A245490 a(n) = n^(n^3).
1, 1, 256, 7625597484987, 340282366920938463463374607431768211456
Offset: 0
Programs
-
Mathematica
Table[n^(n^3), {n, 0, 4}] (* Alonso del Arte, Jul 24 2014 *)
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Pasha Zusmanovich's wiki page.
Pasha Zusmanovich has authored 10 sequences.
Table[n^(n^3), {n, 0, 4}] (* Alonso del Arte, Jul 24 2014 *)
def a(n):
ans = 0
for x in GF(13^n):
if x!=0 and x.trace()==0 and x.multiplicative_order()==13^n-1: ans += 1
return ans # Robin Visser, Jun 01 2024
def a(n):
ans = 0
for x in GF(3^n):
if x!=0 and x.trace()==0 and x.multiplicative_order()==3^n-1: ans += 1
return ans # Robin Visser, Apr 26 2024
def a(n):
ans = 0
for x in GF(5^n):
if x!=0 and x.trace()==0 and x.multiplicative_order()==5^n-1: ans += 1
return ans # Robin Visser, May 10 2024
def a(n):
ans = 0
for x in GF(7^n):
if x!=0 and x.trace()==0 and x.multiplicative_order()==7^n-1: ans += 1
return ans # Robin Visser, Jun 01 2024
def a(n):
ans = 0
for x in GF(11^n):
if x!=0 and x.trace()==0 and x.multiplicative_order()==11^n-1: ans += 1
return ans # Robin Visser, May 10 2024
p := 2;
for n in [1..17] do
F := GF(p^n);
num := 0;
for f in F do
if (f = Zero(F)) then continue; fi;
if (Trace(f) <> Zero(F)) then continue; fi;
if (Order(f) <> Size(F) - 1) then continue; fi;
num := num + 1;
od;
Print (num, ",");
od;
Comments