A046173 -id:A046173 - cp's OEIS frontend

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

0, 1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050, 9127651499, 90354434940, 894416697901, 8853812544070, 87643708742799, 867583274883920, 8588189040096401, 85014307126080090, 841554882220704499, 8330534515080964900, 82463790268588944501, 816307368170808480110

Offset: 0

N. J. A. Sloane

Indices of square numbers which are also generalized pentagonal numbers.
If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008
The n-th term is a(n) = ((5+sqrt(24))^n - (5-sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, May 31 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 10's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) (A001079) are the nonnegative proper solutions of the Pell equation b(n)^2 - 6*(2*a(n))^2 = +1. See the cross reference to A001079 below. - Wolfdieter Lang, Jun 26 2013
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,9}. - Milan Janjic, Jan 25 2015
For n > 1, this also gives the number of (n-1)-decimal-digit numbers which avoid a particular two-digit number with distinct digits. For example, there are a(5) = 9701 4-digit numbers which do not include "39" as a substring; see Wikipedia link. - Charles R Greathouse IV, Jan 14 2016
All possible solutions for y in Pell equation x^2 - 24*y^2 = 1. The values for x are given in A001079. - Herbert Kociemba, Jun 05 2022
Dickson on page 384 gives the Diophantine equation "(20)  24x^2 + 1 = y^2" and later states "... three consecutive sets (x_i, y_i) of solutions of (20) or 2x^2 + 1 = 3y^2 satisfy x_{n+1} = 10x_n - x_{n-1}, y_{n+1} = 10y_n - y_{n-1} with (x_1, y_1) = (0, 1) or (1, 1), (x_2, y_2) = (1, 5) or (11, 9), respectively." The first set of values (x_n, y_n) = (A001079(n-1), a(n-1)). - Michael Somos, Jun 19 2023

			a(2)=10 and (3(-8)^2-(-8))/2=10^2, a(3)=99 and (3(81)^2-(81))/2=99^2. - _Michael Somos_, Sep 05 2006
G.f. = x + 10*x^2 + 99*x^3 + 980*x^4 + 9701*x^5 + 96030*x^6 + ...

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 12.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
D. Fortin, B-spline Toeplitz inverse under corner perturbations, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118. - From _N. J. A. Sloane_, Oct 22 2012
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=10, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=12.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Wikipedia, Curse of 39
Jianqiang Zhao, Finite Multiple zeta Values and Finite Euler Sums, arXiv:1507.04917 [math.NT], 2015.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A001079, A001109, A001353, A001906, A004187, A004254, A018913, A046173, A049310, A054320, A072256, A101950, A105138, A108741 (squares).
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), this sequence (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=5;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[ n eq 1 select 0 else n eq 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..20] ]; // Vincenzo Librandi, Aug 19 2011

A004189 := proc(n)
    option remember;
    if n <= 1 then
        n ;
    else
        10*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A004189(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
seq( simplify(ChebyshevU(n-1, 5)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n-1,1,5], {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008; modified by G. C. Greubel, Jun 06 2019 *)
LinearRecurrence[{10, -1}, {0, 1}, 20] (* Jean-François Alcover, Nov 15 2017 *)
ChebyshevU[Range[21] -2, 5] (* G. C. Greubel, Dec 23 2019 *)

{a(n) = subst(poltchebi(n+1) - 5*poltchebi(n), 'x, 5) / 24}; /* Michael Somos, Sep 05 2006 */

a(n)=([9,1;8,1]^(n-1)*[1;1])[1,1] \\ Charles R Greathouse IV, Jan 14 2016

vector(21, n, n--; polchebyshev(n-1, 2, 5) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number1(n,10,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,5) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = S(2*n-1, sqrt(12))/sqrt(12) = S(n-1, 10); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
A001079(n) = sqrt(24*(a(n)^2)+1), that is a(n) = sqrt((A001079(n)^2-1)/24).
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ( (5+2*sqrt(6))^n - (5-2*sqrt(6))^n )/(4*sqrt(6)).
G.f.: x/(1-10*x+x^2). (End)
a(-n) = -a(n). - Michael Somos, Sep 05 2006
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 10*a(n-1) - a(n-2). (End)
a(n+1) = Sum_{k=0..n} A101950(n,k)*9^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/2*(2 + sqrt(6)).
Product {n >= 2} (1 - 1/a(n)) = 1/5*(2 + sqrt(6)). (End)
a(n) = (A054320(n-1) + A072256(n))/2. - Richard R. Forberg, Nov 21 2013
a(2*n - 1) = A046173(n).
E.g.f.: exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*8^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*12^k. - Peter Bala, Jul 18 2023

A001079 a(n) = 10*a(n-1) - a(n-2); a(0) = 1, a(1) = 5.

Original entry on oeis.org

1, 5, 49, 485, 4801, 47525, 470449, 4656965, 46099201, 456335045, 4517251249, 44716177445, 442644523201, 4381729054565, 43374646022449, 429364731169925, 4250272665676801, 42073361925598085, 416483346590304049

Offset: 0

N. J. A. Sloane

Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(6). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions >1 to the equation x^2=ceiling(x*r*floor(x/r)) where r=sqrt(6). - Benoit Cloitre, Feb 24 2004
a(n) and b(n) (A004189) are the nonnegative proper solutions to the Pell equation a(n)^2 - 6*(2*b(n))^2 = +1, n >= 0. The formula given below by Gregory V. Richardson follows. - Wolfdieter Lang, Jun 26 2013
a(n) are the integer square roots of (A032528 + 1). They are also the values of m where (A032528(m) - 1) has integer square roots. See A122653 for the integer square roots of (A032528 - 1), and see A122652 for the values of m where (A032528(m) + 1) has integer square roots. - Richard R. Forberg, Aug 05 2013
a(n) are also the values of m where floor(2m^2/3) has integer square roots, excluding m = 0. The corresponding integer square roots are given by A122652(n). - Richard R. Forberg, Nov 21 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 24 = 0. - Colin Barker, Feb 09 2014
Dickson on page 384 gives the Diophantine equation "24x^2 + 1 = y^2" and later states "y_{n+1} = 10y_n - y_{n-1}" where y_n is this sequence. - Michael Somos, Jun 19 2023

			Pell equation: n = 0: 1^2 - 24*0^2 = +1, n = 1: 5^2 - 6*(1*2)^2 = 1, n = 2: 49^2 - 6*(2*10)^2 = +1. - _Wolfdieter Lang_, Jun 26 2013
G.f. = 1 + 5*x + 49*x^2 + 485*x^3 + 4801*x^4 + 47525*x^5 + 470449*x^6 + ...

Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.
L. Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 374.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

T. D. Noe, Table of n, a(n) for n=0..200
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Leonhard Euler, Vollstaendige Anleitung zur Algebra, Zweiter Teil.
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, par. 18.
Tanya Khovanova, Recursive Sequences
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
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
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A004189, A001078, A046173, A046172, A036353, A138281, A004189.

I:=[1,5]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 10 2016

A001079 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,5]) ;
    else
        10*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A001079(n),n=0..20) ; # R. J. Mathar, Apr 30 2017

Table[(-1)^n Round[N[Cos[2 n ArcSin[Sqrt[3]]], 50]], {n, 0, 20}] (* Artur Jasinski, Oct 29 2008 *)
a[ n_] := ChebyshevT[n, 5]; (* Michael Somos, Aug 24 2014 *)
CoefficientList[Series[(1-5*x)/(1-10*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
a[n_] := 3^n*Sum[(2/3)^k*Binomial[2*n, 2*k], {k,0,n}]; Flatten[Table[a[n], {n,0,18}]] (* Detlef Meya, May 21 2024 *)

{a(n) = subst(poltchebi(n), 'x, 5)}; /* Michael Somos, Sep 05 2006 */

{a(n) = real((5 + 2*quadgen(24))^n)}; /* Michael Somos, Sep 05 2006 */

{a(n) = n = abs(n); polsym(1 - 10*x + x^2, n)[n+1] / 2}; /* Michael Somos, Sep 05 2006 */

x='x+O('x^30); Vec((1-5*x)/(1-10*x+x^2)) \\ G. C. Greubel, Dec 20 2017

For all members x of the sequence, 6*x^2 -6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 5) = (S(n, 10)-S(n-2, 10))/2 with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 10) = A004189(n+1).
a(n) = sqrt(1+24*A004189(n)^2) (cf. Richardson comment).
a(n)*a(n+3) - a(n+1)*a(n+2) = 240. - Ralf Stephan, Jun 06 2005
Chebyshev's polynomials T(n,x) evaluated at x=5.
G.f.: (1-5*x)/(1-10*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n)= ((5+2*sqrt(6))^n + (5-2*sqrt(6))^n)/2.
a(-n) = a(n).
a(n+1) = 5*a(n) + 2*(6*a(n)^2-6)^(1/2) - Richard Choulet, Sep 19 2007
(sqrt(2)+sqrt(3))^(2*n)=a(n)+A001078(n)*sqrt(6). - Reinhard Zumkeller, Mar 12 2008
a(n+1) = 2*A054320(n) + 3*A138288(n). - Reinhard Zumkeller, Mar 12 2008
a(n) = cosh(2*n* arcsinh(sqrt(2))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^n * cos(2*n* arcsin(sqrt(3))). - Artur Jasinski, Oct 29 2008
a(n) = cos(2*n* arccos(sqrt(3))). - Artur Jasinski, Sep 10 2016
a(n) = A142238(2n-1) = A041006(2n-1) = A041038(2n-1), for all n > 0. - M. F. Hasler, Feb 14 2009
2*a(n)^2 = 3*A122652(n)^2 + 2. - Charlie Marion, Feb 01 2013
E.g.f.: cosh(2*sqrt(6)*x)*exp(5*x). - Ilya Gutkovskiy, Sep 10 2016
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 10*x + sqrt(1 + 96*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 20*x + 4*x^2) is the g.f. of A098270.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - 3/a(n)) = 1/4.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 2/a(n)) = 1/6.
Sum_{n >= 1} 1/(a(n)^2 - 3) = 1/4 - 1/sqrt(24). (End)
a(n) = 3^n*Sum_{k=0..n} (2/3)^k*binomial(2*n, 2*k). - Detlef Meya, May 21 2024

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A108741 Member r=100 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 100, 9801, 960400, 94109401, 9221760900, 903638458801, 88547347201600, 8676736387298001, 850231618608002500, 83314021887196947001, 8163923913326692803600, 799981229484128697805801, 78389996565531285692164900, 7681419682192581869134354401, 752700738858307491889474566400

Offset: 0

Henry Bottomley, Jun 22 2005

Partial sums of A046173. [Joerg Arndt, Jun 10 2013]

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A000217, A004189.

I:=[0,1,100]; [n le 3 select I[n] else 99*Self(n-1)-99*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Feb 02 2016

LinearRecurrence[{99, -99, 1}, {0, 1, 100}, 20] (* Vincenzo Librandi, Feb 02 2016 *)

a(n) = ((49+20*sqrt(6))^n+(49-20*sqrt(6))^n -2)/96 = 98*a(n-1)-a(n-2)+2 = 99*a(n-1)-99*a(n-2)+a(n-3) = (a(n-1)-1)^2/a(n-2) = A004189(n)^2.
G.f.: -x*(x+1)/((x-1)*(x^2-98*x+1)). [Colin Barker, Oct 24 2012]
From Wolfdieter Lang, Feb 01 2016: (Start)
a(n) = (T(n, 49) - 1)/48 = (T(2*n, 5) - 1)/48 with Chebyshev's T polynomials A053120. See the name.
a(n) = A000217((T(n, 5) - 1)/2)/3. n >= 0.
a(n) = S(n-1, 10)^2 = A004189(n)^2, with Chebyshev's S polynomials A049310. This is the triangular number = 3*square number identity. Cf. the famous triangular number = square number identity: A000217((T(n, 3) - 1)/2) = S(n-1, 6)^2. A001109 and A001110. (End)

A036353 Square pentagonal numbers.

Original entry on oeis.org

0, 1, 9801, 94109401, 903638458801, 8676736387298001, 83314021887196947001, 799981229484128697805801, 7681419682192581869134354401, 73756990988431941623299373152801, 708214619789503821274338711878841001, 6800276705461824703444258688161258139001

Offset: 0

Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)

Lim_{n -> oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^8 = 4801 + 1960*sqrt(6). - Ant King, Nov 06 2011

Pentagonal numbers (A000326) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 11 2015

Colin Barker, Table of n, a(n) for n = 0..252
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Byungchan Kim, Eunmi Kim, and Jeremy Lovejoy, On weighted overpartitions related to some q-series in Ramanujan's lost notebook, Int'l J. Number Theory (2021). Also at Université de Paris (France, 2020).
Eric Weisstein's World of Mathematics, Pentagonal Square Number
Index entries for linear recurrences with constant coefficients, signature (9603,-9603,1).

Cf. A000326, A001078, A001079, A001110, A046172, A046173, A248205.

Table[Floor[1/96 ( Sqrt[2] + Sqrt[3] ) ^ ( 8*n - 4 ) ] , {n, 0, 9}] (* Ant King, Nov 06 2011 *)
LinearRecurrence[{9603,-9603,1},{0,1,9801,94109401},20] (* Harvey P. Dale, Apr 14 2019 *)

for(n=0,10^9,g=(n*(3*n-1)/2); if(issquare(g),print(g)))

concat(0, Vec(x*(1+198*x+x^2)/((1-x)*(1-9602*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 24 2015

a(n) = 9602*a(n-1) - a(n-2) + 200; g.f.: x*(1+198*x+x^2)/((1-x)*(1-9602*x+x^2)). - Warut Roonguthai, Jan 05 2001
a(n+1) = 4801*a(n)+100+980*(24*a(n)^2+a(n))^(1/2). - Richard Choulet, Sep 21 2007
From Ant King, Nov 06 2011: (Start)
a(n) = floor(1/96*(sqrt(2) + sqrt(3))^(8*n-4)).
a(n) = 9603*a(n-1) - 9603*a(n-2) + a(n-3).
(End)

More terms from Eric W. Weisstein

A046172 Indices of pentagonal numbers (A000326) that are also squares (A000290).

Original entry on oeis.org

1, 81, 7921, 776161, 76055841, 7452696241, 730288175761, 71560788528321, 7012226987599681, 687126683996240401, 67331402804643959601, 6597790348171111800481, 646516122717964312487521, 63351982236012331511976561, 6207847743006490523861215441

Offset: 1

Eric W. Weisstein

if P_x = y^2 is a pentagonal number that is also a square, the least both pentagonal and square number that is greater as P_x, is P_(49*x + 40*y - 8) = (60*x + 49*y - 10)^2 (in fact, P_(49*x + 40*y - 8) - (60*x + 49*y - 10)^2 = (3/2)*x^2 - (1/2)*x - y^2). - Richard Choulet, Apr 28 2009
a(n)*(3*a(n)-1)/2 = m^2 is equivalent to the Pell equation (6*a(n)-1)^2 - 6*(2*m)^2 = 1 or x(n)^2 - 6*y(n)^2 = 1. - Paul Weisenhorn, May 15 2009
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^4 = 49 + 20*sqrt(6). - Ant King, Nov 07 2011
Numbers k such that the k-th pentagonal number is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014
Indices of pentagonal numbers (A000326) that are also centered octagonal numbers (A016754). - Colin Barker, Jan 11 2015

Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7; http://dx.doi.org/10.1080/0020739X.2016.1164346

Colin Barker, Table of n, a(n) for n = 1..503
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, section 21.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
W. Sierpiński, Sur les nombres pentagonaux, Bull. Soc. Roy. Sci. Liege 33 (1964) 513-517.
Eric Weisstein's World of Mathematics, Pentagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A036353, A046173.

Cf. A000217, A000290, A000326, A251914, A248205.

LinearRecurrence[{99, -99, 1}, {1, 81, 7921}, 13] (* Ant King, Nov 07 2011 *)
Table[Round[(1 + x^(2*n+1))^2 / (12*x^(2*n+1)) /. x->5+2*Sqrt@6],{n,0,99}] (* Federico Provvedi, Apr 24 2023 *)

a(n) = 98*a(n-1) - a(n-2) - 16; g.f.: x*(1 - 18*x + x^2)/((1-x)*(1 - 98*x + x^2)). - Warut Roonguthai Jan 05 2001 - Corrected by Colin Barker, Jan 11 2015
a(n+1) = 49*a(n) - 8 + 10*sqrt(8*(3*a(n)^2 - a(n))) with a(1) = 1. - Richard Choulet, Apr 28 2009
a(n) = 1/6+((5 + 2*sqrt(6))^(2*n+1)/12) + ((5 - 2*sqrt(6))^(2*n+1)/12) for n >= 0. - Richard Choulet, Apr 29 2009
From Paul Weisenhorn, May 15 2009: (Start)
x(n+2) = 98*x(n+1) - x(n) with x(1)=5, x(2)=485;
y(n+2) = 98*y(n+1) - y(n) with y(n)=A046173(n)*2;
m(n+2) = 98*m(n+1) - m(n) with m(n)=A046173(n);
a(n) = A072256(n)^2.
(End)
a(n) = b(n)*b(n), b(n) = 10*b(n-1)- b(n-2), b(1)=1, b(2)=9, b(n)=((5 + sqrt(24))^n - (5 - sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, Sep 21 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
a(n) = ceiling((1/12)*(sqrt(3) + sqrt(2))^(4*n-2)).
(End)
a(n) = (1 + x^(2n+1))^2 / (12*x^(2*n+1)), with x = 5 + 2*sqrt(6). - Federico Provvedi, Apr 24 2023

A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers.

Original entry on oeis.org

1, 1926, 850137, 2564464982, 1132138928657, 3415133918621062, 1507685261236261801, 4547981651299964079126, 2007805569980855008712097, 6056610836775865229750164742, 2673822786819976661810784866297, 8065673443881586606920210924732502

Offset: 1

Richard Choulet, Dec 20 2007

We solve the equation P(p) = P(r) + P(r+1) with unknowns p and r, equivalent to (6*p-1)^2 = 2*(6*r+2)^2 + 17. The Diophantine equation X^2 = 2*Y^2 + 17 whose solutions give p and r are obtained by (x(n), y(n)) such that:
x(1)=5, x(2)=215, x(3)=4517, x(4)=248087 and the same recurrence relation on the odd and even indices x(n+2) = 1154*x(n+1) - x(n);
y(1)=2, y(2)=152, y(3)=3194, y(4)=175424 and the same recurrence relation on the odd and even indices y(n+2) = 1154*y(n+1) - y(n).
The solutions (p,r) are given by the (u(n),v(n)) such that
u(1)=1, u(2)=36, u(3)=753, u(4)=41348 and the same recurrence relation on the odd and even indices u(n+2) = 1154*u(n+1) - u(n) - 192 or u(n+1) = 577*u(n) - 96 + 68*sqrt(72*u(n)^2 - 24*u(n) - 32);
v(1)=0, v(2)=25, v(3)=532, v(4)=29237 and the same recurrence relation on the odd and even indices v(n+2) = 1154*v(n+1) - v(n) + 384 or v(n+1) = 577*v(n) + 192 + 68*sqrt(72*u(n)^2 + 48*u(n) + 15).

			With P(m) = m*(3*m-1)/2,
a(1)=1 because a(1) = P(1) = P(0) + P(1);
a(2)=1926 because P(36) = 1926 = P(25) + P(26) = 925 + 1001;
a(3)=850137 because P(753) = 850137 = P(532) + P(533) = 424270 + 425867 ...

Colin Barker, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (1,1331714,-1331714,-1,1).

Cf. A046173, A144796 (x(n)).

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) )); // G. C. Greubel, Mar 16 2019

a:=proc(m) if type (sqrt(18*m^2-6*m-8)/6-1/3), integer=true then m*(3*m-1)/2 else fi end : seq(a(m),m=1..100000)od; # Emeric Deutsch

# (3 # - 1)/2 &@ Select[Range[10^6], IntegerQ[Sqrt[18 #^2 - 6 # - 8]/6 - 1/3] &] (* or *)
Rest@ CoefficientList[Series[x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/( (1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)), {x, 0, 12}], x] (* Michael De Vlieger, Jul 14 2016 *)

Vec(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) + O(x^20)) \\ Colin Barker, Oct 20 2014

a=(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2))).series(x, 20).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 16 2019

For odd and even indices respectively: a(n+2) = 1331714*a(n+1) - a(n) - 416160; on the odd and the even indices respectively we also have a(n+1) = 665857*a(n) - 208080 + 19618*sqrt(1152*a(n)^2 - 720*a(n) - 32).

The g.f., h(z), is such that h(z) = a(1)*z + a(2)*z^2 + ... and is given by h(z) = z*(1 + 1925*z - 483503*z^2 + 65395*z^3 + 22*z^4)/((1-z)*(1 - 1331714*z^2 + z^4)).

Fixed typo in g.f. in formula, and more terms from Colin Barker, Oct 20 2014

A173205 a(n) = 98*a(n-1)-a(n-2) for n>2, a(1)=1, a(2)=98.

Original entry on oeis.org

1, 98, 9603, 940996, 92208005, 9035443494, 885381254407, 86758327488392, 8501430712608009, 833053451508096490, 81630736817080848011, 7998979154622415008588, 783818326416179589993613

Offset: 1

Sture Sjöstedt, Feb 12 2010

Similar to A046173 and A046172.

Indranil Ghosh, Table of n, a(n) for n = 1..502
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Index entries for linear recurrences with constant coefficients, signature (98,-1).

Cf. A046172, A046173.

CoefficientList[Series[x/(1-98x+x^2),{x,0,20}],x]  (* Harvey P. Dale, Mar 13 2011 *)

a(n) = ((49+20*sqrt(6))^n - (49-20*sqrt(6))^n)/(40*sqrt(6)).

G.f.: x/(1-98*x+x^2). - R. J. Mathar, Feb 13 2010

More terms from R. J. Mathar, Feb 13 2010

Definition rewritten (with Mathar's recurrence) by Bruno Berselli, Nov 10 2011

A251914 Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to the pentagonal number P(m) for some m.

Original entry on oeis.org

0, 98, 9700, 950598, 93149000, 9127651498, 894416697900, 87643708742798, 8588189040096400, 841554882220704498, 82463790268588944500, 8080609891439495856598, 791817305570802005002200, 77590015336047156994359098, 7603029685627050583442189500

Offset: 1

Colin Barker, Dec 11 2014

Also nonnegative integers x in the solutions to 2*x^2-3*y^2+4*x+y+2 = 0, the corresponding values of y being A046172.

			98 is in the sequence because T(98)+T(99) = 4851+4950 = 9801 = P(81).

Colin Barker, Table of n, a(n) for n = 1..503
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A000217, A000326, A046172.

LinearRecurrence[{99, -99, 1}, {0, 98, 9700}, 20] (* Vincenzo Librandi, Mar 03 2016 *)

concat(0, Vec(2*x^2*(x-49) / ((x-1)*(x^2-98*x+1)) + O(x^100)))

a(n) = A046173(n)-1.
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3).
G.f.: 2*x^2*(x-49) / ((x-1)*(x^2-98*x+1)).
a(n) = 2*(-1/2+1/48*(49+20*sqrt(6))^(-n)*(-12-5*sqrt(6)+(-12+5*sqrt(6))*(49+20*sqrt(6))^(2*n))). - Colin Barker, Mar 03 2016

cp's OEIS Frontend

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

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A001079 a(n) = 10*a(n-1) - a(n-2); a(0) = 1, a(1) = 5.

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A108741 Member r=100 of the family of Chebyshev sequences S_r(n) defined in A092184.

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A036353 Square pentagonal numbers.

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A046172 Indices of pentagonal numbers (A000326) that are also squares (A000290).

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A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers.

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