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G.f. satisfies: [x^n] A(x)^(2^m) = 2^m, where m = A007814(n+1) is the highest exponent of 2 dividing n+1, for n>=0.
2

%I #10 Oct 04 2015 22:09:57

%S 1,1,1,-3,1,3,1,21,1,-21,1,255,1,-255,1,478677,1,-478677,1,7152407,1,

%T -7152407,1,-1291535081,1,1291535081,1,-21021866227,1,21021866227,1,

%U 8367123104756933,1,-8367123104756933,1,125486744208053623,1,-125486744208053623,1,-22639240870533272321,1,22639240870533272321,1,-368298497943774746859,1,368298497943774746859,1,-1120119534438107659394201,1,1120119534438107659394201

%N G.f. satisfies: [x^n] A(x)^(2^m) = 2^m, where m = A007814(n+1) is the highest exponent of 2 dividing n+1, for n>=0.

%H Paul D. Hanna, <a href="/A262940/b262940.txt">Table of n, a(n) for n = 0..1024</a>

%F a(2*n) = 1 for n>=0.

%F a(4*n+1) = -a(4*n-1) for n>0.

%F Coefficient of x^k in A(x)^(2^n) equals 2^n at k = m*2^(n+1) + 2^n - 1 for m>=0.

%e G.f.: A(x) = 1 + x + x^2 - 3*x^3 + x^4 + 3*x^5 + x^6 + 21*x^7 + x^8 - 21*x^9 + x^10 + 255*x^11 + x^12 - 255*x^13 + x^14 + 478677*x^15 + x^16 +...

%e The coefficients in A(x)^n begin:

%e n=1: [1, 1, 1, -3, 1, 3, 1, 21, 1, -21, 1, 255, ...];

%e n=2: [1, 2, 3, -4, -3, 2, 19, 44, 29, 2, -153, 512, ...];

%e n=3: [1, 3, 6, -2, -9, -9, 40, 102, 117, 51, -354, 504, ...];

%e n=4: [1, 4, 10, 4, -13, -32, 44, 200, 341, 220, -586, 4, ...];

%e n=5: [1, 5, 15, 15, -10, -64, 10, 310, 775, 755, -679, -1305, ...];

%e n=6: [1, 6, 21, 32, 6, -96, -78, 372, 1443, 2030, -45, -3528, ...];

%e n=7: [1, 7, 28, 56, 42, -112, -224, 302, 2275, 4459, 2520, -5852, ...];

%e n=8: [1, 8, 36, 88, 106, -88, -412, 8, 3075, 8352, 8888, -5568, ...]; ...

%e where the coefficient of x^k in A(x)^(2^m) = 2^m where m = A007814(k+1) for k>=0, like so:

%e [x^0] A(x)^1 = 1;

%e [x^1] A(x)^2 = 2;

%e [x^2] A(x)^1 = 1;

%e [x^3] A(x)^4 = 4;

%e [x^4] A(x)^1 = 1;

%e [x^5] A(x)^2 = 2;

%e [x^6] A(x)^1 = 1;

%e [x^7] A(x)^8 = 8; ...

%o (PARI) {a(n) = local(A=[1,1]); for(k=3,n+1, A=concat(A,0); m=2^valuation(k,2); A[k] = 1 - Vec(Ser(A)^m)[k]/m ); A[n+1]}

%o for(n=0,64,print1(a(n),", "))

%Y Cf. A262939.

%K sign,look

%O 0,4

%A _Paul D. Hanna_, Oct 04 2015