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Search: id:A082630
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| A082630 |
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Start with the sequence S(0)={1,1} and for n>0 define S(n) to be I(S(n-1)) where I denotes the operation of inserting, for i=1,2,3..., the term a(i)+a(i+1) between any two terms for which 7a(i+1)<=11a(i). The listed terms are the initial terms of the limit of this process. |
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7
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| 1, 2, 5, 8, 19, 30, 71, 112, 265, 418, 989, 1560, 3691, 5822, 13775, 21728, 51409, 81090, 191861, 302632, 716035, 1129438, 2672279, 4215120, 9973081, 15731042, 37220045, 58709048, 138907099, 219105150
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OFFSET
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1,2
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COMMENT
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The bisection {1,5,19,265,...} appears to be A001834 and to satisfy the recurrence a(n)=4a(n)-a(n-2) and the condition that 3a(n)^2+6 is a square. The other bisection {2,8,30,112,...} appears to be A052530, and one-half of this bisection, {1,4,15,56,...}, appears to be A001353 and to satisfy a(n)=4a(n-1)-a(n-2) and the condition that 3a(n)^2+1 is a square.
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FORMULA
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The sequence appears to satisfy a(n)=4a(n-2)-a(n-4).
Apparently a(n)a(n+3) = -2 + a(n+1)a(n+2). - R. Stephan, May 29 2004
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EXAMPLE
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Let S(0)={1,1}. Since 7*1<=11*1 we obtain S(1)={1,2,1}. Then since 7*2>11*1 and 7*1<=11*2, we obtain S(2)={1,2,3,1). Continuing, we get S(3)={1,2,5,3,4,1}, S(4)={1,2,5,8,3,7,4,5,1), S(5)={1,2,5,8,11,3,...}, S(6)={1,2,3,5,8,19,11,...}, etc.
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CROSSREFS
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Cf. A001353, A001834, A052530.
Sequence in context: A032063 A037233 A133147 this_sequence A025078 A055614 A076870
Adjacent sequences: A082627 A082628 A082629 this_sequence A082631 A082632 A082633
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KEYWORD
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nonn
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AUTHOR
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John W. Layman (layman(AT)math.vt.edu), May 23 2003
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