A101879 -id:A101879 - cp's OEIS frontend

A005246 a(n) = (1 + a(n-1)*a(n-2))/a(n-3), a(0) = a(1) = a(2) = 1.

1, 1, 1, 2, 3, 7, 11, 26, 41, 97, 153, 362, 571, 1351, 2131, 5042, 7953, 18817, 29681, 70226, 110771, 262087, 413403, 978122, 1542841, 3650401, 5757961, 13623482, 21489003, 50843527, 80198051, 189750626, 299303201, 708158977, 1117014753

Offset: 0

N. J. A. Sloane

For n >= 4 we have the linear recurrence a(n) = 4*a(n-2) - a(n-4). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 04 2001
Integer solutions to the equation floor(sqrt(3)*x^2) = x*floor(sqrt(3)*x). - Benoit Cloitre, Mar 18 2004
For n > 2, a(n) is the smallest integer > a(n-1) such that sqrt(3)*a(n) is closer to and greater than an integer than sqrt(3)*a(n-1). I.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(3)*a(n)) < frac(sqrt(3)*a(n-1)). - Benoit Cloitre, Jan 20 2003
The lower principal and intermediate convergents to 3^(1/2), beginning with 1/1, 3/2, 5/3, 12/7, 19/11, form a strictly increasing sequence; essentially, numerators=A143643 and denominators=A005246. - Clark Kimberling, Aug 27 2008
This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z. The g.f. is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n). The general form of a(n) is given by: a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (b-Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p) and a(2*m+1) = a*Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p). - Richard Choulet, Feb 24 2010
a(n) for n > 1 are the integer square roots of (floor(m^2/3)+1), where the values of m are given by A143643.  Also see A082630. - Richard R. Forberg, Nov 14 2013
The a(n) = (1 + a(n-1)*a(n-2))/a(n-3) recursion has the Laurent property. If a(0), a(1), a(2) are variables, then a(n) is a Laurent polynomial (a rational function with a monomial denominator). - Michael Somos, Feb 27 2019

			G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 7*x^5 + 11*x^6 + 26*x^7 + 41*x^8 + ...
From _Richard Choulet_, Feb 24 2010: (Start)
a(4) = 4^2 - 4^0 - 3*4^1 = 3.
a(7) = 4^3 - 4*binomial(2,1) - 2*(4^2-1) = 26. (End)

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
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..500
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Peter Cameron's Blog, The ADE affair, 3, Posted 23/06/2011.
T. Crilly, Double sequences of positive integers, Math. Gaz., 69 (1985), 263-271.
Enrica Duchi, Andrea Frosini, Renzo Pinzani and Simone Rinaldi, A Note on Rational Succession Rules, J. Integer Seqs., Vol. 6, 2003.
R. K. Guy, Letter to N. J. A. Sloane, Feb 1986
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Valentin Ovsienko, Serge Tabachnikov, Dual numbers, weighted quivers, and extended Somos and Gale-Robinson sequences, arXiv:1705.01623 [math.CO], 2017. See p. 10.
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 linear recurrences with constant coefficients, signature (0,4,0,-1).

Bisections are A001835 and A001075.
Cf. A101265. Row sums of A211956.
Cf. A001353.

a005246 n = a005246_list !! n
a005246_list = 1 : 1 : 1 : map (+ 1) (zipWith div
   (zipWith (*) (drop 2 a005246_list) (tail a005246_list)) a005246_list)
-- Reinhard Zumkeller, Mar 07 2012

A005246:=-(-1-z+2*z**2+z**3)/(1-4*z**2+z**4); # Conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.
for q from 1 to 10 do :a:=1:b:=1:Q:=(a*b^2+q*b+a+q)/(a*b): for m from 0 to 15 do U(m):=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)):od: for m from 0 to 15 do V(m):=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)):od:for m from 0 to 15 do W(2*m):=U(m):od:for m from 0 to 14 do W(2*m+1):=V(m):od:seq(W(m),m=0..30):od; # Richard Choulet, Feb 24 2010

RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(1+a[n-1]a[n-2])/a[n-3]},a,{n,40}] (* Harvey P. Dale, May 28 2013 *)
a[n_] := Cosh[(n-1)*ArcSinh[1/Sqrt[2]]]*If[EvenQ[n], Sqrt[2/3], 1]; Table[a[n] // FunctionExpand, {n, 0, 34}] (* Jean-François Alcover, Dec 10 2014, after Peter Bala *)
a[ n_] := With[{m = If[ n < 0, 2 - n, n]}, SeriesCoefficient[ (1 + x - 3 x^2 - 2 x^3) / (1 - 4 x^2 + x^4), {x, 0, m}]]; (* Michael Somos, Feb 10 2017 *)

{a(n) = if( n<0, n = 2 - n); polcoeff((1 + x - 3*x^2 - 2*x^3) / (1 - 4*x^2 + x^4) + x * O(x^n), n)}; /* Michael Somos, Nov 15 2006 */

{a(n) = real( (2 + quadgen(12))^(n\2) * if( n%2, 1, 1 - 1 / quadgen(12)) )}; /* Michael Somos, May 24 2012 */

G.f.: (1 + x - 3*x^2 - 2*x^3)/(1 - 4*x^2 + x^4).
Limit_{n->oo} a(2n+1)/a(2n) = (3+sqrt(3))/3 = 1.5773502...; lim_{n->oo} a(2n)/a(2n-1) = (3+sqrt(3))/2 = 2.3660254.... - Benoit Cloitre, Aug 07 2002
A101265(n) = a(n)*a(n+1). - Franklin T. Adams-Watters, Apr 24 2006
a(n) = a(2-n) for all n in Z. - Michael Somos, Nov 15 2006
a(2*n + 1) = A001075(n). a(2*n) = A001835(n). a(2*n + 1) - a(2*n) = a(2*n + 2) - a(2*n + 1) = A001353(n). - Michael Somos, May 24 2012
For n > 2: a(n) = a(n-1) + Sum_{k=1..floor((n-1)/2)} a(2*k). - Reinhard Zumkeller, Dec 16 2007
From Richard Choulet, Feb 24 2010: (Start)
a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*4^(m-2*p) - 3*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*4^(m-1-2*p).
a(2*m+1) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*4^(m-2*p) - 2*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*4^(m-1-2*p). (End)
From Tim Monahan, Jul 01 2011: (Start)
Closed form without extra leading 1: ((sqrt(6)+3)*(sqrt(2+sqrt(3))^n+(sqrt(2-sqrt(3))^n))+(3-sqrt(6))*((-sqrt(2+sqrt(3)))^n+(-sqrt(2-sqrt(3)))^n))/12.
Closed form with extra leading 1: ((6+3*sqrt(6)-2*sqrt(3)-3*sqrt(2))*(sqrt(2+sqrt(3))^n)+(6+3*sqrt(6)+2*sqrt(3)+3*sqrt(2))*(sqrt(2-sqrt(3))^n)+(6-3*sqrt(6)-2*sqrt(3)+3*sqrt(2))*((-sqrt(2+sqrt(3)))^n)+(6-3*sqrt(6)+2*sqrt(3)-3*sqrt(2))*((-sqrt(2-sqrt(3)))^n))/24. (End)
a(2*n+2) = Sum_{k = 0..n} 2^k*binomial(n+k,2*k); a(2*n+1) = Sum_{k = 0..n} n/(n+k)*2^k*binomial(n+k,2*k) for n >= 1. Row sums of A211956. - Peter Bala, May 01 2012
a(n) = ((sqrt(2)+sqrt(3)+(-1)^n*(sqrt(2)-sqrt(3)))*sqrt(2+(2-sqrt(3))^n*(2+ sqrt(3))-(-2+sqrt(3))*(2+ sqrt(3))^n))/(4*sqrt(3)). - Gerry Martens, Jun 06 2015
0 = a(n) - 2*a(n+1) + a(n+2) if n is even, 0 = a(n) - 3*a(n+1) + a(n+2) if n is odd for all n in Z. - Michael Somos, Feb 10 2017

More terms from Michael Somos, Aug 01 2001

A032908 One of four 3rd-order recurring sequences for which the first derived sequence and the Galois transformed sequence coincide.

Original entry on oeis.org

2, 2, 3, 6, 14, 35, 90, 234, 611, 1598, 4182, 10947, 28658, 75026, 196419, 514230, 1346270, 3524579, 9227466, 24157818, 63245987, 165580142, 433494438, 1134903171, 2971215074, 7778742050, 20365011075, 53316291174, 139583862446, 365435296163, 956722026042

Offset: 0

Michele Elia (elia(AT)polito.it)

a(n) is also a sequence with the property that the difference between the sum and product of two consecutive terms is equal to the square of the difference between those terms, i.e., a(n)*a(n+1) - (a(n)+ a(n+1)) = (a(n) - a(n + 1))^2. The difference between those two terms, a(n + 1) - a(n) = F(2n -2), the (2n - 2)th Fibonacci number. - John Baker, May 18 2010
Conjecture: consecutive terms of this sequence and consecutive terms of A101265 provide all the positive integer solutions of (a+b)*(a+b+1) == 0 (mod (a*b)). - Robert Israel, Aug 26 2015
Conjecture is true: see Mathematics Stack Exchange link. - Robert Israel, Sep 06 2015
Consecutive terms of this sequence and consecutive terms of A101879 provide all the positive integer pairs for which K = (a+1)/b + (b+1)/a is an integer. For this sequence, K = 3. - Andrey Vyshnevyy, Sep 18 2015

L. E. Dickson, History of the Theory of Numbers, Dover, New York, 1971.

Robert Israel, Table of n, a(n) for n = 0..2154
Michele Elia, A Note on derived linear recurring sequences, pp. 83-92 of Proceedings Seventh Int. Conference on Fibonacci Numbers and their Applications (Austria, 1996), Applications of Fibonacci Numbers, Volume 7.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 919.
Robert Israel, Will Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Mathematics Stack Exchange, Sep 1 2015.
Hieu D. Nguyen and Douglas Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. - From _N. J. A. Sloane_, Mar 16 2014
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A001519, A001622, A049310, A101265, A104457.

[2] cat [n le 1 select 2 else Floor((3*Self(n-1) + Sqrt(5*Self(n-1)^2 - 10*Self(n-1) + 1) - 1)/2): n in [1..30]]; // Vincenzo Librandi, Aug 27 2015

f:= proc(n) option remember; local x;
  x:= procname(n-1); (3*x + sqrt(5*x^2 - 10*x + 1) - 1)/2 end proc:
f(0):= 2: f(1):= 2:
map(f, [$0..30]); # Robert Israel, Aug 26 2015

Table[Fibonacci[2 n - 1] + 1, {n, 0, 27}] (* Michael De Vlieger, Aug 26 2015 *)
LinearRecurrence[{4,-4,1},{2,2,3},40] (* Harvey P. Dale, Apr 11 2018 *)

Vec((2-6*x+3*x^2)/(1-4*x+4*x^2-x^3)+O(x^66)) \\ Joerg Arndt, Jul 02 2013

G.f.: (2 - 6*x + 3*x^2)/((1 - x)*(1 - 3*x + x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n-1) + 1 = A001519(n) + 1. - Vladeta Jovovic, Mar 19 2003
a(n) = 3*a(n - 1) - a(n - 2) - 1. - N. Sato, Jan 21 2010
From  Wolfdieter Lang, Aug 27 2014: (Start)
a(n) = 1 + S(n-1, 3) - S(n-2, 3) =  1 + A001519(n), with Chebyshev S-polynomials (see A049310). For n < 0, we have S(-1, x) = 0 and S(-2, x) = -1.
This follows from the partial fraction decomposition of the g.f., 1/(1 - x) + (1 - 2*x)/ (1 - 3*x + x^2), using the recurrence for S, or from A001519. (End)
From Robert Israel, Aug 26 2015: (Start)
(a(n) + a(n+1))*(a(n) + a(n+1) + 1) = 5*a(n)*a(n+1).
a(n+1) = (3*a(n) + sqrt(5*a(n)^2 - 10*a(n) + 1) - 1)/2 for n >= 1. (End)
a(n) = 1 + (2^(-1-n) * ((3 - sqrt(5))^n * (1 + sqrt(5)) + (-1 + sqrt(5)) * (3 + sqrt(5))^n)) / sqrt(5). - Colin Barker, Nov 02 2016
Sum_{n>=0} 1/a(n) = phi (A001622). - Amiram Eldar, Oct 05 2020
Product_{n>=1} (1 + 1/a(n)) = (3+sqrt(5))/2 = phi^2 (A104457). - Amiram Eldar, Nov 28 2024

More terms from Ralf Stephan, Mar 10 2003

Index for Chebyshev polynomials and cross reference added by Wolfdieter Lang, Aug 27 2014

A275173 a(n) = (a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 6, 9, 22, 36, 51, 82, 129, 321, 529, 753, 1217, 1921, 4786, 7891, 11236, 18166, 28681, 71462, 117828, 167779, 271266, 428289, 1067137, 1759521, 2505441, 4050817, 6395649, 15935586, 26274979, 37413828, 60490982, 95506441, 237966646

Offset: 0

Seiichi Manyama, Jul 19 2016

Inspired by A048736.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,16,0,0,0,0,-16,0,0,0,0,1).

Cf. A101879, A048736, A275174.

Variants: A275175, A275176.

RecurrenceTable[{a[n] == (a[n - 3] + a[n - 1] a[n - 5])/a[n - 6], a[1] == 1, a[2] == 1, a[3] == 1, a[4] == 1, a[5] == 1, a[6] == 1}, a, {n, 42}] (* or *)
CoefficientList[Series[(1 + x + x^2 + x^3 + x^4 - 15 x^5 - 14 x^6 - 13 x^7 - 12 x^8 - 10 x^9 + 9 x^10 + 6 x^11 + 4 x^12 + 3 x^13 + 2 x^14)/((1 - x) (1 + x + x^2 + x^3 + x^4) (1 - 15 x^5 + x^10)), {x, 0, 41}], x] (* Michael De Vlieger, Jul 19 2016 *)
nxt[{a_,b_,c_,d_,e_,f_}]:={b,c,d,e,f,(d+f*b)/a}; NestList[nxt,{1,1,1,1,1,1},50][[;;,1]] (* Harvey P. Dale, Jan 06 2024 *)

Vec((1 +x +x^2 +x^3 +x^4 -15*x^5 -14*x^6 -13*x^7 -12*x^8 -10*x^9 +9*x^10 +6*x^11 +4*x^12 +3*x^13 +2*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -15*x^5 +x^10)) + O(x^50)) \\ Colin Barker, Jul 19 2016

def A(k, l, n)
  a = Array.new(k * 2, 1)
  ary = [1]
  while ary.size < n + 1
    break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
    a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
    ary << a[0]
  end
  ary
end
def A275173(n)
  A(3, 1, n)
end

G.f.: (1 +x +x^2 +x^3 +x^4 -15*x^5 -14*x^6 -13*x^7 -12*x^8 -10*x^9 +9*x^10 +6*x^11 +4*x^12 +3*x^13 +2*x^14) / ((1 -x)*(1 +x +x^2 +x^3 +x^4)*(1 -15*x^5 +x^10)). - Colin Barker, Jul 19 2016

a(n) = 16*a(n-5) - 16*a(n-10) + a(n-15). - G. C. Greubel, Jul 20 2016

A182756 Numbers k > 1 such that are sequences B_k of type: {b(1) = 1, b(2) = k, for n >= 3; b(n) = the smallest number h > b(n-1) such that [[b(n-2) + b(n-1)] * [b(n-2) + h] * [b(n-1) + h]] / [b(n-2) * b(n-1) * h] is an integer}.

Original entry on oeis.org

2, 3, 6, 14, 21, 35, 77, 90, 234, 286, 611

Offset: 1

Jaroslav Krizek, Nov 27 2010

With 1 complement of A182757.

Is this a union/superset of A032908 and A101879? - Ralf Stephan, Nov 29 2010

			For n =1; a(n) = 2; B_2 = A038754(n): 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, …
For n =2; a(n) = 3; B_3 = A182751(n): 1, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, …
For n =3; a(n) = 6; B_6 = A182752(n): 1, 6, 14, 84, 196, 1176, 2744, 16464, 38416, …
For n =4; a(n) = 14; B_14 = A182753(n): 1, 14, 35, 490, 1225, 17150, 42875, …
For n =5; a(n) = 21; B_21 = A182754(n): 1, 21, 77, 1617, 5929, 124509, 456533, …
For n =6; a(n) = 35; B_35 = A182755(n): 1, 35, 90, 3150, 8100, 283500, 729000, …

Cf. A182751, A182752, A182753, A182754, A182755, A182757, A038754.

A276122 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1)^2+a(n-2)^2+a(n-1)+a(n-2))/a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 22, 526, 69427, 219111589, 91273561736491, 119994570874632853695766, 65713991236617279734602790963627271046, 47311933073383646516067037755547920981262829886906923065810924

Offset: 0

Bruno Langlois, Aug 21 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..16

Cf. A064098, A072881, A101879, A165903.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 1] + a[n - 2])/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 11}] (* Michael De Vlieger, Aug 21 2016 *)

a(n) = 6*a(n-1)*a(n-2)-a(n-3)-1.

a(n) ~ 1/6 * c^(phi^n), where c = 2.059783590102273... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 20 2017

a(10) corrected by Seiichi Manyama, Aug 21 2016

A275174 a(n) = (a(n-4) + a(n-1) * a(n-7)) / a(n-8), a(0) = a(1) = ... = a(7) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 33, 53, 74, 96, 141, 209, 300, 714, 1151, 1611, 2094, 3083, 4578, 6579, 15665, 25257, 35355, 45959, 67673, 100497, 144431, 343906, 554491, 776186, 1008991, 1485711, 2206346, 3170896, 7550257, 12173533, 17040724

Offset: 0

Seiichi Manyama, Jul 19 2016

Inspired by A048736.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,23,0,0,0,0,0,0,-23,0,0,0,0,0,0,1).

Cf. A101879, A048736, A275173.

RecurrenceTable[{a[n] == (a[n - 4] + a[n - 1] a[n - 7])/a[n - 8], a[1] == 1, a[2] == 1, a[3] == 1, a[4] == 1, a[5] == 1, a[6] == 1, a[7] == 1, a[8] == 1}, a, {n, 42}] (* Michael De Vlieger, Jul 19 2016 *)

Vec((1 +x +x^2 +x^3 +x^4 +x^5 +x^6 -22*x^7 -21*x^8 -20*x^9 -19*x^10 -18*x^11 -16*x^12 -13*x^13 +14*x^14 +10*x^15 +7*x^16 +5*x^17 +4*x^18 +3*x^19 +2*x^20) / ((1 -x)*(1 +x +x^2 +x^3 +x^4 +x^5 +x^6)*(1 -22*x^7 +x^14)) + O(x^20)) \\ Colin Barker, Jul 19 2016

def A(k, l, n)
  a = Array.new(k * 2, 1)
  ary = [1]
  while ary.size < n + 1
    break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
    a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
    ary << a[0]
  end
  ary
end
def A275174(n)
  A(4, 1, n)
end

G.f.: (1 +x +x^2 +x^3 +x^4 +x^5 +x^6 -22*x^7 -21*x^8 -20*x^9 -19*x^10 -18*x^11 -16*x^12 -13*x^13 +14*x^14 +10*x^15 +7*x^16 +5*x^17 +4*x^18 +3*x^19 +2*x^20) / ((1 -x)*(1 +x +x^2 +x^3 +x^4 +x^5 +x^6)*(1 -22*x^7 +x^14)). - Colin Barker, Jul 19 2016

a(n) = 23*a(n-7) - 23*a(n-14) + a(n-21).

A276271 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)+a(n-2)+a(n-3))/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 6, 46, 2206, 4870846, 3954191749561, 339905052007042640998641, 52373274877565894156748130733610185904753361, 563138297002425210235477817802336090254190075906443582099838858026136728896536841

Offset: 0

Seiichi Manyama, Aug 26 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..15

Cf. A101879, A276122.

nxt[{a_,b_,c_,d_}]:={b,c,d,(b^2+c^2+d^2+b+c+d)/a}; NestList[nxt,{1,1,1,1},12][[All,1]] (* Harvey P. Dale, Mar 10 2017 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(0){|s, i| s + i * i} + a[1..-1].inject(:+)
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276271(n)
  A(4, n)
end

a(n) = 8*a(n-1)*a(n-2)*a(n-3)-a(n-4)-1.

a(n) ~ 2^(-3/2) * c^(d^n), where c = 1.2578918597... and d = A058265 = 1.83928675521416... = (1 + (19 + 3*sqrt(33))^(1/3) + (19 - 3*sqrt(33))^(1/3))/3. - Vaclav Kotesovec, Mar 20 2017

A276453 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = 1, a(2) = 2, a(3) = 6.

Original entry on oeis.org

1, 1, 2, 6, 42, 903, 136052, 881442036, 2581196224947732, 342795531574625708871288171, 5732512385084161208637718426682572229606557631, 5754497648510061274107897581706624823818534711463525598519384262130236399970112

Offset: 0

Seiichi Manyama, Sep 03 2016

nonn

From Antoine de Saint Germain, Dec 30 2024: (Start)
Sequence consists of integers, see Math StackExchange link.
Values of a unitary Y-frieze pattern associated to the linearly oriented quiver K4 (i.e., the quiver whose underlying graph is the complete graph on the vertices {1,2,3,4}, oriented such that i -> j whenever i < j). (End)

Seiichi Manyama, Table of n, a(n) for n = 0..15
Math StackExchange, Is A276175 integer-only?

Cf. A051786, A101879, A276175.

I:=[1, 1, 2, 6]; [n le 4 select I[n] else (Self(n-1)+1)*(Self(n-2)+1)*(Self(n-3)+1)/Self(n-4): n in [1..13]]; // Vincenzo Librandi, Dec 30 2024

RecurrenceTable[{a[n] == (a[n - 1] + 1) (a[n - 2] + 1) (a[n - 3] + 1)/a[n - 4], a[0] == 1, a[1] == 1, a[2] == 2, a[3] == 6}, a, {n, 0, 11}] (* Michael De Vlieger, Sep 03 2016 *)

def A276453(n)
  a = [1, 1, 2, 6]
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(1){|s, i| s * (i + 1)}
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end

a(n) = A051786(n)*A051786(n+1)*A051786(n+2).

A112449 a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 2, 10, 505, 12878813, 4229958765311886322, 5876687051603582015287706866081267480733704277890

Offset: 0

Andrew Hone, Dec 12 2005

nonn

A second-order recurrence with the Laurent property. This property is satisfied by any second-order recurrence of the form a(n+2) = f(a(n+1))/a(n) with f being a polynomial of the form f(x) = x*p(x) where p is a polynomial of degree d with integer coefficients such that p(0)=1 and p has the reciprocal property x^d*p(1/x) = p(x). Hence if a(0) = a(1) = 1 then a(n) is an integer for all n.

As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) or about 0.962 (A202543).

Seiichi Manyama, Table of n, a(n) for n = 0..10
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.

Cf. A101879, A112373, A202543.

a[0]:=1; a[1]:=1; f(x):=x^3+x;
for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od;
s[3]:=ln(10); s[4]:=ln(505);
for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-2*s[n+1]))-s[n]): od: print(evalf(ln(s[10002])/(10002))): evalf(ln((3+sqrt(5))/2));
# s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...
f:=proc(n) option remember; local i,j,k,t1,t2,t3; if n <= 1 then RETURN(1); fi; (f(n-1)^3+f(n-1))/f(n-2); end;
# N. J. A. Sloane

nxt[{a_,b_}]:={b,(b^3+b)/a}; NestList[nxt,{1,1},10][[All,1]] (* Harvey P. Dale, Jun 26 2017 *)

def A(l, m, n)
  a = Array.new(2 * m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(:*) + a[m] ** l
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A112449(n)
  A(3, 1, n)
end # Seiichi Manyama, Nov 20 2016

a(1-n) = a(n). - Seiichi Manyama, Nov 20 2016

A387123 Numbers k such that Sum_{i=1..r} (k-i) and Sum_{i=1..r} (k+i) are both triangular for some r with 1 <= r < k.

Original entry on oeis.org

2, 6, 9, 21, 24, 38, 50, 53, 65, 77, 90, 96, 104, 133, 147, 195, 201, 224, 247, 286, 324, 377, 450, 483, 553, 588, 605, 614, 713, 792, 901, 1014, 1029, 1043, 1066, 1074, 1155, 1274, 1349, 1575, 1784, 1885, 1920, 2034, 2057, 2109, 2279, 2312, 2342, 2622

Offset: 1

Ctibor O. Zizka, Aug 17 2025

For m >= 1, if k = m*(m+1)^2/2 then r = m, thus A006002 is a subsequence. For k >= 286 from A101265 or A101879, r = k-1.

			For k = 6: the least r = 5, T_i = 1 + 2 + 3 + 4 + 5 = 15, T_j = 7 + 8 + 9 + 10 + 11 = 45, both T_i and T_j are triangular numbers, thus k = 6 is a term.

Cf. A000217, A006002, A101265, A101879.

triQ[n_] := IntegerQ[Sqrt[8*n + 1]]; q[k_] := Module[{r = 1, s1 = 0, s2 = 0}, While[s1 += k - r; s2 += k + r; r < k && (! triQ[s1] || ! triQ[s2]), r++]; 1 <= r < k]; Select[Range[3000], q] (* Amiram Eldar, Aug 17 2025 *)

isok(k) = my(sm=0, sp=0); for (r=1, k-1, sm+=k-r; sp+=k+r; if (ispolygonal(sm, 3) && ispolygonal(sp, 3), return(r));); \\ Michel Marcus, Aug 17 2025

from itertools import count, islice
from sympy.ntheory.primetest import is_square
def A387123_gen(startvalue=1): # generator of terms >= startvalue
    for k in count(max(startvalue,1)):
        if any(is_square(((k*r<<1)-r*(r+1)<<2)+1) and is_square(((k*r<<1)+r*(r+1)<<2)+1) for r in range(1,k)):
            yield k
A387123_list = list(islice(A387123_gen(),50)) # Chai Wah Wu, Aug 21 2025

cp's OEIS Frontend

A005246 a(n) = (1 + a(n-1)*a(n-2))/a(n-3), a(0) = a(1) = a(2) = 1.

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A032908 One of four 3rd-order recurring sequences for which the first derived sequence and the Galois transformed sequence coincide.

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A275173 a(n) = (a(n-3) + a(n-1) * a(n-5)) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

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A182756 Numbers k > 1 such that are sequences B_k of type: {b(1) = 1, b(2) = k, for n >= 3; b(n) = the smallest number h > b(n-1) such that [[b(n-2) + b(n-1)] * [b(n-2) + h] * [b(n-1) + h]] / [b(n-2) * b(n-1) * h] is an integer}.

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A276122 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1)^2+a(n-2)^2+a(n-1)+a(n-2))/a(n-3).

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A275174 a(n) = (a(n-4) + a(n-1) * a(n-7)) / a(n-8), a(0) = a(1) = ... = a(7) = 1.

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A276271 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)+a(n-2)+a(n-3))/a(n-4).

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A276453 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = 1, a(2) = 2, a(3) = 6.