A371673 Expansion of g.f. A(x) satisfying [x^(n-1)] A(x)^(n^2) = A000108(n-1) * n^n for n >= 1, where A000108 is the Catalan numbers.

1, 1, 2, 15, 284, 8575, 345460, 17190684, 1012901520, 68810750943, 5291667341342, 454479660308531, 43140290728900554, 4487833959824527910, 508072065566891421336, 62222074620010689986918, 8200304581300850453687880, 1157674985567876068399895997, 174357014524193551292388873190

Offset: 0

Author

Paul D. Hanna, Apr 02 2024

Keywords

• nonn

Comments

Conjecture: a(n) is odd for n > 0 iff n = 2*A003714(k) + 1 for some k, where A003714 is the Fibbinary numbers (integers whose binary representation contains no consecutive ones). See A263075, A263190, and A171791.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 15*x^3 + 284*x^4 + 8575*x^5 + 345460*x^6 + 17190684*x^7 + 1012901520*x^8 + 68810750943*x^9 + 5291667341342*x^10 + ...
The table of coefficients of x^k in A(x)^(n^2) begin:
 n=1: [1,  1,    2,    15,    284,    8575,    345460, ...];
 n=2: [1,  4,   14,    88,   1365,   38304,   1497150, ...];
 n=3: [1,  9,   54,   363,   4410,  105705,   3874824, ...];
 n=4: [1, 16,  152,  1280,  13804,  263408,   8535648, ...];
 n=5: [1, 25,  350,  3875,  43750,  688205,  18352800, ...];
 n=6: [1, 36,  702, 10200, 133389, 1959552,  42189822, ...];
 n=7: [1, 49, 1274, 23863, 376320, 5810763, 108707676, ...];
 ...
where the terms along the main diagonal start as
 [1, 4, 54, 1280, 43750, 1959552, 108707676, ...]
which equals A000108(n-1)*n^n for n >= 1:
 [1, 1*2^2, 2*3^3, 5*4^4, 14*5^5, 42*6^6, 132*7^7, ...].
Compare the above table to the coefficients in 1/(1 - n*x)^n:
 n=1: [1,  1,    1,     1,      1,       1,         1, ...];
 n=2: [1,  4,   12,    32,     80,     192,       448, ...];
 n=3: [1,  9,   54,   270,   1215,    5103,     20412, ...];
 n=4: [1, 16,  160,  1280,   8960,   57344,    344064, ...];
 n=5: [1, 25,  375,  4375,  43750,  393750,   3281250, ...];
 n=6: [1, 36,  756, 12096, 163296, 1959552,  21555072, ...];
 n=7: [1, 49, 1372, 28812, 504210, 7764834, 108707676, ...];
 ...
to see that the main diagonals are equal.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A263075, A263190, A171791.

Cf. A000108, A003714, A118113.

Programs

PARI

{a(n) = my(A=[1], m); for(i=1,n, A=concat(A,0); m=#A;
A[m] = ( m^m*binomial(2*m-1,m-1)/(2*m-1) - Vec( Ser(A)^(m^2) )[m] )/(m^2) );A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) [x^(n-1)] A(x)^(n^2) = n^n * binomial(2*n-1,n-1)/(2*n-1) for n >= 1.

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