A141708 Least positive multiple of 2n-1 which is palindromic in base 2.

1, 3, 5, 7, 9, 33, 65, 15, 17, 513, 21, 2047, 325, 27, 1421, 31, 33, 455, 2553, 195, 1025, 129, 45, 4841, 1421, 51, 3339, 165, 513, 6077, 427, 63, 65, 1273, 2553, 10437, 73, 975, 231, 1501, 891, 3735, 85, 3219, 2047, 273, 93, 2565, 5917, 99, 23533, 4841, 1365, 107

Offset: 1

Author

M. F. Hasler, Jul 17 2008

Keywords

• base
• easy
• nice
• nonn

Comments

Even numbers cannot be palindromic in base 2 (unless leading zeros are considered), that's why we search for odd numbers 2n-1 their smallest multiple k(2n-1) which is palindromic in base 2. Obviously this must always be odd.

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A050782, A141707, A062279.

Programs

Haskell

a141708 n = a141707 n * (2 * n - 1) -- Reinhard Zumkeller, Apr 20 2015

Mathematica

pal2[n_]:=Module[{k=1},While[IntegerDigits[k n,2] != Reverse[ IntegerDigits[ k n,2]],k++];k n]; pal2/@Range[1,121,2] (* Harvey P. Dale, Feb 29 2012 *)

PARI

A141708(n,L=10^9)={ n=2*n-1; forstep(k=1,L,2, binary(k*n)-vecextract(binary(k*n),"-1..1") || return(k*n))}

Python

def binpal(n): b = bin(n)[2:]; return b == b[::-1]
def a(n):
    m = 2*n - 1
    km = m
    while not binpal(km): km += m
    return km
print([a(n) for n in range(1, 55)]) # Michael S. Branicky, Mar 20 2022

Formula

a(n) = (2n-1)*A141707(n).