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Integers m such that 3*m+1 and 7*m+1 are both perfect squares.
4

%I #38 Apr 08 2021 03:24:07

%S 0,5,120,2760,63365,1454640,33393360,766592645,17598237480,

%T 403992869400,9274237758725,212903475581280,4887505700610720,

%U 112199727638465285,2575706229984090840,59129043561995624040,1357392295695915262085,31160893757444055403920

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

%C 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.

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

%C Bisection of A089927. - _R. J. Mathar_, Jul 10 2009

%H Michael De Vlieger, <a href="/A160695/b160695.txt">Table of n, a(n) for n = 1..736</a>

%H Francesca Arici and Jens Kaad, <a href="https://arxiv.org/abs/2012.11186">Gysin sequences and SU(2)-symmetries of C*-algebras</a>, arXiv:2012.11186 [math.OA], 2020.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (24,-24,1).

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

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

%F 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]

%F G.f.: 5*x^2/((1-x)*(x^2-23*x+1)). - _R. J. Mathar_, Jul 10 2009

%F From _Francesca Arici_, Sep 12 2020: (Start)

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

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

%F a(n) = 5*A334673(n-1). - _Hugo Pfoertner_, Apr 07 2021

%p j:=0: for n from 0 to 1000000 do a:=sqrt(3*n+1): b:=sqrt(7*n+1):

%p if (trunc(a)=a) and (trunc(b)=b) then j:=j+1: print(j,n,a,b): end if:

%p end do:

%t LinearRecurrence[{24,-24,1},{0,5,120},30] (* _Harvey P. Dale_, Dec 17 2013 *)

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

%K nonn,easy

%O 1,2

%A _Paul Weisenhorn_, May 24 2009

%E Edited and extended by _R. J. Mathar_, Jul 10 2009

%E Name edited by _Michel Marcus_, Sep 12 2020