%I #11 Sep 13 2024 10:53:35
%S 1,3,15,114,1230,17541,310401,6502368,156119778,4202894190,
%T 124940121966,4056054468282,142598505603138,5393588598473229,
%U 218309634723396555,9413799845705265408,430826174378500622916,20856810739070097320814,1064963403446152512517140,57204475911565427208990300
%N Expansion of g.f. A(x) satisfying [x^n] x*A'(x) / (1 - n*A(x)) = n^(n+1) for n >= 1.
%C Conjecture: for n > 1, a(n) is odd iff n is of the form 2^k-2 or 2^k-1 for some k > 1.
%C Compare to [x^n] x*G'(x) / (1 - n*G(x)) = ((n+1)^n - 1)/n for n >= 1 when G(x) = x/(1-x).
%H Paul D. Hanna, <a href="/A375449/b375449.txt">Table of n, a(n) for n = 1..300</a>
%F a(n) ~ (1 - exp(-1)) * n^n. - _Vaclav Kotesovec_, Sep 13 2024
%e G.f.: A(x) = x + 3*x^2 + 15*x^3 + 114*x^4 + 1230*x^5 + 17541*x^6 + 310401*x^7 + 6502368*x^8 + 156119778*x^9 + 4202894190*x^10 + ...
%e The table of coefficients of x^k in x*A'(x)/(1 - n*A(x)) begins
%e n=1: [1, 7, 55, 547, 7081, 116821, 2351497, ...];
%e n=2: [1, 8, 67, 668, 8356, 132260, 2577541, ...];
%e n=3: [1, 9, 81, 825, 10101, 153387, 2874537, ...];
%e n=4: [1, 10, 97, 1024, 12466, 182698, 3278227, ...];
%e n=5: [1, 11, 115, 1271, 15625, 223481, 3840817, ...];
%e n=6: [1, 12, 135, 1572, 19776, 279936, 4636017, ...];
%e n=7: [1, 13, 157, 1933, 25141, 357295, 5764801, ...];
%e ...
%e in which the main diagonal equals n^(n+1) for n >= 1.
%e RELATED SERIES.
%e x*A'(x)/(1 - A(x)) = x + 7*x^2 + 55*x^3 + 547*x^4 + 7081*x^5 + 116821*x^6 + 2351497*x^7 + 55390315*x^8 + ...
%e Let B(x) satisfy A(x/B(x)) = x, then B(x) begins
%e B(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 + ... + A375448(n)*x^n + ...
%e Below we demonstrate some properties of this related series.
%e The table of coefficients in B(x)^n begins:
%e n=1: [1, 3, 6, 33, 357, 5283, ...];
%e n=2: [1, 6, 21, 102, 948, 13104, ...];
%e n=3: [1, 9, 45, 234, 1935, 24678, ...];
%e n=4: [1, 12, 78, 456, 3561, 41868, ...];
%e n=5: [1, 15, 120, 795, 6150, 67428, ...];
%e ...
%e in which the sum of the first n coefficients in B(x)^n equals the coefficients in x*A'(x)/(1 - A(x)) like so: 1 = 1, 7 = 1 + 6, 55 = 1 + 9 + 45, 547 = 1 + 12 + 78 + 456, 7081 = 1 + 15 + 120 + 795 + 6150, etc.
%e Also, the sum of the first n coefficients in B(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 ...
%o (PARI) {a(n) = my(A=[0, 1], Ax=x, m); for(i=1, n, A = concat(A, 0); Ax = Ser(A); m=#A-1; A[#A] = (m^(m+1) - polcoef( x*Ax' / (1 - m*Ax), m) )/m ); A[n+1]}
%o for(n=1, 20, print1(a(n), ", "))
%Y Cf. A375448, A375452.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Sep 12 2024