A105426 -id:A105426 - cp's OEIS frontend

A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.

1, 4, 31, 244, 1921, 15124, 119071, 937444, 7380481, 58106404, 457470751, 3601659604, 28355806081, 223244789044, 1757602506271, 13837575261124, 108942999582721, 857706421400644, 6752708371622431, 53163960551578804

Offset: 0

N. J. A. Sloane

a(15+30k)-1 and a(15+30k)+1 are consecutive odd powerful numbers. The first pair is 13837575261124 +- 1. See A076445. - T. D. Noe, May 04 2006
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 15*y^2 = 1. The corresponding y values are in A001090. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 04 2014]
The square root of 15*(n^2-1) at those numbers = 5 * A136325. - Richard R. Forberg, Nov 22 2013
For the above Diophantine equation x^2-15*y^2=1, x + y = A105426. - Richard R. Forberg, Nov 22 2013
a(n) solves for x in the Diophantine equation floor(3*x^2/5)= y^2. The corresponding y solutions are provided by A136325(n).  x + y = A070997(n). - Richard R. Forberg, Nov 22 2013
Except for the first term, values of x (or y) in the solutions to x^2 - 8xy + y^2 + 15 = 0. - Colin Barker, Feb 05 2014

Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..200
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A001090, A090965, A098269, A322836 (column 4).

a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 26 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1-4*x)/(1-8*x+x^2) )); // G. C. Greubel, Aug 26 2019

LinearRecurrence[{8,-1},{1,4},20] (* Harvey P. Dale, May 01 2014 *)

PARI
```
a(n)=subst(poltchebi(n),x,4)
```

a(n)=n=abs(n); polcoeff((1-4*x)/(1-8*x+x^2)+x*O(x^n),n) /* Michael Somos, Jun 07 2005 */

def A001091_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-4*x)/(1-8*x+x^2) ).list()
A001091_list(20) # G. C. Greubel, Aug 26 2019

G.f.: A(x) = (1-4*x)/(1-8*x+x^2). - Simon Plouffe in his 1992 dissertation
For all elements x of the sequence, 15*(x^2 -1) is a square. Limit_{n -> infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 11 2002
a(n) = sqrt(15*((A001090(n))^2)+1).
a(n) = ((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
a(n) = 4*S(n-1, 8) - S(n-2, 8) = (S(n, 8) - S(n-2, 8))/2, n>=1; S(n, x) := U(n, x/2) with Chebyshev's polynomials of the 2nd kind, A049310, with S(-1, x) := 0 and S(-2, x) := -1.
a(n) = T(n, 4) with Chebyshev's polynomials of the first kind; see A053120.
a(n)=a(-n). - Ralf Stephan, Jun 06 2005
a(n)*a(n+3) - a(n+1)*a(n+2) = 120. - Ralf Stephan, Jun 06 2005
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*15^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 15*x^2))^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 16*x + 4*x^2) is the g.f. of A098269.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - (5/2)/a(n)) = 1/3.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (3/2)/a(n)) = 1/5.
Sum_{n >= 1} 1/(a(n)^2 - 5/2) =  1/3 - 1/sqrt(15). (End)
a(n) = A001090(n+1)-4*A001090(n). - R. J. Mathar, Dec 12 2023

More terms from Larry Reeves (larryr(AT)acm.org), Aug 25 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A103200 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) for n >= 5.

Original entry on oeis.org

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, 1105706761003, 5132364345954

Offset: 1

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

nonn

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, Mar 29 2005

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.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-1,1).

Cf. A103201, A177187 (first differences).

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

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); # Emeric Deutsch, Apr 13 2005

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] (* Harvey P. Dale, Sep 27 2011 *)
CoefficientList[Series[-x*(1 + x + x^2)/((x - 1)*(x^4 - 8*x^2 + 1)), {x, 0, 40}], x] (* T. D. Noe, Jun 04 2012 *)

Comments from Pierre CAMI and Gerald McGarvey, Apr 20 2005: (Start)
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. (End)
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). - Harvey P. Dale, Sep 27 2011

More terms from Pierre CAMI and Emeric Deutsch, Apr 13 2005

A144479 a(0)=1, a(1)=3, a(n) = 8*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 23, 181, 1425, 11219, 88327, 695397, 5474849, 43103395, 339352311, 2671715093, 21034368433, 165603232371, 1303791490535, 10264728691909, 80814038044737, 636247575665987, 5009166567283159, 39437084962599285, 310487513133511121, 2444463020105489683

Offset: 0

N. J. A. Sloane, Mar 28 2009

A105426 extended backwards.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (8, -1).

I:=[1,3]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 12 2015

LinearRecurrence[{8, -1}, {1, 3}, 25] (* Vincenzo Librandi, Oct 12 2015 *)

Vec((1-5*x)/(1-8*x+x^2) + O(x^40)) \\ Colin Barker, Oct 12 2015

G.f.: (1-5*x)/(1-8*x+x^2). - Philippe Deléham, Mar 28 2009
a(n) = A001090(n+1)-5*A001090(n). - R. J. Mathar, Mar 29 2009
a(n) = (((4-sqrt(15))^n*(1+sqrt(15))+(-1+sqrt(15))*(4+sqrt(15))^n))/(2*sqrt(15)). - Colin Barker, Oct 12 2015

More terms from Philippe Deléham and R. J. Mathar, Mar 28 2009

cp's OEIS Frontend

A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.

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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.

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A144479 a(0)=1, a(1)=3, a(n) = 8*a(n-1) - a(n-2).

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Programs

Magma

Mathematica

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Formula

Extensions

A103200 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) for n >= 5.