A141707 Least k>0 such that (2n-1)k is palindromic in base 2.

1, 1, 1, 1, 1, 3, 5, 1, 1, 27, 1, 89, 13, 1, 49, 1, 1, 13, 69, 5, 25, 3, 1, 103, 29, 1, 63, 3, 9, 103, 7, 1, 1, 19, 37, 147, 1, 13, 3, 19, 11, 45, 1, 37, 23, 3, 1, 27, 61, 1, 233, 47, 13, 1, 21, 23, 59, 525, 5, 1, 93, 23, 41, 1, 1, 49, 27, 13, 187, 87, 269, 15, 111, 13, 29, 7, 1, 13, 3

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 only for odd numbers 2n-1 the k-values such that k(2n-1) is palindromic in base 2. Obviously they are necessarily also odd.

a(A044051(n)) = 1. - Reinhard Zumkeller, Apr 20 2015

Examples

			a(1..5)=1 since 1,3,5,7,9 are already palindromic in base 2.
a(6)=3 since 2*6-1=11 and 2*11=22 are not palindromic in base 2, but 3*11=33 is.
		

Links

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

Crossrefs

Cf. A050782, A141708, A178225, A044051.

Programs

Haskell

a141707 n = head [k | k <- [1, 3 ..], a178225 (k * (2 * n - 1)) == 1]
-- Reinhard Zumkeller, Apr 20 2015

Mathematica

lkp[n_]:=Module[{k=1,n2=2n-1},While[IntegerDigits[k*n2,2]!= Reverse[ IntegerDigits[ k*n2,2]],k++];k]; Array[lkp,80] (* Harvey P. Dale, Mar 19 2016 *)

PARI

A141707(n,L=10^9)={ n=2*n-1; forstep(k=1,L,2, binary(k*n)-vecextract(binary(k*n),"-1..1") || return(k))}

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//m
print([a(n) for n in range(1, 80)]) # Michael S. Branicky, Mar 20 2022