A002315 -id:A002315 - cp's OEIS frontend

A113501 Indices of prime NSW numbers A088165.

1, 2, 3, 9, 14, 23, 29, 81, 128, 210, 468, 473, 746, 950, 3344, 4043, 4839, 14376, 39521, 64563, 72984, 82899, 84338, 85206, 86121, 139160

Offset: 1

Eric W. Weisstein, Jan 09 2006

Very closely related to indices of prime Pell-Lucas numbers (A099088).

a(27) > 221400. - Robert Price, Mar 29 2019

			NSW(1) = 7, NSW(2) = 41, NSW(3) = 239, NSW(9) = 9369319, ...

H. Li, T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 47.
Eric Weisstein's World of Mathematics, NSW Number
Eric Weisstein's World of Mathematics, Integer Sequence Primes

Cf. A002315, A088165, A099088.

nsw = LinearRecurrence[{6, -1}, {7, 41}, 1000]; Position[nsw, ?(PrimeQ[#] &)] // Flatten (* _Amiram Eldar, Dec 07 2018 *)

isok(n) = my(w=3+quadgen(32)); isprime(imag((1+w)*w^n)); \\ Michel Marcus, Dec 07 2018

a(19)-a(20) from Eric W. Weisstein, May 22 2006
a(21) from Eric W. Weisstein, Aug 29 2006
a(22) from Eric W. Weisstein, Nov 11 2006
a(23) from Eric W. Weisstein, Nov 26 2006
a(24) from Eric W. Weisstein, Dec 10 2006
a(25) from Eric W. Weisstein, Jan 25 2007
a(26) from Robert Price, Dec 07 2018

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Joseph L. Pe, Jan 07 2003

Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176.
Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016
From Sergey Pavlov, Mar 14 2017: (Start)
The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

From Benoit Cloitre, Jan 19 2003: (Start)
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
a(n) = A008844(n) - 1. (End)
From R. J. Mathar, Sep 21 2011: (Start)
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = 24*A029546(n-2). (End)
a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020
From Amiram Eldar, Dec 02 2024: (Start)
a(n) = A001542(n-1)*A001542(n).
Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Edited by Bruno Berselli, Nov 17 2016

A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).

Original entry on oeis.org

5, 27, 157, 915, 5333, 31083, 181165, 1055907, 6154277, 35869755, 209064253, 1218515763, 7102030325, 41393666187, 241259966797, 1406166134595, 8195736840773, 47768254910043, 278413792619485, 1622714500806867, 9457873212221717, 55124524772523435, 321289275422918893

Offset: 0

Creighton Dement, Jan 23 2005

A floretion-generated sequence relating to NSW numbers and numbers n such that (n^2 - 8)/2 is a square. It is also possible to label this sequence as the "tesfor-transform of the zero-sequence" under the floretion given in the program code, below. This is because the sequence "vesseq" would normally have been A046184 (indices of octagonal numbers which are also a square) using the floretion given. This floretion, however, was purposely "altered" in such a way that the sequence "vesseq" would turn into A000004. As (a(n)) would not have occurred under "natural" circumstances, one could speak of it as the transform of A000004.
Floretion Algebra Multiplication Program FAMP code: - tesforseq[ + 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e], Note: vesforseq = A000004, lesforseq = A002315, jesforseq = A077445
From Wolfdieter Lang, Feb 05 2015: (Start)
All positive solutions x = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the  fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding y solutions are given by y(n) = A253811(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See, e.g., the Nagell reference on how to find all solutions.
(End)

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, 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 pp. 55, 56.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Tanya Khovanova, Recursive Sequences
Morris Newman, Daniel Shanks, and H. C. Williams, Simple groups of square order and an interesting sequence of primes, Acta Arith., 38 (1980/1981) 129-140. MR82b:20022.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000004, A000129, A001109, A002315, A038761, A038762, A046184, A054490, A164737.

Cf. A077442, A077443(n+1), A077445, A253811.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!((5 - 3*x)/(1-6*x+x^2))); // G. C. Greubel, Jul 26 2018

A101386:= (n) -> simplify(5*ChebyshevU(n, 3) - 3*ChebyshevU(n-1, 3)); seq( A101386(n), n = 0..30); # G. C. Greubel, Mar 17 2020

CoefficientList[ Series[(5-3x)/(1-6x+x^2), {x,0,30}], x] (* Robert G. Wilson v, Jan 29 2005 *)
LinearRecurrence[{6,-1},{5,27},30] (* Harvey P. Dale, Apr 23 2016 *)

Vec((5-3*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Feb 05 2015

[5*chebyshev_U(n,3) -3*chebyshev_U(n-1,3) for n in (0..30)] # G. C. Greubel, Mar 17 2020

a(n) = A002315(n) + A077445(n+1). Note: the offset of A077445 is 1.
a(n+1) - a(n) = 2*A054490(n+1).
a(n) = 6*a(n-1) - a(n-2), a(0)=5, a(1)=27. - Philippe Deléham, Nov 17 2008
From Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009: (Start)
a(n) = ((5+sqrt(18))*(3 + sqrt(8))^n + (5-sqrt(18))*(3 - sqrt(8))^n)/2.
Third binomial transform of A164737. (End)
a(n) = rational part of z(n), with z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions. See the preceding formula. - Wolfdieter Lang, Feb 05 2015
a(n) = 5*A001109(n+1) - 3*A001109(n). - G. C. Greubel, Mar 17 2020
a(n) = Pell(2*n+2) + 3*Pell(2*n+1), where Pell(n) = A000129(n). - G. C. Greubel, Apr 17 2020
E.g.f.: exp(3*x)*(5*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from Robert G. Wilson v, Jan 29 2005

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

Original entry on oeis.org

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset: 0

Paul Weisenhorn, Oct 28 2012

a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.

General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.

			For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
  s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
  if (r=floor(r)) then
    n:=n+1: a(n):=r: x(n):=z:
    b(n):=sqrt((s-110)/11):
    print(n,a(n),b(n),x(n)):
  end if:
end do:

LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* Harvey P. Dale, Aug 15 2022 *)

a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);
a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).
a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).
G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).
With r=sqrt(11); s=10+3*r; t=10-3*r:
a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.
a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.
a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022

A096053 a(n) = (3*9^n - 1)/2.

Original entry on oeis.org

1, 13, 121, 1093, 9841, 88573, 797161, 7174453, 64570081, 581130733, 5230176601, 47071589413, 423644304721, 3812798742493, 34315188682441, 308836698141973, 2779530283277761, 25015772549499853, 225141952945498681

Offset: 0

Benoit Cloitre, Jun 18 2004

Generalized NSW numbers. - Paul Barry, May 27 2005
Counts total area under elevated Schroeder paths of length 2n+2, where area under a horizontal step is weighted 3. Case r=4 for family (1+(r-1)x)/(1-2(1+r)x+(1-r)^2*x^2). Case r=2 gives NSW numbers A002315. Fifth binomial transform of (1+8x)/(1-16x^2), A107906. - Paul Barry, May 27 2005
Primes in this sequence include: a(2) = 13, a(4) = 1093, a(7) = 797161. Semiprimes in this sequence include: a(3) = 121 = 11^2, a(5) = 9841 = 13 * 757, a(6) = 88573 = 23 * 3851, a(9) = 64570081 = 1871 * 34511, a(10) = 581130733 = 1597 * 363889, a(12) = 47071589413 = 47 * 1001523179, a(19) = 225141952945498681 = 13097927 * 17189128703.
Sum of divisors of 9^n. - Altug Alkan, Nov 10 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A083420, A096045, A096046, A096047, A096054.

Cf. A107903, A138894 ((5*9^n-1)/4).

[(3*9^n-1)/2: n in [0..20]]; // Vincenzo Librandi, Nov 01 2011

Table[(3*9^n - 1)/2, {n, 0, 18}] (* L. Edson Jeffery, Feb 13 2015 *)

a(n)=(3*9^n-1)/2 \\ Charles R Greathouse IV, Sep 28 2015

vector(30, n, n--; sigma(9^n)) \\ Altug Alkan, Nov 10 2015

From Paul Barry, May 27 2005: (Start)
G.f.: (1+3*x)/(1-10*x+9*x^2);
a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*4^k;
a(n) = ((1+sqrt(4))*(5+2*sqrt(4))^n+(1-sqrt(4))*(5-2*sqrt(4))^n)/2. (End)
a(n-1) = (-9^n/3)*B(2n,1/3)/B(2n) where B(n,x) is the n-th Bernoulli polynomial and B(k)=B(k,0) is the k-th Bernoulli number.
a(n) = 10*a(n-1) - 9*a(n-2).
a(n) = 9*a(n-1) + 4. - Vincenzo Librandi, Nov 01 2011
a(n) = A000203(A001019(n)). - Altug Alkan, Nov 10 2015
a(n) = A320030(3^n-1). - Nathan M Epstein, Jan 02 2019

Edited by N. J. A. Sloane, at the suggestion of Andrew S. Plewe, Jun 15 2007

A097783 Chebyshev polynomials S(n,11) + S(n-1,11) with Diophantine property.

Original entry on oeis.org

1, 12, 131, 1429, 15588, 170039, 1854841, 20233212, 220710491, 2407582189, 26262693588, 286482047279, 3125039826481, 34088956044012, 371853476657651, 4056299287190149, 44247438682433988, 482665526219583719, 5265073349732986921, 57433141320843272412

Offset: 0

Wolfdieter Lang, Aug 31 2004

All positive integer solutions of Pell equation (3*a(n))^2 - 13*b(n)^2 = -4 together with b(n)=A078922(n+1), n>=0.

			All positive solutions to the Pell equation x^2 - 13*y^2 = -4 are (3=3*1,1), (36=3*12,10), (393=3*131,109), (4287=3*1429,1189 ), ...

Colin Barker, Table of n, a(n) for n = 0..963
Andersen, K., Carbone, L. and Penta, D., 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.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
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
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
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 (11,-1).

Cf. S(n, 11) = A004190(n).

Cf. A000045, A002315, A004146, A006190, A006497, A100047, A113224, A192425.

I:=[1,12]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1 + x) / (1 - 11 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-11*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,11,1)-lucas_number2(n-1,11,1))/9 for n in range(1, 19)] # Zerinvary Lajos, Nov 10 2009

a(n) = S(n, 11) + S(n-1, 11) = S(2*n, sqrt(13)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x) = 0 = U(-1, x).
a(n) = (-2/3)*i*((-1)^n)*T(2*n+1, 3*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-11*x+x^2).
a(n) = L(n,-11)*(-1)^n, where L is defined as in A108299; see also A078922 for L(n,+11). - Reinhard Zumkeller, Jun 01 2005
a(n) = 11*a(n-1) - a(n-2) with a(0)=1 and a(1)=12. - Philippe Deléham, Nov 17 2008
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 12, 0, 131, 0, 1429, 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. See A100047 for the connection with Chebyshev polynomials.
b(n) = 1/2*( (-1)^n - 1 )*F(n,3) + 1/3*( 1 + (-1)^(n+1) )*F(n+1,3), where F(n,x) is the n-th Fibonacci polynomial. The o.g.f. is x*(1 + x^2)/(1 - 11*x^2 + x^4).
Exp( Sum_{n >= 1} 6*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*x^n.
Exp( Sum_{n >= 1} (-6)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = A006497(2n+1)/3. - Adam Mohamed, Aug 22 2024

A123335 a(n) = -2*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=-1.

Original entry on oeis.org

1, -1, 3, -7, 17, -41, 99, -239, 577, -1393, 3363, -8119, 19601, -47321, 114243, -275807, 665857, -1607521, 3880899, -9369319, 22619537, -54608393, 131836323, -318281039, 768398401, -1855077841, 4478554083, -10812186007, 26102926097, -63018038201, 152139002499

Offset: 0

Philippe Deléham, Jun 27 2007

Inverse binomial transform of A077957.
The inverse of the g.f. is 3-x-2/(1+x) which generates 1, 1, -2, +2, -2, +2, ... (-2, +2 periodically continued). - Gary W. Adamson, Jan 10 2011
Pisano period lengths:  1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, ... - R. J. Mathar, Aug 10 2012
a(n) is the rational part of the Q(sqrt(2)) integer (sqrt(2) - 1)^n = a(n) + A077985(n-1)*sqrt(2), with A077985(-1) = 0. - Wolfdieter Lang, Dec 07 2014
3^n*a(n) = A251732(n) gives the rational part of the integer in Q(sqrt(2)) giving the length of a variant of Lévy's C-curve at iteration step n. - Wolfdieter Lang, Dec 07 2014
Define u(0) = 1/0, u(1) = -1/1, and u(n) = -(8 + 3*u(n-1)*u(n-2))/(3*u(n-1) + 2*u(n-2)) for n>1. Then u(n) = a(n)/A000219(n). - Michael Somos, Apr 19 2022

			G.f. = 1 - x + 3*x^2 - 7*x^3 + 17*x^4 - 41*x^5 + 99*x^6 + ... - _Michael Somos_, Apr 19 2022

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,1).

Cf. A000129, A001333, A077985, A251732, A001541 (bisection), A002315 (bisection).

[Round(1/2*((-1-Sqrt(2))^n+(-1+Sqrt(2))^n)): n in [0..30]]; // G. C. Greubel, Oct 12 2017

a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^(-n)):
seq(a(n), n=0..33);  # Alois P. Heinz, Jun 22 2021

LinearRecurrence[{-2,1},{1,-1},40] (* Harvey P. Dale, Nov 03 2011 *)

x='x+O('x^50); Vec((1+x)/(1+2*x-x^2)) \\ G. C. Greubel, Oct 12 2017

{a(n) = real((-1 + quadgen(8))^n)}; /* Michael Somos, Apr 19 2022 */

a(n) = (-1)^n*A001333(n).
G.f.: (1+x)/(1+2*x-x^2).
a(n) = A077985(n) + A077985(n-1). - R. J. Mathar, Mar 28 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 19 2013
G.f.: 1/(1 + x/(1 + 2*x/(1 - x))). - Michael Somos, Apr 19 2022
E.g.f.: exp(-x)*cosh(sqrt(2)*x). - Stefano Spezia, Feb 01 2023

Corrected by N. J. A. Sloane, Oct 05 2008

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

Original entry on oeis.org

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1

Offset: 0

Philippe Deléham, Jun 09 2007

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011
Row sums: A057078. - Philippe Deléham, Jun 11 2007
Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012
If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020

Vincenzo Librandi, Rows n = 1..101, flattened
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry and Aoife Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, example 15.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Cf. A039599, A085478, A123970, A321620.

# The function RiordanSquare is defined in A321620.
RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019

max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)

@CachedFunction
def A129818(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
    return A129818(n-1,k-1) - A129818(n-2,k) - h
for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).
Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009
O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
From Wolfdieter Lang, Dec 20 2010: (Start)
Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).
T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)
T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012
A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019
Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020

A050796 Numbers n such that n^2 + 1 is expressible as the sum of two nonzero squares in at least one way (the trivial solution n^2 + 1 = n^2 + 1^2 is not counted).

Original entry on oeis.org

1, 7, 8, 12, 13, 17, 18, 21, 22, 23, 27, 28, 30, 31, 32, 33, 34, 37, 38, 41, 42, 43, 44, 46, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 64, 67, 68, 70, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 86, 87, 88, 89, 91, 92, 93, 96, 97, 98, 99, 100, 102, 103, 104, 105, 106, 107

Offset: 1

Patrick De Geest, Sep 15 1999

nonn

Analogous solutions exist for the sum of two identical squares z^2 + 1 = 2*r^2 (e.g., 41^2 + 1 = 2*29^2). Values of 'z' are the terms in sequence A002315, values of 'r' are the terms in sequence A001653.
Apart from the first term, numbers n such that (n^2)! == 0 mod (n^2 + 1)^2. - Michel Lagneau, Feb 14 2012
Numbers n such that neither n^2 + 1 nor (n^2 + 1)/2 is prime. - Charles R Greathouse IV, Feb 14 2012

			E.g., 57^2 + 1 = 15^2 + 55^2 = 21^2 + 53^2 = 35^2 + 45^2.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Index entries for sequences related to sums of squares

Cf. A000129, A001653, A002315, A050795, A050798.

t={1}; Do[i=c=2; While[iJayanta Basu, Jun 01 2013 *)

is(n)=!isprime((n^2+1)/if(n%2,2,1)) \\ Charles R Greathouse IV, Feb 14 2012

cp's OEIS Frontend

A113501 Indices of prime NSW numbers A088165.

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A078522 Numbers k such that (k+1)*(2*k+1) is a perfect square.

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A101386 Expansion of g.f.: (5 - 3*x)/(1 - 6*x + x^2).

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A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

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A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

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A096053 a(n) = (3*9^n - 1)/2.

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A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.