Aurifeuillean factorization in Lucas sequences

wpolly · 2018-12-28, 02:24

I have noticed that the Aurifeuillean identity $L_{5k}=L_k(5F_k^2+5F_k+1)(5F_k^2-5F_k+1)$ have analogues in other Lucas sequences, for example

$P=3,Q=-1: \frac{V_{13k}}{V_k}=\left(13^3U_k^6+13^2U_k^4-4\cdot13U_k^2+1\right)^2-\left(13U_k(13U_k^2-1)^2\right)^2, k \text{ odd}$
and
$P=1,Q=2: \frac{U_{7k}}{7U_k}=\left(2^{k-1}V_k\right)^2-\left(7U_{k-1}U_k U_{k+1}\right)^2.$

Are there any known results on such factorization?

science_man_88 · 2018-12-28, 12:55

https://en.m.wikipedia.org/wiki/Lucas_sequence#Examples may give you a way to figure that out.

wpolly · 2019-12-16, 20:42

Something even more strange: certain sequences possess a "double Aurifeullian factorization". For instance: if $P=2$ and $Q=10$ , Then for odd $k$ we have

$V_{10k}=V_{2k}(A+B+C+D)(A+B-C-D)(A-B+C-D)(A-B-C+D),$

where

$<br /> A=6U_{k}V_{k}, B=5\cdot10^{(k-1)/2}V_{k}, C=3\cdot 10^{(k+1)/2}U_{k}, D=3\cdot 10^{k}.<br /> <br />$

2018-12-28, 02:24	#1
wpolly Sep 2002 Vienna, Austria 3×73 Posts	Aurifeuillean factorization in Lucas sequences I have noticed that the Aurifeuillean identity $L_{5k}=L_k(5F_k^2+5F_k+1)(5F_k^2-5F_k+1)$ have analogues in other Lucas sequences, for example $P=3,Q=-1: \frac{V_{13k}}{V_k}=\left(13^3U_k^6+13^2U_k^4-4\cdot13U_k^2+1\right)^2-\left(13U_k(13U_k^2-1)^2\right)^2, k \text{ odd}$ and $P=1,Q=2: \frac{U_{7k}}{7U_k}=\left(2^{k-1}V_k\right)^2-\left(7U_{k-1}U_k U_{k+1}\right)^2.$ Are there any known results on such factorization?

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2018-12-28, 12:55	#2
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	https://en.m.wikipedia.org/wiki/Lucas_sequence#Examples may give you a way to figure that out.

2019-12-16, 20:42	#3
wpolly Sep 2002 Vienna, Austria 3×73 Posts	Something even more strange: certain sequences possess a "double Aurifeullian factorization". For instance: if $P=2$ and $Q=10$ , Then for odd $k$ we have $V_{10k}=V_{2k}(A+B+C+D)(A+B-C-D)(A-B+C-D)(A-B-C+D),$ where $<br /> A=6U_{k}V_{k}, B=5\cdot10^{(k-1)/2}V_{k}, C=3\cdot 10^{(k+1)/2}U_{k}, D=3\cdot 10^{k}.<br /> <br />$