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FORMULA
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Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
a(n) = (1/n) * Sum_{d|n} phi(d) * ([d==0 mod 12] * (4*S2(n/d+3, 4) - 24*S2(n/d+2, 4) + 44*S2(n/d+1, 4) - 24*S2(n/d, 4)) + [d==6 mod 12] * (3*S2(n/d+3, 4) - 18*S2(n/d+2, 4) + 33*S2(n/d+1, 4) - 18*S2(n/d, 4)) + [d==4 mod 12 | d==8 mod 12] * (3*S2(n/d+3, 4) - 19*S2(n/d+2, 4) + 38*S2(n/d+1, 4) - 24*S2(n/d, 4)) + [d==3 mod 12 | d=9 mod 12] * (2*S2(n/d+3, 4) - 13*S2(n/d+2, 4) + 26*S2(n/d+1, 4) - 15*S2(n/d, 4)) + [d==2 mod 12 | d=10 mod 12] * (2*S2(n/d+3, 4) - 13*S2(n/d+2, 4) + 27*S2[n/d+1,4) - 18*S2(n/d, 4)) + [d mod 12 in {1,5,7,11}] * (S2(n/d+3, 4) - 8*S2(n/d+2, 4) + 20*S2(n/d+1, 4) - 15*S2(n/d, 4))), where S2(n, k) is the Stirling subset number, A008277.
G.f.: 1 - Sum_{d>0} (phi(d) / d) * ([d==0 mod 12] * log(1-4x^d) + [d==6 mod 12] * 3*log(1-4x^d) / 4 + [d==4 mod 12 | d==8 mod 12] * (2*log(1-4x^d) + log(1-x^d)) / 3 + [d==3 mod 12 | d=9 mod 12] * (3*log(1-4x^d) + 2*log(1-2x^d)) / 8 + [d==2 mod 12 | d=10 mod 12] * (5*log(1-4x^d) + 4*log(1-x^d)) / 12 + [d mod 12 in {1,5,7,11}] * (log(1-4x^d) + 6*log(1-2x^d) + 8*log(1-x^d)) / 24).
(End)
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MATHEMATICA
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Adn[d_, n_] := Module[{ c, t1, t2}, t2 = 0; For[c = 1, c <= d, c++, If[Mod[d, c] == 0 , t2 = t2 + (x^c/c)*(E^(c*z) - 1)]]; t1 = E^t2; t1 = Series[t1, {z, 0, n+1}]; Coefficient[t1, z, n]*n!]; Pn[n_] := Module[{ d, e, t1}, t1 = 0; For[d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*Adn[d, n/d]/n]]; t1/(1 - x)]; Pnq[n_, q_] := Module[{t1}, t1 = Series[Pn[n], {x, 0, q+1}] ; Coefficient[t1, x, q]]; a[n_] := Pnq[n, 4]; Table[Print[an = a[n]]; an, {n, 1, 25}] (* Jean-François Alcover, Oct 04 2013, after N. J. A. Sloane's Maple code *)
(* This program uses Gilbert and Riordan's recurrence formula, which they recommend for calculations: *)
Adn[d_, n_] := Adn[d, n] = If[1==n, DivisorSum[d, x^# &],
Expand[Adn[d, 1] Adn[d, n-1] + D[Adn[d, n-1], x] x]];
Table[SeriesCoefficient[DivisorSum[n, EulerPhi[#] Adn[#, n/#] &]
Table[(1/n) DivisorSum[n, EulerPhi[#] Which[Divisible[#, 12], 4 StirlingS2[n/#+3, 4] - 24 StirlingS2[n/#+2, 4] + 44 StirlingS2[n/#+1, 4] - 24 StirlingS2[n/#, 4], Divisible[#, 6], 3 StirlingS2[n/#+3, 4] - 18 StirlingS2[n/#+2, 4] + 33 StirlingS2[n/#+1, 4] - 18 StirlingS2[n/#, 4], Divisible[#, 4], 3 StirlingS2[n/#+3, 4] - 19 StirlingS2[n/#+2, 4] + 38 StirlingS2[n/#+1, 4] - 24 StirlingS2[n/#, 4], Divisible[#, 3], 2 StirlingS2[n/#+3, 4] - 13 StirlingS2[n/#+2, 4] + 26 StirlingS2[n/#+1, 4] - 15 StirlingS2[n/#, 4], Divisible[#, 2], 2 StirlingS2[n/#+3, 4] - 13 StirlingS2[n/#+2, 4] + 27 StirlingS2[n/#+1, 4] - 18 StirlingS2[n/#, 4], True, StirlingS2[n/#+3, 4] - 8 StirlingS2[n/#+2, 4] + 20 StirlingS2[n/#+1, 4] - 15 StirlingS2[n/#, 4]] &], {n, 1, 40}]
mx = 40; Drop[CoefficientList[Series[1 - Sum[(EulerPhi[d] / d) Which[
Divisible[d, 12], Log[1 - 4x^d], Divisible[d, 6],
3 Log[1 - 4x^d] / 4, Divisible[d, 4] ,
(2 Log[1 - 4x^d] + Log[1 - x^d]) / 3, Divisible[d, 3],
(3 Log[1 - 4x^d] + 2 Log[1 - 2x^d]) / 8,
Divisible[d, 2], (5 Log[1 - 4x^d] + 4 Log[1 - x^d]) / 12,
True, (Log[1 - 4x^d] + 6 Log[1 - 2x^d] + 8 Log[1 - x^d]) / 24], {d, 1, mx}], {x, 0, mx}], x], 1]
(End)
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