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%I #177 Jul 08 2023 02:37:25
%S 1,2,9,50,289,1682,9801,57122,332929,1940450,11309769,65918162,
%T 384199201,2239277042,13051463049,76069501250,443365544449,
%U 2584123765442,15061377048201,87784138523762,511643454094369,2982076586042450,17380816062160329,101302819786919522
%N Numbers n such that n(n - 1)/2 is a square.
%C Numbers n such that (n-th triangular number - n) is a square.
%C Gives solutions to A007913(2x)=A007913(x-1). - _Benoit Cloitre_, Apr 07 2002
%C Number of closed walks of length 2n on the grid graph P_2 X P_3. - _Mitch Harris_, Mar 06 2004
%C If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - _Bruno Berselli_, Aug 24 2010
%C The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - _Bradley Klee_ & _Bill Gosper_, Jul 22 2015
%C It appears that dividing even terms by two and taking the square root gives sequence A079496. - _Bradley Klee_, Jul 25 2015
%C The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - _Bernard Schott_, Apr 19 2020
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
%D P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.
%H Colin Barker, <a href="/A055997/b055997.txt">Table of n, a(n) for n = 1..100</a>
%H Dario Alpern, <a href="https://www.alpertron.com.ar/SUMPOWER.HTM#4_3_2">a^4+b^3=c^2</a>.
%H Phil Lafer, <a href="https://www.fq.math.ca/Scanned/9-1/lafer.pdf">Discovering the square-triangular numbers</a>, Fib. Quart., 9 (1971), 93-105.
%H Giovanni Lucca, <a href="https://forumgeom.fau.edu/FG2018volume18/FG201808index.html">Integer Sequences and Circle Chains Inside a Circular Segment</a>, Forum Geometricorum, Vol. 18 (2018), 47-55.
%H Kenneth Ramsey, <a href="/A055997/a055997.txt">Generalized Proof re Square Triangular Numbers</a>, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
%F G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
%F a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - _Charlie Marion_, Jul 21 2003; corrected by _Michel Marcus_, Apr 20 2020
%F a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - _Antonio Alberto Olivares_, Jan 31 2004; corrected by _Bernard Schott_, Apr 20 2020
%F A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - _Creighton Dement_, Sep 16 2004
%F a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - _Antonio Alberto Olivares_, Feb 21 2006; corrected by _Michel Marcus_, Apr 20 2020
%F a(n) = A001653(n)-A001652(n-1). - _Charlie Marion_, Apr 10 2006; corrected by _Michel Marcus_, Apr 20 2020
%F a(2k) = A001541(k)^2. - _Alexander Adamchuk_, Nov 24 2006
%F a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with m<n; otherwise, a(n) = 2*A001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - _Kenneth J Ramsey_, Oct 13 2011
%F a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - _Joerg Arndt_, Mar 06 2013
%F a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - _Robert Israel_, Jul 23 2015
%F sqrt(a(n+1)*a(n-1)) = a(n)+1 - _Bradley Klee_ & _Bill Gosper_, Jul 25 2015
%F a(n) = 1 + sum{k=0..n-2} A002315(k). - _David Pasino_, Jul 09 2016; corrected by _Michel Marcus_, Apr 20 2020
%F E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - _Ilya Gutkovskiy_, Jul 09 2016
%F sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - _David Pasino_, Jul 09 2016
%F Limit_{n -> infinity} a(n)/a(n-1) = A156035. - _César Aguilera_, Apr 07 2018
%F a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - _Ridouane Oudra_, Nov 29 2019
%F sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - _Ridouane Oudra_, Nov 29 2019
%F sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - _Bernard Schott_, Apr 18 2020
%p A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
%p map(A,[$1..100]); # _Robert Israel_, Jul 22 2015
%t Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* _Jean-François Alcover_, Mar 06 2013 *)
%t CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Mar 20 2015 *)
%t (1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* _Bill Gosper_, Jul 20 2015 *)
%t a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* _Bradley Klee_, Jul 25 2015 *)
%t LinearRecurrence[{7,-7,1},{1,2,9},30] (* _Harvey P. Dale_, Dec 06 2015 *)
%o (PARI) Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* _Joerg Arndt_, Mar 06 2013 */
%o (PARI) t(n)=(1+sqrt(2))^(n-1);
%o for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ _Hugo Pfoertner_, Nov 29 2019
%o (PARI) a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ _Michel Marcus_, Apr 21 2020
%o (Magma) I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Mar 20 2015
%Y Cf. A007913, A001541.
%Y A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
%Y a(n) = A001108(n-1)+1.
%Y A001110(n-1)=a(n)*(a(n)-1)/2.
%Y Cf. A001652, A001653, A046090.
%Y Identical to A115599, but with additional leading term.
%K easy,nice,nonn
%O 1,2
%A _Barry E. Williams_, Jun 14 2000
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