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 A055997 Numbers n such that n(n-1)/2 is a square. 15
 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 sqrt gives sequence A079496. - Bradley Klee, Jul 25 2015 REFERENCES A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193. 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. 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); e.g., 50-1+(2*50*49)^0.5 = 119. - Charlie Marion, Jul 21 2003 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 even-indexed terms are a(n) = [A001333(n)]^2; the odd-indexed terms are a(n) = 2*[ [A001333(n+1) - A001333(n)]/4 ]^2 = 2*[ [A001333(n+1) - A001333(n)]/4 ]^2 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004 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 + (1/4)*(3-2*sqrt(2))^n. - Antonio Alberto Olivares, Feb 21 2006 a(n) = A001653(n)-A001652(n); e.g., 50=169-119. - Charlie Marion, Apr 10 2006 a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with m infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018 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 */ (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 A274066 * A115599 A047069 A225006 A211789 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 August 19 05:13 EDT 2018. Contains 313844 sequences. (Running on oeis4.)