A036353 -id:A036353 - cp's OEIS frontend

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

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

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

A046173 Indices of square numbers that are also pentagonal.

Original entry on oeis.org

1, 99, 9701, 950599, 93149001, 9127651499, 894416697901, 87643708742799, 8588189040096401, 841554882220704499, 82463790268588944501, 8080609891439495856599, 791817305570802005002201, 77590015336047156994359099, 7603029685627050583442189501

Offset: 1

Eric W. Weisstein

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
a(n)^2 is of the form (2*m-1)*(3*m-2), and the corresponding values of m are 1, 41, 3961, 388081, 38027921, 3726348121, 365144087881, ..., with closed form ((5-2*sqrt(6))^(2n-1)+(5+2*sqrt(6))^(2n-1)+14)/24 (for n>0). - Bruno Berselli, Dec 12 2013
The terms of this sequence satisfy the Diophantine equation m^2 = k * (3k-1)/2, which is equivalent to (6k-1)^2 - 6*(2*m)^2 = 1. Now, with x=6k-1 and y=2*m, we get the Pell-Fermat equation x^2 - 6*y^2 = 1. The solutions (x,y) of this equation are respectively in A046174 and A046175. The indices m=y/2 of the square numbers which are also pentagonal are the terms of this sequence, the indices k=(x+1)/6 of the pentagonal numbers which are also square are in A046172, and the pentagonal square numbers are in A036353. - Bernard Schott, Mar 10 2019
Also, this sequence is related to A302330 by (sqrt(2) + sqrt(3))^(4*n-2) = A302330(n-1)*5 + a(n)*sqrt(24). - Bruno Berselli, Oct 29 2019

			G.f. = x + 99*x^2 + 9701*x^3 + 950599*x^4 + 93149001*x^5 + ...
99 is a term because 99^2 = 9801 = (1/2) * 81 * (3*81 - 1), so 9801 is the 99th square number, also the 81st pentagonal number, and the second pentagonal square number after 1. - _Bernard Schott_, Mar 10 2019

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 35.

Colin Barker, Table of n, a(n) for n = 1..503
M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
L. Euler, De solutione problematum diophanteorum per numeros integros, par. 21
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Pentagonal Square Number
Index entries for linear recurrences with constant coefficients, signature (98,-1).

Cf. A036353 (pentagonal square numbers), A046172 (indices of pentagonal numbers that are also square).
Cf. A046174, A046175 (solutions of x^2 - 6*y^2 = 1).
Cf. A302330.

CoefficientList[Series[(1 + x)/(1 - 98* x + x^2), {x, 0, 30}], x] (* T. D. Noe, Aug 01 2011 *)
LinearRecurrence[{98, -1}, {1, 99}, 30] (* Harvey P. Dale, Jul 31 2017 *)

{a(n) = subst( poltchebi(n) - poltchebi(n-1), 'x, 49) / 48}; /* Michael Somos, Sep 05 2006 */

Vec(x*(x+1)/(x^2-98*x+1) + O(x^30)) \\ Colin Barker, Jun 23 2015

a(n) = 98*a(n-1) - a(n-2); g.f.: (1+x)/(1-98*x+x^2). - Warut Roonguthai, Jan 05 2001
a(1-n) = -a(n) for all n in Z. - Michael Somos, Sep 05 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),5),5). - Marcos Carreira, Dec 27 2006
a(n) = ((12+5*sqrt(6))/24)*(5+2*sqrt(6))^(2*n)+((12-5*sqrt(6))/24)*(5-2*sqrt(6))^(2*n) for n>=0. - Richard Choulet, Apr 29 2009
a(n+1) = 49*a(n) + 10*sqrt(24*a(n)^2+1) for n > =0 with a(0)=1. - Richard Choulet, Apr 29 2009
a(n) = b such that (-1)^n*Integral_{x=-Pi/2..Pi/2} (cos(2*n-1)*x)/(5-sin(x)) dx = c + b*(log(2)-log(3)). - Francesco Daddi, Aug 01 2011
a(n) = floor((1/24) * sqrt(6) * (sqrt(2) + sqrt(3))^(4n-2)). - Ant King, Nov 07 2011
a(n) = A138288(n)*A054320(n). - Gerry Martens, May 13 2024

A342709 12-gonal (dodecagonal) square numbers.

Original entry on oeis.org

1, 64, 3025, 142129, 6677056, 313679521, 14736260449, 692290561600, 32522920134769, 1527884955772561, 71778070001175616, 3372041405099481409, 158414167969674450625, 7442093853169599697984, 349619996931001511354641, 16424697761903901433970161

Offset: 1

Bernard Schott, Mar 19 2021

The 12-gonal square numbers k correspond to the nonnegative integer solutions of the Diophantine equation k = d*(5*d-4) = c^2, equivalent to (5*d-2)^2 - 5*c^2 = 4. Substituting x = 5*d-2 and y = c gives the Pell-Fermat's equation x^2 - 5*y^2 = 4.
The solutions x are in A342710, while corresponding solutions y that are also the indices c of the squares which are 12-gonal are in A033890.
The indices d of the corresponding 12-gonal which are squares are in A081068.

			142129 = 169*(5*169-4) = 377^2, so 142129 is the 169th 12-gonal number and the 377th square, hence 142129 is a term.

Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Intersection of A000290 (squares) and A051624 (12-gonal numbers).

Similar for n-gonal squares: A001110 (triangular), A036353 (pentagonal), A046177 (hexagonal), A036354 (heptagonal), A036428 (octagonal), A036411 (9-gonal), A188896 (there are no 10-gonal squares > 1), A333641 (11-gonal), this sequence (12-gonal).

with(combinat):
seq(fibonacci(4*n-2)^2, n=1..16);

Table[Fibonacci[4*n - 2]^2, {n, 1, 16}] (* Amiram Eldar, Mar 19 2021 *)

a(n) = fibonacci(4*n-2)^2; \\ Michel Marcus, Mar 21 2021

G.f.: x*(1 + 16*x + x^2)/((1 - x)*(1 - 47*x + x^2)). - Stefano Spezia, Mar 20 2021
a(n) = 48*a(n-1) - 48*a(n-2) + a(n-3). - Kevin Ryde, Mar 20 2021
a(n) = 9*A161582(n) + 1. - Hugo Pfoertner, Mar 19 2021
a(n) = A033890(n-1)^2.

A248205 Indices of centered octagonal numbers (A016754) that are also pentagonal numbers (A000326).

Original entry on oeis.org

1, 50, 4851, 475300, 46574501, 4563825750, 447208348951, 43821854371400, 4294094520048201, 420777441110352250, 41231895134294472251, 4040304945719747928300, 395908652785401002501101, 38795007668023578497179550, 3801514842813525291721094751

Offset: 1

Colin Barker, Jan 11 2015

Positive integers y in the solutions to 3*x^2 - 8*y^2 - x + 8*y - 2 = 0, the corresponding values of x being A046172.

			50 is in the sequence because the 50th centered octagonal number is 9801, which is also the 81st pentagonal number.

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

Cf. A000290, A000326, A046172, A036353.

I:=[1,50,4851]; [n le 3 select I[n] else 99*Self(n-1)-99*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jun 13 2015

LinearRecurrence[{99, -99, 1}, {1, 50, 4851}, 20] (* Vincenzo Librandi, Jun 13 2015 *)

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

a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
G.f.: x*(49*x-1) / ((x-1)*(x^2 - 98*x + 1)).
a(n) = (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

A277792 Squares that are also pentagonal pyramidal numbers.

Original entry on oeis.org

0, 1, 196, 2601, 15376, 60025, 181476, 461041, 1032256, 2099601, 3960100, 7027801, 11861136, 19193161, 29964676, 45360225, 66846976, 96216481, 135629316, 187662601, 255360400, 342287001, 452583076, 591024721, 763085376, 975000625, 1233835876, 1547556921, 1925103376, 2376465001, 2912760900

Offset: 0

Ilya Gutkovskiy, Oct 31 2016

Intersection of A000290 and A002411.

			a(2) = 196 because 196 = 14^2 is a perfect square and 196 = 7^2*(7 + 1)/2 is the 7th pentagonal pyramidal number.

Daniel Mondot, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pentagonal Pyramidal Number
Eric Weisstein's World of Mathematics, Square Number
Index to sequences related to polygonal numbers
Index to sequences related to pyramidal numbers
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000290, A001653, A002411, A007588, A036353, A056220, A254333.

[n^2*(2*n^2-1)^2: n in [0..30]]; // Vincenzo Librandi, Nov 01 2016

Table[n^2 (2 n^2 - 1)^2, {n, 0, 30}]
LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,196,2601,15376,60025,181476},40] (* Harvey P. Dale, Nov 01 2024 *)

O.g.f.: x*(1 + 189*x + 1250*x^2 + 1250*x^3 + 189*x^4 + x^5)/(1 - x)^7.
E.g.f.: x*(1 + 97*x + 336*x^2 + 256*x^3 + 60*x^4 + 4*x^5)*exp(x).
a(n) = a(-n).
a(n) = n^2*(2*n^2 - 1)^2.
a(n) = A000290(A007588(n)).
a(n) = A000290(n)*A000290(A056220(n)).
Sum_{n>=1} 1/a(n) = (2*Pi^2+9*sqrt(2)*Pi*cot(Pi/sqrt(2))+3*Pi^2*csc(Pi/sqrt(2))^2-24)/12 = 1.0055779712856...

A333641 11-gonal (or hendecagonal) square numbers.

Original entry on oeis.org

0, 1, 196, 29241, 1755625, 261468900, 38941102225, 2337990844401, 348201795147556, 51858411008887561, 3113535139359330841, 463705205422871375236, 69060571958250748760481, 4146338334574433921200225, 617522713934165528806340100, 91968930524758079223806760025

Offset: 1

Bernard Schott, Mar 31 2020

The 11-gonal square numbers correspond to the nonnegative integer solutions of the Diophantine equation k*(9*k-7)/2 = m^2, equivalent to (18*k-7)^2 - 72*m^2 = 49. Substituting x = 18*k-7 and y = m gives the Pell equation x^2-72*y^2 = 49. The integer solutions (x,y) = (-7,0), (11,1), (119,14), (1451,171), (11243,1325), ... correspond to the following solutions (k,m) = (0,0), (1,1), (7,14), (81,171), (625,1325), ...

			1755625 is a term because 625*(9*625-7)/2 = 1325^2 = 1755625; that means that 1755625 is the 625th 11-gonal number and the square of 1325.

Index entries for linear recurrences with constant coefficients, signature (1,0,1331714,-1331714,0,-1,1).

Intersection of A000290 (squares) and A051682 (11-gonals).
Cf. A106525.
Cf. A001110 (square triangulars), A036353 (square pentagonals), A046177 (square hexagonals), A036354 (square heptagonals), A036428 (square octagonals), A036411 (square 9-gonals), A188896 (only {0,1} are square 10-gonals), this sequence (square 11-gonals), A342709 (square 12-gonals).

for k from 0 to 8000000 do
d:= k*(9*k-7)/2;
if issqr(d) then print(k,sqrt(d),d); else fi; od:

Last /@ Solve[(18*x - 7)^2 - 72*y^2 == 49 && x >= 0 && y >= 0 && y < 10^16, {x, y}, Integers] /. Rule -> (#2^2 &) (* Amiram Eldar, Mar 31 2020 *)

concat(0, Vec(-x*(1 + 195*x + 29045*x^2 + 394670*x^3 + 29045*x^4 + 195*x^5 + x^6)/(-1 + x + 1331714*x^3 - 1331714*x^4 - x^6 + x^7) + O(x^20))) \\ Jinyuan Wang, Mar 31 2020

a(n) = k*(9*k-7)/2 for n > 1, where k = (A106525(4*n-6) + 7)/18. - Jinyuan Wang, Mar 31 2020

More terms from Amiram Eldar, Mar 31 2020

A342300 Least nonnegative number greater than the previous number which is simultaneously an n-gonal and (n+1)-gonal number.

Original entry on oeis.org

0, 1, 3, 36, 9801, 40755, 121771, 297045, 631125, 1212751, 2158695, 3617601, 5773825, 8851275, 13117251, 18886285, 26523981, 36450855, 49146175, 65151801, 85076025, 109597411, 139468635, 175520325, 218664901, 269900415, 330314391, 401087665, 483498225, 578925051, 688851955, 814871421

Offset: 0

Robert G. Wilson v, Jun 04 2021

Also the least nontrivial number simultaneously an n and (n+1)-gonal number for n greater than one.
0 and 1 are always terms of any sequence of polygonal numbers of a particular rank beginning with index 0.
Since the formula for the k-th n-gonal number P(n,k) is k*(4+k*(n-2)-n)/2, one can extrapolate for the non-geometrical terms 0, 1 and 2.
Indices of the n and (n+1)-gonal numbers by pairs: {0, 0} {1, 1}, {3, 2}, {8, 6}, {99, 81}, {165, 143}, {247, 221}, {345, 315}, {459, 425}, {589, 551}, {735, 693}, {897, 851} ..., .
{x, y} of the above are {8n^2 + 10n - 3, 8n^2 - 10n - 7} for n>3 (A303295).
In the first 1000 terms, 1 is congruent to 0 (mod 6), 333 are congruent to 1 (mod 6), and 666 are congruent to 3 (mod 6).

			a(3) is the least triangular and square number > 3, which is 36: A001110(2).
a(4) is the least square and pentagonal number > 36, which is 9801: A036353(2).

Index entries for linear recurrences with constant coefficients, signature (6,-15, 20,-15,6,-1).

Cf. A001110, A036353, A046180, A048903, A048906, A048924, A203627, A189216.

a[n_] := Intersection[ Table[ PolygonalNumber[n, i], {i, 2, 10000}], Table[ PolygonalNumber[n + 1, i], {i, 2, 10000}]][[1]]; a[0] = 0; a[1] = 1; Array[a, 30, 0] (* Or *)
a[n_] := a[n] = 6a[n - 1] -15a[n - 2] +20a[n - 3] -15a[n - 4] +6a[n - 5] -a[n - 6]; a[0] = 0; a[1] = 1; a[2] = 3; a[3] = 36; a[4] = 9801; a[5] = 40755; a[6] = 121771; a[7] = 297045; a[8] = 631125; a[9] = 1212751; Array[a, 30, 0]

a(n) = 32n^5 - 112n^4 + 70n^3 + 93n^2 - 57n - 35 for n > 3; a(0) = 0, a(1) = 1, a(2) = 3, a(3) = 36.

G.f.: x*(1 - 3*x + 33*x^2 + 9610*x^3 - 17556*x^4 + 23575*x^5 - 17753*x^6 + 7122*x^7 - 1189*x^8)/(1 - x)^6. - Stefano Spezia, Jun 08 2021

A378245 Numbers that are both k-gonal and (k+1)-gonal for some k >= 3.

Original entry on oeis.org

1, 36, 1225, 9801, 40755, 41616, 121771, 297045, 631125, 1212751, 1413721, 2158695, 3617601, 5773825, 8851275, 13117251, 18886285, 26523981, 36450855, 48024900, 49146175, 65151801, 85076025, 94109401, 109597411, 139468635, 175520325, 218664901, 269900415, 330314391

Offset: 1

Kelvin Voskuijl, Nov 20 2024

nonn

			a(2) = 36 is both the 8th triangular and the 6th square number.
a(3) = 1225 is both the 49th triangular and the 35th square number.
a(5) = 40755 is both the 165th pentagonal number and the 143th hexagonal number.

Cf. A057145, A342300.
Cf. A001110, A036353, A046180, A048903, A048906, A048924 and A203627 (subsequences).
The subdiagonal of A189216 is also a subsequence.

upto(limit) = my(terms=List(1)); for(k=3, oo, my(found=0); for(n=2, oo, my(a = (2*n - 1)^2, b = (4*n*(3*n - 5) + 6), c = (8*(n-1)^2 + 1), s = (a*k^2 - b*k + c), v = n * (n*k - k - 2*n + 4) / 2); if(issquare(s), my(t = sqrtint(s) + k - 3); if(t % (2*(k-1)) == 0, listput(terms, v); found += 1)); if(v >= limit, break)); if(found == 0, break)); Vec(vecsort(terms)); \\ Daniel Suteu, Dec 08 2024

a(12)-a(30) from Pontus von Brömssen, Dec 07 2024

cp's OEIS Frontend

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

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

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A046173 Indices of square numbers that are also pentagonal.

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A342709 12-gonal (dodecagonal) square numbers.

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A248205 Indices of centered octagonal numbers (A016754) that are also pentagonal numbers (A000326).

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A277792 Squares that are also pentagonal pyramidal numbers.

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