OFFSET
0,2
COMMENTS
For n>=4, A319281(a(n)) == 2^n + [(n mod 4)>0].
It appears that for n>4: a(n)=2*a(n-1)-2*[(n mod 4)==2]; a(n) = ceiling(2^n/15) - [(n mod 4)==0] + 1.
MATHEMATICA
a[n_]:=CountDistinct[Table[PowerMod[x-1, 4, 2^(n-1)], {x, 1, 2^(n-1)}]]; Array[a, 24]
PROG
(PARI) a(n) = #Set(vector(2^(n-1), x, Mod(x-1, 2^(n-1))^4))
(Python)
def A364811(n): return len({pow(x, 4, 1<<n) for x in range(1<<n)}) # Chai Wah Wu, Sep 17 2023
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Albert Mukovskiy, Sep 14 2023
STATUS
approved