A046177 -id:A046177 - cp's OEIS frontend

A046176 Indices of square numbers that are also hexagonal.

1, 35, 1189, 40391, 1372105, 46611179, 1583407981, 53789260175, 1827251437969, 62072759630771, 2108646576008245, 71631910824649559, 2433376321462076761, 82663163018885960315, 2808114166320660573949

Offset: 1

Eric W. Weisstein

Bisection (even part) of Chebyshev sequence with Diophantine property.
(3*b(n))^2 - 2*(2*a(n+1))^2 = 1 with companion sequence b(n) = A077420(n), n >= 0.
Sequence also refers to inradius of primitive Pythagorean triangles with consecutive legs, odd followed by even. - Lekraj Beedassy, Apr 23 2003
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12*sqrt(2). - Ant King, Nov 08 2011
Integers of the form sqrt((m+1)*(2*m+1)). The corresponding values of m form A078522. Subsequence of A284876. - Jonathan Sondow, Apr 07 2017

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=3.
Serge Perrine, About the diophantine equation z^2 = 32y^2 - 16, SCIREA Journal of Mathematics (2019) Vol. 4, Issue 5, 126-139.
Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008) , v_n in Theorem 5.6.
P. H. van der Kamp, Global classification of two-component..., Found. Comput. Math. 9 (5) (2009) 559-597 near Eq. (4.7).
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A000129, A008844, A046177, A078522, A284876.

Cf. A001109, A001110 (partial sums).

List([0..20], n-> Lucas(2,-1, 4*n-2)[1]/2 ); # G. C. Greubel, Jan 13 2020

I:=[1,35]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

seq( simplify(ChebyshevU(2*(n-1), 3)), n = 1..20); # G. C. Greubel, Jan 13 2020

LinearRecurrence[{34, -1}, {1, 35}, 15] (* Ant King, Nov 08 2011 *)
Fibonacci[4*Range[20] -2, 2]/2 (* G. C. Greubel, Jan 13 2020 *)

vector(21, n, polchebyshev(2*(n-1), 2, 3) ) \\ G. C. Greubel, Jan 13 2020

[lucas_number1(4*n-2, 2,-1)/2 for n in (1..20)] # G. C. Greubel, Jan 13 2020

a(n) = 34*a(n-1) - a(n-2); a(0)=-1, a(1)=1.
a(n+1) = S(2*n, 6) = S(n, 34) + S(n-1, 34), n >= 1, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(n, 34) = A029547(n).
G.f.: x*(1+x)/(1-34*x+x^2).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*6^(2*(n-k)), n >= 0.
a(n) = A001109(2n+1). - Lekraj Beedassy, Apr 23 2003
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),3),3). - Marcos Carreira, Dec 27 2006
From Antonio Alberto Olivares, Mar 22 2008: (Start)
a(n) = (sqrt(2)/8)*(3 + 2*sqrt(2))*(17 + 12*sqrt(2))^(n-1) - (sqrt(2)/8)*(3 - 2*sqrt(2))*(17 - 12*sqrt(2))^(n-1).
a(n) = (sqrt(2)/8)*( (17+12*sqrt(2))^(n-1/2) - (17-12*sqrt(2))^(n-1/2) ).
a(n) = (sqrt(2)/8)*( (3+2*sqrt(2))^(2n-1) - (3-2*sqrt(2))^(2n-1) ).
a(n) = (sqrt(2)/8)*( (1+sqrt(2))^(4n-2) - (1-sqrt(2))^(4n-2) ).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). (End)
a(n+1) = 17*a(n) + 6*sqrt(8*a(n)^2+1) for n >= 0. - Richard Choulet, May 01 2009
a(n) = b such that (-1)^(n+1) * Integral_{x=-Pi/2..Pi/2} cos((2*n-1)*x)/(3-sin(x)) dx = c + b*log(2). - Francesco Daddi, Aug 01 2011
a(n) are the nonzero integer square roots of A227970. - Richard R. Forberg, Aug 01 2013
a(n) = y/5, where y are solutions to: y^2 = 2x^2 - x - 3. - Richard R. Forberg, Nov 24 2013
a(n) = sqrt((A078522(n)+1)*(2*A078522(n)+1)). - Jonathan Sondow, Apr 07 2017
a(n) = Pell(4*n-2)/2. - G. C. Greubel, Jan 13 2020
a(n) = A001653(n)*A002315(n). - Gerry Martens, Mar 23 2024

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025

Offset: 0

N. J. A. Sloane

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Muniru A Asiru, Table of n, a(n) for n = 0..100
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes, and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Steven C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018

I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)

a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))

vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Entry edited by N. J. A. Sloane, Sep 14 2007

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.

A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217).

Original entry on oeis.org

1, 18, 595, 20196, 686053, 23305590, 791703991, 26894630088, 913625718985, 31036379815386, 1054323288004123, 35815955412324780, 1216688160731038381, 41331581509442980158, 1404057083160330286975, 47696609245941786776976, 1620280657278860420130193

Offset: 1

Colin Barker, Jan 16 2015

Also positive integers y in the solutions to x^2 - 8*y^2 + x + 8*y - 2 = 0, the corresponding values of x being A008843.

Also the indices of centered octagonal numbers (A016754) which are also hexagonal numbers (A000384). Also positive numbers y in the solutions to 4x^2-8y^2-2x+8y-2=0. - Colin Barker, Jan 25 2015

			18 is in the sequence because the 18th centered octagonal number is 1225, which is also the 49th triangular number.
18 is in the sequence because the 18th centered octagonal number 1225 is also the 25th hexagonal number. - _Colin Barker_, Jan 25 2015

Colin Barker, Table of n, a(n) for n = 1..654
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A008843, A016754, A046177.

Vec(x*(17*x-1)/((x-1)*(x^2-34*x+1)) + O(x^100))

a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
G.f.: x*(17*x-1) / ((x-1)*(x^2-34*x+1)).
a(n) = sqrt((-2-(17-12*sqrt(2))^n-(17+12*sqrt(2))^n)*(2-(17-12*sqrt(2))^(1+n)-(17+12*sqrt(2))^(1+n)))/(8*sqrt(2)). - Gerry Martens, Jun 04 2015

A280181 Indices of centered 9-gonal numbers (A060544) that are also squares (A000290).

Original entry on oeis.org

1, 17, 561, 19041, 646817, 21972721, 746425681, 25356500417, 861374588481, 29261379507921, 994025528680817, 33767606595639841, 1147104598723073761, 38967788749988868017, 1323757712900898438801, 44968794449880558051201, 1527615253583038075302017

Offset: 1

Colin Barker, Dec 28 2016

Also positive integers y in the solutions to 2*x^2 - 9*y^2 + 9*y - 2 = 0, the corresponding values of x being A046176.

Consider all ordered triples of consecutive integers (k, k+1, k+2) such that k is a square and k+1 is twice a square; then the values of k are the squares of the NSW numbers (A002315), the values of k+1 are twice the squares of the odd Pell numbers (A001653), and the values of k+2 are thrice the terms of this sequence. (See the Example section.) - Jon E. Schoenfield, Sep 06 2019

			17 is in the sequence because the 17th centered 9-gonal number is 1225, which is also the 35th square.
From _Jon E. Schoenfield_, Sep 06 2019: (Start)
The following table illustrates the relationship between the NSW numbers (A002315), the odd Pell numbers (A001653), and the terms of this sequence:
.
  |  A002315(n-1)^2  |   2*A001653(n)^2  |
n |   = 3*a(n) - 2   |    = 3*a(n) - 1   |       3*a(n)
--+------------------+-------------------+-------------------
1 |    1^2 =       1 |   1^2*2 =       2 |      1*3 =       3
2 |    7^2 =      49 |   5^2*2 =      50 |     17*3 =      51
3 |   41^2 =    1681 |  29^2*2 =    1682 |    561*3 =    1683
4 |  239^2 =   57121 | 169^2*2 =   57122 |  19041*3 =   57123
5 | 1393^2 = 1940449 | 985^2*2 = 1940450 | 646817*3 = 1940451
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001653, A002315, A046176, A046177, A056771, A060544, A077420.

Cf. A156164.

LinearRecurrence[{35, -35, 1}, {1, 17, 561}, 50] (* G. C. Greubel, Dec 28 2016 *)

Vec(x*(1 - 18*x + x^2) / ((1 - x)*(1 - 34*x + x^2)) + O(x^20))

a(n) = (6 + (3-2*sqrt(2))*(17+12*sqrt(2))^(-n) + (3+2*sqrt(2))*(17+12*sqrt(2))^n) / 12.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
G.f.: x*(1 - 18*x + x^2) / ((1 - x)*(1 - 34*x + x^2)).
a(n) = (A002315(n-1)^2 + 2)/3 = (2*A001653(n)^2 + 1)/3. - Jon E. Schoenfield, Sep 06 2019
a(n) = A077420(floor((n-1)/2)) * A056771(floor(n/2)). - Jon E. Schoenfield, Sep 08 2019
E.g.f.: -1+(1/12)*(6*exp(x)+(3-2*sqrt(2))*exp((17-12*sqrt(2))*x)+(3+2*sqrt(2))*exp((17+12*sqrt(2))*x)). - Stefano Spezia, Sep 08 2019
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2) = A156164. - Andrea Pinos, Oct 07 2022

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

A350712 a(n) is the smallest hexagonal number for which the symmetric representation of sigma(n) has width 2*n, n >= 0, at the diagonal.

Original entry on oeis.org

0, 6, 120, 2016, 7140, 61776, 103740, 738720, 437580, 1185030, 4680270, 4426800, 2031120, 6193440, 4915680, 30728880, 2162160, 48565440, 134734320, 286071240, 163723560, 376902240, 536592420, 137373600, 76576500, 391986000, 214980480, 103672800, 1018606680, 5401294080

Offset: 0

Hartmut F. W. Hoft, Feb 02 2022

nonn

The width of the symmetric representation of sigma for hexagonal numbers at the diagonal is 1 only for number 1. For any hexagonal number h(n) = n*(2*n-1), n>1, the last leg of the Dyck path of h(n)-1 has length 2 and that of h(n) has length 1 (see formula in A237591) so that the width of the symmetric representation of sigma at the diagonal is at least 2 and contains a subpart of size 1 at the diagonal (see A280851).
The geometry of the Dyck paths ensures that a square bisected by the diagonal whose side length equals the width of the symmetric representation of sigma at the diagonal fits between the bounding pair of Dyck paths.
For hexagonal numbers up to h(100000) = 19999900000 only 1225, 1413721, and 1631432881 (the 25th, 841st, and 28561st hexagonal numbers) have width 3 at the diagonal, and none were found of odd width greater than 3.
The next [last] number in the sequence data smaller than h(55000) = 6049945000 is a(42) = 4874349480 [a(49) = 4819214400] with a(31..41) > h(55000).
The numbers [1, 1225, 1413721, 1631432881] mentioned above (in the first comment and in the third comment) are the first four square-hexagonal numbers (A046177). - Omar E. Pol, Feb 04 2022

			a(1) = 6, and a(2) = 120 since all hexagonal numbers k, 6 <= k < 120, have width 2 at the diagonal.

Cf. A000384, A046177, A235791, A237048, A237591, A237593, A249223, A280851, A346875.

(* for function a2[ ] see A237048 and A249223 *)
(* parameter bw is an upper bound estimate for how many values will be returned *)
a350712[n_, bw_] := Module[{widthL=Table[0, bw], wL, cL, i, w}, wL=Map[#(2#-1)&, Range[n]]; cL=Map[Last[a2[#]]&, wL]; For[i=1, i<=n, i++, w=cL[[i]]; If[EvenQ[w]&&widthL[[w/2]]==0, widthL[[w/2]]=wL[[i]]]]; Join[{0}, widthL]]
Take[a350712[55000, 50], 37]

cp's OEIS Frontend

A046176 Indices of square numbers that are also hexagonal.

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A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

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A342709 12-gonal (dodecagonal) square numbers.

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A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217).

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A280181 Indices of centered 9-gonal numbers (A060544) that are also squares (A000290).

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A333641 11-gonal (or hendecagonal) square numbers.

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