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Left half of Pascal's triangle (A034868) modulo 2.
2

%I #28 Aug 15 2017 11:09:26

%S 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,

%T 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,

%U 1,1,1,1,1,0,0,0,0,0,0

%N Left half of Pascal's triangle (A034868) modulo 2.

%C 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).

%H G. C. Greubel, <a href="/A130047/b130047.txt">Table of n, a(n) for the first 100 rows, flattened</a>

%F T(n,k) = mod(binomial(n, k), 2), 0 <= k <= floor(n/2). - _G. C. Greubel_, Aug 12 2017

%e Triangle begins:

%e 1,

%e 1,

%e 1, 0,

%e 1, 1,

%e 1, 0, 0,

%e 1, 1, 0,

%e 1, 0, 1, 0,

%e 1, 1, 1, 1,

%e 1, 0, 0, 0, 0,

%e 1, 1, 0, 0, 0,

%e 1, 0, 1, 0, 0, 0,

%e 1, 1, 1, 1, 0, 0,

%e 1, 0, 0, 0, 1, 0, 0,

%e 1, 1, 0, 0, 1, 1, 0,

%e 1, 0, 1, 0, 1, 0, 1, 0,

%e 1, 1, 1, 1, 1, 1, 1, 1,

%e 1, 0, 0, 0, 0, 0, 0, 0, 0,

%e ...

%e Triangle (right aligned) begins:

%e 1,

%e 1,

%e 1, 0,

%e 1, 1,

%e 1, 0, 0,

%e 1, 1, 0,

%e 1, 0, 1, 0,

%e 1, 1, 1, 1,

%e 1, 0, 0, 0, 0,

%e 1, 1, 0, 0, 0,

%e 1, 0, 1, 0, 0, 0,

%e 1, 1, 1, 1, 0, 0,

%e 1, 0, 0, 0, 1, 0, 0,

%e 1, 1, 0, 0, 1, 1, 0,

%e 1, 0, 1, 0, 1, 0, 1, 0,

%e 1, 1, 1, 1, 1, 1, 1, 1,

%e 1, 0, 0, 0, 0, 0, 0, 0, 0,

%e 1, 1, 0, 0, 0, 0, 0, 0, 0,

%e ...

%p # From _N. J. A. Sloane_, Mar 22 2015:

%p for n from 0 to 20 do

%p lprint(seq(binomial(n,k) mod 2, k=0..floor(n/2))); od:

%p # For row sums:

%p f:=n->add(binomial(n,k) mod 2, k=0..floor(n/2));

%p [seq(f(n),n=0..60)];

%t Table[Mod[Binomial[n, k], 2], {n, 0, 10}, {k, 0, Floor[n/2]}] (* _G. C. Greubel_, Aug 12 2017 *)

%Y Cf. A007318, A034868, A048896, A133179.

%K nonn,tabf

%O 0,1

%A _Philippe Deléham_, Oct 10 2007

%E Corrected by _N. J. A. Sloane_, Mar 22 2015 at the suggestion of Kevin Ryde