technical question for oeis-entries

[N. J. A. Sloane]

Gottfried

The simplest way is for you to take the internal format
for the sequence

%I A096773
%S A096773 0,3,1,13,5,53,21,213,85,853,341,3413,1365,13653,5461,54613,21845,
%T A096773 218453,87381,873813,349525,3495253,1398101,13981013,5592405,55924053,
%U A096773 22369621,223696213,89478485,894784853,357913941,3579139413,1431655765
%N A096773 a(n+2) = 4*a(n) + 1; a(1)=0, a(2) = 3.
%C A096773 Remainders for classes m of integers n (mod 2^(m+1)) which divide the integers in 2 groups after one Collatz (3x+1)-transformation.
%C A096773 With one 3x+1-transformation T(x;p) := x'= (3x+1)/2^p all numbers of the form x = i*2^N +a(N) -> x' = i*6 + 1 for odd N>2, -> x' = i*6 + 5 for even N. x = 4*i + 3 -> x' = 6*i + 5, x = 8*i + 1 -> x' = 6*i + 1, x =16*i +13 -> x' = 6*i + 5, x = 32*i + 1 -> x' = 6*i + 1, x =64*i +53 -> x' = 6*i + 5, x =128*i + 1 -> x' = 6*i + 1, with i a free parameter >=0
%F A096773 a(2m) = (5*2^(2m-1) - 1)/3, a(2m-1) = (2^(2m-2)-1)/3.
%e A096773 a(1) = (2^0-1)/3 = 0, a(2) = (5*2^1 - 1) / 3 = 3, a(3) = (2^2-1)/3 = 1, a(4) = (5*2^3 - 1) / 3 =13, a(5) = (2^4-1)/3 = 5, a(6) = (5*2^5 - 1) / 3 =53, a(7) = (2^6-1)/3 =21.
%t A096773 a[1] = 0; a[2] = 3; a[n_] := a[n] = 4a[n - 2] + 1; Table[ a[n], {n, 35}] (from Robert G. Wilson v Aug 20 2004)
%Y A096773 Bisections are A002450 & A072197.
%K A096773 easy,nonn
%O A096773 1,2
%A A096773 Gottfried Helms (helms(AT)uni-kassel.de), Aug 15 2004
%E A096773 More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 20 2004


and to reformat it yourself - and then email me the result

Neil