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

%I #8 Mar 01 2022 07:29:14

%S 1,1,4,99,4308,283575,25673506,3037201300,454292561764,83892724766919,

%T 18770366957242024,5009355553568692667,1573355916010921955326,

%U 574838333958720246640420,241825797204483146358016660,116088470252288250153307625160

%N G.f. A(x) satisfies: [x^n] A(x)^(2*n+1) = [x^n] 2*A(x)^(n+1), for n > 1.

%e G.f.: A(x) = 1 + x + 4*x^2 + 99*x^3 + 4308*x^4 + 283575*x^5 + 25673506*x^6 + 3037201300*x^7 + 454292561764*x^8 + ...

%e where the coefficient of x^n in A(x)^(2*n+1) equals twice the coefficient of x^n in A(x)^(n+1), for n > 1.

%e To illustrate, we may inspect the table of coefficients of x^k in A(x)^n, which begins:

%e n=1: [1, 1, 4, 99, 4308, 283575, 25673506, ...];

%e n=2: [1, 2, 9, 206, 8830, 576558, 51958427, ...];

%e n=3: [1, 3, 15, 322, 13578, 879294, 78870127, ...];

%e n=4: [1, 4, 22, 448, 18565, 1192144, 106424462, ...];

%e n=5: [1, 5, 30, 585, 23805, 1515486, 134637800, ...];

%e n=6: [1, 6, 39, 734, 29313, 1849716, 163527042, ...];

%e n=7: [1, 7, 49, 896, 35105, 2195249, 193109644, ...];

%e n=8: [1, 8, 60, 1072, 41198, 2552520, 223403640, ...];

%e n=9: [1, 9, 72, 1263, 47610, 2921985, 254427666, ...];

%e n=10: [1, 10, 85, 1470, 54360, 3304122, 286200985, ...];

%e n=11: [1, 11, 99, 1694, 61468, 3699432, 318743513, ...];

%e n=12: [1, 12, 114, 1936, 68955, 4108440, 352075846, ...];

%e n=13: [1, 13, 130, 2197, 76843, 4531696, 386219288, ...];

%e ...

%e to see the following pattern:

%e [x^2] A(x)^5 = 30 = [x^2] 2*A(x)^3 = 2*15;

%e [x^3] A(x)^7 = 896 = [x^3] 2*A(x)^4 = 2*448;

%e [x^4] A(x)^9 = 47610 = [x^4] 2*A(x)^5 = 2*23805;

%e [x^5] A(x)^11 = 3699432 = [x^5] 2*A(x)^6 = 2*1849716;

%e [x^6] A(x)^13 = 386219288 = [x^6] 2*A(x)^7 = 2*193109644;

%e ...

%o (PARI) {a(n) = my(A=[1,1],m); for(i=1,n, A = concat(A,0); m = #A-1;

%o A[#A] = polcoeff( Ser(A)^(2*m+1) - 2*Ser(A)^(m+1), m) );A[n+1]}

%o for(n=0,20,print1(a(n),", "))

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 28 2022