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A141707 Least k>0 such that (2n-1)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 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

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

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Last modified July 28 17:07 EDT 2021. Contains 346335 sequences. (Running on oeis4.)