Timeline for integral transform of Fibonacci polynomials is integral
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 26, 2020 at 8:50 | answer | added | Leucippus | timeline score: 1 | |
| Jun 26, 2017 at 17:58 | vote | accept | T. Amdeberhan | ||
| Jun 26, 2017 at 6:17 | answer | added | Gjergji Zaimi | timeline score: 14 | |
| Jun 26, 2017 at 2:52 | comment | added | Christian Remling | The first few $F_n=\sum F(n,k)x^k$ at least have the stronger property that every term of $\sum \int \ldots$ is an integer separately. If this is true in general, one can perhaps be optimistic about a proof since there is an explicit formula for the coefficients $F(n,k)$. | |
| Jun 26, 2017 at 2:10 | comment | added | T. Amdeberhan | $F_3(1+3z)=9z^2+6z+2$, so $\int_0^1F_3(1+3z)dz=8$. | |
| Jun 26, 2017 at 2:08 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 124 characters in body
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| Jun 26, 2017 at 0:08 | comment | added | T. Amdeberhan | Perhaps you were misled by a typo: $F_0(x)=0$. | |
| Jun 26, 2017 at 0:07 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
edited body
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| Jun 26, 2017 at 0:05 | comment | added | Cherng-tiao Perng | Did you check the case n=3 and k=1? | |
| Jun 25, 2017 at 22:20 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |