A054320 -id:A054320 - cp's OEIS frontend

A006061 Star numbers (A003154) that are squares.

1, 121, 11881, 1164241, 114083761, 11179044361, 1095432263641, 107341182792481, 10518340481399521, 1030690025994360601, 100997104206965939401, 9896685522256667700721, 969774184076946468731281

Offset: 1

N. J. A. Sloane

			a(2)=121 because this is the 2nd star number (A003154) that is a square.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 121, p. 42, Ellipses, Paris 2008.
M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 1..500
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Star Number
Index entries for linear recurrences with constant coefficients, signature (99, -99, 1).

A007667 is 3*a(n)+2, sqrt(a(n)) is A054320.

Cf. A003154.

a:=[1,121,11881];; for n in [4..20] do a[n]:=99*a[n-1]-99*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1+22*x+x^2)/((1-x)*(1-98*x+x^2)) )); // G. C. Greubel, Jul 23 2019

Digits := 1000:q := seq(floor(evalf(( (5+2*sqrt(6))^n*(sqrt(6)-2)-(5-2*sqrt(6))^n*(sqrt(6)+2))^2/16)),n=1..100);
A006061:=-(1+22*z+z**2)/(z-1)/(z**2-98*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

CoefficientList[Series[(1+22*x+x^2)/((1-x)*(1-98*x+x^2)), {x,0,20}], x] (* or *) LinearRecurrence[{99,-99,1}, {1,121,11881}, 20] (* G. C. Greubel, Jul 23 2019 *)

my(x='x+O('x^20)); Vec((1+22*x+x^2)/((1-x)*(1-98*x+x^2))) \\ G. C. Greubel, Jul 23 2019

((1+22*x+x^2)/((1-x)*(1-98*x+x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

A007667 = 3*square star numbers (A006061) + 2.
a(n) = denominator of kappa(sqrt(6)/A054320(n)) where kappa(x) is the sum of successive remainders by computing the Euclidean algorithm for (1, x). - Thomas Baruchel, Nov 29 2003
From Ignacio Larrosa Cañestro, Feb 27 2000: (Start)
a(n) = 99*(a(n-1) - a(n-2)) + a(n-3).
a(n) = (5 - 2*sqrt(6))/8*(sqrt(3) + sqrt(2))^(4*n) + (5 + 2*sqrt(6))/8*(sqrt(3) - sqrt(2))^(4*n) - 1/4. (End)
a(n) = 98*a(n-1) - a(n-2) + 24. - Lekraj Beedassy, Jul 14 2008

More terms from Eric W. Weisstein and Sascha Kurz, Mar 24 2002

A007667 The sum of both two and three consecutive squares.

Original entry on oeis.org

5, 365, 35645, 3492725, 342251285, 33537133085, 3286296790925, 322023548377445, 31555021444198565, 3092070077983081805, 302991312620897818205, 29690056566770003102165

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

			a(2) = 365 = 13^2+14^2 = 10^2+11^2+12^2.

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for sequences related to sums of squares
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A003154, A031138, A006061, A054320.

a:=[5, 365, 35645];; for n in [4..20] do a[n]:=99*a[n-1]-99*a[n-2] + a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( 5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[5*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)),{x,0,20}],x] (* Vincenzo Librandi, Apr 16 2012 *)
LinearRecurrence[{99,-99,1},{5,365,35645},20] (* Harvey P. Dale, Dec 10 2024 *)

my(x='x+O('x^20)); Vec(5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2))) \\ G. C. Greubel, Jul 23 2019

(5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

From Ignacio Larrosa Cañestro, Feb 27 2000: (Start)
a(n) = (b(n)-1)^2 + b(n)^2 + (b(n)+1)^2 = c(n)^2 + (c(n)+1)^2, where b(n) = A054320(n) and c(n) = A031138(n).
a(n) = 3*A006061(n) + 2.
a(n) = 99*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 3*(5 - 2*sqrt(6))/8*(sqrt(3) + sqrt(2))^(4*n) + 3*(5 + 2*sqrt(6))/8*(sqrt(3) - sqrt(2))^(4*n) + 5/4. (End)
G.f.: 5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)). - Colin Barker, Apr 14 2012

Corrected by T. D. Noe, Nov 07 2006

A123479 Coefficients of series giving the best rational approximations to sqrt(6).

Original entry on oeis.org

20, 1980, 194040, 19013960, 1863174060, 182572043940, 17890197132080, 1753056746899920, 171781670999060100, 16832850701160989900, 1649447587042777950120, 161629030679491078121880, 15837995559003082877994140, 1551961935751622630965303860

Offset: 1

Gene Ward Smith, Sep 28 2006

The partial sums of the series 5/2 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(6), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2;2], [2;2,4,2], [2;2,4,2,4,2], [2;2,4,2,4,2,4,2] and so forth.

Sequence of numbers x=a(n) such 4*x+1 and 6*x+1 are both square, and their square roots are A138288(n) and A054320(n). - Paul Cleary, Jun 23 2014

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A123478, A123480, A029549, A123482.

LinearRecurrence[{99,-99,1},{0,20,1980},{2, 25}] (* Paul Cleary, Jun 23 2014 *)

Vec(-20*x/((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 99*a(n+2) - 99*a(n+1) + a(n).
a(n) = -5/24 + (( + 2*6^(1/2))/48)*(49 + 20*6^(1/2))^n + ((5 - 2*6^(1/2))/48)*(49 - 20*6^(1/2))^n.
G.f.: -20*x / ((x-1)*(x^2-98*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

A138281 a(n) = floor((sqrt(2) + sqrt(3))^n).

Original entry on oeis.org

1, 3, 9, 31, 97, 308, 969, 3051, 9601, 30210, 95049, 299052, 940897, 2960313, 9313929, 29304086, 92198401, 290080547, 912670089, 2871501385, 9034502497, 28424933309, 89432354889, 281377831710, 885289046401, 2785353383794, 8763458109129, 27572156006234

Offset: 0

Reinhard Zumkeller, Mar 12 2008

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A089078, A110117, A121503, A135611, A135798.

[Floor((Sqrt(2) + Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Jan 27 2018

Table[Floor[(Sqrt[2] + Sqrt[3])^n], {n, 0, 50}] (* G. C. Greubel, Jan 27 2018 *)

for(n=0,50, print1(floor((sqrt(2) + sqrt(3))^n), ", ")) \\ G. C. Greubel, Jan 27 2018

a(2*n) = floor(A001079(n) + A001078(n)*sqrt(6));
(sqrt(2) + sqrt(3))^(2*n) = A001079(n) + A001078(n)*sqrt(6);
a(2*n+1) = floor(A054320(n)*sqrt(2) + A138288(n)*sqrt(3));
(sqrt(2)+sqrt(3))^(2*n+1) = A054320(n)*sqrt(2) + A138288(n)*sqrt(3).

Terms a(16) and a(18) corrected, terms a(19) onward added by G. C. Greubel, Jan 27 2018

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

A157084 Consider all consecutive integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 10, 108, 1078, 10680, 105730, 1046628, 10360558, 102558960, 1015229050, 10049731548, 99482086438, 984771132840, 9748229241970, 96497521286868, 955226983626718, 9455772314980320

Offset: 0

Charlie Marion, Mar 12 2009

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=3, then first(2) = 312 = 14*21 - 0 + 18.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=3 and n=2, first(2) = 312 = 7*21 + 6*27 + 3 and last(2) = 363 = 8*21 + 7*27 + 6. a(n) = (2^(n+1)((1+sqrt(3/2))^(2n+1) + (1-sqrt(3/2))^(2n+1)) - 2*2)/4; e.g., 108 = (2^3((1+sqrt(3/2))^5 + (1-sqrt(3/2))^5) - 2*2)/4.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=3 and n=2, then first(2) = 312 = (3^3((1+sqrt((4/3))^(5) + (1-sqrt(4/3)^5) - 2*3)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(3) = 1078 = 2*11*49 and a(4) = 10680 = 3*40*89.
In general, if first(n) is the first term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim n->inf first(n+1)/first(n) = k*(1 + sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2) = 108 since 108^2 + 109^2 + 110^2 = 133^2 + 134^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (11, -11, 1).

Cf. A001652, A157088, A157092, A157096.

[(1/2 + Sqrt(6)/4)*(5 + 2*Sqrt(6))^n - (Sqrt(6)/4 - 1/2)*(5 - 2*Sqrt(6))^n - 1: n in [0..50]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[2*x*(-5 + x)/((x - 1)*(x^2 - 10*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)

x='x+O('x^50); concat([0], Vec(2*(x - 5)/((x-1)*(x^2 - 10*x + 1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 10*a(n-1) - a(n-2) + 8.
For n > 0, a(n) = 5*a(n-1) + 4*A157085(n-1) + 2.
Lim_{n->inf} a(n+1)/a(n) = 2(1 + sqrt(3/2))^2 = 5 + 2*sqrt(6).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 2*x*(x - 5)/((x-1)*(x^2 - 10*x + 1)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3).
a(n) = 2*A087125(n) = A054320(n) - 1. (End)
From Sergei N. Gladkovskii, Jan 12 2012: (Start)
G.f.: 1/(x-1) + (x+1)/(x^2-10*x+1).
a(n) = (1/2 + sqrt(6)/4)*(5 + 2*sqrt(6))^n - (sqrt(6)/4 - 1/2)*(5 - 2*sqrt(6))^n - 1. (End)

A004291 Expansion of (1 + 2x + x^2)/(1 - 10x + x^2).

Original entry on oeis.org

1, 12, 120, 1188, 11760, 116412, 1152360, 11407188, 112919520, 1117788012, 11064960600, 109531817988, 1084253219280, 10733000374812, 106245750528840, 1051724504913588, 10410999298607040, 103058268481156812, 1020171685512961080, 10098658586648453988

Offset: 0

N. J. A. Sloane

J. M. Alonso, Growth functions of amalgams, in Alperin, ed., Arboreal Group Theory, Springer, pp. 1-34, esp. p. 32.
P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 160, middle display.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Hacène Belbachir, Soumeya Merwa Tebtoub, László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Pairwise sums of A054320.

CoefficientList[Series[(1+2*x+x^2)/(1-10*x+x^2),{x,0,30}],x] (* Vincenzo Librandi, Jun 13 2012 *)

Vec((1+2*x+x^2)/(1-10*x+x^2) + O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = 12*A004189(n), n> 0. - R. J. Mathar, Oct 29 2012
a(n) = sqrt(3/2)*(-(5 - 2*sqrt(6))^n + (5 + 2*sqrt(6))^n) for n > 0. - Colin Barker, Jan 25 2016
For n > 0: a(n) = 10*a(n-1) - a(n-2) with a(0) = 0, a(1) = 12. - A.H.M. Smeets, Jul 25 2017

A165293 Inverse of A038303, and generalization of A130595.

Original entry on oeis.org

1, 10, -1, 100, -20, 1, 1000, -300, 30, -1, 10000, -4000, 600, -40, 1, 100000, -50000, 10000, -1000, 50, -1, 1000000, -600000, 150000, -20000, 1500, -60, 1, 10000000, -7000000, 2100000, -350000, 35000, -2100, 70

Offset: 1

Mark Dols, Sep 13 2009

Rows sum up to A001019 (powers of 9), diagonals to A004189, a generalization of A010892 (the inverse Fibonacci). Ratio of diagonal sums converges to a decimal sequence: A000108 (Catalan numbers), which is the squared difference of sqrt(2) and sqrt(3), or 5-sqrt(24). Ratio between first binomial transform (A054320 and A138288)of A004189, converges to sqrt(2/3). 1/(2*sqrt(24)) gives A000984 (central binomial coefficients) as a decimal sequence.

Triangle T(n,k), read by rows, given by [10,0,0,0,0,0,0,0,...] DELTA [ -1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 15 2009

			Triangle begins:
      1;
     10,    -1;
    100,   -20,   1;
   1000,  -300,  30,  -1;
  10000, -4000, 600, -40, 1;

Cf. A007318, A130595, A038303, A004189, A010892, A001079, A054320, A138288, A041041, A000108.

Sum_{k=0..n} T(n,k)*x^k = (10-x)^n. - Philippe Deléham, Dec 15 2009

G.f.: x*y/(1-10*x+x*y). - R. J. Mathar, Aug 11 2015

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

cp's OEIS Frontend

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A138281 a(n) = floor((sqrt(2) + sqrt(3))^n).

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A157084 Consider all consecutive integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives X values.

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

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