A266195 Match-making permutation: start with a(1) = 1, then always choose for a(n) the least unused number such that multiplying a(n) by a(n-1) does not produce any carries when performed in base 2.
1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 17, 13, 32, 14, 18, 20, 19, 33, 15, 34, 22, 64, 21, 24, 36, 28, 65, 23, 66, 25, 40, 35, 72, 42, 48, 37, 68, 26, 128, 27, 129, 29, 130, 30, 132, 31, 256, 38, 80, 49, 73, 56, 136, 41, 96, 69, 144, 67, 84, 97, 137, 112, 145, 134, 160, 50, 133, 76, 161, 100, 257, 39, 258, 43, 260, 44
Offset: 1
Examples
For n=11, we first note that a(10) = 10, and the least unused number after a(1) .. a(10) is 11. Trying to multiply 10 (= 1010_2) and 11 (= 1011_2), in the binary system results in 1011 * 1010 ------- c1011 1011 ------- 1101110 = 110, and we see that there's a carry-bit (marked c) affecting the result, thus A048720(10,11) < 10*11 and A061858(10,10) > 0, thus we cannot select 11 for a(11). The next unused number is 12, and indeed, for numbers 10 and 12 (= 1100_2), the binary multiplication results in 1100 * 1010 ------- 1100 1100 ------- 1111000 = 120, which is a clean product without carries (i.e., A061858(10,12) = 0), thus 12 is selected to be a match for 10, and we set a(11) = 12. For a(49) = 31 (= 11111_2) and a(50) = 256 (= 100000000_2) the multiplication results in 100000000 * 11111 ------------- 100000000 100000000 100000000 100000000 100000000 ------------- 1111100000000 = 7936, and we see that the carryless product is this time obtained almost trivially, as the other number is so much larger and more spacious than the other that they can easily avoid any clashing bits that would produce carries.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..2694
- Eric Angelini, a(n)*a(n+1) shows at least twice the same digit, Posting on SeqFan-list Dec 21 2015. [Source of inspiration for this sequence.]
- Rémy Sigrist, Logarithmic scatterplot of the first 500000 terms
- Index entries for sequences that are permutations of the natural numbers
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