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A045502 Numbers k such that 2*k+1 and 3*k+1 are squares.

%I A045502 #36 Dec 02 2023 13:50:46
%S A045502 0,40,3960,388080,38027920,3726348120,365144087880,35780394264160,
%T A045502 3506113493799840,343563341998120200,33665701402321979800,
%U A045502 3298895174085555900240,323258061358982156243760,31675991118006165755988280,3103923871503245261930607720
%N A045502 Numbers k such that 2*k+1 and 3*k+1 are squares.
%C A045502 Problem 1 for the 3rd grade of the 38th Mathematics Competition of the Republic of Slovenia (1998) was to prove that if k is a natural number such that 2*k+1 and 3*k+1 are perfect squares, then k is divisible by 40 (see link with solution Crux Mathematicorum and formula Mar 25 2021). - _Bernard Schott_, Mar 25 2021
%H A045502 Colin Barker, <a href="/A045502/b045502.txt">Table of n, a(n) for n = 0..500</a>.
%H A045502 John Albert, <a href="http://www2.math.ou.edu/~jalbert/putnam/pell.pdf">Pell Equations</a>, Putnam practice, November 17, 2004 (1-2).
%H A045502 R. S. Luthar, <a href="https://www.jstor.org/stable/2322070">Problem E2606</a>, Amer. Math. Monthly, 84 (1977), 823-824.
%H A045502 R. E. Woodrow, <a href="https://cms.math.ca/crux/v25/n4/page196-211.pdf">Problem 1 for the third grade of the 38th Mathematics competition of the Republic of Slovenia (1998)</a>, Crux Mathematicorum, The Olympiad Corner, p. 208, April 1999, Vol. 25, No. 4.
%H A045502 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (99,-99,1).
%H A045502 <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical Competitions</a>.
%F A045502 From _Colin Barker_, Mar 23 2017: (Start)
%F A045502 O.g.f.: 40*x / ((1 - x)*(1 - 98*x + x^2)).
%F A045502 a(n) = 99*a(n-1)- 99*a(n-2) + a(n-3) for n>2.
%F A045502 a(n) = (-10 + (5 - 2*sqrt(6))*(49 + 20*sqrt(6))^(-n) + (5 + 2*sqrt(6))*(49 + 20*sqrt(6))^n)/24. (End)
%F A045502 From _G. C. Greubel_, Jan 13 2020: (Start)
%F A045502 a(n) = 5*(ChebyshevT(n, 49) + 48*ChebyshevU(n-1, 48) - 1)/12.
%F A045502 a(n) = 4*ChebyshevU(n-1, 5)*ChebyshevU(n, 5). (End)
%F A045502 a(n) = 40*A278620(n). - _Bernard Schott_, Mar 25 2021
%p A045502 seq(coeff(series(40*x/((1-x)*(x^2-98*x+1)), x,n+1),x,n),n=0..15); # _Muniru A Asiru_, Jul 17 2018
%t A045502 f[0]=0; f[1]=2; f[n_]:= f[n]= 10*f[n-1] -f[n-2]; a[n_]:= f[n]*f[n+1];
%t A045502 CoefficientList[Series[40x/((1-x)(1-98x+x^2)), {x,0,15}], x] (* _Michael De Vlieger_, Jul 20 2018 *)
%t A045502 Table[5*(ChebyshevT[n, 49] +48*ChebyshevU[n-1, 49] -1)/12, {n,0,15}] (* _G. C. Greubel_, Jan 13 2020 *)
%t A045502 LinearRecurrence[{99,-99,1},{0,40,3960},20] (* _Harvey P. Dale_, Dec 02 2023 *)
%o A045502 (PARI) concat(0, Vec(40*x/((1-x)*(1-98*x+x^2))+O(x^20))) \\ _Colin Barker_, Mar 23 2017
%o A045502 (GAP) a:=[0,40,3960];; for n in [4..15] do a[n]:=99*a[n-1]-99*a[n-2]+a[n-3]; od; a; # _Muniru A Asiru_, Jul 17 2018
%o A045502 (Magma) I:=[0,40,3960]; [n le 3 select I[n] else 99*Self(n-1) -99*Self(n-2) + Self(n-3): n in [1..15]]; // _G. C. Greubel_, Jan 13 2020
%o A045502 (Sage) [4*chebyshev_U(n-1,5)*chebyshev_U(n,5) for n in (0..15)] # _G. C. Greubel_, Jan 13 2020
%Y A045502 Cf. A001078, A245031, A278620.
%K A045502 nonn,easy
%O A045502 0,2
%A A045502 Fred Schwab (fschwab(AT)nrao.edu)