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A261594 G.f.: sqrt( Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k) * 2^(k*(k-1)) ).

%I A261594 #6 Aug 26 2015 03:19:53
%S A261594 1,1,3,37,2149,532611,539508291,2202251249193,36044200375109487,
%T A261594 2361471528989758715269,618991271516919971774301613,
%U A261594 649043297118583276751832395970903,2722266074808493870871954767765560237289,45671958833739479081570180837023756023304348531,3064991675467024774224369897734145197065069681513495767
%N A261594 G.f.: sqrt( Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k) * 2^(k*(k-1)) ).
%C A261594 a(k) = 1 (mod 3) iff k = 9*A005836(n) + [0,1,3,4] for n>=0, with a(k) = 0 (mod 3) elsewhere, where A005836 lists numbers n whose base 3 representation contains no 2 (conjecture).
%e A261594 G.f.: A(x) = 1 + x + 3*x^2 + 37*x^3 + 2149*x^4 + 532611*x^5 + 539508291*x^6 +...
%e A261594 where
%e A261594 A(x)^2 = 1 + 2*x + 7*x^2 + 80*x^3 + 4381*x^4 + 1069742*x^5 + 1080096067*x^6 +...+ A135756(n)*x^n +...
%e A261594 such that
%e A261594 A135756(n) = Sum_{k=0..n} binomial(n,k) * 2^(k*(k-1)).
%e A261594 The residue of the terms modulo 3 begin:
%e A261594 [1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, ...]
%e A261594 where a(k) appears to be congruent to 1 (mod 3) at k = 9*A005836(n) + [0,1,3,4] for n>=0, and congruent to zeros elsewhere.
%o A261594 (PARI) {a(n) = polcoeff( sum(m=0,n, x^m*sum(k=0, m, binomial(m, k)*2^(k*(k-1))) +x*O(x^n))^(1/2), n)}
%o A261594 for(n=0,20,print1(a(n),", "))
%Y A261594 Cf. A135756.
%K A261594 nonn
%O A261594 0,3
%A A261594 _Paul D. Hanna_, Aug 25 2015