A002310 -id:A002310 - cp's OEIS frontend

A049685 a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).

1, 6, 41, 281, 1926, 13201, 90481, 620166, 4250681, 29134601, 199691526, 1368706081, 9381251041, 64300051206, 440719107401, 3020733700601, 20704416796806, 141910183877041, 972666870342481, 6666757908520326, 45694638489299801, 313195711516578281

Offset: 0

Clark Kimberling

In general, Sum_{k=0..n} binomial(2*n-k,k)j^(n-k) = (-1)^n*U(2n, I*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,7), where L is defined as in A108299; see also A033890 for L(n,-7). - Reinhard Zumkeller, Jun 01 2005
Take 7 numbers consisting of 5 ones together with any two successive terms from this sequence. This set has the property that the sum of their squares is 7 times their product. (R. K. Guy, Oct 12 2005.) See also A111216.
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
From Wolfdieter Lang, Feb 09 2021: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 7*x*y = -5 are given by [x(n) = S(n, 7) - S(n-1, 7), y(n) = x(n-1)], for all integer numbers n, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-n, x) = -S(n-2, x), for n >= 2. x(n) = a(n), for n >= 0.
This indefinite binary quadratic form has discriminant D = +45. There is only this family representing -5 properly with x and y positive, and there are no improper solutions.
All proper and improper solutions of the generalized Pell equation X^2 - 45*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n) from the preceding comment, by X(n) = x(n) + x(n-1) = S(n-1, 7) - S(n-2, 7) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 7), for all integer numbers n. For positive integers X(n) = A056854(n) and Y(n) = A004187(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper family of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer numbers n.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

			a(3) = L(4*3 + 2)/3 = 843/3 = 281. - _Indranil Ghosh_, Feb 06 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1193
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Row 7 of array A094954. First differences of A004187.
Cf. A002310, A002320, A049310, A049685, A056854.
Cf. similar sequences listed in A238379.

[Lucas(4*n+2)/3: n in [0..30]]; // G. C. Greubel, Dec 17 2017

Table[LucasL[4*n+2]/3, {n,0,50}] (* or *) LinearRecurrence[{7,-1}, {1,6}, 50] (* G. C. Greubel, Dec 17 2017 *)

a(n)=(fibonacci(4*n+1)+fibonacci(4*n+3))/3 \\ Charles R Greathouse IV, Jun 16 2014

[lucas_number1(n,7,1)-lucas_number1(n-1,7,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i); then q(n, 5)=a(n); a(n) = 7a(n-1) - a(n-2). - Benoit Cloitre, Nov 10 2002
From Ralf Stephan, May 29 2004: (Start)
a(n+2) = 7a(n+1) - a(n).
G.f.: (1-x)/(1-7x+x^2).
a(n)*a(n+3) = 35 + a(n+1)*a(n+2). (End)
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*5^k. - Paul Barry, Aug 30 2004
If another "1" is inserted at the beginning of the sequence, then A002310, A002320 and A049685 begin with 1, 2; 1, 3; and 1, 1; respectively and satisfy a(n+1) = (a(n)^2+5)/a(n-1). - Graeme McRae, Jan 30 2005
a(n) = (-1)^n*U(2n, i*sqrt(5)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
[a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = S(n, 7) - S(n-1, 7) with Chebyshev S polynomials S(n-1, 7) = A004187(n), for n >= 0. - Wolfdieter Lang, Feb 09 2021
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/3. - Stefano Spezia, Apr 14 2025
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/9) * sqrt(1 - T(2*n+1, -7/2)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 6, 0, 41, 0, 281, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -9, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/5 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A290903(n-1) - 1/A290903(n).) (End)

A002320 a(n) = 5*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 14, 67, 321, 1538, 7369, 35307, 169166, 810523, 3883449, 18606722, 89150161, 427144083, 2046570254, 9805707187, 46981965681, 225104121218, 1078538640409, 5167589080827, 24759406763726, 118629444737803

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Together with A002310 these are the two sequences satisfying the requirement that (a(n)^2 + a(n-1)^2)/(1 - a(n)*a(n-1)) be an integer; in both cases this integer is -5. - Floor van Lamoen, Oct 26 2001

From a posting to Netnews group sci.math by ksbrown(AT)seanet.com (K. S. Brown) on Aug 15 1996.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See p. 44.
Tanya Khovanova, Recursive Sequences
MathPages, N = (x^2 + y^2)/(1+xy) is a Square
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A054477.

a002320 n = a002320_list !! n
a002320_list = 1 : 3 :
   (zipWith (-) (map (* 5) (tail a002320_list)) a002320_list)
-- Reinhard Zumkeller, Oct 16 2011

LinearRecurrence[{5,-1},{1,3},30] (* Harvey P. Dale, Nov 13 2014 *)

Sequences A002310, A002320 and A049685 have this in common: each one satisfies a(n+1) = (a(n)^2+5)/a(n-1) - Graeme McRae, Jan 30 2005
G.f.: (1-2x)/(1-5x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*2^k. - _Philippe Deléham, Mar 05 2014
E.g.f.: exp(5*x/2)*(sqrt(21)*cosh(sqrt(21)*x/2) + sinh(sqrt(21)*x/2))/sqrt(21). - Stefano Spezia, Jul 07 2025
From Peter Bala, Jul 07 2025: (Start)
a(n) = ( (4 + sqrt(21))*(5 - sqrt(21))^(n+1) - (4 - sqrt(21))*(5 + sqrt(21))^(n+1) )/(2^(n+1)*sqrt(21)).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 5/a(2*n)) = 1/15, since 5/(a(2*n) + 5/a(2*n)) = 1/a(2*n-1) + 1/a(2*n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n-1) + 5/a(2*n-1)) = 1/5, since 5/(a(2*n-1) + 5/a(2*n-1)) = 1/a(2*n-2) + 1/a(2*n). (End)

A207606 Triangle of coefficients of polynomials u(n,x) jointly generated with A207607; see the Formula section.

Original entry on oeis.org

1, 2, 3, 2, 4, 7, 2, 5, 16, 11, 2, 6, 30, 36, 15, 2, 7, 50, 91, 64, 19, 2, 8, 77, 196, 204, 100, 23, 2, 9, 112, 378, 540, 385, 144, 27, 2, 10, 156, 672, 1254, 1210, 650, 196, 31, 2, 11, 210, 1122, 2640, 3289, 2366, 1015, 256, 35, 2, 12, 275, 1782, 5148, 8008

Offset: 1

Clark Kimberling, Feb 19 2012

As triangle T(n,k) with 0 <= k <= n, it is (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 03 2012

			First five rows:
  1;
  2;
  3,  2;
  4,  7,  2;
  5, 16, 11,  2;
Triangle (2, -1/2, 1/2, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, ...), 0 <= k <= n, begins:
  1;
  2,   0;
  3,   2,   0;
  4,   7,   2,   0;
  5,  16,  11,   2,   0;
  6,  30,  36,  15,   2,   0;
  7,  50,  91,  64,  19,   2,   0;
  8,  77, 196, 204, 100,  23,   2,   0;

G. C. Greubel, Rows n = 1..101 of the triangle, flattened

Cf. A207607.

T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=0 then n+2
    elif k=n then 2
    else 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k)
      fi; end:
1, seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Mar 15 2020

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207606 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A207607 *)
(* Second program *)
T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, n+2, If[k==n, 2, 2*T[n-1, k] - T[n-2, k] + T[n-1, k-1] ]]]; Join[{1}, Table[T[n, k], {n, 0, 10}, {k, 0, n}]]//Flatten (* G. C. Greubel, Mar 15 2020 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else x*u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(u(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==1): return n+1
    elif (k==n): return 2
    else: return 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k)
[1]+[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Mar 15 2020

u(n,x) = u(n-1,x) + v(n-1,x), v(n,x) = x*u(n-1,x) + (x+1)v(n-1,x), where u(1,x)=1, v(1,x)=1.
As triangle T(n,k) with 0 <= k <= n: g.f.: (1-y*x)/(1-(2+y)*x+x^2). - Philippe Deléham, Mar 03 2012
As triangle T(n,k) with 0 <= k <= n: Sum_{k=0..n} T(n,k)*x^k = A132677(n), A000034(n)*A057077(n), A057079(n), A000027(n+1), A001519(n+1), A001075(n), A002310(n), A038725(n), A172968(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Mar 03 2012
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Mar 03 2012
T(n,k) = C(n+k-1,2*k+1) + 2*C(n+k-1,2*k), where C is binomial. - Yuchun Ji, May 23 2019
T(n,k) = T(n-1,k) + A207607(n-1,k). - Yuchun Ji, May 28 2019

A101463 Expansion of g.f. (x^3+x^2+2x+1)/(x^4+5x^2+1).

Original entry on oeis.org

1, 2, -4, -9, 19, 43, -91, -206, 436, 987, -2089, -4729, 10009, 22658, -47956, -108561, 229771, 520147, -1100899, -2492174, 5274724, 11940723, -25272721, -57211441, 121088881, 274116482, -580171684, -1313370969, 2779769539, 6292738363, -13318676011

Offset: 0

Creighton Dement, Jan 20 2005

A floretion-generated sequence relating to Pythagoras' theorem generalized.

Floretion Algebra Multiplication Program. FAMP code: em[J* ]sigcycseq[ + .75'i + .5'k + .25i' + .5j' + .5k' - .25'ii' + .25'jj' - .25'kk' - .75'jk' + .5'ki' - .25'kj' + .25e]

F. A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971.

James A. Sellers, Domino Tilings and Products of Fibonacci and Pell Numbers, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.2.
Index entries for linear recurrences with constant coefficients, signature (0,-5,0,-1)

Elements of even index in the sequence gives A004253. Elements of odd index in the sequence gives A002310.

Cf. A004253, A002310.

CoefficientList[Series[(x^3+x^2+2x+1)/(x^4+5x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[{0,-5,0,-1},{1,2,-4,-9},31] (* Harvey P. Dale, Apr 15 2012 *)

Let b(1)=1, b(2)=2, b(3)=4 and b(n)=(b(n-1)*b(n-2)+(3+(-1)^n)/2)/b(n-3) then b(n)=abs(a(n)) - Benoit Cloitre, Mar 03 2007
a(n) = -5*a(n-2)-a(n-4), n>3. [Harvey P. Dale, Apr 15 2012]
G.f.: ( 1+2*x+x^2+x^3 ) / ( 1+5*x^2+x^4 ). - R. J. Mathar, Jun 18 2014
a(n) = -3a(n-1)+2a(n-2) if n even. a(n) = (5*a(n-1)+a(n-2))/2 if n odd. - R. J. Mathar, Jun 18 2014

cp's OEIS Frontend

A049685 a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).

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A002320 a(n) = 5*a(n-1) - a(n-2).

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A207606 Triangle of coefficients of polynomials u(n,x) jointly generated with A207607; see the Formula section.

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A101463 Expansion of g.f. (x^3+x^2+2*x+1)/(x^4+5*x^2+1).

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A101463 Expansion of g.f. (x^3+x^2+2x+1)/(x^4+5x^2+1).