0,2

Define an "XOR difference triangle" for a sequence A by the following process. Start with A in the left-most column. Generate the next column by performing the XOR operation between adjacent terms of the prior column. Repeat this process to generate the XOR difference triangle for A. Further, we define the "XOR BINOMIAL transform" of A as the main diagonal in the XOR difference triangle for A. The XOR BINOMIAL transform is its self-inverse. Let a sequence B be the XOR BINOMIAL transform of A, then we may express B by: B(n) = SumXOR_{k=0..n} A047999(n,k)*A(k), which is equivalent to: B(n) = (C(n,0)mod 2)*A(0) XOR (C(n,1)mod 2)*A(1) XOR (C(n,2)mod 2)*A(2) XOR ... XOR (X(n,n)mod 2)*A(n), where the coefficients are C(n,k)(mod 2) = A047999(n,k).

This sequence is a rearrangement of the numbers which are 2^k times distinct Fermat numbers (numbers of the form 2^(2^m) + 1). This matches the sizes of polygons constructible with compass and straightedge (A003401) up to 2^32+1, which is the first nonprime Fermat number. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Jun 16 2006

T(n, k) = 2^(n-k)*A001317(k). T(n, n) = A001317(n) = SumXOR_{k=0..n} A047999(n, k)*2^k, where SumXOR is the analogue of summation under the binary XOR operation.

The main diagonal equals A001317 (Pascal's triangle mod 2 in decimal):

{1,3,5,15,17,51,85,255,257,771,1285,3855,...},

and defines the XOR BINOMIAL transform of the powers of 2.

Rows begin:

[1],

[2,3],

[4,6,5],

[8,12,10,15],

[16,24,20,30,17],

[32,48,40,60,34,51],

[64,96,80,120,68,102,85],

[128,192,160,240,136,204,170,255],

[256,384,320,480,272,408,340,510,257],

[512,768,640,960,544,816,680,1020,514,771],

[1024,1536,1280,1920,1088,1632,1360,2040,1028,1542,1285],...

(PARI) T(n, k)=local(B); B=0; for(i=0, k, B=bitxor(B, binomial(k, i)%2*2^(n-i))); B

Cf. A047999, A001317.

Cf. A000215, A003401.

Sequence in context: A080997 A151942 A054582 this_sequence A118315 A075159 A095424

Adjacent sequences: A099881 A099882 A099883 this_sequence A099885 A099886 A099887

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Oct 28 2004