A227324 Result of changing both the prime indices and the exponents in the prime factorization of n: increment odd values, decrement even values.
1, 9, 4, 3, 49, 36, 25, 81, 2, 441, 169, 12, 121, 225, 196, 27, 361, 18, 289, 147, 100, 1521, 841, 324, 7, 1089, 16, 75, 529, 1764, 1369, 729, 676, 3249, 1225, 6, 961, 2601, 484, 3969, 1849, 900, 1681, 507, 98, 7569, 2809, 108, 5, 63, 1444, 363, 2209, 144
Offset: 1
Examples
n = 2^3 => a(n) = 3^4 = 81. n = 3^2 => a(n) = 2^1 = 2.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
Programs
-
Maple
a:= n-> mul(ithprime(i[1])^i[2], i=map(x->map(y->y-1+2*irem(y, 2), [numtheory[pi](x[1]), x[2]]), ifactors(n)[2])): seq(a(n), n=1..100); # Alois P. Heinz, Jul 17 2013
-
Mathematica
a[n_] := If[n == 1, 1, Product[{p, e} = pe; Prime[BitXor[PrimePi[p] - 1, 1] + 1]^(BitXor[e - 1, 1] + 1), {pe, FactorInteger[n]}]]; Array[a, 100] (* Jean-François Alcover, May 31 2019, after Andrew Howroyd *)
-
PARI
a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); prime( bitxor( primepi(p)-1, 1)+1)^(bitxor(e-1, 1)+1))} \\ Andrew Howroyd, Jul 23 2018
-
Python
primes = [2]*2 primes[1] = 3 def addPrime(k): for p in primes: if k%p==0: return if p*p > k: break primes.append(k) for n in range(5, 1000000, 6): addPrime(n) addPrime(n+2) for n in range(1,99): p = 1 j = n i = 0 while j>1: e = 0 while j % primes[i] == 0: j /= primes[i] e+=1 if e: e = ((e-1)^1) + 1 p*= primes[i^1]**e i += 1 print(str(p), end=', ')
Formula
Sum_{k=1..n} a(k) ~ c * n^3, where c = (1/3) * Product_{p prime} ((p-1)*(p^6 + q(p) +(p^3-1)*q(p)^2))/(p^7 - p*q(p)^2) = 0.3120270364..., where q(p) = nextprime(p) = A151800(p) if p has an odd index, and q(p) = prevprime(p) = A151799(p) otherwise. - Amiram Eldar, Sep 17 2023
Extensions
Keyword:mult added by Andrew Howroyd, Jul 23 2018
Comments