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A100735
Inverse modulo 2 binomial transform of 2^n.
4
1, 1, 3, 3, 15, 15, 45, 45, 255, 255, 765, 765, 3825, 3825, 11475, 11475, 65535, 65535, 196605, 196605, 983025, 983025, 2949075, 2949075, 16711425, 16711425, 50134275, 50134275, 250671375, 250671375, 752014125, 752014125, 4294967295, 4294967295
OFFSET
0,3
COMMENTS
The modulo 2 binomial transform and its inverse are defined by
B(n) = Sum_{k=0..n} (binomial(n,k) mod 2)*A(k),
A(n) = Sum_{k=0..n} (-1)^A010060(n-k)*(binomial(n, k) mod 2)*B(k). - N. J. A. Sloane, Dec 20 2019
2^n may be retrieved as Sum_{k=0..n} mod(binomial(n,k),2)*a(k).
FORMULA
a(n) = Sum_{k=0..n} (-1)^A010060(n-k)*mod(binomial(n, k), 2)*2^k.
MATHEMATICA
Table[Sum[(-1)^ThueMorse[n - k]*Mod[Binomial[n, k], 2]*2^k, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Apr 17 2018 *)
PROG
(PARI) for(n=0, 50, print1(abs(sum(k=0, n, (-1)^(hammingweight(k)%2)* lift(Mod(binomial(n, k), 2))*2^k)), ", ")) \\ G. C. Greubel, Apr 17 2018
CROSSREFS
Sequence in context: A333862 A172087 A086116 * A362531 A129356 A290344
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Dec 06 2004
STATUS
approved