A072256 -id:A072256 - cp's OEIS frontend

A054318 a(n)-th star number (A003154) is a square.

1, 5, 45, 441, 4361, 43165, 427285, 4229681, 41869521, 414465525, 4102785725, 40613391721, 402031131481, 3979697923085, 39394948099365, 389969783070561, 3860302882606241, 38213059042991845, 378270287547312205

Offset: 1

Ignacio Larrosa Cañestro, Feb 27 2000

A two-way infinite sequence which is palindromic.
Also indices of centered hexagonal numbers (A003215) which are also centered square numbers (A001844). - Colin Barker, Jan 02 2015
Also positive integers y in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0. - Colin Barker, Jan 02 2015

			a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.

Colin Barker, Table of n, a(n) for n = 1..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

A031138 is 3*a(n)-2. Cf. A003154, A006061, A182432, A211955.
Quintisection of column k=2 of A233427.
Cf. A001844, A003215, A253475.

a:=[1,5,45];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x(1-6x+x^2)/((1-x)(1-10x+x^2)), {x,0,30}], x] (* Michael De Vlieger, Aug 11 2016 *)
LinearRecurrence[{11,-11,1},{1,5,45},30] (* Harvey P. Dale, Nov 05 2016 *)

a(n)=if(n<1,a(1-n),1/2+subst(poltchebi(n)+poltchebi(n-1),x,5)/12)

Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jan 02 2015

(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 1/2 + (3 - sqrt(6))/12*(5 + 2*sqrt(6))^n + (3 + sqrt(6))/12*(5 - 2*sqrt(6))^n.
From Michael Somos, Mar 18 2003: (Start)
G.f.: x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)).
12*a(n)*a(n-1) + 4 = (a(n) + a(n-1) + 2)^2.
a(n) = a(1-n) = 10*a(n-1) - a(n-2) - 4.
a(n) = 12*a(n-1)^2/(a(n-1) + a(n-2)) - a(n-1).
a(n) = (a(n-1) + 4)*a(n-1)/a(n-2). (End)
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + (1/2)*Sum_{k = 1..n} 8^k*binomial(n+k,2*k).
a(n+1) = R(n,4), where R(n,x) is the n-th row polynomial of A211955.
a(n+1) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(3) and T(n,x) the Chebyshev polynomial of the first kind.
Sum {k>=0} 1/a(k) = sqrt(3/2). (End)
A003154(a(n)) = A006061(n). - Zak Seidov, Oct 22 2012
a(n) = (4*a(n-1) + a(n-1)^2) / a(n-2), n >= 3. - Seiichi Manyama, Aug 11 2016
2*a(n) = 1+A072256(n). - R. J. Mathar, Feb 07 2022

More terms from James Sellers, Mar 01 2000

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

Original entry on oeis.org

0, 15, 1485, 145530, 14260470, 1397380545, 136929032955, 13417647849060, 1314792560174940, 128836253249295075, 12624638025870742425, 1237085690282083462590, 121221773009618308591410, 11878496669252312158495605, 1163971451813716973223977895

Offset: 0

Charlie Marion, Feb 15 2012

For n > 1, a(n) = 98*a(n-1) - a(n-2) + 15. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1) = A014105(m) and for n > 1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m). Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).

			2*0 = 3*0.
2*15 = 3*10.
2*1485 = 3*990.
2*145530 = 3*97020.

Colin Barker, Table of n, a(n) for n = 0..500
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008.

A080872 a(n)a(n+3) - a(n+1)a(n+2) = 4, given a(0)=a(1)=1, a(2)=5.

Original entry on oeis.org

1, 1, 5, 9, 49, 89, 485, 881, 4801, 8721, 47525, 86329, 470449, 854569, 4656965, 8459361, 46099201, 83739041, 456335045, 828931049, 4517251249, 8205571449, 44716177445, 81226783441, 442644523201, 804062262961, 4381729054565, 7959395846169, 43374646022449, 78789896198729, 429364731169925

Offset: 0

Paul D. Hanna, Feb 22 2003

Seiichi Manyama, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0, 10, 0, -1).

Cf. A080871, A080873, A080874, A080875.

Bisections are A001079 and A072256.

CoefficientList[Series[(-x^3-5 x^2+x+1)/(x^4-10 x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[{0,10,0,-1},{1,1,5,9},30] (* Harvey P. Dale, May 06 2012 *)

Vec( (-x^3 - 5*x^2 + x + 1)/(x^4 - 10*x^2 + 1) + O(x^66) ) \\ Joerg Arndt, Jan 29 2016

G.f.: (-x^3 - 5*x^2 + x + 1)/(x^4 - 10*x^2 + 1).
a(n) = (3+sqrt(3))/12*(sqrt(3)-sqrt(2))^n+(3-sqrt(3))/12*(-sqrt(3)+sqrt(2))^n+(3+sqrt(3))/12*(sqrt(3)+sqrt(2))^n+(3-sqrt(3))/12*(-sqrt(3)-sqrt(2))^n. [Richard Choulet, Dec 03 2008]
a(n+4) = 10*a(n+2)-a(n). [Richard Choulet, Dec 04 2008]

A260810 a(n) = n^2(3n^2 - 1)/2.

Original entry on oeis.org

0, 1, 22, 117, 376, 925, 1926, 3577, 6112, 9801, 14950, 21901, 31032, 42757, 57526, 75825, 98176, 125137, 157302, 195301, 239800, 291501, 351142, 419497, 497376, 585625, 685126, 796797, 921592, 1060501, 1214550, 1384801, 1572352, 1778337, 2003926, 2250325, 2518776

Offset: 0

Bruno Berselli, Jul 31 2015

Pentagonal numbers with square indices.

After 0, a(k) is a square if k is in A072256.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Subsequence of A001318 and A245288 (see Formula field).
Cf. A000326, A193218 (first differences).
Cf. A000583 (squares with square indices), A002593 (hexagonal numbers with square indices).
Cf. A232713 (pentagonal numbers with pentagonal indices), A236770 (pentagonal numbers with triangular indices).

Magma
```
[n^2*(3*n^2-1)/2: n in [0..40]];
```

I:=[0,1,22,117,376]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Aug 23 2015

A260810:=n->n^2*(3*n^2 - 1)/2: seq(A260810(n), n=0..50); # Wesley Ivan Hurt, Apr 25 2017

Table[n^2 (3 n^2 - 1)/2, {n, 0, 40}]
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 22, 117, 376}, 40] (* Vincenzo Librandi, Aug 23 2015 *)

PARI
```
vector(40, n, n--; n^2*(3*n^2-1)/2)
```
Sage
```
[n^2*(3*n^2-1)/2 for n in (0..40)]
```

G.f.: x*(1 + x)*(1 + 16*x + x^2)/(1 - x)^5.
a(n) = a(-n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A245288(2*n^2).
a(n) = A001318(2*n^2-1) with A001318(-1) = 0.
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi^2/3 - sqrt(3)*Pi*cot(Pi/sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi*cosec(Pi/sqrt(3)) - Pi^2/6 - 3. (End)

A153111 Solutions of the Pell-like equation 1 + 6AA = 7BB, with A, B integers.

Original entry on oeis.org

1, 25, 649, 16849, 437425, 11356201, 294823801, 7654062625, 198710804449, 5158826853049, 133930787374825, 3477041644892401, 90269151979827601, 2343520909830625225, 60841274503616428249, 1579529616184196509249, 41006928746285492812225

Offset: 1

Ctibor O. Zizka, Dec 18 2008

B is of the form B(i) = 26*B(i-1) - B(i-2) for B(0) = 1, B(1) = 25 (this sequence).
A is of the form A(i) = 26*A(i-1) - A(i-2) for A(0) = 1, A(1) = 27.
In general a Pell-like equation of the form 1 + X*A*A = (X + 1)*B*B has the solution A(i) = (4*X + 2)*A(i-1) - A(i-2), for A(0) = 1 and A(1) = (4*X + 3), and B(i) = (4*X + 2)*B(i-1) - B(i-2) for B(0) = 1 and B(1) = (4*X + 1).
Examples in the OEIS:
  X = 1 gives A002315 for A(i) and A001653 for B(i);
  X = 2 gives A054320 for A(i) and A072256 for B(i);
  X = 3 gives A028230 for A(i) and A001570 for B(i);
  X = 4 gives A049629 for A(i) and A007805 for B(i);
  X = 5 gives A133283 for A(i) and A157014 for B(i);
  X = 6 gives A157461 for A(i) and this sequence for B(i).
Positive values of x (or y) satisfying x^2 - 26*x*y + y^2 + 24 = 0. - Colin Barker, Feb 20 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Luigi Cimmino, Algebraic relations for recursive sequences, arXiv:math/0510417 [math.NT], 2005-2008.
Jeroen Demeyer, Diophantine sets of polynomials over number fields, arXiv:0807.1970 [math.NT], 2008.
Franz Lemmermeyer, Conics - a Poor Man's Elliptic Curves, arXiv:math/0311306 [math.NT], 2003.
Index entries for linear recurrences with constant coefficients, signature (26,-1).

Cf. A002315, A001653, A054320, A072256, A001078, A028230, A001570, A049629, A007805, A133283, A140480.

Cf. similar sequences listed in A238379.

I:=[1,25]; [n le 2 select I[n] else 26*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 22 2014

CoefficientList[Series[(1 - x)/(x^2 - 26 x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 22 2014 *)
LinearRecurrence[{26, -1}, {1, 25}, 20] (* Jean-François Alcover, Jan 07 2019 *)

Vec(-x*(x-1)/(x^2-26*x+1) + O(x^100)) \\ Colin Barker, Feb 20 2014

a(n) = 26*a(n-1) - a(n-2). - Colin Barker, Feb 20 2014
G.f.: -x*(x - 1) / (x^2 - 26*x + 1). - Colin Barker, Feb 20 2014
a(n) = (1/14)*(7 - sqrt(42))*(1 + (13 + 2*sqrt(42))^(2*n - 1))/(13 + 2*sqrt(42))^(n - 1). - Bruno Berselli, Feb 25 2014
E.g.f.: (1/7)*(7*cosh(2*sqrt(42)*x) - sqrt(42)*sinh(2*sqrt(42)*x))*exp(13*x) - 1. - Franck Maminirina Ramaharo, Jan 07 2019

More terms from Philippe Deléham, Sep 19 2009; corrected by N. J. A. Sloane, Sep 20 2009

Additional term from Colin Barker, Feb 20 2014

A253175 Indices of hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 7, 67, 661, 6541, 64747, 640927, 6344521, 62804281, 621698287, 6154178587, 60920087581, 603046697221, 5969546884627, 59092422149047, 584954674605841, 5790454323909361, 57319588564487767, 567405431320968307, 5616734724645195301, 55599941815130984701

Offset: 1

Colin Barker, Jan 08 2015

Also positive integers x in the solutions to 4*x^2-6*y^2-2*x+6*y-2 = 0, the corresponding values of y being A253475.

			7 is in the sequence because the 7th hexagonal number is 91, which is also the 6th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A003215, A253475, A006244, A350923.

LinearRecurrence[{11, -11, 1}, {1, 7, 67}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(-x*(x^2-4*x+1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: -x*(x^2-4*x+1) / ((x-1)*(x^2-10*x+1)).
a(n) = (2+(5-2*sqrt(6))^n*(3+sqrt(6))-(-3+sqrt(6))*(5+2*sqrt(6))^n)/8. - Colin Barker, Mar 05 2016
4*a(n) = 1+3*A072256(n). - R. J. Mathar, Feb 07 2022
a(n) = A350923(n)/2. - Paolo Xausa, May 30 2025

A165516 Perfect squares (A000290) that can be expressed as the sum of three consecutive triangular numbers (A000217).

Original entry on oeis.org

4, 64, 361, 6241, 35344, 611524, 3463321, 59923081, 339370084, 5871850384, 33254804881, 575381414521, 3258631508224, 56381506772644, 319312633001041, 5524812282304561, 31289379402593764, 541375222159074304, 3066039868821187801, 53049246959306977201, 300440617765073810704, 5198284826789924691364

Offset: 1

Ant King, Sep 25 2009, Oct 01 2009

Those perfect squares that can be expressed as the sum of three consecutive triangular numbers correspond to integer solutions of the equation T(k)+T(k+1)+T(k+2)=s^2, or equivalently to 3k^2+9k+8=2s^2. Hence solutions occur whenever 1/2 (3k^2+9k+8) is a perfect square, or equivalently when s>=2 and sqrt(24s^2-15) is congruent to 3 mod 6. Furthermore, with the exception of the first term, the members of this sequence are precisely those perfect squares that are also centered triangular numbers (A005448). For s>=2, the values of s are in A129445, and the corresponding indices of the smallest of the 3 triangular numbers are given in A165517.

			The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 (=T63+T64+T65), and hence a(4)=6241.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000290, A129445, A005448, A000290, A000217, A165517.

I:=[4, 64, 361, 6241, 35344]; [n le 5 select I[n] else Self(n-1) + 98*Self(n-2) - 98*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // G. C. Greubel, Oct 21 2018

Select[Range[2,1.8 10^7],Mod[Sqrt[24#^2-15],6]==3 &]^2
CoefficientList[Series[(4 + 60 x - 95 x^2 + x^4)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 14 2014 *)
LinearRecurrence[ {1,98,-98,-1,1}, {4, 64, 361, 6241, 35344}, 50] (* G. C. Greubel, Oct 21 2018 *)

Vec(O(x^66)+x*(4+60*x-95*x^2+x^4)/((1-x)*(1-10*x+x^2)*(1+10*x+x^2))) \\ Joerg Arndt, Mar 13 2014

a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
a(n) = 98*a(n-2) - a(n-4) - 30. - Ant King, Dec 09 2010
a(n) = (1/32)*(10 -3*(sqrt(6)-3) * (5-2*sqrt(6))^n + (2+ sqrt(6)) * (-5-2*sqrt(6))^n -(sqrt(6)-2) *(2*sqrt(6)-5)^n + 3*(3+sqrt(6)) *(5+2*sqrt(6))^n).
G.f.: x*(4+60*x-95*x^2+x^4)/((1-x)*(1-10*x+x^2)*(1+10*x+x^2)).
16*a(n) = 5 +9*A072256(n+1) +2*(-1)^n*A054320(n). - R. J. Mathar, Apr 28 2020

a(1) = 4 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander R. Povolotsky

a(16)-a(21) added by Alex Ratushnyak, Mar 12 2014

A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 6, 55, 540, 5341, 52866, 523315, 5180280, 51279481, 507614526, 5024865775, 49741043220, 492385566421, 4874114620986, 48248760643435, 477613491813360, 4727886157490161, 46801248083088246, 463284594673392295, 4586044698650834700, 45397162391834954701

Offset: 1

Colin Barker, Jan 02 2015

Also positive integers x in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0, the corresponding values of y being A054318.
Also indices of centered hexagonal numbers (A003215) which are also hexagonal numbers (A000384).
Also indices of terms in sequence A193218 which are the square root of a sum of 5th powers (A000539). - Daniel Poveda Parrilla, Jun 10 2017

			6 is in the sequence because the 6th centered square number is 61, which is also the 5th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A000539, A001844, A003215, A054318, A193218, A253175.

LinearRecurrence[{11, -11, 1}, {1, 6, 55}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(x*(5*x-1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: x*(5*x-1) / ((x-1)*(x^2-10*x+1)).
a(n) = sqrt((-2-(5-2*sqrt(6))^n-(5+2*sqrt(6))^n)*(2-(5-2*sqrt(6))^(1+n)-(5+2*sqrt(6))^(1+n)))/(4*sqrt(2)). - Gerry Martens, Jun 04 2015
2*a(n) = 1+A054320(n-1). - R. J. Mathar, Feb 07 2022

A087125 Indices k of hex numbers H(k) that are also triangular.

Original entry on oeis.org

0, 5, 54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 0

Eric W. Weisstein, Aug 14 2003

From the law of cosines, the non-Pythagorean triple {a(n), a(n)+1=A253475(n+1), A072256(n+1)} forms a near-isosceles triangle with the angle bounded by the consecutive sides equal to the regular tetrahedron's central angle (see A156546 and A247412). This implies also that a(n) are those numbers k such that (16/3)*A000217(k)+1 is a perfect square. - Federico Provvedi, Apr 04 2023

Colin Barker, Table of n, a(n) for n = 0..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A006244, A031138, A072256, A054318.

[Round((-4-(5-2*Sqrt(6))^n*(-2+Sqrt(6)) + (2+Sqrt(6))*(5 + 2*Sqrt(6))^n)/8): n in [0..25]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[(-x^2+5*x)/((1-x)*(1-10*x+x^2)), {x, 0, 25}], x] (* G. C. Greubel, Nov 04 2017 *)
LinearRecurrence[{11,-11,1},{0,5,54},30] (* Harvey P. Dale, Jun 14 2022 *)
Table[(x Sqrt[z^(2 n + 1) + z^-(2 n + 1) - 2] - 4) / 8 //. {x -> Sqrt[2], y -> Sqrt[3], z -> (5 + 2 x y)}, {n, 0, 100}] // Round (* Federico Provvedi, Apr 16 2023 *)

concat(0, Vec(x*(x-5)/((x-1)*(x^2-10*x+1)) + O(x^50))) \\ Colin Barker, Jun 23 2015

G.f.: (-x^2+5*x)/((1-x)*(1-10*x+x^2)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n > 2. - Colin Barker, Jun 23 2015
a(n) = (-4 - (5-2*sqrt(6))^n*(-2 + sqrt(6)) + (2+sqrt(6))*(5+2*sqrt(6))^n)/8. - Colin Barker, Mar 05 2016
a(n) = 10*a(n-1) - a(n-2) + 4 for n > 1. - Charlie Marion, Feb 14 2023
a(n) = ((x^(n+1)+1)*(x^n-1))/(2*x^n*(x-1)), with x=5+2*sqrt(6). - Federico Provvedi, Apr 04 2023
a(n) = sqrt(3*A161680(A054318(n+1)) + 1/4) - 1/2 = floor(sqrt(3*A000217(A054318(n+1)-1) + 1/4)). - Federico Provvedi, Apr 16 2023

A098308 Unsigned member r=-8 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 8, 81, 800, 7921, 78408, 776161, 7683200, 76055841, 752875208, 7452696241, 73774087200, 730288175761, 7229107670408, 71560788528321, 708378777612800, 7012226987599681, 69413891098384008, 687126683996240401

Offset: 0

Wolfdieter Lang, Oct 18 2004

((-1)^(n+1))*a(n) = S_{-8}(n), n>=0, defined in A092184.

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,9,-1).

LinearRecurrence[{9,9,-1},{0,1,8},40] (* Harvey P. Dale, Aug 11 2013 *)

a(n)= (T(n, 5)-(-1)^n)/6, with Chebyshev's polynomials of the first kind evaluated at x=5: T(n, 5)=A001079(n)=((5+2*sqrt(6))^n + (5-2*sqrt(6))^n)/2.
a(n)= 10*a(n-1)-a(n-2)+2*(-1)^(n+1), n>=2, a(0)=0, a(1)=1.
a(n)= 9*a(n-1) + 9*a(n-2) - a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=8.
G.f.: x*(1-x)/((1+x)*(1-10*x+x^2)) = x*(1-x)/(1-9*x-9*x^2+x^3) (from the Stephan link, see A092184).
a(x)=1/12(2*(-1)^x+(5+2*Sqrt[6])(5-2*Sqrt[6])^x+(5-2*Sqrt[6])(5+2*Sqrt[6])^x). - Harvey P. Dale, Aug 11 2013
a(n-1)+a(n) = A072256(n). - R. J. Mathar, Feb 19 2017

cp's OEIS Frontend

A054318 a(n)-th star number (A003154) is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

PARI

Sage

Formula

Extensions

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2*T(m) = 3*T(k) for some k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A080872 a(n)*a(n+3) - a(n+1)*a(n+2) = 4, given a(0)=a(1)=1, a(2)=5.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A260810 a(n) = n^2*(3*n^2 - 1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Magma

Maple

Mathematica

PARI

Sage

Formula

A153111 Solutions of the Pell-like equation 1 + 6*A*A = 7*B*B, with A, B integers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A253175 Indices of hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

A080872 a(n)a(n+3) - a(n+1)a(n+2) = 4, given a(0)=a(1)=1, a(2)=5.

A260810 a(n) = n^2(3n^2 - 1)/2.

A153111 Solutions of the Pell-like equation 1 + 6AA = 7BB, with A, B integers.