A164126 First differences of A006995.

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

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Aug 10 2009

Keywords

• nonn
• base
• easy

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

Examples

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

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

See A178225.

Programs

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

Python

def A164126(n):
    if n == 1: return 1
    m = (a:=1<<(l:=n.bit_length()-2))|(n&a-1)
    k = (m<Chai Wah Wu, Jun 11 2024

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)

Extensions

a(1) changed to 1 and keyword:base added by R. J. Mathar, Aug 26 2009