Skip to main content
15 events
when toggle format what by license comment
May 16, 2019 at 8:35 review Reopen votes
May 16, 2019 at 14:20
Apr 2, 2019 at 20:36 comment added Sam Hopkins I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
Apr 2, 2019 at 12:27 history closed user44191
Stopple
Max Alekseyev
Zhi-Wei Sun
Jan-Christoph Schlage-Puchta
Not suitable for this site
Apr 2, 2019 at 4:07 answer added Alexey Ustinov timeline score: 13
Apr 2, 2019 at 2:55 comment added Somos The identity is with arguments renamed the same as $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. Read my essay "In the elliptic realm" to see connection.
Apr 2, 2019 at 1:19 history became hot network question
Apr 2, 2019 at 1:16 answer added Ira Gessel timeline score: 7
Apr 2, 2019 at 0:32 vote accept user918212
Apr 2, 2019 at 0:08 answer added Cherng-tiao Perng timeline score: 12
Apr 1, 2019 at 23:25 comment added user44191 This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
Apr 1, 2019 at 23:10 review Close votes
Apr 2, 2019 at 12:27
Apr 1, 2019 at 22:52 comment added user44191 Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
Apr 1, 2019 at 22:49 comment added user918212 It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
Apr 1, 2019 at 22:47 comment added user44191 This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
Apr 1, 2019 at 22:34 history asked user918212 CC BY-SA 4.0