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A253414
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G.f. satisfies (1+x^2)*g(x) = 1 + x*g(x^2).
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1
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1, 1, -1, 0, 1, -1, -1, 1, 1, 0, -1, -1, 1, 0, -1, 1, 1, 0, -1, 0, 1, -1, -1, 0, 1, 1, -1, -1, 1, 0, -1, 1, 1, 0, -1, 0, 1, -1, -1, 1, 1, 0, -1, -1, 1, 0, -1, 0, 1, 1, -1, 0, 1, -1, -1, 0, 1, 1, -1, -1, 1, 0, -1, 1, 1, 0, -1, 0, 1, -1, -1, 1, 1, 0, -1, -1, 1, 0, -1, 1, 1, 0, -1, 0, 1, -1, -1, 0, 1, 1, -1, -1, 1, 0, -1, 0, 1, 1, -1, 0, 1
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OFFSET
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0
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COMMENTS
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The set of all terms is {-1,0,1}.
Proof. It is not difficult to verify that the function g(x) defined by g(x) = (1 - x^2)*( 1/(1 - x^4) + x/(1 - x^8) + x^3/(1 - x^16) + x^7/(1 - x^32) + ...), for |x| < 1, satisfies the functional equation (1 + x^2)*g(x) = 1 + x*g(x^2).
We thus have g(x) = (1 - x^2)*S(x), where S(x) = sum {n>=0} ( x^(4*n) + x^(8*n + 1) + x^(16*n + 3) + x^(32*n + 7) + ...). It is easy to see that the arithmetic progressions {4*n}, {8*n + 1}, {16*n + 3}, {32*n + 7}, ... are disjoint and hence S(x) has coefficients in {0,1}. It follows that g(x) = (1 - x^2)*S(x) has coefficients in {-1,0,1}. (End)
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LINKS
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FORMULA
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a(2*k) = (-1)^k.
a(1) = 1.
a(2*k+1)=a(k)-a(2*k-1) for k >= 1.
G.f.: g(x) = 1/(1 + x^2) + x/((1 + x^2)*(1 + x^4)) + x^3/((1 + x^2)*(1 + x^4)*(1 + x^8)) + ... = (1 - x^2)*( 1/(1 - x^4) + x/(1 - x^8) + x^3/(1 - x^16) + ... ). - Peter Bala, Jan 05 2015
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EXAMPLE
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(1+x^2)*(1 + x - x^2 + x^4 + ...) = 1 + x*(1 + x^2 - x^4 + ...)
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MAPLE
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N:= 100: # to get a(0) to a(N)
A:= Array(0..N):
for i from 0 to floor(N/2) do
A[2*i] := (-1)^i
od:
A[1]:= 1:
for i from 1 to floor((N-1)/2) do
A[2*i+1]:= A[i] - A[2*i-1]
od:
seq(A[i], i=0..N);
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MATHEMATICA
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nmax = 100; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[(1 + x^2) A[x] - (1 + x A[x^2]) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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