A397594 - OEIS

%I #15 Jul 11 2026 03:53:33

%S 1,3,27,528,16512,705195,37887834,2442170091,183244178880,

%T 15675764998821,1505681872695420,160476900178362804,

%U 18799367838421680984,2401670597893104602064,332382395344108056499278,49549320572895140393543289,7916911857698298814547857788,1349889314907247276753745833113

%N G.f. A(x) such that the sum of the first n coefficients in A(x/n)^n equals 4^(n-1) for n >= 1.

%C Conjecture: for n > 3, a(n) is odd iff n is of the form 2^k-1 or 2^k+1 (k > 1).

%H Paul D. Hanna, <a href="/A397594/b397594.txt">Table of n, a(n) for n = 0..400</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n has the following properties.

%F (1) 4^(n-1) = Sum_{k=0..n-1} [x^k] A(x/n)^n for n >= 1.

%F (2) (4*n)^(n-1) = [x^(n-1)] A(x)^n / (1 - n*x) for n >= 1.

%F (3) (4*n)^(n-1) = [x^(n-1)] B'(x) / (1 - n*B(x)) for n >= 1 where B(x) = x*A(B(x)) and B(x) = Series_Reversion(x/A(x)).

%F (4) [x^(n-1)] B'(x)/(1 - B(x)) = Sum_{k=0..n-1} [x^k] A(x)^n for n >= 1 where B(x) = x*A(B(x)).

%F a(n) = Sum_{k=0..n} 4^k * A397590(n,k) for n >= 0.

%F a(n) ~ (exp(1) - 1/4) * exp(-3*exp(-1)/4) * 4^n * n^(n-1). - _Vaclav Kotesovec_, Jul 11 2026

%e G.f.: A(x) = 1 + 3*x + 27*x^2 + 528*x^3 + 16512*x^4 + 705195*x^5 + 37887834*x^6 + 2442170091*x^7 + 183244178880*x^8 + ...

%e where 4^(n-1) = Sum_{k=0..n-1} [x^k] A(x/n)^n for n >= 1, as illustrated below.

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

%e   n = 1: [1,  3,  27,  528,  16512,  705195, ...];

%e   n = 2: [1,  6,  63, 1218,  36921, 1537974, ...];

%e   n = 3: [1,  9, 108, 2097,  61956, 2519154, ...];

%e   n = 4: [1, 12, 162, 3192,  92427, 3672468, ...];

%e   n = 5: [1, 15, 225, 4530, 129225, 5024808, ...];

%e   n = 6: [1, 18, 297, 6138, 173322, 6606468, ...]; ...

%e The sums of the first n coefficients in A(x/n)^n are

%e   n = 1: 4^0 = 1;

%e   n = 2: 4^1 = 1 + 6/2;

%e   n = 3: 4^2 = 1 + 9/3 + 108/3^2;

%e   n = 4: 4^3 = 1 + 12/4 + 162/4^2 + 3192/4^3;

%e   n = 5: 4^4 = 1 + 15/5 + 225/5^2 + 4530/5^3 + 129225/5^4;

%e   n = 6: 4^5 = 1 + 18/6 + 297/6^2 + 6138/6^3 + 173322/6^4 + 6606468/6^5;

%e   etc.

%e RELATED SERIES.

%e Let B(x) = Series_Reversion(x/A(x)) then

%e B(x) = x + 3*x^2 + 36*x^3 + 798*x^4 + 25845*x^5 + 1101078*x^6 + 58218699*x^7 + 3681384168*x^8 + ...

%e where B(x) = x*A(B(x)) and

%e B'(x)/(1 - B(x)) = 1 + 7*x + 118*x^2 + 3367*x^3 + 133996*x^4 + 6786244*x^5 + 416216494*x^6 + ...

%e the coefficients of which are partial sums of [x^k] A(x)^n:

%e 1 = 1, 7 = (1 + 6), 118 = (1 + 9 + 108), 3367 = (1 + 12 + 162 + 3192), ...

%e The table of coefficients in B'(x)/(1 - n*B(x)) begins

%e   n = 1: [1,  7, 118, 3367, 133996, ...];

%e   n = 2: [1,  8, 130, 3572, 139321, ...];

%e   n = 3: [1,  9, 144, 3813, 145326, ...];

%e   n = 4: [1, 10, 160, 4096, 152161, ...];

%e   n = 5: [1, 11, 178, 4427, 160000, ...]; ...

%e in which the main diagonal equals

%e [1, 8, 144, 4096, 160000, ..., (4*n)^(n-1), ...].

%o (PARI) \\ by definition

%o {a(n,y=4) = my(A=[1], m, V); for(i=1, n, A = concat(A, 0); m=#A; V=Vec( subst(Ser(A)^m, x, x/m) );

%o A[m] = (y^(m-1) - sum(k=1, #V, V[k]) )*m^(m-2) ); GF=Ser(A); polcoef(GF,n)}

%o {upto(n,y=4) = a(n,y); Vec(GF)}

%o upto(25)

%o (PARI) \\ from (4*n)^(n-1) = [x^(n-1)] A(x)^n / (1 - n*x)

%o {a(n,y=4) = my(A=[1], m, V); for(i=1, n, A = concat(A, 0); m=#A;

%o A[m] = ((y*m)^(m-1) - polcoef(Ser(A)^m/(1-m*x),m-1))/m ); GF=Ser(A); polcoef(GF,n)}

%o {upto(n,y=4) = a(n,y); Vec(GF)}

%o upto(25)

%Y Cf. A397590, A397591, A397592, A397593, A397595.

%Y Cf. A180747, A375457.

%K nonn,new

%O 0,2

%A _Paul D. Hanna_, Jul 10 2026