A055997 Numbers k such that k*(k - 1)/2 is a square.

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

Offset: 1

Author

Barry E. Williams, Jun 14 2000

Keywords

• easy
• nice
• nonn

Comments

Numbers k such that (k-th triangular number - k) is a square.

Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002

Number of closed walks of length 2k 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.
• Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 46, 56.
• 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).

Crossrefs

Cf. A007913, A001541.

A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.

a(n) = A001108(n-1)+1.

A001110(n-1)=a(n)*(a(n)-1)/2.

Cf. A001652, A001653, A046090.

Identical to A115599, but with additional leading term.

Programs

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

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 *)

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

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

A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004

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 mA001653(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

a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015

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