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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, 1, 2, 15, 284, 8575, 345460, 17190684, 1012901520, 68810750943, 5291667341342, 454479660308531, 43140290728900554, 4487833959824527910, 508072065566891421336, 62222074620010689986918, 8200304581300850453687880, 1157674985567876068399895997, 174357014524193551292388873190
OFFSET
0,3
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.
LINKS
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.
EXAMPLE
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.
PROG
(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), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Apr 02 2024
STATUS
approved