A078522 Numbers k such that (k+1)*(2*k+1) is a perfect square.

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Author

Joseph L. Pe, Jan 07 2003

Keywords

• nonn
• easy

Comments

Equivalently, both k+1 and 2*k+1 are perfect squares.

The square roots of (k+1)*(2*k+1) are in A046176.

Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016

From Sergey Pavlov, Mar 14 2017: (Start)

The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).

Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).

(End)

Links

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

Crossrefs

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

Programs

GAP

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

Magma

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

Maple

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

Mathematica

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

PARI

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

Sage

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

Formula

From Benoit Cloitre, Jan 19 2003: (Start)

a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.

a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).

a(n) = A008844(n) - 1. (End)

From R. J. Mathar, Sep 21 2011: (Start)

G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).

a(n) = 24*A029546(n-2). (End)

a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019

a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020

From Amiram Eldar, Dec 02 2024: (Start)

a(n) = A001542(n-1)*A001542(n).

Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Extensions

Edited by Bruno Berselli, Nov 17 2016