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%I #16 Apr 11 2024 03:25:26
%S 1,1,2,15,284,8575,345460,17190684,1012901520,68810750943,
%T 5291667341342,454479660308531,43140290728900554,4487833959824527910,
%U 508072065566891421336,62222074620010689986918,8200304581300850453687880,1157674985567876068399895997,174357014524193551292388873190
%N Expansion of g.f. A(x) satisfying [x^(n-1)] A(x)^(n^2) = A000108(n-1) * n^n for n >= 1, where A000108 is the Catalan numbers.
%C Conjecture: a(n) is odd for n > 0 iff n = 2*A003714(k) + 1 for some k, where A003714 is the Fibbinary numbers (integers whose binary representation contains no consecutive ones). See A263075, A263190, and A171791.
%H Paul D. Hanna, <a href="/A371673/b371673.txt">Table of n, a(n) for n = 0..300</a>
%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F (1) [x^(n-1)] A(x)^(n^2) = n^n * binomial(2*n-1,n-1)/(2*n-1) for n >= 1.
%F (2) [x^(n-1)] A(x)^(n^2) = [x^(n-1)] 1/(1 - n*x)^n for n >= 1.
%e G.f.: A(x) = 1 + x + 2*x^2 + 15*x^3 + 284*x^4 + 8575*x^5 + 345460*x^6 + 17190684*x^7 + 1012901520*x^8 + 68810750943*x^9 + 5291667341342*x^10 + ...
%e The table of coefficients of x^k in A(x)^(n^2) begin:
%e n=1: [1, 1, 2, 15, 284, 8575, 345460, ...];
%e n=2: [1, 4, 14, 88, 1365, 38304, 1497150, ...];
%e n=3: [1, 9, 54, 363, 4410, 105705, 3874824, ...];
%e n=4: [1, 16, 152, 1280, 13804, 263408, 8535648, ...];
%e n=5: [1, 25, 350, 3875, 43750, 688205, 18352800, ...];
%e n=6: [1, 36, 702, 10200, 133389, 1959552, 42189822, ...];
%e n=7: [1, 49, 1274, 23863, 376320, 5810763, 108707676, ...];
%e ...
%e where the terms along the main diagonal start as
%e [1, 4, 54, 1280, 43750, 1959552, 108707676, ...]
%e which equals A000108(n-1)*n^n for n >= 1:
%e [1, 1*2^2, 2*3^3, 5*4^4, 14*5^5, 42*6^6, 132*7^7, ...].
%e Compare the above table to the coefficients in 1/(1 - n*x)^n:
%e n=1: [1, 1, 1, 1, 1, 1, 1, ...];
%e n=2: [1, 4, 12, 32, 80, 192, 448, ...];
%e n=3: [1, 9, 54, 270, 1215, 5103, 20412, ...];
%e n=4: [1, 16, 160, 1280, 8960, 57344, 344064, ...];
%e n=5: [1, 25, 375, 4375, 43750, 393750, 3281250, ...];
%e n=6: [1, 36, 756, 12096, 163296, 1959552, 21555072, ...];
%e n=7: [1, 49, 1372, 28812, 504210, 7764834, 108707676, ...];
%e ...
%e to see that the main diagonals are equal.
%o (PARI) {a(n) = my(A=[1], m); for(i=1,n, A=concat(A,0); m=#A;
%o A[m] = ( m^m*binomial(2*m-1,m-1)/(2*m-1) - Vec( Ser(A)^(m^2) )[m] )/(m^2) );A[n+1]}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A263075, A263190, A171791.
%Y Cf. A000108, A003714, A118113.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Apr 02 2024