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 A164126 First differences of A006995. 3

%I

%S 1,2,2,2,2,6,2,4,6,4,2,12,6,12,2,8,12,8,6,8,12,8,2,24,12,24,6,24,12,

%T 24,2,16,24,16,12,16,24,16,6,16,24,16,12,16,24,16,2,48,24,48,12,48,24,

%U 48,6,48,24,48,12,48,24,48,2,32,48,32,24,32,48,32,12,32,48,32,24,32,48,32,6

%N First differences of A006995.

%C Contribution from _Hieronymus Fischer_, Feb 18 2012: (Start)

%C From the formula section it follows that a(2^m - 1 + 2^(m-1) - k) = a(2^m - 1 + k) for 0 <= k <= 2^(m-1), as well as a(2^m - 1 + 2^(m-1) - k) = 2 for k=0, 2^(m-1) and a(2^m - 1 + 2^(m-1) - k) = 6 for k=2^(m-2), hence, starting from positions n=2^m-1, the following 2^(m-1) terms form symmetric tuples limited on the left and on the right by a '2' and always having a '6' as the center element.

%C Example: for n = 15 = 2^4 - 1, we have the (2^3+1)-tuple (2,8,12,8,6,8,12,8,2).

%C Further on, since a(2^m - 1 + 2^(m-1) + k) = a(2^(m+1) - 1 - k) for 0 <= k <= 2^(m-1) an analogous statement holds true for starting positions n = 2^m + 2^(m-1) - 1.

%C Example: for n = 23 = 2^4 + 2^3 - 1, we find the (2^3+1)-tuple (2,24,12,24,6,24,12,24,2).

%C If we group the sequence terms according to the value of m=floor(log_2(n)), writing those terms together in separate lines and opening each new line for n >= 2^m + 2^(m-1), then a kind of a 'logarithmic shaped' cone end will be formed, where both the symmetry and the calculation rules become obvious. The first 63 terms are depicted below:

%C 1

%C 2

%C 2

%C 2 2

%C 6 2

%C 4 6 4 2

%C 12 6 12 2

%C 8 12 8 6 8 12 8 2

%C 24 12 24 6 24 12 24 2

%C 16 24 16 12 16 24 16 6 16 24 16 12 16 24 16 2

%C 48 24 48 12 48 24 48 6 48 24 48 12 48 24 48 2

%C .

%C (End)

%C Decremented by 1, also the sequence of run lengths of 0's in A178225. - _Hieronymus Fischer_, Feb 19 2012

%H Vincenzo Librandi, <a href="/A164126/b164126.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A006995(n+1) - A006995(n).

%F Contribution from _Hieronymus Fischer_, Feb 17 2012: (Start)

%F a(4*2^m - 1) = a(6*2^m - 1) = 2;

%F a(5*2^m - 1) = a(7*2^m - 1) = 6 (for m > 0);

%F Let m = floor(log_2(n)), then

%F Case 1: a(n) = 2, if n+1 = 2^(m+1) or n+1 = 3*2^(m-1);

%F Case 2: a(n) = 2^(m-1), if n = 0(mod 2) and n < 3*2^(m-1);

%F Case 3: a(n) = 3*2^(m-1), if n = 0(mod 2) and n >= 3*2^(m-1);

%F Case 4: a(n) = 3*2^(m-1)/gcd(n+1-2^m, 2^m), otherwise.

%F Cases 2-4 above can be combined as

%F Case 2': a(n) = (2 - (-1)^(n-(n-1)*floor(2*n/(3*2^m))))*2^(m-1)/gcd(n+1-2^m, 2^m).

%F Recursion formula:

%F Let m = floor(log_2(n)); then

%F Case 1: a(n) = 2*a(n-2^(m-1)), if 2^m <= n < 2^m + 2^(m-2) - 1;

%F Case 2: a(n) = 6, if n = 2^m + 2^(m-2) - 1;

%F Case 3: a(n) = a(n-2^(m-2)), if 2^m + 2^(m-2) <= n < 2^m + 2^(m-1) - 1;

%F Case 4: a(n) = 2, if n = 2^m + 2^(m-1) - 1;

%F Case 5: a(n) = (2 + (-1)^n)*a(n-2^(m-1)), otherwise (which means 2^m + 2^(m-1) <= n < 2^(m+1)).

%F (End)

%e a(1) = A006995(2) - A006995(1) = 1 - 0 = 1.

%t f[n_]:=FromDigits[RealDigits[n,2][]]==FromDigits[Reverse[RealDigits[n, 2][]]]; a=1;lst={};Do[If[f[n],AppendTo[lst,n-a];a=n],{n,1,8!,1}]; lst

%Y Cf. A006995, A164124, A029971, A016041, A164125.

%Y See A178225.

%K nonn,base,easy

%O 1,2

%A _Vladimir Joseph Stephan Orlovsky_, Aug 10 2009

%E a(1) changed to 1 and keyword:base added by _R. J. Mathar_, Aug 26 2009

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Last modified November 15 19:18 EST 2019. Contains 329149 sequences. (Running on oeis4.)