A375448 - OEIS

%I #12 Sep 13 2024 06:36:56

%S 1,3,6,33,357,5283,96534,2067312,50345955,1367512761,40875976152,

%T 1331343423234,46892513148468,1775323414999818,71885746640828286,

%U 3100014000785085216,141857882269044077865,6866221878372182554395,350521791594556907681202,18824690900373744731703396

%N Expansion of the g.f. A(x) with the property that the sum of the first n coefficients in A(x/n)^n equals n^2 for n >= 1.

%H Paul D. Hanna, <a href="/A375448/b375448.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) satisfies [x^n] x*B'(x/n) / (1 - n*B(x/n)) = n^2 for n >= 1, where B(x/A(x)) = x and B(x) is the g.f. of A375449.

%F a(n) ~ c * n^(n+1), where c = 0.5698891168602..., conjecture: c = (exp(1)-1)/exp(3*exp(-1)). - _Vaclav Kotesovec_, Sep 13 2024

%e G.f.: A(x) = 1 + 3*x + 6*x^2 + 33*x^3 + 357*x^4 + 5283*x^5 + 96534*x^6 + 2067312*x^7 + 50345955*x^8 + 1367512761*x^9 + 40875976152*x^10 + ...

%e The defining property of g.f. A(x) is described below.

%e The table of coefficients in A(x)^n begins:

%e   n=1: [1,  3,   6,   33,   357,   5283,   96534, ...];

%e   n=2: [1,  6,  21,  102,   948,  13104,  230139, ...];

%e   n=3: [1,  9,  45,  234,  1935,  24678,  414234, ...];

%e   n=4: [1, 12,  78,  456,  3561,  41868,  667746, ...];

%e   n=5: [1, 15, 120,  795,  6150,  67428, 1017540, ...];

%e   n=6: [1, 18, 171, 1278, 10107, 105246, 1501578, ...];

%e   n=7: [1, 21, 231, 1932, 15918, 160587, 2172807, ...];

%e   ...

%e in which the sum of the first n coefficients in A(x/n)^n equals n^2, as illustrated by

%e   1 = 1;

%e   4 = 1 + 6/2;

%e   9 = 1 + 9/3 + 45/3^2;

%e   16 = 1 + 12/4 + 78/4^2 + 456/4^3;

%e   25 = 1 + 15/5 + 120/5^2 + 795/5^3 + 6150/5^4;

%e   36 = 1 + 18/6 + 171/6^2 + 1278/6^3 + 10107/6^4 + 105246/6^5;

%e   49 = 1 + 21/7 + 231/7^2 + 1932/7^3 + 15918/7^4 + 160587/7^5 + 2172807/7^6;

%e   ...

%e RELATED SERIES.

%e Let B(x) be the series reversion of x/A(x), B(x/A(x)) = x, then

%e B(x) = x + 3*x^2 + 15*x^3 + 114*x^4 + 1230*x^5 + 17541*x^6 + 310401*x^7 + 6502368*x^8 + ... + A375449(n)*x^n + ...

%e Further, let C(x) = x*B'(x)/(1 - B(x)) = x + 7*x^2 + 55*x^3 + 547*x^4 + 7081*x^5 + 116821*x^6 + 2351497*x^7 + 55390315*x^8 + ...

%e then the coefficient of x^n in C(x) equals the sum of the initial n terms of A(x)^n for n >= 1; 1 = 1, 7 = 1 + 6, 55 = 1 + 9 + 45, 547 = 1 + 12 +  78 + 456, 7081 = 1 + 15 + 120 + 795 + 6150, etc.

%o (PARI) {a(n) = my(A=[1], m, V); for(i=0, n, A = concat(A, 0); m=#A; V=Vec( subst(Ser(A)^m, x, x/m) );

%o A[m] = (m^2 - sum(k=1, #V, V[k]) )*m^(m-2) ); H=A; A[n+1]}

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

%Y Cf. A375449, A375457.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Sep 12 2024