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A171791 G.f. A(x) satisfies: [x^n] A(x)^((n+1)^2) = 0 for n>1 with a(0)=a(1)=1.

%I A171791 #28 Oct 22 2020 16:54:36
%S A171791 1,1,-4,25,-194,1603,-15264,122316,-1897710,-8845133,-1169435932,
%T A171791 -52853978047,-3193246498792,-205347570309000,-14534295599537024,
%U A171791 -1115833257773950536,-92445637289048967654,-8219735646409095418617
%N A171791 G.f. A(x) satisfies: [x^n] A(x)^((n+1)^2) = 0 for n>1 with a(0)=a(1)=1.
%C A171791 It appears that for k>0, a(k) is odd iff k = 2*A003714(n)+1 for n>=0, where A003714 is the fibbinary numbers (integers whose binary representation contains no consecutive ones); this is true for at least the first 520 terms. [See also A263190 and A263075.] - _Paul D. Hanna_, Oct 09 2013
%C A171791 Observation of Paul D. Hanna is true for at least the first 1028 terms. - _Sean A. Irvine_, Apr 25 2014
%H A171791 Paul D. Hanna, <a href="/A171791/b171791.txt">Table of n, a(n) for n = 0..200</a>
%F A171791 The g.f. A(x) satisfies the following relations.
%F A171791 (1) 0 = [x^(n-1)] A(x)^(n^2), for n > 1.
%F A171791 (2) 0 = [x^(n-1)] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), for n > 1. - _Paul D. Hanna_, Oct 22 2020
%F A171791 (3) 0 = [x^n] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), for n > 0. - _Paul D. Hanna_, Oct 22 2020
%e A171791 G.f.: A(x) = 1 + x - 4*x^2 + 25*x^3 - 194*x^4 + 1603*x^5 +...
%e A171791 The coefficients in the square powers of g.f. A(x) begin:
%e A171791 A^1:  [1,  1,   -4,    25,   -194,    1603,   -15264,    122316, ...];
%e A171791 A^4:  [1,  4,  -10,    56,   -427,    3360,   -33546,    218880, ...];
%e A171791 A^9:  [1,  9,    0,    21,   -252,    1701,   -25992,     -2970, ...];
%e A171791 A^16: [1, 16,   56,     0,    -84,    -784,   -18656,   -384896, ...];
%e A171791 A^25: [1, 25,  200,   525,      0,   -2695,   -38600,   -878150, ...];
%e A171791 A^36: [1, 36,  486,  3000,   7821,       0,  -101322,  -1916352, ...];
%e A171791 A^49: [1, 49,  980, 10241,  58898,  170079,        0,  -4515000, ...];
%e A171791 A^64: [1, 64, 1760, 27136, 256048, 1500352,  4979712,         0, ...];
%e A171791 A^81: [1, 81, 2916, 61425, 838026, 7720839, 48097152, 184870512, 0,...]; ...
%e A171791 Note how the coefficient of x^n in A(x)^((n+1)^2) = 0 for n>1.
%e A171791 ALTERNATE RELATION.
%e A171791 The coefficients in A(x)^(n^2) * (1 - n*x*A(x)'/A(x)) begin:
%e A171791 n=1: [1, 0, 4, -50, 582, -6412, 76320, -733896, 13283970, ...];
%e A171791 n=2: [1, 2, 0, -28, 427, -5040, 67092, -547200, 15539502, ...];
%e A171791 n=3: [1, 6, 0, 0, 84, -1134, 25992, 3960, 13172355, ...];
%e A171791 n=4: [1, 12, 28, 0, 0, 196, 9328, 288672, 13426530, ...];
%e A171791 n=5: [1, 20, 120, 210, 0, 0, 7720, 351260, 15775425, ...];
%e A171791 n=6: [1, 30, 324, 1500, 2607, 0, 0, 319392, 17452530, ...];
%e A171791 n=7: [1, 42, 700, 5852, 25242, 48594, 0, 0, 15518020, ...];
%e A171791 n=8: [1, 56, 1320, 16960, 128024, 562632, 1244928, 0, 0, ...];
%e A171791 n=9: [1, 72, 2268, 40950, 465570, 3431484, 16032384, 41082336, 0, 0, ...]; ...
%e A171791 in which the two adjacent diagonals above the main diagonal are all zeros after initial terms, illustrating that
%e A171791 (1) 0 = [x^(n-1)] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), and
%e A171791 (2) 0 = [x^n] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), for n > 0.
%o A171791 (PARI) {a(n) = local(A=[1,1]); for(m=3,n+1, A=concat(A,0); A[ #A]=-Vec(Ser(A)^(m^2))[m]/m^2); A[n+1]}
%o A171791 for(n=0, 20, print1(a(n), ", "))
%Y A171791 Cf. A263190, A263075, A229044, A185072, A229041, A230218, A292877.
%K A171791 sign
%O A171791 0,3
%A A171791 _Paul D. Hanna_, Jan 24 2010