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Square root of coefficients of power series: A083352(x)^2 + A083352(x) - 1; term-by-term square root of A083353.
3

%I #21 Nov 19 2017 14:20:37

%S 1,3,6,9,12,15,18,24,33,36,42,54,66,78,87,99,129,144,159,180,198,231,

%T 264,276,300,333,387,420,429,456,531,582,603,591,675,735,816,879,864,

%U 945,1122,1173,1251,1260,1335,1503,1605,1650,1728,1785,1959,2070,2220,2268,2340,2514,2700,2883,2919,2967,3294,3552,3447,3744,3633,4110,4251,4338,4221,4851,4962,5226,5217,5487,5718,6243,6033,6534,6753,6678,7158

%N Square root of coefficients of power series: A083352(x)^2 + A083352(x) - 1; term-by-term square root of A083353.

%C After the first term, each term seems to be a multiple of 3.

%H Paul D. Hanna, <a href="/A083354/b083354.txt">Table of n, a(n) for n = 0..300</a>

%F a(n) = sqrt(A083353(n)).

%e A083352(x) = 1 + 3x + 9x^2 + 9x^3 + 3x^4 + 15x^5 + 33x^6 + ...; thus,

%e A083353(x) = A083352(x)^2 + A083352(x) - 1 = 1 + 9x + 36x^2 + 81x^3 + 144x^4 + 225x^5 + ...

%o (PARI) {for(i=1, 30, A=[1]; print1(1, ", "); for(i=1, 200, A0=concat(A, 0); for(n=1, 100*A[#A], A0[#A0]=n; B=Vec(Ser(A0)^2 + Ser(A0) - 1); if(issquare(B[#B]), print1(sqrtint(B[#B]), ", "); A0[#A0]=n; A=A0; break)))); C=vector(#B,n,sqrtint(B[n]))}

%Y Cf. A083352, A083353.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Apr 26 2003

%E Extended by _Paul D. Hanna_, Nov 19 2017