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A103200
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a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n)=a(n-4)+sqrt(60*a(n-2)^2+60*a(n-2)+1) for n>=5.
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10
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1, 2, 11, 19, 90, 153, 712, 1208, 5609, 9514, 44163, 74907, 347698, 589745, 2737424, 4643056, 21551697, 36554706, 169676155, 287794595, 1335857546, 2265802057, 10517184216, 17838621864, 82801616185, 140443172858, 651895745267
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| The original version of this question was as follows: Let a(1) = 1, a(2) = 2, a(3) = 11, a(4) = 19; for n = 1..4 let b(n) = sqrt(60 a(n)^2 + 60 a(n) + 1); for n >= 5 let a(n) = a(n-4) + b(n-2), b(n) = sqrt(60 a(n)^2 + 60 a(n) +1). Bhanu and Deshpande ask for a proof that a(n) and b(n) are always integers. The b(n) sequence is A103201.
This sequence is also the interleaving of two sequences c and d that can be extended backwards: c(0) = c(1) = 0, c(n) = sqrt(60 c(n-1)^2 + 60 c(n-1) +1) + c(n-2) giving 0,0,1,11,90,712,5609,... d(0) = 1, d(1) = 0, d(n) = sqrt(60 d(n-1)^2 + 60 d(n-1) +1) + d(n-2) giving 1,0,2,19,153,1208,9514,... and interleaved: 0,1,0,0,1,2,11,19,90,153,712,1208,5609,9514,... lim n->infinity a(n)/a(n-2) = 1/(4 - sqrt(15)), (1/(4-sqrt(15)))^n approaches an integer as n->infinity. - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Mar 29 2005
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REFERENCES
| K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, An interesting sequence of quadruples and related open problems, Institute of Sciences, Nagpur, India, Preprint, 2005.
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index to sequences with linear recurrences with constant coefficients, signature (1,8,-8,-1,1).
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FORMULA
| Comments from Pierre CAMI (pierre-cami(AT)bbox.fr) and Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Apr 20 2005: Sequence satisfies a(0)=0, a(1)=1, a(2)=2, a(3)=11; for n>3, a(n) = 8*a(n-2)-a(n-4)+3. G.f.: -x*(1+x+x^2) / ( (x-1)*(x^4-8*x^2+1) ). Note that the 3 = the sum of the coefficients in the numerator of the g.f., 8 appears in the denominator of the g.f. and 8 = 2*3 + 2. Similar relationships hold for other series defined as nonnegative n such that m*n^2 + m*n + 1 is a square, here m=60. Cf. A001652, A001570, A049629, A105038, A105040, A104240, A077288, A105036, A105037.
a(2n) = (A105426(n)-1)/2, a(2n+1) = (A001090(n+2)-5*A001090(n+1)-1)/2. - Ralf Stephan, May 18 2007
a(1)=1, a(2)=2, a(3)=11, a(4)=19, a(5)=90, a(n)=a(n-1)+8*a(n-2)- 8*a(n-3)- a(n-4)+a(n-5) [From Harvey P. Dale, Sep 27 2011]
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MAPLE
| a[1]:=1: a[2]:=2:a[3]:=11: a[4]:=19: for n from 5 to 31 do a[n]:=a[n-4]+sqrt(60*a[n-2]^2+60*a[n-2]+1) od:seq(a[n], n=1..31); (Deutsch)
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MATHEMATICA
| RecurrenceTable[{a[1]==1, a[2]==2, a[3]==11, a[4]==19, a[n]==a[n-4]+ Sqrt[60a[n-2]^2+60a[n-2]+1]}, a[n], {n, 40}] (* or *) LinearRecurrence[ {1, 8, -8, -1, 1}, {1, 2, 11, 19, 90}, 40] (* From Harvey P. Dale, Sep 27 2011 *)
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PROG
| (MAGMA) I:=[1, 2, 11, 19, 90]; [n le 5 select I[n] else Self(n-1)+8*Self(n-2)-8*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Sep 28 2011
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CROSSREFS
| Cf. A103201, A177187 (first differences).
Sequence in context: A186267 A067660 * A105076 A067670 A159879 A017185
Adjacent sequences: A103197 A103198 A103199 * A103201 A103202 A103203
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KEYWORD
| nonn
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AUTHOR
| K. S. Bhanu and M. N. Deshpande, Mar 24 2005
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EXTENSIONS
| More terms from Pierre CAMI (pierre-cami(AT)bbox.fr) and Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 13 2005
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