A246161 Permutation of positive integers: a(1) = 1, a(A014580(n)) = A000069(1+a(n)), a(A091242(n)) = A001969(1+a(n)), where A000069 and A001969 are the odious and evil numbers, and A014580 resp. A091242 are the binary coded irreducible resp. reducible polynomials over GF(2).
1, 2, 4, 3, 5, 9, 8, 6, 10, 18, 7, 17, 11, 12, 20, 36, 15, 34, 19, 23, 24, 40, 72, 30, 16, 68, 39, 46, 48, 80, 13, 144, 60, 33, 136, 78, 21, 92, 96, 160, 37, 27, 288, 120, 66, 272, 14, 156, 43, 184, 192, 320, 75, 54, 35, 576, 240, 132, 22, 544, 25, 29, 312, 86, 368, 384, 41
Offset: 1
Keywords
Examples
Consider n=21. In binary it is 10101, encoding for polynomial x^4 + x^2 + 1, which factorizes as (x^2 + x + 1)(x^2 + x + 1) over GF(2), in other words, 21 = A048720(7,7). As such, it occurs as the 14th term in A091242, reducible polynomials over GF(2), coded in binary. By definition of this permutation, a(21) is thus obtained as A001969(1+a(14)). 14 in turn is 8th term in A091242, thus a(14) = A001969(1+a(8)). In turn, 8 = A091242(4), thus a(8) = A001969(1+a(4)), and 4 = A091242(1). By working the recursion back towards the toplevel, the result is a(21) = A001969(1+A001969(1+A001969(1+A001969(1+1)))) = 24. Consider n=35. In binary it is 100011, encoding for polynomial x^5 + x + 1, which factorizes as (x^2 + x + 1)(x^3 + x^2 + 1) over GF(2), in other words, 35 = A048720(7,13). As such, it occurs as the 26th term in A091242, thus a(35) = A001969(1+a(26)), and as 26 = A091242(18) and 18 = A091242(12) and 12 = A091242(7) and 7 = A014580(3) [the polynomial x^2 + x + 1 is irreducible over GF(2)], and 3 = A014580(2) and 2 = A014580(1), we obtain the result as a(35) = A001969(1+A001969(1+A001969(1+A001969(1+A000069(1+A000069(1+A000069(2))))))) = 136.
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