A397593 - OEIS

%I #14 Jul 11 2026 03:49:14

%S 1,2,14,212,5060,163612,6627896,321431160,18124740188,1164431516480,

%T 83964054624948,6716502875164072,590440591362465196,

%U 56598091771350294496,5876895519035724906276,657270233781923280277160,78784141824808570443009244,10077258172565896291161041472,1370190694994211436654204353668

%N G.f. A(x) such that the sum of the first n coefficients in A(x/n)^n equals 3^(n-1) for n >= 1.

%H Paul D. Hanna, <a href="/A397593/b397593.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) 3^(n-1) = Sum_{k=0..n-1} [x^k] A(x/n)^n for n >= 1.

%F (2) (3*n)^(n-1) = [x^(n-1)] A(x)^n / (1 - n*x) for n >= 1.

%F (3) (3*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} 3^k * A397590(n,k) for n >= 0.

%F a(n) ~ (exp(1) - 1/3) * exp(-2*exp(-1)/3) * 3^n * n^(n-1). - _Vaclav Kotesovec_, Jul 11 2026

%e G.f.: A(x) = 1 + 2*x + 14*x^2 + 212*x^3 + 5060*x^4 + 163612*x^5 + 6627896*x^6 + 321431160*x^7 + 18124740188*x^8 + ...

%e where 3^(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,  2,  14,  212,  5060,  163612, ...];

%e   n = 2: [1,  4,  32,  480, 11164,  353400, ...];

%e   n = 3: [1,  6,  54,  812, 18480,  573084, ...];

%e   n = 4: [1,  8,  80, 1216, 27192,  826832, ...];

%e   n = 5: [1, 10, 110, 1700, 37500, 1119292, ...];

%e   n = 6: [1, 12, 144, 2272, 49620, 1455624, ...]; ...

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

%e   n = 1: 3^0 = 1;

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

%e   n = 3: 3^2 = 1 + 6/3 + 54/3^2;

%e   n = 4: 3^3 = 1 + 8/4 + 80/4^2 + 1216/4^3;

%e   n = 5: 3^4 = 1 + 10/5 + 110/5^2 + 1700/5^3 + 37500/5^4;

%e   n = 6: 3^5 = 1 + 12/6 + 144/6^2 + 2272/6^3 + 49620/6^4 + 1455624/6^5;

%e   etc.

%e RELATED SERIES.

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

%e B(x) = x + 2*x^2 + 18*x^3 + 304*x^4 + 7500*x^5 + 242604*x^6 + 9707376*x^7 + 463312384*x^8 + ...

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

%e B'(x)/(1 - B(x)) = 1 + 5*x + 61*x^2 + 1305*x^3 + 39321*x^4 + 1507673*x^5 + 69860085*x^6 + ...

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

%e 1 = 1, 5 = (1 + 4), 61 = (1 + 6 + 54), 1305 = (1 + 8 + 80 + 1216), ...

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

%e   n = 1: [1, 5,  61, 1305, 39321, ...];

%e   n = 2: [1, 6,  70, 1416, 41436, ...];

%e   n = 3: [1, 7,  81, 1555, 43941, ...];

%e   n = 4: [1, 8,  94, 1728, 46956, ...];

%e   n = 5: [1, 9, 109, 1941, 50625, ...]; ...

%e in which the main diagonal equals

%e [1, 6, 81, 1728, 50625, ..., (3*n)^(n-1), ...].

%o (PARI) \\ by definition

%o {a(n,y=3) = 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=3) = a(n,y); Vec(GF)}

%o upto(25)

%o (PARI) \\ from (3*n)^(n-1) = [x^(n-1)] A(x)^n / (1 - n*x)

%o {a(n,y=3) = 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=3) = a(n,y); Vec(GF)}

%o upto(25)

%Y Cf. A397590, A397591, A397592, A397594, A397595.

%Y Cf. A180747, A375457.

%K nonn,new

%O 0,2

%A _Paul D. Hanna_, Jul 10 2026