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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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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.

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), 236-242

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

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Last modified August 19 21:19 EDT 2019. Contains 326133 sequences. (Running on oeis4.)