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A130047
Left half of Pascal's triangle (A034868) modulo 2.
2
1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0
OFFSET
0,1
COMMENTS
Row sums yield: 1, 1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, ...(see A048896).
FORMULA
T(n,k) = mod(binomial(n, k), 2), 0 <= k <= floor(n/2). - G. C. Greubel, Aug 12 2017
EXAMPLE
Triangle begins:
1,
1,
1, 0,
1, 1,
1, 0, 0,
1, 1, 0,
1, 0, 1, 0,
1, 1, 1, 1,
1, 0, 0, 0, 0,
1, 1, 0, 0, 0,
1, 0, 1, 0, 0, 0,
1, 1, 1, 1, 0, 0,
1, 0, 0, 0, 1, 0, 0,
1, 1, 0, 0, 1, 1, 0,
1, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0,
...
Triangle (right aligned) begins:
1,
1,
1, 0,
1, 1,
1, 0, 0,
1, 1, 0,
1, 0, 1, 0,
1, 1, 1, 1,
1, 0, 0, 0, 0,
1, 1, 0, 0, 0,
1, 0, 1, 0, 0, 0,
1, 1, 1, 1, 0, 0,
1, 0, 0, 0, 1, 0, 0,
1, 1, 0, 0, 1, 1, 0,
1, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 0, 0, 0, 0, 0, 0, 0,
...
MAPLE
# From N. J. A. Sloane, Mar 22 2015:
for n from 0 to 20 do
lprint(seq(binomial(n, k) mod 2, k=0..floor(n/2))); od:
# For row sums:
f:=n->add(binomial(n, k) mod 2, k=0..floor(n/2));
[seq(f(n), n=0..60)];
MATHEMATICA
Table[Mod[Binomial[n, k], 2], {n, 0, 10}, {k, 0, Floor[n/2]}] (* G. C. Greubel, Aug 12 2017 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Philippe Deléham, Oct 10 2007
EXTENSIONS
Corrected by N. J. A. Sloane, Mar 22 2015 at the suggestion of Kevin Ryde
STATUS
approved