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LSpice
LSpice
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We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answersanswers to your previous questionyour previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "-1"$-1$" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using the generating function for Stirling numbers and Binet's formula for Fibonacci numbers, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}(2\phi-3)+(1+z)^{2-\phi}(2\phi+1)}{\sqrt{5}}-1,$$ where $\phi:=\frac{1+\sqrt{5}}2$$\phi\mathrel{:=}\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "-1" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using the generating function for Stirling numbers and Binet's formula for Fibonacci numbers, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}(2\phi-3)+(1+z)^{2-\phi}(2\phi+1)}{\sqrt{5}}-1,$$ where $\phi:=\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "$-1$" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using the generating function for Stirling numbers and Binet's formula for Fibonacci numbers, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}(2\phi-3)+(1+z)^{2-\phi}(2\phi+1)}{\sqrt{5}}-1,$$ where $\phi\mathrel{:=}\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).

deleted 4 characters in body; added 71 characters in body
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Max Alekseyev
Max Alekseyev
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We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "-1" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using the generating function for Stirling numbers and Binet's formula for Fibonacci numbers, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}\phi^{-3}-(1+z)^{2-\phi}(1-\phi)^{-3}}{\sqrt{5}}-1,$$$$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}(2\phi-3)+(1+z)^{2-\phi}(2\phi+1)}{\sqrt{5}}-1,$$ where $\phi:=\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "-1" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using Binet's formula, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}\phi^{-3}-(1+z)^{2-\phi}(1-\phi)^{-3}}{\sqrt{5}}-1,$$ where $\phi:=\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "-1" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using the generating function for Stirling numbers and Binet's formula for Fibonacci numbers, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}(2\phi-3)+(1+z)^{2-\phi}(2\phi+1)}{\sqrt{5}}-1,$$ where $\phi:=\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
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  • 152

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "-1" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using Binet's formula, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}\phi^{-3}-(1+z)^{2-\phi}(1-\phi)^{-3}}{\sqrt{5}}-1,$$ where $\phi:=\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).