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.
%I A251690 #38 Jan 05 2015 23:01:20
%S A251690 1,-1,-2,-2,0,-1,0,-3,0,-3,-3,0,-3,-2,-3,-1,-2,0,-1,-2,0,0,-2,0,0,0,
%T A251690 -2,0,-3,0,-2,0,-1,0,-3,-2,-1,-1,-3,-1,0,-2,-2,-3,-1,-3,-1,-1,0,0,-1,
%U A251690 -1,-3,-3,-1,0,-1,0,-2,0,-3,-3,-3,-2,-1,-2,-1,-2,-2,-2,-3,-1,-3,-1,-3,-1,0,-2,-2,-2,-1,-1,-2,-2,0,-3,-3,-2,-3,-1,-3,-2,0,0,0,-2,-2,-2,-2,-3,-3,0,-2,0,-3,-1,0,-2,-3,-1,-3,0,-1,0,-2,-1,-1,-3,-3,-1,-3,-3,0,-3,-2,-3,-2
%N A251690 G.f. A(x) satisfies the condition that G(A(x)) is a power series in x consisting entirely of positive integer coefficients such that G(A(x) - x^k) has negative coefficients for k>0, where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
%C A251690 Compare to the similar series F(x) for the Catalan function C(x) = 1 + x*C(x)^2, where C(F(x)) consists entirely of positive integer coefficients such that C(F(x) - x^k) has negative coefficients for k>0; in which case F(x) = (x+x^2) - (x+x^2)^2, and C(F(x)) = 1/(1-x-x^2).
%C A251690 It seems that a(n) lies in [-3,0] for n>1; does this continue to hold?
%H A251690 Paul D. Hanna, <a href="/A251690/b251690.txt">Table of n, a(n) for n = 1..200</a>
%e A251690 G.f.: A(x) = x - x^2 - 2*x^3 - 2*x^4 - x^6 - 3*x^8 - 3*x^10 - 3*x^11 - 3*x^13 - 2*x^14 - 3*x^15 - x^16 - 2*x^17 - x^19 - 2*x^20 - 2*x^23 - 2*x^27 - 3*x^29 +...
%e A251690 Given G(x) = 1 + x*G(x)^3, which begins
%e A251690 G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 +...
%e A251690 then
%e A251690 G(A(x)) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 17*x^5 + 36*x^6 + 78*x^7 + 169*x^8 + 370*x^9 + 813*x^10 + 1793*x^11 + 3971*x^12 +...+ A251691(n)*x^n +...
%e A251690 consists entirely of positive integer coefficients such that
%e A251690 G(A(x) - x^k) has negative coefficients for k>0, as illustrated by:
%e A251690 k=1: G(A(x) - x) = 1 - x^2 - 2*x^3 + x^4 + 12*x^5 + 11*x^6 - 48*x^7 +...
%e A251690 k=2: G(A(x) - x^2) = 1 + x + x^2 - 2*x^3 - 19*x^4 - 83*x^5 - 267*x^6 +...
%e A251690 k=3: G(A(x) - x^3) = 1 + x + 2*x^2 + 3*x^3 + 2*x^4 - 13*x^5 - 97*x^6 - 471*x^7 +...
%e A251690 k=4: G(A(x) - x^4) = 1 + x + 2*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 6*x^6 - 58*x^7 - 413*x^8 +...
%e A251690 k=5: G(A(x) - x^5) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 30*x^6 + 48*x^7 + 33*x^8 - 215*x^9 - 1632*x^10 +...
%e A251690 k=6: G(A(x) - x^6) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 17*x^5 + 35*x^6 + 72*x^7 + 139*x^8 + 234*x^9 + 228*x^10 - 655*x^11 - 6102*x^12 +...
%e A251690 etc.
%e A251690 The position of the first negative term in G(A(x) - x^n), n>=1, begins:
%e A251690 [2, 3, 5, 7, 9, 11, 13, 16, 18, 20, 23, 25, 28, 30, 32, 35, 37, 40, 42, 45, 47, 50, 52, 55, 57, 59, 62, 64, 67, 69, 72, 74, 77, 79, 82, 84, 87, 89, 92, 94, 97, 99, 102, 104, 106, 109, 111, 114, 116, 119, ...];
%e A251690 does the ratio of the position of the first negative term in G(A(x)-x^n) divided by n approach some constant?
%o A251690 (PARI) /* Prints initial N terms: */
%o A251690 N=100;
%o A251690 /* G(x) = 1 + x*G(x)^3 is the g.f. of A001764: */
%o A251690 {G=1+serreverse(x/(1+x +x*O(x^(3*N+10)))^3);}
%o A251690 /* Print terms as you build vector A, then print A at the end: */
%o A251690 {A=[1,-1];print1("1, -1, ");
%o A251690 for(l=1,N,A=concat(A,-4);
%o A251690 for(i=1,4,A[#A]=A[#A]+1;
%o A251690 V=Vec(subst(G,x,x*truncate(Ser(A)) +O(x^floor(3*#A+1)) ));
%o A251690 if((sign(V[3*#A])+1)/2==1, print1(A[#A],", "); break));); A}
%Y A251690 Cf. A251691, A001764.
%K A251690 sign
%O A251690 1,3
%A A251690 _Paul D. Hanna_, Dec 31 2014