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 A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x. 17

%I

%S 1,25,841,28561,970225,32959081,1119638521,38034750625,1292061882721,

%T 43892069261881,1491038293021225,50651409893459761,

%U 1720656898084610641,58451683124983302025,1985636569351347658201,67453191674820837076801,2291422880374557112953025

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

%C 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

%C Solutions to A007913(x)=A007913(2x-1). - _Benoit Cloitre_, Apr 07 2002

%C From _Ant King_, Nov 09 2011: (Start)

%C Indices of positive hexagonal numbers that are also perfect squares.

%C 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).

%C (End)

%C Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - _Colin Barker_, Jan 25 2015

%C 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

%C 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

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

%H Muniru A Asiru, <a href="/A008844/b008844.txt">Table of n, a(n) for n = 0..100</a>

%H Daniel C. Fielder, <a href="http://www.fq.math.ca/Scanned/6-3/fielder.pdf">Special integer sequences controlled by three parameters</a>, Fibonacci Quarterly 6, 1968, 64-70.

%H M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, <a href="http://www.jstor.org/stable/2968551">Problem 47</a>, Amer. Math. Monthly, 4 (1897), 25-28.

%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2018volume18/FG201808index.html">Integer Sequences and Circle Chains Inside a Circular Segment</a>, Forum Geometricorum, Vol. 18 (2018), 47-55.

%H S. C. Schlicker, <a href="http://www.jstor.org/stable/10.4169/math.mag.84.5.339">Numbers Simultaneously Polygonal and Centered Polygonal</a>, Mathematics Magazine, Vol. 84, No. 5, December 2011

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexagonalSquareNumber.html">Hexagonal Square Number.</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).

%F From _Benoit Cloitre_, Jan 19 2003: (Start)

%F a(n) = A078522(n) + 1.

%F a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)

%F G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).

%F a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000

%F a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - _Richard Choulet_, Sep 14 2007

%F 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

%F From _Ant King_, Nov 09 2011: (Start)

%F a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).

%F a(n) = 34*a(n-1) - a(n-2) - 8.

%F a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).

%F a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)).

%F (End)

%F From _Ravi Kumar Davala_, May 26 2013: (Start)

%F a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).

%F a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).

%F a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).

%F a(n+1)*a(n-1) = (a(n)+4)^2.

%F (End)

%F a(n) = A001652(n)^2 + A046090(n)^2. - _César Aguilera_, Jan 15 2018

%F Lim_{n -> infinity} a(n)/a(n-1) = A156164. - _César Aguilera_, Jan 28 2018

%F sqrt(2*a(n))-1 = A002315(n). - _Ezhilarasu Velayutham_, Apr 05 2019

%e From _Ravi Kumar Davala_, May 26 2013: (Start)

%e A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly

%e 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.

%e A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly

%e 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

%e 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.

%e (End)

%e 1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

%p 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

%t LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* _Ant King_, Nov 09 2011 *)

%t CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* _Vincenzo Librandi_, Jan 20 2018 *)

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

%o (PARI) vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ _Altug Alkan_, Nov 11 2015

%o (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

%o (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

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

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E Entry edited by _N. J. A. Sloane_, Sep 14 2007

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Last modified April 15 07:59 EDT 2021. Contains 342975 sequences. (Running on oeis4.)