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XOR BINOMIAL transform of A038712.
3

%I #18 Dec 30 2016 16:04:48

%S 1,2,0,4,0,0,0,8,0,0,0,0,0,0,0,16,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,32,0,

%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,64,0,0,0,

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

%N XOR BINOMIAL transform of A038712.

%C See A099884 for the definitions of the XOR BINOMIAL transform and the XOR difference triangle.

%C a(n) = A062383(n+1) - A062383(n). - _Reinhard Zumkeller_, Aug 06 2009

%C A038712 has offset 1, but we need to use offset 0 for the XOR BINOMIAL. - _Michael Somos_, Dec 30 2016

%F a(2^n-1) = 2^n for n>=0 and a(k)=0 otherwise. a(n) = SumXOR_{i=0..n} (C(n, i)mod 2)*A038712(n-i) and SumXOR is summation under XOR.

%F a(n) = A048298(n+1). - _Michael Somos_, Dec 30 2016

%e G.f. = 1 + 2*x + 4*x^3 + 8*x^7 + 16*x^15 + 32*x^31 + 64*x^63 + 128*x^127 + ...

%e XOR difference triangle of A038712 begins:

%e [1],

%e [3,2],

%e [1,2,0],

%e [7,6,4,4],

%e [1,6,0,4,0],

%e [3,2,4,4,0,0],

%e [1,2,0,4,0,0,0],

%e [15,14,12,12,8,8,8,8],...

%e where A038712 is in the leftmost column and A099894 (this sequence) forms the main diagonal.

%e a(1) = 1*1 XOR 0*1 = 1, a(2) = 1*1 XOR 0*3 XOR 1*1 = 0, a(3) = 1*1 XOR 1*3 XOR 1*1 XOR 1*7 = 4 where (1, 3, 1, 7) are the first four terms of A038712. - _Michael Somos_, Dec 30 2016

%t a[ n_] := With[ {m = n+1}, If[ m >=0 && Total[ IntegerDigits[ m, 2]] == 1, m, 0]]; (* _Michael Somos_, Dec 30 2016 *)

%o (PARI) {a(n)=local(B);B=0;for(i=0,n,B=bitxor(B,binomial(n,i)%2*A038712(n-i) ));B}

%o (PARI) {a(n) = my(m = n+1); m * ( m>=0 && hammingweight(m) == 1)}; /* _Michael Somos_, Dec 30 2016 */

%Y Cf. A099884, A038712, A099895, A062383, A048298.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Oct 29 2004