A160695 Integers m such that 3*m+1 and 7*m+1 are both perfect squares.

0, 5, 120, 2760, 63365, 1454640, 33393360, 766592645, 17598237480, 403992869400, 9274237758725, 212903475581280, 4887505700610720, 112199727638465285, 2575706229984090840, 59129043561995624040, 1357392295695915262085, 31160893757444055403920

Offset: 1

Author

Paul Weisenhorn, May 24 2009

Keywords

• nonn
• easy

Comments

The ansatz 3*a(n)+1=A^2, 7*a(n)+1=B^2 is equivalent to the Pell equation x^2-21*y^2=1 (see A077232 for d=21), with x=(21*a(n)+5)/2 and y=A*B/2.

The associated A are in A004253, the B in A030221.

Bisection of A089927. - R. J. Mathar, Jul 10 2009

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..736
• Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
• Index entries for linear recurrences with constant coefficients, signature (24,-24,1).

Crossrefs

Cf. A004253, A004254, A030221, A089927, A160682, A245031, A334673.

Programs

Maple

j:=0: for n from 0 to 1000000 do a:=sqrt(3*n+1): b:=sqrt(7*n+1):
if (trunc(a)=a) and (trunc(b)=b) then j:=j+1: print(j,n,a,b): end if:
end do:

Mathematica

LinearRecurrence[{24,-24,1},{0,5,120},30] (* Harvey P. Dale, Dec 17 2013 *)

Formula

a(n) = 24*a(n-1) - 24*a(n-2) + a(n-3).

a(n) = (A004253(n)^2 - 1)/3 = (A030221(n)^2 - 1)/7.

a(n) = ((5+w)/2*((23+5*w)/2)^(n-1) + (5-w)/2*((23-5*w)/2)^(n-1) - 5)/21; where w=sqrt(21). [Corrected by Kevin Ryde, Sep 11 2020]

G.f.: 5*x^2/((1-x)*(x^2-23*x+1)). - R. J. Mathar, Jul 10 2009

From Francesca Arici, Sep 12 2020: (Start)

a(n) = 23*a(n-1) - a(n-2) + 5.

a(n) = A004254(n)* A004254(n+1). (End)

a(n) = 5*A334673(n-1). - Hugo Pfoertner, Apr 07 2021

Extensions

Edited and extended by R. J. Mathar, Jul 10 2009

Name edited by Michel Marcus, Sep 12 2020