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
KEYWORD
base,easy,nice,nonn
AUTHOR
M. F. Hasler, Jul 17 2008
STATUS
approved