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A103200 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. +0
10
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; listen)
OFFSET

1,2

COMMENT

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

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.

FORMULA

Comments from Pierre CAMI (pierrecami(AT)tele2.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)/(-1+x+8*x^2-8*x^3-x^4+x^5). 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

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)

CROSSREFS

Cf. A103201.

Sequence in context: A163997 A067931 A067660 this_sequence A105076 A067670 A159879

Adjacent sequences: A103197 A103198 A103199 this_sequence A103201 A103202 A103203

KEYWORD

nonn

AUTHOR

K. S. Bhanu and M. N. Deshpande, Mar 24 2005

EXTENSIONS

More terms from Pierre CAMI (pierrecami(AT)tele2.fr) and Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 13 2005

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Last modified March 20 09:10 EDT 2010. Contains 173642 sequences.


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