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 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))…). 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS If n is not a binary palindrome, then a(n)=n. For n>3: a(n)3: The number of iterations such that A006995(n)= A006995(A006995(A006995(…(…(r))…) is given by A206921(n). LINKS FORMULA 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)). EXAMPLE 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)). PROG /* C program fragment, omitting formal details, n!=3 */ k=0; p=A006995(n); while A178225(p)==1 {   k++;   p=A206915(p); } return p; CROSSREFS Cf. A006995, A206921, A178225, A206915, A154809. Sequence in context: A289677 A113967 A205386 * A276775 A271169 A224979 Adjacent sequences:  A206919 A206920 A206921 * A206923 A206924 A206925 KEYWORD nonn,base AUTHOR Hieronymus Fischer, Mar 12 2012 STATUS approved

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Last modified March 23 14:17 EDT 2019. Contains 321431 sequences. (Running on oeis4.)