A099902 Multiplies by 2 and shifts right under the XOR BINOMIAL transform (A099901).

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

Author

Paul D. Hanna, Oct 30 2004

Keywords

• eigen
• nonn

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

• Robert Israel, Table of n, a(n) for n = 0..3290

Crossrefs

Cf. A099884, A099900, A099901.

Programs

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

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)

Python

def A099902(n): return sum(int(not ~((n<<1)-k)&k)<Chai Wah Wu, Jul 30 2025

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