with 5 numbers divisible by 2^0,
%e and 3 numbers divisible by 2^1,
%e and 2 numbers divisible by 2^2,
%e and 0 numbers divisible by 2^3,
%e and 0 numbers divisible by 2^4.
%t Flatten[Table[b = Binomial[n, Range[0, n]]; Table[Count[b, _?(Mod[#, 2^k] == 0 &)], {k, 0, n}], {n, 0, 12}]] (* _T. D. Noe_, Dec 18 2012 *)
%Y Cf. A000079, A007318.
%K nonn,tabl
%O 0,2
%A _Michel Marcus_, Dec 17 2012
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