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

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Examples

			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.
		

References

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

Links

• 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).

Crossrefs

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

Programs

GAP

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

Magma

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

Maple

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

Mathematica

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 *)

PARI

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

PARI

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

Formula

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

Extensions

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