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Search: id:A081097
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| A081097 |
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Numbers n such that n^2= (1/5)*(n+floor(sqrt(5)*n*floor(sqrt(5)*n))). |
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1
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| 1, 6, 23, 40, 273, 1870, 7343, 12816, 87841, 602070, 2364359, 4126648, 28284465, 193864606, 761316191, 1328767776, 9107509825, 62423800998, 245141449079, 427859097160, 2932589879121, 20100270056686, 78934785287183
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Conjecture : if m is an integer and sqrt(m) is irrational, the sequence of n such that n^2= (1/m)*(n+floor(sqrt(m)*n*floor(sqrt(m)*n))) always satisfies a recurrence of order m. Ex : if m=6 sequence n=b(k) satisfies : b(6k)=4*b(6k-1)+4*b(6k-2)-b(6k-3)-1; b(6k+1)=.... etc.
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FORMULA
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a(1)=1; a(2)=6; a(3)=23; a(4)=40; a(4n)=2*a(4n-1)-a(4n-2); a(4n+1)=7*a(4n)-a(4n-2)-1; a(4n+2)=7*a(4n+1)-a(4n-1)-1; a(4n+3)=4*a(4n+2)-a(4n+1)/2-1/2
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PROGRAM
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(PARI) x=1; y=6; z=23; u=40; for(n=5, 50, v=if((n%4-1)*(n%4-2), if(n%4, 4*u-z/2-1/2, 2*u-z), if(n%4-1, 7*u-z-1, 7*u-y-1)); x=y; y=z; z=u; u=v; print1(v, ", "))
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CROSSREFS
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Cf. A046090.
Sequence in context: A062818 A012327 A012523 this_sequence A031293 A022269 A026817
Adjacent sequences: A081094 A081095 A081096 this_sequence A081098 A081099 A081100
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 15 2003
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