cp's OEIS Frontend

A220645 T(n,k): number of binomial coefficients C(n,r), for 0 <= r <= n, divisible by 2^k.

%I A220645 #23 Feb 07 2023 05:57:37
%S A220645 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 A220645 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 A220645 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 A220645 T(n,k): number of binomial coefficients C(n,r), for 0 <= r <= n, divisible by 2^k.
%C A220645 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 A220645 The number A119387(n) gives the position of the last positive number in each row. - _T. D. Noe_, Dec 18 2012
%H A220645 T. D. Noe, <a href="/A220645/b220645.txt">Rows n = 0..100 of triangle, flattened</a>
%H A220645 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 A220645 Triangle starts:
%e A220645 0: 1
%e A220645 1: 2 0
%e A220645 2: 3 1 0
%e A220645 3: 4 0 0 0
%e A220645 4: 5 3 2 0 0
%e A220645 5: 6 2 0 0 0 0
%e A220645 For n=4, the corresponding Pascal's triangle row is:
%e A220645 1 4 6 4 1,
%e A220645 with 5 numbers divisible by 2^0,
%e A220645 and 3 numbers divisible by 2^1,
%e A220645 and 2 numbers divisible by 2^2,
%e A220645 and 0 numbers divisible by 2^3,
%e A220645 and 0 numbers divisible by 2^4.
%t A220645 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 A220645 Cf. A000079, A007318.
%K A220645 nonn,tabl
%O A220645 0,2
%A A220645 _Michel Marcus_, Dec 17 2012