IntersectionNumbers

Let $F=\mathrm{GF}(q)$ where $q=p^\ell$ is a prime power. Let $\Phi=\Phi_k$ be a subgroup of order $k$ of the multiplicative group $F^*=F\setminus \{0\}=\langle\zeta\rangle$. For each $a\in F^*$, we want to know the number $t_a$ of elements in the intersection $\Phi\cap \Phi a+1$. Note that we are interested only when $\Phi$ is circular. The noncircular Ferrero pair numbers can be found in Modisett Primes.

Assume that $\Phi$ is generated by $\phi=\zeta^{(q-1)/k}$ and $a=\zeta^s$. In the following, under each $k$, the prime power $q$ and the integers $s$ are listed. Since we are interested in the case when $(F,\Phi)$ is circular, we use red to indicate $s$ for those $a=\zeta^s$ with $t_a=1$, black ones for those with $t_a=2$. Those $s$ not listed means that if $a=\zeta^s$, then $t_a=0$.

For any $\psi\in\Phi$, we have $\Phi \psi a=\Phi a$. So, we may assume that $1\leq s\leq(q-1)/k$.

• ${k=4}$
  • $q= 13$, $s = {\color{red}1}, 2$
  • $q= 17$, $s= 1, {\color{red}2}$
  • $q= 29$, $s= {\color{red}1}, 4$
  • $q= 37$, $s= {\color{red}1}, 5$
  • $q= 41$, $s= {\color{red}6}, 8$
  • $q= 53$, $s= {\color{red}1}, 7$
  • $q= 61$, $s= {\color{red}1}, 8$
  • $q= 73$, $s= {\color{red}8}, 13$
  • $q= 89$, $s= {\color{red}16}, 19$
  • $q= 97$, $s= 5, {\color{red}10}$
  • $q= 101$, $s= {\color{red}1}, 13$
  • $q=13^2=2197$, $s={\color{red}183}, 366$
  • $q=17^2=4913$, $s=307, {\color{red}614}$
  • $q=29^2=24389$, $s={\color{red}871}, 3484$
  • $q=37^2=50653$, $s={\color{red}1407}, 7035$
  • $q=3^4=81$, $s=10, {\color{red}20}$
  • $q=7^4=2401$, $s=100, {\color{red}200}$
• ${k=5}$
• ${k=6}$
  • $q=19$, $s= {\color{red}1}, 2, 3$
  • $q=31$, $s= 3, {\color{red}4}, 5$
  • $q=37$, $s= {\color{red}1}, 4, 6$
  • $q=43$, $s= 4, {\color{red}6}, 7$
  • $q=61$, $s={\color{red}1}, 8, 10$
  • $q=67$, $s={\color{red}1}, 3, 11$
  • $q=73$, $s={\color{red}8}, 9, 12$
  • $q=79$, $s={\color{red}4}, 7, 13$
  • $q=5^2=25$, $s={\color{red}2}, 3, 4$
  • $q=11^2=121$, $s={\color{red}12}, 18, 20$
  • $q=17^2=289$, $s={\color{red}12}, 33, 48$
  • $q=23^2=529$, $s={\color{red}48}, 60, 88$
  • $q=29^2=841$, $s=5, {\color{red}30}, 140$
  • $q=31^3=29791$, $s= 2979, {\color{red}3972}, 4965$
  • $q=37^3=50653$, $s= {\color{red}1407}, 5628, 8442$
  • $q=43^3=79507$, $s=7572, {\color{red}11358}, 13251$
  • $q=61^3=226981$, $s={\color{red}3783}, 30264, 37830$
• ${k=8}$
  • $q=73$, $s= 1, 3, 4, {\color{red}8}$
  • $q=89$, $s= 1, {\color{red}5}, 7, 8$
  • $q=97$, $s= 2, 3, 5, {\color{red}10}$
  • $q=113$, $s= 1, 5, 6, {\color{red}12}$
  • $q=137$, $s= 2, 3, 5, {\color{red}10}$
  • $q=193$, $s= 6, {\color{red}10}, 11, 17$
  • $q=233$, $s= 7, 13, {\color{red}14}, 23$
  • $q=241$, $s= 1, 4, 5, {\color{red}10}$
  • $q=19^2=361$, $s=10, 14, {\color{red}20}, 41$
  • $q=23^2=529$, $s=24, 27, {\color{red}48},63$
  • $q=29^2=841$, $s=15, {\color{red}30}, 39, 81$
  • $q=31^2=961$, $s=24, 44, {\color{red}48}, 100$
  • $q=73^3=389017$, $s= 5403, 16209, 21612, {\color{red}43224}$
  • $q=89^3=704969$, $s=8011, {\color{red}40055}, 56077, 64088$
  • $q=97^3=912673$, $s=19014, 28521, 47535, {\color{red}95070}$
• The GAP program used to generated the above lists. Change the values of $k$ and $l$ and ignore the unbound global variables warning):

=python k:= 6; p:= 1; l:=1; nextPrime:= function(p,k) local p1, q; p1:= NextPrimeInt(p); q:= p1^l; while (q mod k) <> 1 do p1:= NextPrimeInt(p1); q:= p1^l; od; return(p1); end; intersect:= function(A,B) local C; C:= ShallowCopy(A); IntersectSet(C,B); return(C); end; for i in [1..10] do p:= nextPrime(p,k); q:= p^l; z:= Z(q); F:= Field(z); ph:= z^((q-1)/k); B:= List([1..k], i-> ph^i); I:= Filtered( List([1..(q-1)/k], i-> [i, Size(intersect(Set(B*z^i+z^0),B))]), u-> u[2]<>0); Print(p, " / ", q, "\n", I, "\n\n"); od;