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a(n) = (A172126(n) - 1)/3.
2

%I #17 Sep 01 2024 09:37:19

%S 0,1,3,5,13,21,28,53,85,113,213,227,341,453,853,909,1365,1813,1820,

%T 3413,3637,5461,7253,7281,13653,14549,14563,21845,29013,29125,54613,

%U 58197,58253,87381,116053,116501,116508,218453,232789,233013,349525,464213,466005,466033

%N a(n) = (A172126(n) - 1)/3.

%C 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

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

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

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

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

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

%H Amiram Eldar, <a href="/A172143/b172143.txt">Table of n, a(n) for n = 1..10000</a>

%t 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 *)

%o (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, ", ")))

%Y Cf. A002450, A072197, A172126, A172241.

%K nonn

%O 1,3

%A _Ralf Stephan_, Nov 19 2010

%E More terms from _Amiram Eldar_, Sep 01 2024