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A268489
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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.
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0
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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
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OFFSET
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1,1
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LINKS
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EXAMPLE
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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...
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PROG
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(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}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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