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Search: id:A036840
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| A036840 |
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a(n) = average of the repeating terms of {T(n,k)} rounded to the nearest integer (rounding up if there's a choice), if {T(n,k)} is eventually periodic; = 0 otherwise. Here T(n,k) is the "phi/sigma tug-of-war sequence with seed n" defined by T(n,1) = phi(n), T(n,2) = sigma(phi(n)), T(n,3) = phi(sigma(phi(n))), ..., T(n,k) = phi(T(n,k-1)) if k is odd and = sigma(T(n,k-1)) if k is even. |
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+0
3
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| 1, 1, 3, 3, 7, 3, 7, 7, 7, 7, 7, 7, 20, 7, 12, 12, 39, 7, 39, 12, 20, 7, 20, 12, 20, 20, 39, 20, 39, 12, 39, 39, 20, 39, 39, 20, 154, 39, 39, 39, 39, 20, 154, 20, 39, 20, 39, 39, 154, 20, 154, 39, 154, 39, 39, 39, 154, 39, 39, 39, 100, 39, 154, 154, 100, 20, 100, 154, 39, 39
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Conjecture: a(n) is never 0; i.e. the sequence {T(n,k)} is eventually periodic for every n.
a(n) - n can be thought of as the final score in the phi/sigma tug-of-war with seed n. For example a(5) - 5 = 7 - 5 = 2, so sigma wins by "2 points" over phi at 5. a(8) - 8 = 7 - 8 = -1, so phi wins by "1 point" over sigma at 8. a(3) - 3 = 3 - 3 = 0, so it is a tie at 3. Are sigma's margins of victory over phi bounded? Are phi's bounded?
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EXAMPLE
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The sequence {T(5,k)} is 4, 7, 6, 12, 4, 7, .... The average of the repeating numbers is 7.25 which rounds off to 7. So a(5) = 7. The sequence {T(37,k)} is 36, 91, 72, 195, 96, 252, 72, 195, .... The average of the repeating numbers is 153.75, which rounds off to 154. So a(37) = 154.
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MATHEMATICA
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a[ n_ ] := Module[ {}, For[ m=n; seq={}, !MemberQ[ seq, m ], m=DivisorSigma[ 1, EulerPhi[ m ] ], AppendTo[ seq, m ] ]; rp=Drop[ seq, Position[ seq, m ][ [ 1, 1 ] ]-1 ]; Floor[ 1/2+(Plus@@Join[ rp, EulerPhi/@rp ])/2/Length[ rp ] ] ]
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CROSSREFS
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Cf. A000010, A000203, A036845, A066437.
Sequence in context: A076560 A096915 A100803 this_sequence A159913 A030316 A034257
Adjacent sequences: A036837 A036838 A036839 this_sequence A036841 A036842 A036843
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KEYWORD
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nonn
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AUTHOR
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Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Jan 09 2002
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EXTENSIONS
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Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Jan 18 2002
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