A379127 a(1) = 1; for n > 1, a(n) is the largest proper divisor d of 2n-1 such that A048720(A065621(d),(2n-1)/d) is equal to 2n-1.

1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 11, 5, 1, 1, 1, 1, 3, 1, 7, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 43, 1, 19, 9, 1, 1, 1, 1, 1, 1, 1, 1, 3, 5, 1, 1, 23, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Antti Karttunen, Dec 21 2024

Keywords

• nonn

Comments

The position of the first occurrence of odd numbers k = 1, 3, 5, 7, 9, ... in this sequence is given by (1/2) * (A379128(2*k-1)+1).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences related to binary expansion of n
• Index entries for sequences operating on polynomials in ring GF(2)[X]

Crossrefs

Odd bisection of A325567.

Cf. A048720, A065621, A277320, A379128.

Programs

PARI

A048720(b, c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2);
A065621(n) = bitxor(n-1, n+n-1);
A325567(n) = if(1==n, n, fordiv(n, d, if((d>1)&&A048720(A065621(n/d), d)==n, return(n/d))));
A379127(n) = A325567(2*n-1);

Formula

a(n) = A325567(2*n-1).