A154809 Numbers whose binary expansion is not palindromic.
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
Offset: 1
Examples
a(1)=2, since 2 = 10_2 is not binary palindromic.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Programs
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Haskell
a154809 n = a154809_list !! (n-1) a154809_list = filter ((== 0) . a178225) [0..]
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Magma
[n: n in [0..600] | not (Intseq(n, 2) eq Reverse(Intseq(n, 2)))]; // Vincenzo Librandi, Jul 05 2015
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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
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Mathematica
palQ[n_Integer, base_Integer]:=Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], ! palQ[#, 2] &] (* Vincenzo Librandi, Jul 05 2015 *) -
PARI
isok(n) = binary(n) != Vecrev(binary(n)); \\ Michel Marcus, Jul 05 2015
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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<
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).
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:
(End)
Extensions
Extended by Ray Chandler, Mar 14 2010
Comments