cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A264228 G.f. A(x) satisfies: A(x)^3 = A( x^3/(1-3*x) ), with A(0) = 0.

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%I A264228 #29 Mar 11 2024 10:01:36
%S A264228 1,1,2,5,13,35,97,274,785,2275,6656,19630,58295,174175,523238,1579584,
%T A264228 4789919,14584723,44577799,136732988,420784888,1298937282,4021383654,
%U A264228 12483820395,38853994422,121220646116,379062880051,1187912517953,3730305167438,11736596024002,36994041916973,116807229667919,369415244627269,1170113816365089
%N A264228 G.f. A(x) satisfies: A(x)^3 = A( x^3/(1-3*x) ), with A(0) = 0.
%C A264228 Radius of convergence is r = (sqrt(13) - 3)/2, where r = r^3/(1-3*r), with A(r) = 1.
%C A264228 Compare to a g.f. M(x) of Motzkin numbers: M(x)^2 = M(x^2/(1-2*x)) where M(x) = (1-x - sqrt(1-2*x-3*x^2))/(2*x).
%H A264228 Paul D. Hanna, <a href="/A264228/b264228.txt">Table of n, a(n) for n = 1..600</a>
%F A264228 G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.
%F A264228 (1) A(x)^3 = A( x^3/(1-3*x) ).
%F A264228 (2) A( x/(1+3*x) )^3 = A( x^3/(1+3*x)^2 ). - _Paul D. Hanna_, Mar 25 2023
%F A264228 (3) A( x/(1+x+x^2) )^3 = A( x^3/(1-x^3)^2 ). - _Paul D. Hanna_, Mar 11 2024
%e A264228 G.f.: A(x) = x + x^2 + 2*x^3 + 5*x^4 + 13*x^5 + 35*x^6 + 97*x^7 + 274*x^8 + 785*x^9 + 2275*x^10 + 6656*x^11 + 19630*x^12 + 58295*x^13 + 174175*x^14 + ...
%e A264228 where A(x)^3 = A( x^3/(1-3*x) ).
%e A264228 RELATED SERIES.
%e A264228 A(x)^3 = x^3 + 3*x^4 + 9*x^5 + 28*x^6 + 87*x^7 + 270*x^8 + 839*x^9 + 2610*x^10 + 8127*x^11 + 25331*x^12 + 79035*x^13 + 246852*x^14 + 771808*x^15 + ...
%e A264228 A( x/(1+x+x^2) ) = x + x^4 + 2*x^7 + 6*x^10 + 22*x^13 + 88*x^16 + 367*x^19 + 1570*x^22 + 6843*x^25 + 30271*x^28 + 135530*x^31 + 612852*x^34 + 2794412*x^37 + 12832472*x^40 + ...
%e A264228 Let B(x) = x/Series_Reversion(A(x)), then A(x) = x*B(A(x)), where
%e A264228 B(x) = 1 + x + x^2 + x^3 - x^5 - x^6 + 2*x^8 + 3*x^9 - 6*x^11 - 9*x^12 + 20*x^14 + 30*x^15 - 71*x^17 - 110*x^18 + 267*x^20 + 419*x^21 - 1041*x^23 + ...
%e A264228 Let C0(x) and C2(x) be series trisections of B(x), B(x) = C0(x) + x + C2(x):
%e A264228 C0(x) = 1 + x^3 - x^6 + 3*x^9 - 9*x^12 + 30*x^15 - 110*x^18 + 419*x^21 - 1648*x^24 + 6652*x^27 - 27369*x^30 + 114384*x^33 - 484276*x^36 + ...
%e A264228 C2(x) = x^2 - x^5 + 2*x^8 - 6*x^11 + 20*x^14 - 71*x^17 + 267*x^20 - 1041*x^23 + 4168*x^26 - 17047*x^29 + 70902*x^32 + ... + (-1)^(n-1)*A370446(n)*x^(3*n-1) + ...
%e A264228 then C0(x) = x^2/C2(x).
%o A264228 (PARI)  {a(n) = my(A=x); for(i=1, n, A = ( subst(A, x, x^3/(1-3*x +x*O(x^n))) )^(1/3) ); polcoeff(A, n)}
%o A264228 for(n=1, 40, print1(a(n), ", "))
%Y A264228 Cf. A264229, A264230, A361763, A370440, A370446.
%K A264228 nonn
%O A264228 1,3
%A A264228 _Paul D. Hanna_, Nov 08 2015