A367014 Let q be the n-th prime power (A246655), then a(n) = q^3 + q^2 - q; number of solutions to x*y = z*w in the finite field F_q.
10, 33, 76, 145, 385, 568, 801, 1441, 2353, 4336, 5185, 7201, 12673, 16225, 20385, 25201, 30721, 33760, 51985, 70561, 81313, 105985, 120001, 151633, 208801, 230641, 266176, 305185, 362881, 394273, 499201, 537921, 578593, 712801, 921985, 1040401, 1103233, 1236385, 1306801
Offset: 1
Keywords
Examples
For q = A246655(3) = 4, we see that in F_4 = F_2(t), where t^2 + t + 1 = 0: - x*y = z*w = 0 has 7 solutions for the pair (x,y) and 7 solutions for the pair (z,w); - x*y = z*w = 1 has 3 solutions for the pair (x,y) and 3 solutions for the pair (z,w); - x*y = z*w = t has 3 solutions for the pair (x,y) and 3 solutions for the pair (z,w); - x*y = z*w = 1+t has 3 solutions for the pair (x,y) and 3 solutions for the pair (z,w), so a(4) = 7*7 + 3*3*3 = 76.
Links
- Jianing Song, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
Map[#^3+#^2-#&,Select[Range[200],PrimePowerQ]] (* Paolo Xausa, Nov 26 2023 *)
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PARI
lim_A367014(N) = for(n=2, N, if(isprimepower(n), print1(n^3 + n^2 - n, ", ")))
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Python
from sympy import primepi, integer_nthroot def A367014(n): def bisection(f,kmin=0,kmax=1): while f(kmax) > kmax: kmax <<= 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax def f(x): return int(n+x-sum(primepi(integer_nthroot(x,k)[0]) for k in range(1,x.bit_length()))) return (m:=bisection(f,n,n))*(m*(m+1)-1) # Chai Wah Wu, Jan 19 2025
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