

A141707


Least k>0 such that (2n1)k is palindromic in base 2.


4



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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 2n1 the kvalues such that k(2n1) 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*61=11 and 2*11=22 are not palindromic in base 2, but 3*11=33 is.


MATHEMATICA

lkp[n_]:=Module[{k=1, n2=2n1}, 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*n1; 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


CROSSREFS

Cf. A050782, A141708, A178225, A044051.
Sequence in context: A091084 A016610 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



