A011943 -id:A011943 - cp's OEIS frontend

A102344 Numbers n such that the Diophantine equation (x+2)^3-x^3=2*n^2 has solutions.

2, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367

Offset: 1

Richard Choulet, Sep 08 2008

n^2 = 3*(2*x+4)^2+16.

Essentially the same as A011943. - Chris Boyd, Apr 18 2015

			The first examples are 2^3-0^3=2*2^2 ; 5^3-3^3=2*7^2 ; 57^3-55^3=2*97^2 ; 781^3-779^3=2*1351^2 ; 10865^3-10863^3=2*18817^2

Index entries for linear recurrences with constant coefficients, signature (14,-1).

Cf. A011943.

I:=[2,7,97]; [n le 3 select I[n] else 14*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 19 2015

2, seq(othopoly[T](n,7),n=1..50); # Robert Israel, Apr 19 2015

a[1]=2; a[2]=7; a[3]=97; a[n_] := a[n] = 14*a[n-1]-a[n-2]; Table[a[n], {n, 1, 17}] (* Jean-François Alcover, Dec 17 2013 *)
LinearRecurrence[{14,-1},{2,7,97},20] (* Harvey P. Dale, Sep 26 2016 *)

Vec(x*(2-21*x+x^2)/(1-14*x+x^2) + O(x^30)) \\ Michel Marcus, Apr 19 2015

a(n+2) = 14*a(n+1)-a(n) for n>=2.

G.f.: x*(2-21*x+x^2)/(1-14*x+x^2). a(n)=7*A007655(n+2)-97*A007655(n+1), n>1. - R. J. Mathar, Sep 11 2008

More terms from Vincenzo Librandi, Apr 19 2015

A110294 a(2n) = A028230(n), a(2n+1) = -A067900(n+1).

Original entry on oeis.org

1, -8, 15, -112, 209, -1560, 2911, -21728, 40545, -302632, 564719, -4215120, 7865521, -58709048, 109552575, -817711552, 1525870529, -11389252680, 21252634831, -158631825968, 296011017105, -2209456310872, 4122901604639, -30773756526240, 57424611447841

Offset: 0

Creighton Dement, Jul 18 2005

See A110293.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A001353, A001570, A011943, A028230, A067900, A110293.

[(3*(-1)^n-1)*Evaluate(ChebyshevSecond(n+1), 2)/2: n in [0..40]]; // G. C. Greubel, Jan 04 2023

seriestolist(series((1-8*x+x^2)/((x^2-4*x+1)*(x^2+4*x+1)), x=0,25));

CoefficientList[Series[(1-8x+x^2)/((1-4x+x^2)(1+4x+x^2)), {x, 0, 24}], x] (* Michael De Vlieger, Nov 01 2016 *)
LinearRecurrence[{0,14,0,-1},{1,-8,15,-112},30] (* Harvey P. Dale, Dec 16 2024 *)

Vec((1-8*x+x^2)/((1-4*x+x^2)*(1+4*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 01 2016

[(3*(-1)^n-1)*chebyshev_U(n,2)/2 for n in range(41)] # G. C. Greubel, Jan 04 2023

G.f.: (1-8*x+x^2) / ((1-4*x+x^2)*(1+4*x+x^2)).
a(n) = 14*a(n-2) - a(n-4) for n>3. - Colin Barker, Nov 01 2016
a(n) = (3*(-1)^n - 1)*A001353(n+1)/2. - R. J. Mathar, Sep 11 2019

A232765 Values of y solving x^2 = floor(y^2/3 + y).

Original entry on oeis.org

0, 1, 4, 9, 28, 73, 144, 409, 1036, 2025, 5716, 14449, 28224, 79633, 201268, 393129, 1109164, 2803321, 5475600, 15448681, 39045244, 76265289, 215172388, 543830113, 1062238464, 2996964769, 7574576356, 14795073225, 41742334396, 105500238889, 206068786704, 581395716793, 1469428768108

Offset: 1

Richard R. Forberg, Nov 29 2013

nonn

The corresponding values of x are given by A232771.
a(n) + 3 gives the values of y solving x^2 = floor(y^2/3 - y), and yields the same values for x.
a(3n+1) are squares whose square roots are given by A005320.
Let b(n) equal the second differences of a(n) where b(1) = 2. Then, for n>0, b(3n-1) = b(3n-2) = 2* A001570(n+1);  b(3n)= 2*A011943(n);  and b(3n) = (b(3n+1) + b(3n-1))/2.

Cf. A005320, A011943, A001570, A110293.

is(n)=issquare(n^2\3+n)
print1("0, 1");for(x=3,99,y=round(sqrt(3)*x-3/2);if(is(y),print1(", "y))) \\ Charles R Greathouse IV, Dec 09 2013

Empirical g.f.: -x^2*(x+1)*(x^2+x+1)^2 / ((x-1)*(x^6-14*x^3+1)). - Colin Barker, Dec 30 2014

a(23) corrected by Colin Barker, Dec 30 2014

A099270 Unsigned member r=-12 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 12, 169, 2352, 32761, 456300, 6355441, 88519872, 1232922769, 17172398892, 239180661721, 3331356865200, 46399815451081, 646266059449932, 9001325016847969, 125372284176421632, 1746210653453054881

Offset: 0

Wolfdieter Lang, Oct 18 2004

((-1)^(n+1))*a(n) = S_{-12}(n), n>=0, defined in A092184.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for two-way infinite sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (13, 13, -1).

a[n_] := (ChebyshevT[n, 7] - (-1)^n)/8; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Jun 21 2013, from 1st formula *)
CoefficientList[Series[x (1 - x) / ((1 + x) (1 - 14 x + x^2)), {x, 0, 20}], x] (* Vincenzo Librandi, Jun 21 2013 *)

a(n)=real(((7+4*quadgen(12))^n-(-1)^n)/8) /* Michael Somos, Apr 30 2005 */

a(n)=n=abs(2*n); round(2^(n-4)*prod(k=1,n,2-sin(2*Pi*k/n)))

a(n) = (T(n, 7)-(-1)^n)/8, with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7)=A011943(n)=((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.
a(n) = 13*a(n-1) + 13*a(n-2) - a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=12.
G.f.: x*(1-x)/((1+x)*(1-14*x+x^2)) = x*(1-x)/(1-13*x-13*x^2+x^3) (from the Stephan link, see A092184).
a(n) = 14*a(n-1)-a(n-2)-2*(-1)^n, a(0)=0, a(1)=1. a(-n)=a(n).

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]Sqrt[Prime[m]])^n + (Prime[n] - gap[n]Sqrt[Prime[m]])^n)/2]].

Original entry on oeis.org

1, 2, 2, 13, 17, 21, 185, 245, 305, 425, 7361, 12833, 18817, 32321, 47873, 215171, 271051, 328691, 449251, 576851, 853171, 12334505, 21164697, 31341961, 55836009, 86013257, 164203785, 212610281, 532365557, 659940697, 793109789, 1076412613

Offset: 1

Roger L. Bagula, Aug 18 2008

General Lucas-like Binet sequences
where Prime[m]starts at 1:
a(n)=((Prime[n]+gap[n]*Sqrt[Prime[m])^n+(Prime[n]-gap[n]*Sqrt[Prime[m])^n)/2.
Row sums are:
{1, 4, 51, 1160, 119205, 2694186, 583504495, 12222749556, 4868938911913,
3621654266405174, 21636046625243691}

			{1},
{2, 2},
{13, 17, 21},
{185, 245, 305, 425},
{7361, 12833, 18817, 32321, 47873},
{215171, 271051, 328691, 449251, 576851, 853171},
{12334505, 21164697, 31341961, 55836009, 86013257, 164203785, 212610281},
{532365557, 659940697, 793109789, 1076412613, 1382639597, 2065328317, 2442521189, 3270431797},
{40436937953, 68810349217, 102354570337, 185966400481, 293310073697, 587469359713, 778486092257, 1259085279457, 1553019848801},
{7312866926183, 15217609281335, 25813998655559, 56317915837223,
101380456546055, 246072307427783, 351480840333479, 643872497781095,
837435900955463, 1336749872660999}, {512759709537725, 608866569299409,
709085196658213, 922088454409101, 1152233212894709, 1665820807145925,
1950209769575213, 2576571400365309, 2919512658836837, 3667365684348213,
4951533162173037}

Cf. A011943, A081336, A034478.

gap[n_] := Prime[n + 1] - Prime[n]; t[n_, m_] := If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]]; Table[Table[FullSimplify[t[n, m]], {m, 0, n}], {n, 0, 10}]; Flatten[%]

gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]].

cp's OEIS Frontend

A102344 Numbers n such that the Diophantine equation (x+2)^3-x^3=2*n^2 has solutions.

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A110294 a(2*n) = A028230(n), a(2*n+1) = -A067900(n+1).

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A232765 Values of y solving x^2 = floor(y^2/3 + y).

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A099270 Unsigned member r=-12 of the family of Chebyshev sequences S_r(n) defined in A092184.

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A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]].

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A110294 a(2n) = A028230(n), a(2n+1) = -A067900(n+1).

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]Sqrt[Prime[m]])^n + (Prime[n] - gap[n]Sqrt[Prime[m]])^n)/2]].