The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A154809 Numbers whose binary expansion is not palindromic. 14
 2, 4, 6, 8, 10, 11, 12, 13, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 29, 30, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 64, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Complement of A006995. The (a(n)-n+1)-th binary palindrome equals the greatest binary palindrome <= a(n). The corresponding formula identity is: A006995(a(n)-n+1)=A206913(a(n)). - Hieronymus Fischer, Mar 18 2012 A145799(a(n)) < a(n). - Reinhard Zumkeller, Sep 24 2015 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 FORMULA A030101(n) != n. - David W. Wilson, Jun 09 2009 A178225(a(n)) = 0. - Reinhard Zumkeller, Oct 21 2011 From Hieronymus Fischer, Feb 19 2012 and Mar 18 2012: (Start) Inversion formula: If d is any number from this sequence, then the index number n for which a(n)=d can be calculated by n=d+1-A206915(A206913(d)). Explicitly: Let p=A206913(d), m=floor(log_2(p)) and p>2, then: a(n)=d+1+(((5-(-1)^m)/2) + sum(k=1...floor(m/2)|(floor(p/2^k) mod 2)/2^k))*2^floor(m/2). Example 1: d=1000, A206913(d)=975, A206915(975)=62, hence n=1001-62=939. Example 2: d=10^6, A206913(d)=999471, A206915(999471)=2000, hence n=1000001-2000=998001. Recursion formulas: a(n+1)=a(n)+1+A178225(a(n)+1) Also: Case 1: a(n+1)=a(n)+2, if A206914(a(n))=a(n)+1; Case 2: a(n+1)=a(n)+1, else. Also: Case 1: a(n+1)=a(n)+1, if A206914(a(n))>a(n)+1; Case 2: a(n+1)=a(n)+2, else. Iterative calculation formula: Let f(0):=n+1, f(j):=n-1+A206915(A206913(f(j-1)) for j>0; then a(n)=f(j), if f(j)=f(j-1). The number of necessary steps is typically <4 and is limited by O(log_2(n)). Example 3: n=1000, f(0)=1001, f(1)=1061, f(2)=1064=f(3), hence a(1000)=1064. Example 4: n=10^6, f(0)=10^6+1, f(1)=1001999, f(2)=1002001=f(3), hence a(10^6)=1002001. Formula identity: a(n) = n-1 + A206915(A206913(a(n))). (End) EXAMPLE a(1)=2, since 2 = 10_2 is not binary palindromic. MAPLE ispali:= proc(n) local L; L:= convert(n, base, 2); ListTools:-Reverse(L)=L end proc: remove(ispali, [\$1..1000]); # Robert Israel, Jul 05 2015 MATHEMATICA palQ[n_Integer, base_Integer]:=Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], ! palQ[#, 2] &] (* Vincenzo Librandi, Jul 05 2015 *) PROG (Haskell) a154809 n = a154809_list !! (n-1) a154809_list = filter ((== 0) . a178225) [0..] (Magma) [n: n in [0..600] | not (Intseq(n, 2) eq Reverse(Intseq(n, 2)))]; // Vincenzo Librandi, Jul 05 2015 (PARI) isok(n) = binary(n) != Vecrev(binary(n)); \\ Michel Marcus, Jul 05 2015 (Python) def A154809(n): def f(x): return n+(x>>(l:=x.bit_length())-(k:=l+1>>1))-(int(bin(x)[k+1:1:-1], 2)>(x&(1<

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified August 10 05:36 EDT 2024. Contains 375044 sequences. (Running on oeis4.)