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A099902
Multiplies by 2 and shifts right under the XOR BINOMIAL transform (A099901).
5
1, 3, 7, 11, 23, 59, 103, 139, 279, 827, 1895, 2955, 5655, 14395, 24679, 32907, 65815, 197435, 460647, 723851, 1512983, 3881019, 6774887, 9142411, 18219287, 54002491, 123733863, 192940939, 369104407, 939538491, 1610637415, 2147516555
OFFSET
0,2
COMMENTS
Equals the XOR BINOMIAL transform of A099901. Also, equals the main diagonal of the XOR difference triangle A099900, in which the central terms of the rows form the powers of 2.
Bisection of A101624. - Paul Barry, May 10 2005
LINKS
FORMULA
a(n) = SumXOR_{k=0..n} (binomial(n-k+floor(k/2), floor(k/2)) mod 2)*2^k for n >= 0.
a(n) = SumXOR_{i=0..n} (C(n, i) mod 2)*A099901(n-i), where SumXOR is the analog of summation under the binary XOR operation and C(i, j) mod 2 = A047999(i, j).
a(n) = Sum_{k=0..n} A047999(n-k+floor(k/2), floor(k/2)) * 2^k.
From Paul Barry, May 10 2005: (Start)
a(n) = Sum_{k=0..2n} (binomial(k, 2n-k) mod 2)*2^(2n-k);
a(n) = Sum_{k=0..n} (binomial(2n-k, k) mod 2)*2^k. (End)
a(n) = Sum_{k=0..2n} A106344(2n,k)*2^(2n-k). - Philippe Deléham, Dec 18 2008
MAPLE
a:= n -> add((binomial(n-k+floor(k/2), floor(k/2)) mod 2)*2^k, k=0..n):
map(a, [$0..100]); # Robert Israel, Jan 24 2016
PROG
(PARI) {a(n)=local(B); B=0; for(k=0, n, B=bitxor(B, binomial(n-k+k\2, k\2)%2*2^k)); B}
(PARI) a(n)=sum(k=0, n, binomial(n-k+k\2, k\2)%2*2^k)
CROSSREFS
KEYWORD
eigen,nonn
AUTHOR
Paul D. Hanna, Oct 30 2004
STATUS
approved