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A032202
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Sequence (a(n): n >= 1) that shifts left 2 places under the "CIK" (necklace, indistinct, unlabeled) transform and satisfies a(1) = a(2) = 1.
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1
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1, 1, 1, 2, 3, 6, 10, 22, 41, 92, 193, 435, 963, 2215, 5051, 11754, 27375, 64381, 151898, 360661, 859149, 2055804, 4934428, 11883930, 28699336, 69497354, 168691424, 410399073, 1000486306, 2443761830, 5979742904, 14656709518
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OFFSET
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1,4
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COMMENTS
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a(n+2) = (1/n)*Sum_{d|n} phi(n/d)*c(d), where c(n) = n*a(n) + Sum_{s=1..n-1} c(s)*a(n-s) with a(1) = a(2) = 1, c(1) = 1, and c(2) = 3.
G.f.: If A(x) = Sum_{n>=1} a(n)*x^n, then Sum_{n>=1} a(n+2)*x^n = -Sum_{n>=1} (phi(n)/n)*log(1-A(x^n)).
The g.f. of the auxiliary sequence (c(n): n>=1) is C(x) = Sum_{n>=1} c(n)*x^n = x*(dA(x)/dx)/(1-A(x)) = x + 3*x^2 + 7*x^3 + 19*x^4 + 46*x^5 + 117*x^6 + 281*x^7 + 707*x^8 + 1717*x^9 + 4288*x^10 + 10583*x^11 + 26401*x^12 + ...
(End)
The first two terms of the sequence must be specified. In general, if the sequence (b(n): n >= 1) is such that (b(n+2): n >= 1) = CIK((b(n): n >= 1)), then b(3) = b(1), b(4) = (1/2)*(b(1)^2 + 2*b(2) + b(1)), b(5) = (b(1)/3)*(b(1)^2 + 3*b(2) + 5), and so on. - Petros Hadjicostas, Jan 01 2019
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LINKS
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MATHEMATICA
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m = 33; a[1] = a[2] = 1; A[_] = 0;
Do[A[x_] = x(a[1] + x a[2] - x Sum[EulerPhi[n] Log[1-A[x^n]]/n, {n, 1, m}]) + O[x]^m // Normal, {m}];
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PROG
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(PARI)
CIK(p, n)={sum(d=1, n, eulerphi(d)/d*log(subst(1/(1+O(x*x^(n\d))-p), x, x^d)))}
seq(n)={my(p=1+O(x)); for(i=1, n\2, p=1+x+x*CIK(x*p, 2*i)); Vec(p+O(x^n))} \\ Andrew Howroyd, Jun 20 2018
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CROSSREFS
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KEYWORD
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nonn,eigen
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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