A172143 a(n) = (A172126(n) - 1)/3.

0, 1, 3, 5, 13, 21, 28, 53, 85, 113, 213, 227, 341, 453, 853, 909, 1365, 1813, 1820, 3413, 3637, 5461, 7253, 7281, 13653, 14549, 14563, 21845, 29013, 29125, 54613, 58197, 58253, 87381, 116053, 116501, 116508, 218453, 232789, 233013, 349525, 464213, 466005, 466033

Offset: 1

Author

Ralf Stephan, Nov 19 2010

Keywords

• nonn

Comments

Conjecture: sequence consists of an infinite number of subsequences S(m,0) = A172241(n) = (1/18)[8^n-(-1)^n-9], m>0, S(m,n+1) = 4*S(m,n)+1. The first subsequences

S(1,n) = A002450(n) = (4^n-1)/3 = 0, 1, 5, 21, 85, ...,

S(2,n) = A072197(n) = (10*4^n-1)/3 = 3, 13, 53, 213, ...,

S(3,n) = (85*4^n-1)/3 = 28, 113, 453, ...,

S(4,n) = (682*4^n-1)/3 = 227, 909, 3637, ..., and generally,

S(m,n) = [(3*A172241(m) + 1) * 4^n - 1]/3.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002450, A072197, A172126, A172241.

Programs

Mathematica

seq[max_] := Module[{kmax = Floor[Log[4, 3*max+1]], s = {}, s1, odd},Do[odd = (4^k-1)/3; s1 = 2^Range[0, Floor[Log2[max/odd]]] * odd; s = Join[s, s1], {k, 1, kmax}]; Select[(Union[s] - 1)/3, IntegerQ]]; seq[10^7] (* Amiram Eldar, Sep 01 2024 *)

PARI

for(n=1, 300000, o=3*n/2^valuation(n, 2)+1; b=ispower(o); if(b&&b%2==0&&round(sqrtn(o, b/2))==4&&(n-1)%3==0, print1((n-1)/3, ", ")))

Extensions

More terms from Amiram Eldar, Sep 01 2024