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A005563 a(n) = n*(n+2) = (n+1)^2 - 1.
(Formerly M2720)
247

%I M2720

%S 0,3,8,15,24,35,48,63,80,99,120,143,168,195,224,255,288,323,360,399,

%T 440,483,528,575,624,675,728,783,840,899,960,1023,1088,1155,1224,1295,

%U 1368,1443,1520,1599,1680,1763,1848,1935,2024,2115,2208,2303,2400,2499,2600

%N a(n) = n*(n+2) = (n+1)^2 - 1.

%C Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).

%C Second-order linear recurrences y(m) = 2y(m-1) + A005563(n)y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - _Len Smiley_, Dec 08 2001

%C Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - _Roberto E. Martinez II_, Jan 07 2002

%C Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - _Benoit Cloitre_, May 28 2002

%C Also numbers n such that 4n + 4 is a square. - _Cino Hilliard_, Dec 18 2003

%C The function sqrt(x^2 + 1), starting with 1, produces an integer after n(n+2) iterations. - _Gerald McGarvey_, Aug 19 2004

%C a(n) mod 3 = 0 if and only if n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - _Reinhard Zumkeller_, Oct 16 2006

%C A129296(n) = number of divisors of a(n+1) that are not greater than n. - _Reinhard Zumkeller_, Apr 09 2007

%C Nonnegative X values of solutions to the equation X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 06 2007

%C Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007

%C From _R. K. Guy_, Feb 01 2008: (Start)

%C Toads and Frogs puzzle:

%C This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by

%C T T - F F

%C T - T F F

%C T F T - F

%C T F T F -

%C T F - F T

%C - F T F T

%C F - T F T

%C F F T - T

%C F F - T T

%C I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.

%C Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)

%C For n > 0: A143053(a(n)) = A000290(n+1). - _Reinhard Zumkeller_, Jul 20 2008

%C a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - _Paul Curtz_, Oct 28 2008

%C A053186(a(n)) = 2*n. - _Reinhard Zumkeller_, May 20 2009

%C Row 3 of array A163280, n >= 1. - _Omar E. Pol_, Aug 08 2009

%C Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009 [Comment edited by _N. J. A. Sloane_, Sep 24 2009]

%C Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod(f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - _A.K. Devaraj_, Sep 18 2009

%C For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). - _Rick L. Shepherd_, Sep 27 2009

%C For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - _Gary W. Adamson_, Jul 15 2010

%C Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g. a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - _Gary W. Adamson_, Jul 30 2010

%C a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - _Peter Luschny_, Jul 08 2011

%C a(n) = A002378(n) + floor(sqrt(A002378(n))); Pronic number + its root. - _Fred Daniel Kline_, Sep 16 2011

%C a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - _Aldo González Lorenzo_, Oct 12 2011

%C a(n-1) = A008833(n) * A068310(n) for n > 1. - _Reinhard Zumkeller_, Nov 26 2011

%C Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - _J. M. Bergot_, May 03 2012

%C Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to <S^2> = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - _Stanislav Sykora_, May 26 2012

%C Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - _Raphie Frank_, Sep 28 2012

%C Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - _Takumi Sato_, Oct 10 2012

%C The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427.. = A156035. - _Kival Ngaokrajang_, Sep 07 2013

%C The integers in the closed form solution of a(n) = 2*a(n-1) + A005563(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08, 2001, are m and -m + 2 where m >= 3 is a positive integer. - _Felix P. Muga II_, Mar 18 2014

%C Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + A005563(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim a(n+1)/a(n) = m as n approaches infinity. - _Felix P. Muga II_, Mar 18 2014

%C For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - _Emeric Deutsch_, Aug 07 2014

%C If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - _Bruno Berselli_, Sep 04 2014

%C A253607(a(n)) = 1. - _Reinhard Zumkeller_, Jan 05 2015

%C For n >= 1, a(n) is the dimension of the simple Lie Algebra A_n. - _Wolfdieter Lang_, Oct 21 2015

%C Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - _David Covert_, Jan 15 2016

%C n > 0, a(n) % (n+1) = a(n) / (n+1) = n - _Torlach Rush_, Apr 04 2016

%C Conjecture: When using the Sieve of Eratosthenes and sieving (n+1...a(n)), with divisors (1...n) and n>0, there will be no more than a(n-1) composite numbers. - _Fred Daniel Kline_, Apr 08 2016

%C a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - _Andres Cicuttin_, Jun 02 2016

%C Also for n>0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - _R. J. Cano_, Dec 21 2016

%C The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kusniec link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - _Charles Kusniec_, Jul 03 2017

%C Also the number of independent vertex sets in the n-barbell graph. - _Eric W. Weisstein_, Aug 16 2017

%D E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see index under Toads and Frogs Puzzle.

%D Martin Gardner, Perplexing Puzzles and Tantalizing Teasers, page 21 (for "The Dime and Penny Switcheroo")

%D R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.

%D Derek Holton, Math in School, 37 #1 (Jan 2008) 20-22.

%D Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585

%D Edouard Lucas, Récréations Mathématiques, Gauthier-Villars, Vol. 2 (1883) 141-143.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H T. D. Noe, <a href="/A005563/b005563.txt">Table of n, a(n) for n = 0..1000</a>

%H Jeremiah Bartz, Bruce Dearden, Joel Iiams, <a href="https://arxiv.org/1810.07895">Classes of Gap Balancing Numbers</a>, arXiv:1810.07895 [math.NT], 2018.

%H R. K. Guy, <a href="/A005555/a005555.pdf">Letter to N. J. A. Sloane, May 1990</a>

%H R. K. Guy, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/GUY/catwalks.html">Catwalks, sandsteps and Pascal pyramids</a>, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.

%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>

%H Charles Kusniec, <a href="https://oeis.org/A000027/a000027_3.pdf">Klauber triangle</a>

%H F. P. Muga II, <a href="https://www.researchgate.net/publication/267327689">Extending the Golden Ratio and the Binet-de Moivre Formula</a>, March 2014; Preprint on ResearchGate.

%H Simon Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/MasterThesis.pdf">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992.

%H Simon Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf">1031 Generating Functions and Conjectures</a>, Université du Québec à Montréal, 1992.

%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014.

%H K. V. S. Sarma, I. V. N. Uma, <a href="http://www.ijma.info/index.php/ijma/article/view/1540">On Szeged index of standard graphs</a>, International J. of Math. Archive, 3(8), 2012, 3129-3135. - _Emeric Deutsch_, Aug 07 2014

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BarbellGraph.html">Barbell Graph</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/IndependentVertexSet.html">Independent Vertex Set</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Near-SquarePrime.html">Near-Square Prime</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: x*(3-x)/(1-x)^3. - _Simon Plouffe_ in his 1992 dissertation

%F A002378(a(n)) = A002378(n)*A002378(n+1); e.g., A002378(15)=240=12*20. - _Charlie Marion_, Dec 29 2003

%F a(n) = A067725(n)/3. - _Zerinvary Lajos_, Mar 06 2007

%F a(n) = sum_{k=1..n} A144396(k). - _Zerinvary Lajos_, May 11 2007

%F a(n) = A134582(n+1)/4. - _Zerinvary Lajos_, Feb 01 2008

%F a(n) = Real((n+1+i)^2). - _Gerald Hillier_, Oct 12 2008

%F a(n) = (n!+(n+1)!)/(n-1)!..n>0. - _Gary Detlefs_, Aug 10 2009

%F a(n) = floor(n^5/(n^3+1)) with offset 1..a(1)=0. - _Gary Detlefs_, Feb 11 2010

%F a(n) = a(n-1) + 2*n+1 (with a(0)=0). - _Vincenzo Librandi_, Nov 18 2010

%F Sum_{n>=1} 1/a(n) = 3/4. - _Mohammad K. Azarian_, Dec 29 2010

%F For n>0 a(n)=2/(Integral_{x=0..Pi/2} (sin(x))^(n-1)*(cos(x))^3). - _Francesco Daddi_, Aug 02 2011

%F G.f.: U(0) where U(k)= -1 + (k+1)^2/(1 - x/(x + (k+1)^2/U(k+1))) ; (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Oct 19 2012

%F a(n) = 15*C(n+4,3)*C(n+4,5)/(C(n+4,2)*C(n+4,4)). - _Gary Detlefs_, Aug 05 2013

%F a(n) = (n+2)!/((n-1)!+n!), n>0. - _Ivan N. Ianakiev_, Nov 11 2013

%F a(n) = 3*C(n+1,2)-C(n,2) for n >=0. - _Felix P. Muga II_, Mar 11 2014

%F a(n) = ( A016742(n+1) - 4)/4 for n>=0. - _Felix P. Muga II_, Mar 11 2014

%F a(-2 - n) = a(n) for all n in Z. - _Michael Somos_, Aug 07 2014

%F E.g.f.: x*(x + 3)*exp(x). - _Ilya Gutkovskiy_, Jun 03 2016

%F For n >= 1, a(n^2+n-2) = a(n-1) * a(n). - _Miko Labalan_, Oct 15 2017

%e G.f. = 3*x + 8*x^2 + 15*x^3 + 24*x^4 + 35*x^5 + 48*x^6 + 63*x^7 + 80*x^8 + ...

%t Table[n^2 - 1, {n, 42}] (* _Zerinvary Lajos_, Mar 21 2007 *)

%t ListCorrelate[{1, 2}, Range[-1, 50], {1, -1}, 0, Plus, Times] (* _Harvey P. Dale_, Aug 29 2015 *)

%t Range[20]^2 - 1 (* _Eric W. Weisstein_, Aug 16 2017 *)

%t LinearRecurrence[{3, -3, 1}, {3, 8, 15}, {0, 20}] (* _Eric W. Weisstein_, Aug 16 2017 *)

%t CoefficientList[Series[((-3 + x) x)/(-1 + x)^3, {x, 0, 20}], x] (* _Eric W. Weisstein_, Aug 16 2017 *)

%o (PARI) a(n)=n*(n+2) \\ _Charles R Greathouse IV_, Dec 22 2011

%o (PARI) concat(0, Vec(x*(3-x)/(1-x)^3 + O(x^90))) \\ _Altug Alkan_, Oct 22 2015

%o (Maxima) makelist(n*(n+2), n, 0, 56); /* _Martin Ettl_, Oct 15 2012 */

%o (Haskell)

%o a005563 n = n * (n + 2)

%o a005563_list = zipWith (*) [0..] [2..] -- _Reinhard Zumkeller_, Dec 16 2012

%Y A column of triangle A102537.

%Y a(n+1), n>=2, first column of triangle A120070.

%Y Cf. A013468, A007531, A062196, A002378, A000290(n) = a(n-1) + 1.

%Y Cf. A046092, A067725, A123865, A123866, A123867, A123868, A028560, A000217, A067728, A140091, A140681, A212331, A069817, A253607.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E Partially edited by _Joerg Arndt_, Mar 11 2010

%E More terms from _N. J. A. Sloane_, Aug 01 2010

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Last modified February 22 10:06 EST 2019. Contains 320390 sequences. (Running on oeis4.)