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A055997 Numbers n such that n(n - 1)/2 is a square. 16
1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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

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

Number of closed walks of length 2n on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004

a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006

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.

P. 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

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

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(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

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

Lim_{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

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.

Sequence in context: A138416 A328741 A274066 * A115599 A047069 A225006

Adjacent sequences:  A055994 A055995 A055996 * A055998 A055999 A056000

KEYWORD

easy,nice,nonn

AUTHOR

Barry E. Williams, Jun 14 2000

STATUS

approved

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Last modified July 6 05:26 EDT 2020. Contains 335475 sequences. (Running on oeis4.)