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Search: id:A038122
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| A038122 |
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Start with {1,2,...,n}, replace any two numbers a,b by |a^2-b^2|, repeat until single number k remains; a(n) = minimal value of k. |
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+0
1
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| 1, 3, 0, 16, 15, 63, 8, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0, 3, 1, 0, 0, 1, 3, 0, 4, 3, 3, 4, 0
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Due mostly to the efforts of Dean Hickerson (dean.hickerson(AT)yahoo.com), supported by David W. Wilson and Michael Kleber, it is now known that this has period 12 beginning at n=8.
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LINKS
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Dean Hickerson and Michael Kleber, "Reducing a Set by Subtracting Squares", J. Integer Sequences, Vol. 2, 1999, #4.
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FORMULA
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For n<4 and n>7, a(n) = n*(n+1)/2 mod 6 = A010875(A000217(n)). - Henry Bottomley (se16(AT)btinternet.com), Feb 24 2003
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EXAMPLE
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a(2) = 3 from (1,2); a(3) = 0 from ((1,2),3); a(4) = 16 from (((1,2),3),4); a(5) = 15 from ((((2,3),5),1),4)
a(6) = 63 from (((1,4),(3,5)),(2,6)) [ Michael Kleber (michael.kleber(AT)gmail.com) ]
a(7) = 8 from (((((4,5),6),(2,7)),1),3) [ Kleber ]
a(8) = 0 from ((((4,5),7)(2,6))((1,3),8)) [ Guy ]
a(9) = 3 from (2,(1,(((6,7),((3,4),8)),(5,9)))) [ Kleber ]
a(10)= 1 from ((((((((4,5),9),6),(8,10)),2),3),7),1) [ This and the following are due to Dean Hickerson (dean.hickerson(AT)yahoo.com) ]
a(11)= 0 from ((((((3,7),(9,11)),6),(8,10)),(1,2)),(4,5))
a(12)= 0 from ((((((1,3),7),(8,10)),(((5,6),9),(11,12))),2),4)
a(13)= 1 from (((((((((3,7),(9,11)),6),(8,10)),5),(12,13)),2),4),1) ...
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CROSSREFS
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Sequence in context: A013351 A013407 A013494 this_sequence A143779 A036968 A024040
Adjacent sequences: A038119 A038120 A038121 this_sequence A038123 A038124 A038125
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KEYWORD
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nonn,nice,easy
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AUTHOR
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R. K. Guy (rkg(AT)cpsc.ucalgary.ca)
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