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

1,6

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

Links

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

Example

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.

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 *)

Prog

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

Crossrefs

Cf. A050782, A141708, A178225, A044051.

Sequence in context: A016610 A356551 A305470 * A329593 A263490 A190180

Adjacent sequences: A141704 A141705 A141706 * A141708 A141709 A141710

Keyword

base,easy,nice,nonn

Author

M. F. Hasler, Jul 17 2008

Status

approved