

A220645


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


2



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, 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, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 9, 7, 2, 0, 0, 0, 0
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OFFSET

0,2


COMMENTS

Since the sum of binomial coefficients of the nth row of Pascal's triangle is 2^n, T(n, k)=0 for k >= n. So only n elements, from k=0 to n1, will be displayed at row n, giving a triangle instead of a table.
The number A119387(n) gives the position of the last positive number in each row.  T. D. Noe, Dec 18 2012


LINKS

T. D. Noe, Rows n = 0..100 of triangle, flattened
F. T. Howard, The number of binomial coefficients divisible by a fixed power of 2, Proc. Amer. Math. Soc. 29 (1971), 236242


EXAMPLE

Triangle starts:
0: 1
1: 2 0
2: 3 1 0
3: 4 0 0 0
4: 5 3 2 0 0
5: 6 2 0 0 0 0
For n=4, the corresponding Pascal's triangle row is:
1 4 6 4 1,
with 5 numbers divisible by 2^0,
and 3 numbers divisible by 2^1,
and 2 numbers divisible by 2^2,
and 0 numbers divisible by 2^3,
and 0 numbers divisible by 2^4.


MATHEMATICA

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 *)


CROSSREFS

Cf. A000079, A007318.
Sequence in context: A070679 A177443 A176919 * A127374 A098862 A003988
Adjacent sequences: A220642 A220643 A220644 * A220646 A220647 A220648


KEYWORD

nonn,tabl


AUTHOR

Michel Marcus, Dec 17 2012


STATUS

approved



