OFFSET
1,2
COMMENTS
Numbers n such that (n-th triangular number - n) is a square.
Number of closed walks of length 2n on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
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
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
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.
LINKS
Colin Barker, Table of n, a(n) for n = 1..100
Dario Alpern, a^4+b^3=c^2.
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
FORMULA
a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
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
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
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
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
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
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020
MAPLE
A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
map(A, [$1..100]); # Robert Israel, Jul 22 2015
MATHEMATICA
Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
(1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
a[1]=1; a[2]=2; a[n_]:=(a[n-1]+1)^2/a[n-2]; a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
LinearRecurrence[{7, -7, 1}, {1, 2, 9}, 30] (* Harvey P. Dale, Dec 06 2015 *)
PROG
(PARI) Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */
(PARI) t(n)=(1+sqrt(2))^(n-1);
for(k=1, 24, print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)), ", ")) \\ Hugo Pfoertner, Nov 29 2019
(PARI) a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020
(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
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
Barry E. Williams, Jun 14 2000
STATUS
approved