A232637 - OEIS

A232637

Odious numbers of order 2: a(n) = A000069(A000069(n)).

1, 2, 7, 13, 14, 21, 25, 26, 31, 37, 41, 42, 49, 50, 55, 61, 62, 69, 73, 74, 81, 82, 87, 93, 97, 98, 103, 109, 110, 117, 121, 122, 127, 133, 137, 138, 145, 146, 151, 157, 161, 162, 167, 173, 174, 181, 185, 186, 193, 194, 199, 205, 206, 213, 217, 218, 223, 229, 233, 234, 241, 242, 247, 253

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OFFSET

1,2

COMMENTS

Odious numbers with odious subscripts.

From Antti Karttunen, Nov 29 2013: (Start)

Starting from 4 and iterating A000069(4), A000069(A000069(4)), A000069(A000069(A000069(4))), etc. gives A004119 from its second term onward: 4, 7, 13, 25, 49, 97, 193, ..., which is thus a subsequence of this sequence from the term 7 onward.

Proof: All of the terms A004119(n) are odious although A004119(n)-1 is evil, and the formula for A000069(n) reduces to a(n) = 2n - 1 when n-1 is evil, and iterating that formula starting from 4 gives A004119 from 7 onward (cf. Philippe Deléham's formula there dated Feb 20 2004).

(End)

These numbers are never multiples of 4. Probably there are infinitely many multiples of m in this sequence for any m not divisible by 4. Equivalently, A233419(n) > 0 for all n. - Charles R Greathouse IV, Dec 05 2013

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..8193

FORMULA

a(n) = A000069(A000069(n)).

4n-6 <= a(n) <= 4n-3, see PARI script. - Charles R Greathouse IV, Dec 05 2013

EXAMPLE

The first odious number, A000069(1) = 1, and A000069(1) = 1, so a(1) = 1.

The second odious number, A000069(2) = 2, and A000069(2) = 2, so a(2) = 2.

Those were the only fixed points of A000069, and after that, we have:

The third odious number, A000069(3) = 4, and A000069(4) = 7, thus a(3) = 7.

The fourth odious number, A000069(4) = 7, and A000069(7) = 13, thus a(4) = 13.

PROG