A206915 -id:A206915 - cp's OEIS frontend

A206919 Sum of binary palindromes <= n.

0, 1, 1, 4, 4, 9, 9, 16, 16, 25, 25, 25, 25, 25, 25, 40, 40, 57, 57, 57, 57, 78, 78, 78, 78, 78, 78, 105, 105, 105, 105, 136, 136, 169, 169, 169, 169, 169, 169, 169, 169, 169, 169, 169, 169, 214, 214, 214, 214, 214, 214, 265, 265, 265, 265, 265, 265, 265, 265

Offset: 0

Hieronymus Fischer, Feb 18 2012

Sum of binary palindromes A006995(k) <= n.

Different from A206920.

			a(2)=1, since the only binary palindromes <= 1 are p=0 and p=1;
a(5)=9, since the sum of all binary palindromes <= 5 is 9 = 0 + 1 + 3 + 5.

Cf. A006995, A206913, A206915.

a(n) = sum(k=1, n, my(b=binary(k)); if (b==Vecrev(b), k)); \\ Michel Marcus, Sep 09 2018

a(n) = Sum_{k=1..A206915(A206913(n))} A006995(k).
a(n) = A206920(A206915(A206913(n))).
Let p = A206913(n) > 3, m = floor(log_2(p)), then
a(n) = (8/7)*((3/4)*(4-(-1)^m)/(3+(-1)^m)*2^(3*floor(m/2))-1) + (floor(p/2^floor(m/2)) mod 2)*p + 2^m + 1 + Sum_{k=1..floor(m/2)-1} (floor(p/2^k) mod 2)*(2^k+2^(m-k)+2^(m-floor(m/2)+1)*(4^(floor(m/2)-k-1)-1)+(2-(-1)^m)*2^floor(m/2)+2^(floor(m/2)-k)*(p-floor((p mod (2^(m-k+1)))/2^k)*2^k)). - [Corrected; missing factor to the sum term (2-(-1)^m) pasted by the author, Sep 08 2018]

A206922 Root of the n-th binary palindrome. Least number r > 1 such that A006995(n) can be represented by a finite or infinite number of iterations A006995(A006995(A006995(...(...(r))...).

Original entry on oeis.org

2, 2, 3, 4, 4, 6, 4, 8, 6, 10, 11, 12, 13, 14, 4, 16, 8, 18, 19, 20, 6, 22, 23, 24, 25, 26, 10, 28, 29, 30, 11, 32, 12, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 13, 46, 47, 48, 49, 50, 14, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 4, 64, 16, 66, 67, 68

Offset: 1

Hieronymus Fischer, Mar 12 2012

If n is not a binary palindrome, then a(n)=n.

For n>3: a(n)

For n<>3: The number of iterations such that A006995(n)= A006995(A006995(A006995(...(...(r))...) is given by A206921(n).

			a(1)=2, since A006995(1) = 0 = A006995(A006995(2)).
a(3)=3, since A006995(3) = 3 = A006995(A006995(A006995(...(3)...).
a(7)=4, since A006995(7) = 15 = A006995(A006995(A006995(4)).
a(9)=6, since A006995(9) = 21 = A006995(A006995(6)).

Cf. A006995, A206921, A178225, A206915, A154809.

/* C program fragment, omitting formal details, n!=3 */
k=0;
p=A006995(n);
while A178225(p)==1
{
  k++;
  p=A206915(p);
}
return p;

a(n) <= n for n > 1.
a(n)=p(k), where p(k) can be determined by the following iteration: set k=0, p(0)=A006995(n). Repeat while A178225(p(k))==1, set k=k+1, p(k)=A206915(p(k-1)) end repeat [for n<>3].
Recursion for n<>3:
  Case 1: a(n)=n, if n is not a binary palindrome;
  Case 2: a(n)=a(A206915(n)), else.
Formally: a(n)=if (A178225(n)==0) then n else a(A206915(n)).

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A206919 Sum of binary palindromes <= n.

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