0,1

Numbers k such that 2*k^2 + 14*k + 25 is a square. - James R. Buddenhagen, Jul 19 2008

If k satisfies the previous condition, i.e., P(k) = 2*k^2 + 14*k + 25 is a square, then there exists an integer m such that (k+3)^2 + (k+4)^2 = m^2; finding such numbers k is equivalent to finding Pythagorean triples (X,X+1,Z) with X in A001652 (1st formula). For k >= 0, P(k) = A001844(k+3), and the successive values of P(k) that are squares are in A008844. - Bernard Schott, Mar 24 2019

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, New-York, 1964, pp. 122-124.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

a(n) = A001652(n) - 3.

a(n) = ( (1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 14 )/4.

a(n) - a(n-1) = A001541(n), n > 0. - R. J. Mathar, Apr 23 2009

a(n) = (Q(2*n+1) -14)/4 = (4*P(n)*P(n+1) + (-1)^n - 7)/2, where P(n) = A000129(n) (Pell) and Q(n) = A002203(n) (Pell-Lucas). - G. C. Greubel, Apr 04 2019

CoefficientList[Series[(3-21x+4x^2)/((x-1)(x^2-6x+1)), {x, 0, 30}], x] (* or *) LinearRecurrence[{7, -7, 1}, {-3, 0, 17}, 30] (* Harvey P. Dale, Dec 12 2016 *)

(PARI) my(x='x+O('x^30)); Vec((3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))) \\ G. C. Greubel, Feb 12 2018

(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))) ); // G. C. Greubel, Feb 12 2018

(SageMath) ((3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 04 2019

Cf. A001652 (X values), A001653 (Z values), A001844, A008844.

Sequence in context: A024040 A357811 A338489 * A127187 A057398 A361455

Adjacent sequences: A009756 A009757 A009758 * A009760 A009761 A009762

N. J. A. Sloane, Nick Baxter (nick(AT)visigenic.com)

G.f. and Binet formula corrected by R. J. Mathar, Aug 24 2016

approved