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

0,2

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

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).

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

Example

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.

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 *)

Prog

(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
(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

Crossrefs

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

Sequence in context: A246761 A122142 A151557 * A251925 A181892 A274469

Adjacent sequences: A008841 A008842 A008843 * A008845 A008846 A008847

Keyword

nonn,easy,changed

Author

N. J. A. Sloane

Extensions

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

Status

approved