

A005563


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


246



0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600
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OFFSET

0,2


COMMENTS

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).
Secondorder linear recurrences y(m) = 2y(m1) + A005563(n)y(m2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers.  Len Smiley, Dec 08 2001
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
Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(ij) then det(M_n) = (1)^(n1)*a(k1)^(n1).  Benoit Cloitre, May 28 2002
Also numbers n such that 4n + 4 is a square.  Cino Hilliard, Dec 18 2003
The function sqrt(x^2 + 1), starting with 1, produces an integer after n(n+2) iterations.  Gerald McGarvey, Aug 19 2004
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*(1A079978(n)).  Reinhard Zumkeller, Oct 16 2006
A129296(n) = number of divisors of a(n+1) that are not greater than n.  Reinhard Zumkeller, Apr 09 2007
Sequence allows us to find X values of 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
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
From R. K. Guy, Feb 01 2008: (Start)
Toads and Frogs puzzle:
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
T T  F F
T  T F F
T F T  F
T F T F 
T F  F T
 F T F T
F  T F T
F F T  T
F F  T T
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.
Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)
For n > 0: A143053(a(n)) = A000290(n+1).  Reinhard Zumkeller, Jul 20 2008
a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15).  Paul Curtz, Oct 28 2008
A053186(a(n)) = 2*n.  Reinhard Zumkeller, May 20 2009
Row 3 of array A163280, n >= 1.  Omar E. Pol, Aug 08 2009
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]
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 noninteger 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
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
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
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
a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k2)*i(k3)). Thus P_0(n) = 2*nn^2 and a(n) = P_0(n+2). See also A067998 and for the case k=1 A080956.  Peter Luschny, Jul 08 2011
a(n) = A002378(n) + floor(sqrt(A002378(n))); Pronic number + its root.  Fred Daniel Kline, Sep 16 2011
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
a(n1) = A008833(n) * A068310(n) for n > 1.  Reinhard Zumkeller, Nov 26 2011
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
Given a particle with spin S = n/2 (always a halfinteger value), the quantummechanical 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
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
Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m)))  2 for m > 0.  Takumi Sato, Oct 10 2012
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
The integers in the closed form solution of a(n) = 2*a(n1) + A005563(m2)*a(n2), 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
Let m >= 3 be a positive integer. If a(n) = 2*a(n1) + A005563(m2) * a(n2), 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
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
If P_{k}(n) is the nth kgonal number, then a(n) = t*P_{s}(n+2)  s*P_{t}(n+2) for s=t+1.  Bruno Berselli, Sep 04 2014
A253607(a(n)) = 1.  Reinhard Zumkeller, Jan 05 2015
For n >= 1, a(n) is the dimension of the simple Lie Algebra A_n.  Wolfdieter Lang, Oct 21 2015
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
n > 0, a(n) % (n+1) = a(n) / (n+1) = n  Torlach Rush, Apr 04 2016
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(n1) composite numbers.  Fred Daniel Kline, Apr 08 2016
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
Also for n>0, a(n) is the number of times that n1 occurs among the first (n+1)! terms of A055881.  R. J. Cano, Dec 21 2016
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
Also the number of independent vertex sets in the nbarbell graph.  Eric W. Weisstein, Aug 16 2017


REFERENCES

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.
Martin Gardner, Perplexing Puzzles and Tantalizing Teasers, page 21 (for "The Dime and Penny Switcheroo")
R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
Derek Holton, Math in School, 37 #1 (Jan 2008) 2022.
HsienKuei 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/wpcontent/files/2016/12/aathhrr1.pdf. Also Exact and Asymptotic Solutions of a DivideandConquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
Edouard Lucas, Récréations Mathématiques, GauthierVillars, Vol. 2 (1883) 141143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n = 0..1000
Jeremiah Bartz, Bruce Dearden, Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
R. K. Guy, Letter to N. J. A. Sloane, May 1990
R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Charles Kusniec, Klauber triangle
F. P. Muga II, Extending the Golden Ratio and the Binetde Moivre Formula, March 2014; Preprint on ResearchGate.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
K. V. S. Sarma, I. V. N. Uma, On Szeged index of standard graphs, International J. of Math. Archive, 3(8), 2012, 31293135.  Emeric Deutsch, Aug 07 2014
Eric Weisstein's World of Mathematics, Barbell Graph
Eric Weisstein's World of Mathematics, Independent Vertex Set
Eric Weisstein's World of Mathematics, NearSquare Prime
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

G.f.: x*(3x)/(1x)^3.  Simon Plouffe in his 1992 dissertation
A002378(a(n)) = A002378(n)*A002378(n+1); e.g., A002378(15)=240=12*20.  Charlie Marion, Dec 29 2003
a(n) = A067725(n)/3.  Zerinvary Lajos, Mar 06 2007
a(n) = sum_{k=1..n} A144396(k).  Zerinvary Lajos, May 11 2007
a(n) = A134582(n+1)/4.  Zerinvary Lajos, Feb 01 2008
a(n) = Real((n+1+i)^2).  Gerald Hillier, Oct 12 2008
a(n) = (n!+(n+1)!)/(n1)!..n>0.  Gary Detlefs, Aug 10 2009
a(n) = floor(n^5/(n^3+1)) with offset 1..a(1)=0.  Gary Detlefs, Feb 11 2010
a(n) = a(n1) + 2*n+1 (with a(0)=0).  Vincenzo Librandi, Nov 18 2010
Sum_{n>=1} 1/a(n) = 3/4.  Mohammad K. Azarian, Dec 29 2010
For n>0 a(n)=2/(Integral_{x=0..Pi/2} (sin(x))^(n1)*(cos(x))^3).  Francesco Daddi, Aug 02 2011
G.f.: U(0) where U(k)= 1 + (k+1)^2/(1  x/(x + (k+1)^2/U(k+1))) ; (continued fraction, 3step).  Sergei N. Gladkovskii, Oct 19 2012
a(n) = 15*C(n+4,3)*C(n+4,5)/(C(n+4,2)*C(n+4,4)).  Gary Detlefs, Aug 05 2013
a(n) = (n+2)!/((n1)!+n!), n>0.  Ivan N. Ianakiev, Nov 11 2013
a(n) = 3*C(n+1,2)C(n,2) for n >=0.  Felix P. Muga II, Mar 11 2014
a(n) = ( A016742(n+1)  4)/4 for n>=0.  Felix P. Muga II, Mar 11 2014
a(2  n) = a(n) for all n in Z.  Michael Somos, Aug 07 2014
E.g.f.: x*(x + 3)*exp(x).  Ilya Gutkovskiy, Jun 03 2016
For n >= 1, a(n^2+n2) = a(n1) * a(n).  Miko Labalan, Oct 15 2017


EXAMPLE

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 + ...


MATHEMATICA

Table[n^2  1, {n, 42}] (* Zerinvary Lajos, Mar 21 2007 *)
ListCorrelate[{1, 2}, Range[1, 50], {1, 1}, 0, Plus, Times] (* Harvey P. Dale, Aug 29 2015 *)
Range[20]^2  1 (* Eric W. Weisstein, Aug 16 2017 *)
LinearRecurrence[{3, 3, 1}, {3, 8, 15}, {0, 20}] (* Eric W. Weisstein, Aug 16 2017 *)
CoefficientList[Series[((3 + x) x)/(1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Aug 16 2017 *)


PROG

(PARI) a(n)=n*(n+2) \\ Charles R Greathouse IV, Dec 22 2011
(PARI) concat(0, Vec(x*(3x)/(1x)^3 + O(x^90))) \\ Altug Alkan, Oct 22 2015
(Maxima) makelist(n*(n+2), n, 0, 56); /* Martin Ettl, Oct 15 2012 */
(Haskell)
a005563 n = n * (n + 2)
a005563_list = zipWith (*) [0..] [2..]  Reinhard Zumkeller, Dec 16 2012


CROSSREFS

A column of triangle A102537.
a(n+1), n>=2, first column of triangle A120070.
Cf. A013468, A007531, A062196, A002378, A000290(n) = a(n1) + 1.
Cf. A046092, A067725, A123865, A123866, A123867, A123868, A028560, A000217, A067728, A140091, A140681, A212331, A069817, A253607.
Sequence in context: A258837 A131386 A132411 * A067998 A066079 A185079
Adjacent sequences: A005560 A005561 A005562 * A005564 A005565 A005566


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

Partially edited by Joerg Arndt, Mar 11 2010
More terms from N. J. A. Sloane, Aug 01 2010


STATUS

approved



