A002315 -id:A002315 - cp's OEIS frontend

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

1, 10, 89, 791, 7030, 62479, 555281, 4935050, 43860169, 389806471, 3464398070, 30789776159, 273643587361, 2432002510090, 21614379003449, 192097408520951, 1707262297685110, 15173263270645039, 134852107138120241, 1198495700972437130, 10651609201613813929

Offset: 0

Wolfdieter Lang, Aug 04 2000

This is the m=11 member of the m-family of sequences S(n,m-2)+S(n-1,m-2) = S(2*n,sqrt(m)) (for S(n,x) see Formula). The m=4..10 instances are A005408, A002878, A001834, A030221, A002315, A033890 and A057080, resp. The m=1..3 (signed) sequences are: A057078, A057077 and A057079, resp.
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim_{n->oo} a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 89, 389806471, 192097408520951, 7477414486269626733119, ... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 10, 0, 89, 0, 791, 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 = -7, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
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
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=11.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A005408, A002878, A001834, A030221, A002315, A033890, A057080.
Cf. A057078, A057077, A057079.
Cf. A018913, A049310.
Cf. A100047.

A057081 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,10]);
    else
        9*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1 + x)/(1 - 9*x + x^2), {x,0,50}], x] (* or *) LinearRecurrence[{9,-1}, {1,10}, 50] (* G. C. Greubel, Apr 12 2017 *)

Vec((1+x)/(1-9*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,9,1)-lucas_number2(n-1,9,1))/7 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

a(n) = 9*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 9) + S(n-1, 9) = S(2*n, sqrt(11)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 9) = A018913(n).
G.f.: (1+x)/(1-9*x+x^2).
Let q(n, x) = Sum{i=0..n} x^(n-i)*binomial(2*n-i, i), a(n) = (-1)^n*q(n, -11). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-9)*(-1)^n, where L is defined as in A108299; see also A070998 for L(n,+9). - Reinhard Zumkeller, Jun 01 2005
From Peter Bala, Jun 08 2025: (Start)
a(n) = (1/sqrt(7)) * ( ((sqrt(11) + sqrt(7))/2)^(2*n+1) - ((sqrt(11) - sqrt(7))/2)^(2*n+1) ).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/11 (telescoping series: 11/(a(n) - 1/a(n)) = 1/A018913(n+1) + 1/A018913(n)).
Conjecture: for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(11/7) [telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (1 - 4/b(n+1))/(1 - 4/b(n)), where b(n) = 2 + A056918(n)]. (End)

A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

Original entry on oeis.org

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350

Offset: 1

Benoit Cloitre, May 10 2003

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have
  a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
  a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ).
It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)
Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 1..2608
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Bisections are A001542 and A002315.

Cf. A084069, A084070, A133566, A079496, A001541, A001653, A008844, A055792, A049629, A108412, A143608.

a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
From Paul Barry, Jun 06 2006: (Start)
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007
Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007
From Peter Bala, Mar 23 2018: (Start)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
sqrt(1 + 2*a(2*n)^2) = A001541(n).
1 + 2*a(2*n)^2 = A055792(n+1).
a(2*n) - a(2*n-1) = A001653(n).
(1 + a(2*n+1)^2)/2 = A008844(n). (End)
a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021

A084158 a(n) = A000129(n) * A000129(n+1)/2.

Original entry on oeis.org

0, 1, 5, 30, 174, 1015, 5915, 34476, 200940, 1171165, 6826049, 39785130, 231884730, 1351523251, 7877254775, 45912005400, 267594777624, 1559656660345, 9090345184445, 52982414446326, 308804141493510, 1799842434514735, 10490250465594899, 61141660359054660, 356359711688733060

Offset: 0

Paul Barry, May 18 2003

May be called Pell triangles.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A001652, A001654, A002203, A002315, A003499.

Cf. A041011, A046729, A079291, A084159, A084175, A163960, A383720.

[Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n)/16): n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

with(combinat): a:=n->fibonacci(n,2)*fibonacci(n-1,2)/2: seq(a(n), n=1..22); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{5,5,-1},{0,1,5},30] (* Harvey P. Dale, Sep 07 2011 *)

Pell(n)=([2, 1; 1, 0]^n)[2, 1];
a(n)=Pell(n)*Pell(n+1)/2 \\ Charles R Greathouse IV, Mar 21 2016

a(n)=([0,1,0; 0,0,1; -1,5,5]^n*[0;1;5])[1,1] \\ Charles R Greathouse IV, Mar 21 2016

[(lucas_number2(2*n+1,2,-1) -2*(-1)^n)/16 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = ((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/16.
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3). - Mohamed Bouhamida, Sep 02 2006; corrected by Antonio Alberto Olivares, Mar 29 2008
a(n) = (-1/8)*(-1)^n + (( sqrt(2)+1)/16)*(3+2*sqrt(2))^n + ((-sqrt(2)+1)/16)*(3-2*sqrt(2))^n. - Antonio Alberto Olivares, Mar 30 2008
sqrt(a(n) - a(n-1)) = A000129(n). - Antonio Alberto Olivares, Mar 30 2008
O.g.f.: x/((1+x)(1-6*x+x^2)). - R. J. Mathar, May 18 2008
a(n) = A041011(n)*A041011(n+1). - R. K. Guy, May 18 2008
From Mohamed Bouhamida, Aug 30 2008: (Start)
a(n) = 6*a(n-1) - a(n-2) - (-1)^n.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3) - 2*(-1)^n. (End)
In general, for n>k+1, a(n+k) = A003499(k+1)*a(n-1) - a(n-k-2) - (-1)^n A000129(k+1)^2. - Charlie Marion, Jan 04 2012
For n>0, a(2n-1)*a(2n+1) = oblong(a(2n)); a(2n)*a(2n+2) = oblong(a(2n+1)-1). - Charlie Marion, Jan 09 2012
a(n) = A046729(n)/4. - Wolfdieter Lang, Mar 07 2012
a(n) = sum of squares of first n Pell numbers A000129 (A079291). - N. J. A. Sloane, Jun 18 2012
a(n) = (A002315(n) - (-1)^n)/8. - Adam Mohamed, Sep 05 2024
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} Pell(3*k)/Pell(k) * x^k/k ). - Seiichi Manyama, May 07 2025

A055997 Numbers k such that k*(k - 1)/2 is a square.

Original entry on oeis.org

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

Offset: 1

Barry E. Williams, Jun 14 2000

Numbers k such that (k-th triangular number - k) is a square.
Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002
Number of closed walks of length 2k on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - Bradley Klee & Bill Gosper, Jul 22 2015
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.

Colin Barker, Table of n, a(n) for n = 1..100
Dario Alpern, a^4+b^3=c^2.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 46, 56.
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A007913, A001541.
A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
a(n) = A001108(n-1)+1.
A001110(n-1)=a(n)*(a(n)-1)/2.
Cf. A001652, A001653, A046090.
Identical to A115599, but with additional leading term.

I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
map(A,[$1..100]); # Robert Israel, Jul 22 2015

Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
(1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
LinearRecurrence[{7,-7,1},{1,2,9},30] (* Harvey P. Dale, Dec 06 2015 *)

Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */

t(n)=(1+sqrt(2))^(n-1);
for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ Hugo Pfoertner, Nov 29 2019

a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020

a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - Charlie Marion, Jul 21 2003; corrected by Michel Marcus, Apr 20 2020
a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004; corrected by Bernard Schott, Apr 20 2020
A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004
a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Feb 21 2006; corrected by Michel Marcus, Apr 20 2020
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with mA001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020

A077444 Numbers k such that (k^2 + 4)/2 is a square.

Original entry on oeis.org

2, 14, 82, 478, 2786, 16238, 94642, 551614, 3215042, 18738638, 109216786, 636562078, 3710155682, 21624372014, 126036076402, 734592086398, 4281516441986, 24954506565518, 145445522951122, 847718631141214, 4940866263896162, 28797478952235758, 167844007449518386

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003
Equivalently, 2*n^2 + 8 is a square.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 2 + n^2/2. - Ctibor O. Zizka, Nov 09 2009
The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013
a((p+1)/2) == 2 (mod p) where p is an odd prime. - Altug Alkan, Mar 17 2016

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Amiram Eldar, Table of n, a(n) for n = 1..1306
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Tanya Khovanova, Recursive Sequences
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
J. J. O'Connor and E. F. Robertson, Pell's Equation. [Broken link]
Eric Weisstein's World of Mathematics, Pell Equation.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

(A077445(n))^2 - 2*a(n) = 8.
First differences of A001541.
Pairwise sums of A001542.
Bisection of A002203 and A080039.
Cf. A001653.

[n: n in [0..10^8] | IsSquare((n^2 + 4) div 2)]; // Vincenzo Librandi, Jun 20 2015

LinearRecurrence[{6,-1},{2,14},30] (* Harvey P. Dale, Jul 25 2018 *)

for(n=1,20,q=(1+sqrt(2))^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec(2*x*(1+x)/(1-6*x+x^2) + O(x^100)) \\ Altug Alkan, Mar 17 2016

a(n) = (((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + ((3 + 2*sqrt(2))^(n-1) - (3 - 2*sqrt(2))^(n-1))) / (2*sqrt(2)).
a(n) = 2*A002315(n-1).
Recurrence: a(n) = 6*a(n-1) - a(n-2), starting 2, 14.
Offset 0, with a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = a^((2n+1)/2) - b^((2n+1)/2). a(n) = 2*(A001109(n+1) + A001109(n)) = (A003499(n+1) - A003499(n))/2 = 2*sqrt(A001108(2n+1)) = sqrt(A003499(2n+1)-2). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Limit_{n->oo} a(n)/a(n-1) = 5.82842712474619009760... = 3 + 2*sqrt(2). See A156035.
From R. J. Mathar, Nov 16 2007: (Start)
G.f.: 2*x*(1+x)/(1-6*x+x^2).
a(n) = 2*(7*A001109(n) - A001109(n+1)). (End)
a(n) = (1+sqrt(2))^(2*n-1) - (1+sqrt(2))^(1-2*n). - Gerson Washiski Barbosa, Sep 19 2010
a(n) = floor((1 + sqrt(2))^(2*n-1)). - Thomas Ordowski, Jun 07 2013
a(n) = sqrt(2*A075870(n)^2-4). - Derek Orr, Jun 18 2015
a(n) = 2*sqrt((2*A001653(n)^2)-1). - César Aguilera, Jul 13 2023
E.g.f.: 2*(1 + exp(3*x)*(sqrt(2)*sinh(2*sqrt(2)*x) - cosh(2*sqrt(2)*x))). - Stefano Spezia, Aug 27 2024

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))n^(k-j), n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 5, -1, 1, 5, 11, 7, -2, 1, 6, 19, 29, 9, -1, 1, 7, 29, 71, 76, 11, 1, 1, 8, 41, 139, 265, 199, 13, 2, 1, 9, 55, 239, 666, 989, 521, 15, 1, 1, 10, 71, 377, 1393, 3191, 3691, 1364, 17, -1, 1, 11, 89, 559, 2584, 8119, 15289, 13775, 3571, 19, -2

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Oct 22 2017

This array is used to compute A269254: A269254(n) = least k such that A(n,k) is a prime, or -1 if no such k exists.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
  1   2    1    -1     -2      -1        1         2          1          -1
  1   3    5     7      9      11       13        15         17          19
  1   4   11    29     76     199      521      1364       3571        9349
  1   5   19    71    265     989     3691     13775      51409      191861
  1   6   29   139    666    3191    15289     73254     350981     1681651
  1   7   41   239   1393    8119    47321    275807    1607521     9369319
  1   8   55   377   2584   17711   121393    832040    5702887    39088169
  1   9   71   559   4401   34649   272791   2147679   16908641   133121449
  1  10   89   791   7030   62479   555281   4935050   43860169   389806471
  1  11  109  1079  10681  105731  1046629  10360559  102558961  1015229051

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Rows: A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, ...
Columns: A000012, A000027, A028387, ...

(* Array: *)
Grid[Table[LinearRecurrence[{n, -1}, {1, 1 + n}, 10], {n, 10}]]
(* Array antidiagonals flattened (gives this sequence): *)
A294099[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] n^(k - j), {j, 0, k}]; Flatten[Table[A294099[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*n^(k-j))}; /* Michael Somos, Jun 19 2023 */

A(n,0) = 1, A(n,1) = n + 1, A(n,k) = n*A(n,k-1) - A(n,k-2), n >= 1, k >= 2.
G.f. for row n: (1 + x)/(1 - n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A299045. - Michael Somos, Jun 19 2023

A075870 Numbers k such that 2*k^2 - 4 is a square.

Original entry on oeis.org

2, 10, 58, 338, 1970, 11482, 66922, 390050, 2273378, 13250218, 77227930, 450117362, 2623476242, 15290740090, 89120964298, 519435045698, 3027489309890, 17645500813642, 102845515571962, 599427592618130, 3493720040136818, 20362892648202778, 118683635849079850

Offset: 1

Gregory V. Richardson, Oct 16 2002

Lim_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2).
Also gives solutions to the equation x^2-2 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators = A075870, denominators = A002315. - Clark Kimberling, Aug 27 2008
Numbers n such that sqrt(floor(n^2/2 - 1)) is an integer. The integer square roots are given by A002315. - Richard R. Forberg, Aug 01 2013
a(n) are the integer square roots of m^2 + (m+2)^2. The values of m are given by A065113 (except for m = 0). The values of this expression are given by A165518. - Richard R. Forberg, Aug 15 2013
Values of x (or y) in the solutions to x^2 - 6*x*y + y^2 + 16 = 0. - Colin Barker, Feb 04 2014
Also integers k such that k^2 is equal to the sum of four consecutive triangular numbers. - Colin Barker, Dec 20 2014
Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = (y-1)^2 + (y+1)^2. y-values are listed in A002315. Example: for x=58 and y=41, 57*58/2 + 58*59/2 = 40^2 + 42^2. - Bruno Berselli, Mar 19 2018

			From _Muniru A Asiru_, Mar 19 2018: (Start)
For k=2, 2*2^2 - 4 = 8 - 4 = 4 = 2^2.
For k=10, 2*10^2 - 4 = 200 - 4 =  196 = 14^2.
For k=58, 2*58^2 - 4 = 6728 - 4 =  6724 = 82^2.
... (End)

A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Colin Barker, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A000217, A000290, A002315.

Twice A001653.

a:=[2,10];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018

a:= proc(n) option remember: if n = 1 then 2 elif n = 2 then 10 elif  n >= 3 then 6*procname(n-1) - procname(n-2) fi; end: seq(a(n), n = 0..25); # Muniru A Asiru, Mar 19 2018

LinearRecurrence[{6,-1},{2,10},30] (* Harvey P. Dale, Sep 27 2018 *)

Vec(2*x*(1-x)/(1-6*x+x^2) + O(x^100)) \\ Colin Barker, Dec 20 2014

a(n) = 2 * A001653(n).
a(n) = (1/sqrt(2))*((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1)) = 6*a(n-1) - a(n-2).
G.f.: 2*x*(1-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/2). - Paul Weisenhorn, Jun 11 2020
From Peter Bala, Aug 21 2022: (Start)
a(n) = 2*Pell(2*n-1).
1/a(n) - 1/a(n+1) = 1/(Pell(2*n) + 1/Pell(2*n)), where Pell(n) = A000129(n). (End)

More terms from Colin Barker, Dec 20 2014

A097775 Pell equation solutions (14a(n))^2 - 197b(n)^2 = -1 with b(n) = A097776(n), n >= 0.

Original entry on oeis.org

1, 787, 618581, 486203879, 382155630313, 300373839222139, 236093455472970941, 185569155627915937487, 145857120230086453893841, 114643510931692324844621539, 90109653735189937241418635813, 70826073192348358979430203127479, 55669203419532074967894898239562681

Offset: 0

Wolfdieter Lang, Aug 31 2004

			(x,y) = (14*1=14;1), (11018=14*787;785), (8660134=14*618581;617009), ... give the positive integer solutions to x^2 - 197*y^2 =-1.

Colin Barker, Table of n, a(n) for n = 0..345
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (786,-1).

Cf. A097774 for S(n, 2*393).

Cf. similar sequences of the type (1/k)*sinh((2*n + 1)*arcsinh(k)): A002315 (k=1), A049629 (k=2), A097314 (k=3), A078989 (k=4), A097726 (k=5), A097729 (k=6), A097732 (k=7), A097735 (k=8), A097738 (k=9), A097741 (k=10), A097766 (k=11), A097769 (k=12), A097772 (k=13), this sequence (k=14).

LinearRecurrence[{786, -1}, {1, 787}, 20] (* Harvey P. Dale, Dec 12 2017 *)

Vec((1+x)/(1-2*393*x+x^2) + O(x^100)) \\ Colin Barker, Apr 04 2015

G.f.: (1 + x)/(1 - 2*393*x + x^2).
a(n) = S(n, 2*393) + S(n-1, 2*393) = S(2*n, 2*sqrt(197)), with Chebyshev polynomials of the 2nd kind. See A049310 for the triangle of S(n, x) = U(n, x/2) coefficients. S(-1, x) = 0 = U(-1, x).
a(n) = ((-1)^n)*T(2*n+1, 14*i)/(14*i) with the imaginary unit i and Chebyshev polynomials of the first kind. See the T-triangle A053120.
a(n) = 786*a(n-1) - a(n-2) for n > 1; a(0)=1, a(1)=787. - Philippe Deléham, Nov 18 2008
a(n) = (1/14)*sinh((2*n + 1)*arcsinh(14)). - Bruno Berselli, Apr 05 2018

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

Original entry on oeis.org

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461

Offset: 1

Paul Weisenhorn, Feb 21 2009

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n)+1 = a(n)^2,
   (A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
  A   a-sequence    b-sequence     c-sequence
  1     A001653       A002315(n-1)   A078522
  2     A072256       A054320(n-1)   A045502(n-1)
  3     A001570       A028230        A059989(n-1)
  4     A007805       A049629(n-1)   A157459
  5  -> A157014 <-    A133283        A157460
  6     A153111       A157461        A157874
  7     A157877       A157878        A157879
  8     A077420(n-1)  A046176        A157880
  9     A097315(n-1)  A097314(n-1)   A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. similar sequences listed in A238379.

a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014

seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020

CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041049 *)
a[30, 20] (* Gerry Martens, Jun 07 2015 *)
Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)

Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012

[chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Edited by Alois P. Heinz, Sep 09 2011

A046729 Expansion of 4x/((1+x)(1-6*x+x^2)).

Original entry on oeis.org

0, 4, 20, 120, 696, 4060, 23660, 137904, 803760, 4684660, 27304196, 159140520, 927538920, 5406093004, 31509019100, 183648021600, 1070379110496, 6238626641380, 36361380737780, 211929657785304, 1235216565974040, 7199369738058940, 41961001862379596, 244566641436218640

Offset: 0

N. J. A. Sloane

Related to Pythagorean triples: alternate terms of A001652 and A046090.
Even-valued legs of nearly isosceles right triangles: legs differ by 1. 0 is smaller leg of degenerate triangle with legs 0 and 1 and hypotenuse 1. - Charlie Marion, Nov 11 2003
The complete (nearly isosceles) primitive Pythagorean triple is given by {a(n), a(n)+(-1)^n, A001653(n)}. - Lekraj Beedassy, Feb 19 2004
Note also that A046092 is the even leg of this other class of nearly isosceles Pythagorean triangles {A005408(n), A046092(n), A001844(n)}, i.e., {2n+1, 2n(n+1), 2n(n+1)+1} where longer sides (viz. even leg and hypotenuse) are consecutive. - Lekraj Beedassy, Apr 22 2004
Union of even terms of A001652 and A046090. Sum of legs of primitive Pythagorean triangles is A002315(n) = 2*a(n) + (-1)^n. - Lekraj Beedassy, Apr 30 2004

			[1,0,1]*[1,2,2; 2,1,2; 2,2,3]^0 gives (degenerate) primitive Pythagorean triple [1, 0, 1], so a(0) = 0. [1,0,1]*[1,2,2; 2,1,2; 2,2,3]^7 gives primitive Pythagorean triple [137903, 137904, 195025] so a(7) = 137904.
G.f. = 4*x + 20*x^2 + 120*x^3 + 696*x^4 + 4060*x^5 + 23660*x^6 + ...

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, p. 17. MR2002669.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).
Index entries for two-way infinite sequences

Cf. A000129, A001109, A001333, A001652, A002315, A029549, A046090.

Cf. A046727, A047235, A084158, A084159, A089499, A114336.

[4*Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n) / 16): n in [0..30]]; // Vincenzo Librandi, Jul 29 2019

LinearRecurrence[{5,5,-1}, {0,4,20}, 25] (* Vincenzo Librandi, Jul 29 2019 *)

a(n)=n%2+(real((1+quadgen(8))^(2*n+1))-1)/2

a(n)=if(n<0,-a(-1-n),polcoeff(4*x/(1+x)/(1-6*x+x^2)+x*O(x^n),n))

[(lucas_number2(2*n+1,2,-1) -2*(-1)^n)/4 for n in range(41)] # G. C. Greubel, Feb 11 2023

a(n) = ((1+sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) + 2*(-1)^(n+1))/4.
a(n) = A089499(n)*A089499(n+1).
a(n) = 4*A084158(n). - Lekraj Beedassy, Jul 16 2004
a(n) = ceiling((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/4. - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 12 2004
a(n) is the k-th entry among the complete near-isosceles primitive Pythagorean triple A114336(n), where k = (3*(2n-1) - (-1)^n)/2, i.e., a(n) = A114336(A047235(n)), for positive n. - Lekraj Beedassy, Jun 04 2006
a(n) = A046727(n) - (-1)^n = 2*A114620(n). - Lekraj Beedassy, Aug 14 2006
From George F. Johnson, Aug 29 2012: (Start)
2*a(n)*(a(n) + (-1)^n) + 1 = (A000129(2*n+1))^2;
n > 0, 2*a(n)*(a(n) + (-1)^n) + 1 = ((a(n+1) - a(n-1))/4)^2, a perfect square.
a(n+1) = (3*a(n) + 2*(-1)^n) + 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n-1) = (3*a(n) + 2*(-1)^n) - 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n+1) = 6*a(n) - a(n-1) + 4*(-1)^n.
a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
a(n+1) *a(n-1) = a(n)*(a(n) + 4*(-1)^n).
a(n) = (sqrt(1 + 8*A029549(n)) - (-1)^n)/2.
a(n) = A002315(n) - A084159(n) = A084159(n) - (-1)^n.
a(n)  = A001652(n) + (1 - (-1)^n)/2 = A046090(n) - (1 + (-1)^n)/2.
Limit_{n->oo} a(n)/a(n-1) =  3 + 2*sqrt(2).
Limit_{n->oo} a(n)/a(n-2) = 17 + 12*sqrt(2).
Limit_{n->oo} a(n)/a(n-r) = (3 + 2*sqrt(2))^r.
Limit_{n->oo} a(n-r)/a(n) = (3 - 2*sqrt(2))^r. (End)
From G. C. Greubel, Feb 11 2023: (Start)
a(n) = (A001333(2*n+1) - 2*(-1)^n)/4.
a(n) = (1/2)*(A001109(n+1) + A001109(n) - (-1)^n). (End)
E.g.f.: exp(-x)*(exp(4*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 1)/2. - Stefano Spezia, Aug 03 2024

cp's OEIS Frontend

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

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A084068 a(1) = 1, a(2) = 2; a(2*k) = 2*a(2*k-1) - a(2*k-2), a(2*k+1) = 4*a(2*k) - a(2*k-1).

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A084158 a(n) = A000129(n) * A000129(n+1)/2.

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A055997 Numbers k such that k*(k - 1)/2 is a square.

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A077444 Numbers k such that (k^2 + 4)/2 is a square.

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A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*n^(k-j), n >= 1, k >= 0.

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A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))n^(k-j), n >= 1, k >= 0.

A097775 Pell equation solutions (14a(n))^2 - 197b(n)^2 = -1 with b(n) = A097776(n), n >= 0.

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

A046729 Expansion of 4x/((1+x)(1-6*x+x^2)).