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A363560 Expansion of g.f. A(x) satisfying A(x)^3 = 1 + x*(A(x) + A(x)^2 + A(x)^9).

%I A363560 #17 Aug 16 2023 10:26:23
%S A363560 1,1,3,18,126,966,7863,66696,583111,5217513,47547405,439777242,
%T A363560 4117802109,38956162023,371795456373,3575401032544,34611064585803,
%U A363560 336998629754631,3298200003722997,32428037256038775,320151289224740949,3172536384239678856,31544584654878015766
%N A363560 Expansion of g.f. A(x) satisfying A(x)^3 = 1 + x*(A(x) + A(x)^2 + A(x)^9).
%C A363560 Compare to: G(x)^3 = 1 + x*(G(x) + G(x)^2 + G(x)^3) holds when G(x) = 1/(1-x).
%C A363560 Conjecture: a(n) == 0 (mod 3) for n > 0 except when n == 1 (mod 7).
%H A363560 Paul D. Hanna, <a href="/A363560/b363560.txt">Table of n, a(n) for n = 0..500</a>
%F A363560 G.f. A(x) = Sum_{n>=0} a(n)*x^n may be defined by the following formulas.
%F A363560 (1) A(x) = 1 + x*(A(x) - A(x)^3 + A(x)^4 - A(x)^6 + A(x)^7).
%F A363560 (2) A(x)^2 = 1 + x*(A(x) + A(x)^2 - A(x)^3 + A(x)^5 - A(x)^6 + A(x)^8) .
%F A363560 (3) A(x)^3 = 1 + x*(A(x) + A(x)^2 + A(x)^9).
%F A363560 (4) A(x)^4 = 1 + x*(A(x) + A(x)^2 + A(x)^4 - A(x)^6 + A(x)^7 + A(x)^10).
%F A363560 (5) A(x)^5 = 1 + x*(A(x) + A(x)^2 + A(x)^4 + A(x)^5 - A(x)^6 + A(x)^8 + A(x)^11).
%F A363560 (6) A(x)^6 = 1 + x*(A(x) + A(x)^2 + A(x)^4 + A(x)^5 + A(x)^9 + A(x)^12).
%F A363560 (7) A(x) = (1/x) * Series_Reversion( x/(1 + Series_Reversion( x/(1 + x*(1+x)^2 + x*(1+x)^5) ) ) ).
%e A363560 G.f.: A(x) = 1 + x + 3*x^2 + 18*x^3 + 126*x^4 + 966*x^5 + 7863*x^6 + 66696*x^7 + 583111*x^8 + 5217513*x^9 + 47547405*x^10 + ...
%e A363560 such that
%e A363560 A(x)^3 = 1 + x*(A(x) + A(x)^2 + A(x)^9).
%e A363560 Also,
%e A363560 A(x) = 1 + x*(A(x) - A(x)^3 + A(x)^4 - A(x)^6 + A(x)^7).
%e A363560 RELATED TABLE.
%e A363560 The table of coefficients in A(x)^n begins:
%e A363560 n=1: [1, 1,  3,  18,  126,   966,   7863,   66696, ...];
%e A363560 n=2: [1, 2,  7,  42,  297,  2292,  18738,  159450, ...];
%e A363560 n=3: [1, 3, 12,  73,  522,  4059,  33354,  284886, ...];
%e A363560 n=4: [1, 4, 18, 112,  811,  6360,  52566,  450888, ...];
%e A363560 n=5: [1, 5, 25, 160, 1175,  9301,  77370,  666780, ...];
%e A363560 n=6: [1, 6, 33, 218, 1626, 13002, 108919,  943524, ...];
%e A363560 n=7: [1, 7, 42, 287, 2177, 17598, 148540, 1293937, ...];
%e A363560 n=8: [1, 8, 52, 368, 2842, 23240, 197752, 1732928, ...];
%e A363560 n=9: [1, 9, 63, 462, 3636, 30096, 258285, 2277756, ...];
%e A363560 ...
%e A363560 from which one can verify the formulas involving powers of A(x).
%e A363560 RELATED SERIES.
%e A363560 Let G(x) = 1 + Series_Reversion( x/(1 + x*(1+x)^2 + x*(1+x)^5) )
%e A363560 where
%e A363560 G(x) = 1 + x + 2*x^2 + 11*x^3 + 61*x^4 + 380*x^5 + 2502*x^6 + 17163*x^7 + 121312*x^8 + 877370*x^9 + 6461765*x^10 + ...
%e A363560 then
%e A363560 A(x) = G(x*A(x)),
%e A363560 and so
%e A363560 A(x) = (1/x) * Series_Reversion( x/G(x) );
%e A363560 thus,
%e A363560 x*A(x) = (A(x) - 1) / (1 + (A(x) - 1)*(A(x)^2 + A(x)^5) )
%e A363560 which is equivalent to
%e A363560 A(x) = 1 + x*(A(x) - A(x)^3 + A(x)^4 - A(x)^6 + A(x)^7).
%e A363560 TERMS MODULO 3.
%e A363560 It appears that a(n) == 0 (mod 3) for n > 0 except when n == 1 (mod 7).
%e A363560 The residues of a(7*k + 1) modulo 3, for k >= 0, begin
%e A363560 a(7*k + 1) (mod 3) = [1, 1, 1, 1, 0, 2, 1, 0, 0, 1, 1, 2, 0, 0, 2, 2, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, ...].
%o A363560 (PARI) {a(n) = my(A=1+x); for(i=1, n, A = (1 + x*(A + A^2 + A^9) +x*O(x^n))^(1/3) ); polcoeff(A, n)}
%o A363560 for(n=0, 30, print1(a(n), ", "))
%Y A363560 Cf. A300048, A161634.
%K A363560 nonn
%O A363560 0,3
%A A363560 _Paul D. Hanna_, Aug 12 2023