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 A164126 First differences of A006995. 3
 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, 24, 2, 16, 24, 16, 12, 16, 24, 16, 6, 16, 24, 16, 12, 16, 24, 16, 2, 48, 24, 48, 12, 48, 24, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Contribution from Hieronymus Fischer, Feb 18 2012: (Start) 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. Example: for n = 15 = 2^4 - 1, we have the (2^3+1)-tuple (2,8,12,8,6,8,12,8,2). 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. 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). 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:                           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 24 2    16 24 16 12 16 24 16 6 16 24 16 12 16 24 16 2    48 24 48 12 48 24 48 6 48 24 48 12 48 24 48 2 . (End) Decremented by 1, also the sequence of run lengths of 0's in A178225. - Hieronymus Fischer, Feb 19 2012 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 FORMULA a(n) = A006995(n+1) - A006995(n). Contribution from Hieronymus Fischer, Feb 17 2012: (Start) a(4*2^m - 1) = a(6*2^m - 1) = 2; a(5*2^m - 1) = a(7*2^m - 1) = 6 (for m > 0); Let m = floor(log_2(n)), then   Case 1: a(n) = 2, if n+1 = 2^(m+1) or n+1 = 3*2^(m-1);   Case 2: a(n) = 2^(m-1), if n = 0(mod 2) and n < 3*2^(m-1);   Case 3: a(n) = 3*2^(m-1), if n = 0(mod 2) and n >= 3*2^(m-1);   Case 4: a(n) = 3*2^(m-1)/gcd(n+1-2^m, 2^m), otherwise. Cases 2-4 above can be combined as   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). Recursion formula: Let m = floor(log_2(n)); then   Case 1: a(n) = 2*a(n-2^(m-1)), if 2^m <= n < 2^m + 2^(m-2) - 1;   Case 2: a(n) = 6, if n = 2^m + 2^(m-2) - 1;   Case 3: a(n) = a(n-2^(m-2)), if 2^m + 2^(m-2) <= n < 2^m + 2^(m-1) - 1;   Case 4: a(n) = 2, if n = 2^m + 2^(m-1) - 1;   Case 5: a(n) = (2 + (-1)^n)*a(n-2^(m-1)), otherwise (which means 2^m + 2^(m-1) <= n < 2^(m+1)). (End) EXAMPLE a(1) = A006995(2) - A006995(1) = 1 - 0 = 1. MATHEMATICA f[n_]:=FromDigits[RealDigits[n, 2][[1]]]==FromDigits[Reverse[RealDigits[n, 2][[1]]]]; a=1; lst={}; Do[If[f[n], AppendTo[lst, n-a]; a=n], {n, 1, 8!, 1}]; lst CROSSREFS Cf. A006995, A164124, A029971, A016041, A164125. See A178225. Sequence in context: A198889 A329814 A130754 * A261902 A163368 A151948 Adjacent sequences:  A164123 A164124 A164125 * A164127 A164128 A164129 KEYWORD nonn,base,easy AUTHOR Vladimir Joseph Stephan Orlovsky, Aug 10 2009 EXTENSIONS a(1) changed to 1 and keyword:base added by R. J. Mathar, Aug 26 2009 STATUS approved

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Last modified August 8 19:29 EDT 2020. Contains 336298 sequences. (Running on oeis4.)