<think>
Alright, so I've come across this problem where I'm supposed to find the determinant of an n-by-n matrix A. Each entry of the matrix is defined by c_{i+j-1}, where c_k is a term from a sequence given by a generating function.

First, let me parse the problem statement:

We have a sequence c_0, c_1, c_2, ... such that the generating function is (1 - 3x - sqrt(1 - 14x + 9x^2)) / 4. This equals the sum from k=0 to infinity of c_k x^k for sufficiently small x.

Then, we're given an n-by-n matrix A where each entry at position (i,j) is c_{i+j-1}. So, it's a Hankel matrix because each anti-diagonal is constant. I remember that Hankel matrices have some special properties and sometimes their determinants can be found using generating functions or continued fractions.

The goal is to find the determinant of this matrix A for any positive integer n.

Hmm. Let me think about how generating functions relate to determinants of Hankel matrices. There's something called the Hankel determinant, which is exactly this kind of determinant. For sequences defined by a generating function, sometimes there are formulas or recurrence relations that can help compute these determinants.

Given that the generating function is (1 - 3x - sqrt(1 - 14x + 9x^2)) / 4, I should first try to understand this generating function. Maybe it's related to some known combinatorial structure or orthogonal polynomials?

Let me compute the first few terms of the sequence c_k to see if I recognize a pattern.

First, let's write down the generating function:

G(x) = (1 - 3x - sqrt(1 - 14x + 9x^2)) / 4

I can try expanding this square root using a binomial expansion or maybe find a recursive formula for c_k. Alternatively, perhaps I can express G(x) as a continued fraction or relate it to known generating functions.

Wait, another thought: The expression under the square root is 1 - 14x + 9x^2. Let me see if that's a perfect square or something close.

Let’s compute (a - bx)^2 = a^2 - 2abx + b^2 x^2. Comparing with 1 - 14x + 9x^2, we have:

- The constant term is 1, so a^2 = 1 ⇒ a = 1 or -1.

Let's take a=1 for simplicity.

Then, the coefficient of x is -2ab = -14 ⇒ ab = 7.

And the coefficient of x² is b² = 9 ⇒ b = 3 or -3.

So let's test with b=3. Then ab = 7 ⇒ 3a =7 ⇒ a=7/3, which conflicts because earlier we had a^2=1 ⇒ a=±1. So that doesn't work.

Alternatively, maybe it's not a perfect square but can be expressed as (something)^2. Alternatively, perhaps I made an error in interpreting the expression under sqrt.

Wait, actually, 1 -14x +9x² is already quadratic; let me check if it factors:

Looking for two numbers p and q such that pq = 9*1=9 and p + q = -14. Hmm, factors of 9 are (3,3), but 3+3=6 ≠ -14. So maybe it doesn't factor nicely.

Alternatively, perhaps completing the square:

Let me write 1 -14x +9x² as 9x² -14x +1.

Compute discriminant: D = (-14)^2 -4*9*1=196-36=160. Hmm, sqrt(160) is 4*sqrt(10), not an integer. So that might complicate things.

But maybe I can write it as 9(x² - (14/9)x) +1.

Completing the square inside:

x² - (14/9)x = x² - (14/9)x + (7/9)^2 - (7/9)^2 = (x - 7/9)^2 - (49/81).

Therefore, 9(x² - (14/9)x) +1 = 9[(x -7/9)^2 - 49/81] +1 = 9(x -7/9)^2 - 49/9 +1=9(x -7/9)^2 - 40/9.

So, 1 -14x +9x² = 9(x -7/9)^2 - 40/9. Hmm, not sure if that's helpful.

Alternatively, perhaps I can write the square root as sqrt(1 -14x +9x^2). Maybe using generating function techniques, like expressing G(x) in terms of continued fractions or looking for a recursive relation.

Let’s try to find an expression for c_k. Since we have G(x) = (1 -3x - sqrt(1-14x+9x²))/4, let me denote H(x)=sqrt(1 -14x +9x^2). Then,

G(x) = (1 - 3x - H(x)) /4

Multiply both sides by 4:

4 G(x) = 1 -3x - H(x)

Then, rearrange: H(x) = 1 -3x -4 G(x)

But since H(x)=sqrt(1-14x +9x^2), squaring both sides gives:

H(x)^2 = (1 -14x +9x²)

Also,

From H(x) = 1 -3x -4 G(x), let's square both sides:

(1 -3x -4 G(x))² = 1 -14x +9x²

Expanding the left side:

= (1)^2 + (-3x)^2 + (-4G(x))^2 + 2*(1*(-3x) + 1*(-4G(x)) + (-3x)*(-4G(x)))

= 1 +9x² +16 G(x)^2 -6x -8 G(x) +24 x G(x)

So, set equal to right side:

1 +9x² +16 G(x)^2 -6x -8 G(x) +24 x G(x) = 1 -14x +9x²

Subtracting 1 from both sides:

9x² +16 G(x)^2 -6x -8 G(x) +24 x G(x) = -14x +9x²

Now, subtract the right side:

(9x² -9x²) + (16 G(x)^2) + (-6x +14x) + (-8 G(x)) +24 x G(x)=0

Simplify:

16 G(x)^2 +8x -8 G(x)+24 x G(x)=0

Divide both sides by 8:

2 G(x)^2 +x -G(x)+3x G(x)=0

Bring all terms to one side:

2 G(x)^2 + (3x -1) G(x) +x =0

So, this is a quadratic equation in G(x):

2 G(x)^2 + (3x -1) G(x) + x=0

Let me solve for G(x):

G(x)= [ -(3x -1) ± sqrt{(3x -1)^2 -8x} ] /4

But we know that G(x) is given as (1 -3x - sqrt(1-14x+9x²))/4, so the negative sign in front of the square root must be taken. So,

G(x)= [ -(3x -1) - sqrt{(3x -1)^2 -8x} ] /4

Wait, let's compute discriminant D=(3x-1)^2 -8x=9x² -6x +1 -8x=9x² -14x +1. Which is exactly the same as inside H(x). So sqrt(D)=sqrt(9x² -14x +1)=H(x).

Therefore, G(x)= [ -3x +1 - H(x) ] /4 which matches the given expression.

So now we have a quadratic equation satisfied by G(x):

2 G(x)^2 + (3x -1) G(x) +x =0

This suggests that c_k satisfies a recurrence relation. Let me express this as:

2 G(x)² + (3x -1)G(x) + x=0.

Let’s write the coefficients in terms of generating functions:

Multiply through by 1/ (since we have G², linear term, and constant).

But maybe we can find a recurrence relation for c_k. Let me consider expressing G(x)^2 as a convolution.

Expressed as: G(x) = sum_{k=0}^\infty c_k x^k

Therefore,

G(x)^2 = (sum_{k=0}^\infty c_k x^k)(sum_{m=0}^\infty c_m x^m) = sum_{n=0}^\infty (sum_{i=0}^n c_i c_{n-i}) )x^n.

Now, let's plug into the quadratic equation:

2 G(x)^2 + (3x -1)G(x) + x=0

Express term by term:

- The constant term: 2c_0² + (-1)c_0 + 0 =0 ?

Wait, no. Let me think.

Let me compute each coefficient for powers of x in the equation above:

Left side is 2 G² + (3x -1)G +x.

Express as sum_{n=0}^\infty [2 c_n^{(2)}] x^n + sum_{n=0}^\infty [ if n >=1 then 3c_{n-1} else 0 ] x^n + sum_{n=0}^\infty (-c_n) x^n +x.

Wait, perhaps it's better to write each term:

2 G(x)^2: coefficient of x^n is 2 times the convolution sum_{i=0}^n c_i c_{n-i}

(3x -1)G(x): expanding this gives (3x)(sum c_k x^k ) + (-1)(sum c_k x^k )

So, that's sum 3c_k x^{k+1} - sum c_k x^k.

Which can be rewritten as: sum_{n=0}^\infty [if n >=1 then 3c_{n-1} else 0]x^n - sum_{n=0}^\infty c_n x^n

Similarly, the term +x is just a single term at x^1.

So altogether:

Left side: sum_{n=0}^\infty [2 c_n^{(2)} + (3 delta_{n>=1} c_{n-1}) - c_n]x^n +x=0

But wait, actually when we write all terms together, the equation is 2G² + (3x -1)G + x=0.

So:

Left side = 2 G(x)^2 + (3x -1) G(x) + x = sum_{n=0}^\infty [2 (sum_{i=0}^n c_i c_{n-i}) ] x^n + sum_{n=0}^\infty [ if n >=1 then 3 c_{n-1} else 0 - c_n]x^n + x.

Wait, perhaps it's better to reindex and write all coefficients for each x^n:

For each n >=0, the coefficient is:

2 * sum_{i=0}^n c_i c_{n-i} + [3 c_{n-1} if n>=1 else 0] - c_n + [1 if n=1 else 0].

Set all these coefficients to zero since left side equals zero.

So for each n:

If n =0: coefficient is 2c_0² + 0 (since no term from 3x*G) - c_0 + 0=0

Thus, equation at n=0: 2c_0² -c_0=0 ⇒ c_0(2c_0 -1)=0. Since c_0 is the constant term in G(x), let's compute it by setting x=0.

From G(0) = (1 -0 - sqrt(1 -0 +0))/4=(1-1)/4=0, so c_0=0? Wait, but from n=0 equation: 2c_0² -c_0=0. If c_0=0, then it's satisfied.

Wait, let me compute G(0). The generating function is (1 -3x - sqrt(1 -14x +9x²))/4. At x=0, that's (1 -0 -sqrt(1 -0 +0))/4=(1-1)/4=0, so c_0=0.

But in the equation above for n=0: 2c_0² -c_0 =0 ⇒ if c_0=0, then 0 -0=0 which is okay.

Wait, but from G(x)=sum_{k=0}^\infty c_k x^k, so c_0=G(0)=0. So that's consistent.

Then for n >=1:

Coefficient of x^n: 2 * sum_{i=0}^n c_i c_{n-i} +3c_{n-1} -c_n +delta_{n,1}=0

Wait, because the term from 3x*G(x) gives coefficient 3c_{n-1} when n >=1, and then we have subtracting c_n. Additionally, the last x adds a delta_{n,1}=1 only at n=1.

Therefore, for each n>=0:

If n=0: 2c_0² -c_0 =0 ⇒ already satisfied with c_0=0

For n >=1:

2 * sum_{i=0}^n c_i c_{n-i} +3c_{n-1} -c_n + [if n=1, add 1] =0.

But let me write for n>=1:

For general n >=1,

Equation: 2C(n) + 3c_{n-1} -c_n + delta_{n,1}=0

Where C(n)=sum_{i=0}^n c_i c_{n-i}

Wait, this seems complex because of the convolution term. Maybe we can find a recurrence relation for c_n.

But if I denote D(n) = 2C(n) +3c_{n-1} -c_n

Then,

For n >=1:

D(n) + delta_{n,1}=0

So, for n>=2: D(n)=0

And for n=1: D(1)+1=0 ⇒ D(1)=-1.

Let me see if I can write a recurrence.

Alternatively, perhaps we can rearrange the equation:

For n >=1,

c_n = 3c_{n-1} +2 sum_{i=0}^n c_i c_{n-i}

But wait, that seems messy because of the convolution. Hmm.

Wait, but for n=1:

From D(1)+1=0,

D(1)= -1

Where D(1) = 2C(1) +3c_0 -c_1=2(c_0 c_1 +c_1 c_0) +3c_0 -c_1

But since c_0=0, this simplifies to:

D(1)=2*(0*c_1 + c_1*0) +0 -c_1 = -c_1.

Therefore,

From n=1: D(1)+1=-c_1+1=0 ⇒ c_1=1.

So, c_1=1.

Now for n=2:

D(2)= 2C(2) +3c_1 -c_2

Where C(2)=sum_{i=0}^2 c_i c_{2-i}=c_0 c_2 +c_1 c_1 +c_2 c_0 =0* c_2 +1*1 +c_2 *0=1.

Thus,

D(2)= 2*1 +3*c_1 -c_2 =2 +3*1 -c_2=5 -c_2

But since for n>=2, D(n)=0 ⇒ 5 - c_2=0 ⇒c_2=5.

Similarly, for n=3:

C(3)=sum_{i=0}^3 c_i c_{3-i}=c_0 c_3 +c_1 c_2 +c_2 c_1 +c_3 c_0 =0*c_3 +1*5 +5*1 +c_3 *0=5+5=10

Then D(3)=2*C(3) +3c_2 -c_3=2*10 +3*5 -c_3=20+15 -c_3=35 -c_3.

Since n=3>=2, set D(3)=0 ⇒ 35 -c_3=0 ⇒ c_3=35.

Wait a second. Let's list out the terms we have so far:

c_0=0

c_1=1

c_2=5

c_3=35

Let me see if I can spot a pattern here. These numbers look familiar: 1, 5, 35...

Wait, they're similar to Catalan numbers? But the nth Catalan number is C_n = (1/(n+1))binomial(2n, n). Let's compute:

C_0=1

C_1=1

C_2=2

C_3=5

C_4=14

Hmm, close but not exactly matching. c_1=1=C_1, c_2=5=C_3, c_3=35= which is 35 vs C_5=42? No, doesn't match.

Alternatively, maybe it's related to the number of some trees or another combinatorial structure.

Wait, let me compute more terms. Let me find c_4.

For n=4:

C(4)= sum_{i=0}^4 c_i c_{4-i}=c_0c_4 +c_1c_3 +c_2c_2 +c_3c_1 +c_4c_0

= 0*c_4 +1*35 +5*5 +35*1 +0*c_4

=0 +35 +25 +35 +0 =95.

Then D(4)=2*C(4) +3c_3 -c_4=2*95 +3*35 -c_4=190+105 -c_4=295 -c_4

Set to zero: c_4=295.

So, c_4=295. Hmm.

Looking at OEIS for sequence 0,1,5,35,295... Let's see:

Wait, I remember that the number of certain trees or labeled structures might give this.

Alternatively, perhaps these are the numbers related to generating functions with such a form.

Wait, the generating function is (1 -3x - sqrt(1 -14x +9x^2))/4. Maybe it's similar to the generating function for some combinatorial objects, perhaps 3 trees or something else?

Alternatively, maybe it's the number of ternary trees or something? But I don't recall exactly.

Alternatively, since we have a quadratic equation satisfied by G(x), perhaps we can express c_n in terms of hypergeometric functions or using continued fractions for Hankel determinants.

Wait, another approach: The determinant of an n x n Hankel matrix with moments c_k is equal to the product of the first n elements of some orthogonal polynomial system evaluated at specific points. But I might need more information on that front.

Alternatively, perhaps there's a recursive formula for the determinant.

Given that the problem states "for a positive integer n", we're supposed to find a general expression. So maybe it's 1 or something factorial?

Wait, but let's think about small n and see if we can compute determinants manually to spot a pattern.

First, when n=1:

A is [c_{1+1-1}] = [c_1] = [1]. Determinant is 1.

n=2: Matrix [[c_1, c_2],[c_2,c_3]] = [[1,5],[5,35]]

Determinant: (1)(35) - (5)(5)=35 -25=10

Wait, determinant is 10 for n=2.

n=3: Matrix:

[c1 c2 c3]

[c2 c3 c4]

[c3 c4 c5]

So entries:

c1=1, c2=5, c3=35, c4=295, c5 needs to be computed. But let's compute determinant.

But wait, computing for n=3 would require knowing c5 and so on. Maybe it's getting too cumbersome.

Wait, but perhaps the determinants follow a simple pattern like 1,10, something else?

Alternatively, maybe all determinants are 1? But no, because det(A) for n=2 is 10.

Alternatively, perhaps they are n! ?

But for n=1:1!=1

n=2:2!=2 ≠10

So that doesn't hold.

Wait, let's compute determinant for n=3:

First, need c_5. Let me calculate it.

From our earlier approach, for n=4, we got c_4=295. For n=5:

C(5)=sum_{i=0}^5 c_i c_{5-i}=c0*c5 +c1*c4 +c2*c3 +c3*c2 +c4*c1 +c5*c0

Again, since c0=0,

C(5)=0 +1*295 +5*35 +35*5 +295*1 +0=295 +175 +175 +295.

Compute: 295+175=470, 470+175=645, 645+295=940

Then D(5)=2*C(5)+3c_4 -c_5=2*940 +3*295 -c_5=1880 +885 -c5=2765 -c5.

Set equal to zero (since n=5 >=2), so c5=2765

Thus, c_5=2765

So the matrix for n=3 is:

Row 1: [c1,c2,c3] = [1,5,35]

Row 2: [c2,c3,c4]= [5,35,295]

Row 3: [c3,c4,c5]= [35,295,2765]

Compute determinant:

| 1   5    35 |

| 5   35  295|

|35  295 2765|

This will be tedious. Let's compute using expansion by minors.

First, expand along the first row:

det = 1 * det([[35,295],[295,2765]]) - 5 * det([[5,295],[35,2765]]) +35*det([[5,35],[35,295]])

Compute each minor:

First minor: det([[35,295],[295,2765]) = (35)(2765) - (295)(295)

= 35*2765: Let's compute that.

35 * 2000 =70,000

35 *765=26,775

So total:70,000 +26,775=96,775

Then subtract (295)^2:

295^2: Let's compute 300^2=90,000; subtract 2*300*5 +5^2= 90,000 -3000 -25=86,975

Wait, no. Wait (a - b)^2=a² -2ab +b² where a=300, b=5.

So, 295=300-5

Thus,

295²=(300 -5)²=90,000 - 2*300*5 +25=90,000 -3,000 +25=87,025.

So, minor1 determinant:96,775 -87,025=9,750

Second minor:

det([[5,295],[35,2765])=5*2765 -295*35

Compute 5*2765=13,825

Compute 295*35: Let's do 300*35=10,500 minus 5*35=175 ⇒10,500 -175=10,325.

Thus, determinant=13,825 -10,325=3,500

Third minor:

det([[5,35],[35,295])=5*295 -35*35=1,475 -1,225=250

Therefore, determinant is:

1 * 9,750 -5 *3,500 +35*250 =9,750 -17,500 +8,750.

Compute:9,750 -17,500= -7,750; then -7,750 +8,750=1,000

So det(A)=1,000 for n=3.

Hmm. So:

n | determinant

1 | 1

2 | 10

3 | 1,000

Wait a minute. These numbers: 1, 10, 1000... that's 10^{0}, 10^{1}, 10^{3}? Hmm, not exactly.

Wait, n=1:1, n=2:10=1*10^1, n=3:1000=1*10^3. So it seems like det(A)=10^{n(n-1)/2} or something?

For n=1: 10^{0}=1

n=2:10^{1}=10

n=3:10^{3}=1,000

Wait, but n=4 would then be 10^6? Let me see if the pattern holds.

But before that, maybe let's try n=4 determinant. It might take a long time to compute c_5, c_6, etc., but perhaps I can proceed.

Alternatively, note that for n=1:1=1!, n=2:10≈ factorial-like? 3! is 6≠10; 10 is not a factorial. Wait, 1, 10, 1000...

Wait, if you look at exponents:

n=1: exponent 0

n=2: exponent 1 (since 10^1)

n=3: exponent 3 (since 10^3)

Hmmm. That doesn't form an arithmetic progression.

Alternatively, think about the exponents as triangular numbers? For n=1: T(0)=0, n=2:T(1)=1, n=3:T(2)=3. So perhaps det(A)=10^{T(n-1)} where T(k) is the kth triangular number.

Let me check:

For n=1: 10^{T(0)}=10^0=1 ✓

n=2:10^{T(1)}=10^1=10 ✓

n=3:10^{T(2)}=10^3=1,000 ✓

So the pattern seems to hold. Then for general n, det(A)=10^{n(n-1)/2}.

But let's test this hypothesis. Let me see if I can find a reference or some reasoning behind it.

Wait, in problems involving Hankel determinants and generating functions, especially when dealing with orthogonal polynomials or moment sequences, the determinant often relates to products of terms from continued fractions or other structures.

Given that our generating function is G(x)=(1-3x -sqrt(1-14x +9x²))/4, perhaps it's related to a known combinatorial class whose Hankel determinants are powers of 10. Alternatively, the sequence c_k might be such that det(A)=10^{n(n-1)/2}.

But how can we verify this? Perhaps by induction or using properties of continued fractions.

Alternatively, maybe consider the continued fraction representation of G(x). Since G(x) satisfies a quadratic equation, perhaps its continued fraction is periodic or has some regular structure. If so, then the Hankel determinants might follow a geometric progression, leading to exponential terms like 10^{something}.

Wait, another thought: Perhaps c_k can be expressed as (35)^{k-1} for k>=1? Let's check:

c_1=1, which would need to be (35)^{0}=1 ✓

c_2=5 vs 35^{1}=35≠5 ✗. So that doesn't work.

Alternatively, maybe c_k satisfies a linear recurrence. From the quadratic equation earlier:

We had 2 G(x)^2 + (3x -1)G(x)+x =0.

If we can write this as G(x)=rational function involving sqrt(1-14x+9x²), perhaps we can relate it to some generating function that's a solution to a Riccati equation. Maybe the terms c_k are related to Bessel numbers or something similar.

Alternatively, think about the generating function expression:

G(x) = (1 -3x - sqrt(1-14x+9x²))/4

Let me compute G(x) in terms of continued fraction. Since it's a solution to a quadratic equation, perhaps express as a continued fraction with periodic terms.

Alternatively, note that the generating function resembles those for certain types of trees or graphs where each edge contributes a factor, leading to multiplicative factors in determinants.

But maybe stepping back: The Hankel determinant of order n is known to satisfy det(A_n)=det(A_{n-1}) * something. If we see from our examples:

For n=1: det=1

n=2: 10=1*10

n=3:1,000=10*100=10*10²

Wait, if det(A_{n})=det(A_{n-1})*10^{n-1}, then for n=1, det=1.

For n=2, 1*10^1=10 ✓

n=3:10*10^2=1,000 ✓

Then for n=4, it would be 1,000 *10^3=1,000,000. So det(A_4)=10^{6}=1,000,000.

Therefore, the general formula seems to be det(A_n)=10^{n(n-1)/2}

Because for n=1: 10^{0}=1

For n=2:10^{1}=10

n=3:10^{3}=1,000

Wait no, wait n=4 would be 10^{6}, which is n(n-1)/2 where n=4: 4*3/2=6 ✓.

Yes, so det(A_n)=10^{T(n)}=10^{n(n-1)/2}

Thus, the determinant of A is 10^{n(n-1)/2}

So boxed answer: \boxed{10^{\frac{n(n-1)}{2}}}

**Final Answer**
The determinant of matrix \(A\) is \boxed{10^{\frac{n(n-1)}{2}}}.
</think>



To find the determinant of the \(n\)-by-\(n\) matrix \(A\) with entries \(c_{i+j-1}\) where the sequence \(c_0, c_1, c_2, \ldots\) is defined by the generating function

\[
\frac{1 - 3x - \sqrt{1 - 14x + 9x^2}}{4} = \sum_{k=0}^\infty c_k x^k,
\]

we start by examining the sequence \(c_k\) and its properties. The generating function satisfies a quadratic equation, suggesting that the sequence might be related to known combinatorial structures or orthogonal polynomials.

By calculating the initial terms of the sequence \(c_k\), we find:
- \(c_0 = 0\)
- \(c_1 = 1\)
- \(c_2 = 5\)
- \(c_3 = 35\)
- \(c_4 = 295\)
- \(c_5 = 2765\)

We then compute the determinants of small Hankel matrices formed by these terms:
- For \(n=1\), the determinant is 1.
- For \(n=2\), the matrix \(\begin{pmatrix} 1 & 5 \\ 5 & 35 \end{pmatrix}\) has a determinant of 10.
- For \(n=3\), the matrix \(\begin{pmatrix} 1 & 5 & 35 \\ 5 & 35 & 295 \\ 35 & 295 & 2765 \end{pmatrix}\) has a determinant of 1000.

Observing the pattern, we hypothesize that the determinant of the \(n\)-by-\(n\) Hankel matrix is given by \(10^{n(n-1)/2}\). This hypothesis is supported by the calculations for small \(n\) and fits the observed pattern.

Thus, the determinant of the matrix \(A\) is

\[
\boxed{10^{\frac{n(n-1)}{2}}}
\]