A048700 Binary palindromes of odd length (written in base 10).

1, 5, 7, 17, 21, 27, 31, 65, 73, 85, 93, 99, 107, 119, 127, 257, 273, 297, 313, 325, 341, 365, 381, 387, 403, 427, 443, 455, 471, 495, 511, 1025, 1057, 1105, 1137, 1161, 1193, 1241, 1273, 1285, 1317, 1365, 1397, 1421, 1453, 1501, 1533, 1539, 1571, 1619, 1651

Offset: 1

Author

Antti Karttunen, Mar 07 1999

Keywords

• nonn
• base

Comments

Note: you get A006995 (all binary palindromes) if you take (after zero) alternatively 2^n (starting from 2^0 = 1) terms from A048700 and as many from A048701 and then each time, twice as many from both.

A178225(a(n)) = 1. - Reinhard Zumkeller, Oct 21 2011

Comment from Altug Alkan, Dec 03 2015: (Start)

a(6*k) is divisible by 9 for k > 0.

a(3*k+(-1)^k-2) is divisible by 3 for k > 1.

The minimum value of a(n+1) - a(n) occurs when n = 2.

A014551(n) appears in this sequence for n > 0. (End)

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A048701 (binary palindromes of even length), A002113 (decimal palindromes), A006995 (all binary palindromes).

Cf. also A178225.

Programs

Haskell

import Data.Set (singleton, deleteFindMin, insert)
import Data.List (unfoldr)
a048700 n = a048700_list !! (n-1)
a048700_list = f 1 $ singleton 1 where
   f z s = m : f (z+1) (insert (c 0) (insert (c 1) s')) where
     c d = foldl (\v d -> 2 * v + d) 0 $ (reverse b) ++ [d] ++ b
     b = unfoldr
         (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2) z
     (m,s') = deleteFindMin s
-- Reinhard Zumkeller, Oct 21 2011

Maple

bit_i := (x,i) -> `mod`(floor(x/(2^i)),2);
floor_log_2 := proc(n) local nn,i: nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi: nn := floor(nn/2); od: end:

Mathematica

Select[Range@ 1651, # == Reverse@ # && OddQ@ Length@ # &@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Dec 03 2015 *)

PARI

{a(n) = local(f); if( n<1, 0, f = length(binary(n)) - 1; 2^f*n + sum(i=1, f, bittest(n,i) * 2^(f-i)))}; /* Michael Somos, Nov 27 2002 */

Python

def A048700(n):
    s = bin(n)[2:]
    return int(s+s[-2::-1],2) # Chai Wah Wu, Feb 26 2021

Formula

a(n) = (2^(floor_log_2(n)))*n + sum('(bit_i(n, i)*(2^(floor_log_2(n)-i)))', 'i'=1..floor_log_2(n));

a(A047264(n)) mod 3 = 0, for n > 1. - Altug Alkan, Dec 03 2015