A268489 a(n) + f(a(n)) is always odd, where f is the sequence obtained by replacing each term with the corresponding row of A027746 (list of its prime factors). Lexicographic first such sequence without duplicates.

2, 1, 5, 7, 4, 10, 11, 13, 8, 17, 14, 16, 15, 19, 22, 28, 27, 30, 34, 35, 36, 38, 39, 42, 43, 44, 46, 47, 48, 50, 51, 56, 57, 59, 64, 66, 67, 69, 70, 72, 75, 77, 79, 82, 84, 88, 89, 91, 92, 94, 96, 98, 103, 104, 107, 110, 112, 113, 119, 121, 126, 129, 132

Offset

1,1

Links

Table of n, a(n) for n=1..63.

Example

The reasoning below leads to the following list of indices n, prime factors f(n) and terms a(n):
n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11,12,13,14,15,16,...
F = 2, 1, 5, 7, 2, 2, 2, 5,11, 13, 2, 2, 2,17, 2, 7,...
a = 2, 1, 5, 7, 4,10,11,13, 8, 17,14,...
a(1)=1=f(1) is not possible because 1+1 is not odd, but a(1)=2=f(1) does not lead to a contradiction.
Then, a(2)=1=f(2) is also possible, because a(2)+f(a(2)) = 1+2 is odd.
a(3)=3=f(3) not possible because 3+3 is even. a(3)=4=f(3) is not possible because this leads to f(4)=2. But a(3)=5=f(3) is possible, leading to the constraint f(5) = 2, the only possible even prime factor.
We note that 3 can never occur because 3+f(3) = 3+3 is not odd.
a(4)=4 => f(4)=2 is not possible because 4+2 is not odd. a(4)=6 => f(5)=3 is not possible. a(4)=7=f(4) is possible, but requires that f(7) = 2.
a(5)=2q because f(5)=2, a(5)=4, i.e., q=2=f(6) is ok since f(4)=7 is odd.
We note that 6 can never occur because 6+f(6) = 6+2 is not odd.
a(6)=2q because f(7)=2. The number 6 is excluded. a(6)=8 is not possible because then f(8)=2 is not odd. a(6)=10, f(8)=5 is possible, and must have f(10)=odd.
a(7)=8 => f(9,10,11)=2 is not possible because f(10)=odd. a(7)=9 => f(9,10)=3 is not possible (9+3 is even). a(7)=11= f(9) is possible, we must have f(11)=2.
Now f(10) must be odd and f(11)=2 so a(8) must be an odd prime, a(8)=13=f(10) is the smallest available. Thus f(13)=2.
a(9)=2x because f(11)=2 and we can use a(9)=8, f(11,12,13)=2, which is possible because 8+f(8)=13. And so on...

Prog

(PARI) {list(n, s=[], check=(s, f=concat(apply(A027746_row, s)))->!for(i=1, #s, s[i]<=#f&&(bittest(s[i]+f[s[i]], 0)||return)))=for(n=2, n, S=Set(s); s=concat(s, 1); for(k=1, S[#S]+99, !setsearch(S, k)&&(s[n]=k)&&check(s)&&break)); s}

Crossrefs

Sequence in context: A249265 A259972 A193762 * A198422 A105459 A227048

Adjacent sequences: A268486 A268487 A268488 * A268490 A268491 A268492

Keyword

nonn

Author

M. F. Hasler, Feb 06 2016

Status

approved