A028560 -id:A028560 - cp's OEIS frontend

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

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

Offset: 0

N. J. A. Sloane

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).
Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), 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(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002
Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003
For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k 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*(1-A079978(n)). - Reinhard Zumkeller, Oct 16 2006
a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007
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, 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, 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)
a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - Paul Curtz, Oct 28 2008
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, 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 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
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} ((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
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
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 half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to  = 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(n-1) + a(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
Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - 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 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
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
For n > 0, a(n) mod (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(n-1) 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 n-1 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 Kival Ngaokrajang 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 n-barbell graph. - Eric W. Weisstein, Aug 16 2017
Interleaving of A000466 and A033996. - Bruce J. Nicholson, Nov 08 2019
a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020
From Muge Olucoglu, Jan 19 2021: (Start)
For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).
.
          0 1 2   5 6 7
        +---------------
  *(0,1)| X X
   (0,2)| X   X
   (1,5)|   X     X
  *(2,6)|     X     X
  *(5,7)|         X   X
   (6,7)|           X X
.
*: represents the elements that are covered. (End)
1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022
Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023
For n >= 1, also the number of minimum cyclic edge cuts in the (n+2)-trapezohedron graph. - Eric W. Weisstein, Nov 21 2024
For n >= 1, a(n) is the sum of the interior angles of a polygon with n+2 sides, in radians, multiplied by (n+2)/Pi. - Stuart E Anderson, Aug 06 2025

			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 + ...

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, p. 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) 20-22.
Edouard Lucas, Récréations Mathématiques, Gauthier-Villars, Vol. 2 (1883) 141-143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024. See p. 22.
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.
Hsien-Kuei Hwang, S. Janson, and T.-H. 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.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, 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.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
Kival Ngaokrajang, Klauber triangle
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
K. V. S. Sarma and I. V. N. Uma, On Szeged index of standard graphs, International J. of Math. Archive, 3(8), 2012, 3129-3135. - _Emeric Deutsch_, Aug 07 2014
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Leo Tavares, Illustration: Bounded Squares.
Leo Tavares, Illustration: Trapezoids.
Leo Tavares, Illustration: Trapagons.
Eric Weisstein's World of Mathematics, Barbell Graph.
Eric Weisstein's World of Mathematics, Cyclic Edge Cut.
Eric Weisstein's World of Mathematics, Independent Vertex Set.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Eric Weisstein's World of Mathematics, Trapezohedral Graph.
Wikipedia, Quine-McCluskey Algorithm.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

A column of triangle A102537.
a(n+1), n>=2, first column of triangle A120070.
Cf. A000035, A000217, A000290, A000466, A001651, A002378, A005563.
Cf. A007531, A008585, A008833, A013468, A016742, A019973, A028560.
Cf. A033996, A046092, A053186, A054000, A055881, A056108, A062196.
Cf. A068310, A067725, A067728, A067998, A069817, A079978, A080956.
Cf. A085692, A120072, A123865, A123866, A123867, A123868, A134582.
Cf. A140091, A140681, A143053, A144396, A156035, A163280, A212331.
Cf. A253607, A367204.

a005563 n = n * (n + 2)
a005563_list = zipWith (*) [0..] [2..]  -- Reinhard Zumkeller, Dec 16 2012

[n*(n+2): n in [0..60]]; // G. C. Greubel, Mar 29 2024

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 *)
Table[n (n + 2), {n, 20}] (* Eric W. Weisstein, Nov 21 2024 *)
CoefficientList[Series[(-3 + x)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Nov 21 2024 *)
LinearRecurrence[{3, -3, 1}, {3, 8, 15}, 20] (* Eric W. Weisstein, Nov 21 2024 *)

makelist(n*(n+2), n, 0, 56); /* Martin Ettl, Oct 15 2012 */

a(n)=n*(n+2) \\ Charles R Greathouse IV, Dec 22 2011

concat(0, Vec(x*(3-x)/(1-x)^3 + O(x^90))) \\ Altug Alkan, Oct 22 2015

[n*(n+2) for n in range(61)] # G. C. Greubel, Mar 29 2024

G.f.: x*(3-x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = A000290(n+1) - 1.
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
A143053(a(n)) = A000290(n+1), for n > 0. - Reinhard Zumkeller, Jul 20 2008
a(n) = Real((n+1+i)^2). - Gerald Hillier, Oct 12 2008
A053186(a(n)) = 2*n. - Reinhard Zumkeller, May 20 2009
a(n) = (n! + (n+1)!)/(n-1)!, 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(n-1) + 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
a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^(n-1)*(cos(x))^3), for n > 0. - Francesco Daddi, Aug 02 2011
a(n) = A002378(n) + floor(sqrt(A002378(n))); pronic number + its root. - Fred Daniel Kline, Sep 16 2011
a(n-1) = A008833(n) * A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011
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
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)!/((n-1)! + 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
A253607(a(n)) = 1. - Reinhard Zumkeller, Jan 05 2015
E.g.f.: x*(x + 3)*exp(x). - Ilya Gutkovskiy, Jun 03 2016
For n >= 1, a(n^2 + n - 2) = a(n-1) * a(n). - Miko Labalan, Oct 15 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/4. - Amiram Eldar, Nov 04 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 2.
Product_{n>=1} (1 - 1/a(n)) = -sqrt(2)*sin(sqrt(2)*Pi)/Pi. (End)
a(n) = A000290(n+2) - n*2. See Bounded Squares illustration. - Leo Tavares, Oct 05 2021
From Leo Tavares, Oct 10 2021: (Start)
a(n) = A008585(n) + 2*A000217(n-1). See Trapezoids illustration.
2*A005563 = A054000(n+1). See Trapagons illustration.
a(n) = 2*A000217(n) + n. (End)
a(n) = (n+2)!!/(n-2)!! for n > 1. - Jacob Szlachetka, Jan 02 2022

Partially edited by Joerg Arndt, Mar 11 2010

More terms from N. J. A. Sloane, Aug 01 2010

A117950 a(n) = n^2 + 3.

Original entry on oeis.org

3, 4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364, 403, 444, 487, 532, 579, 628, 679, 732, 787, 844, 903, 964, 1027, 1092, 1159, 1228, 1299, 1372, 1447, 1524, 1603, 1684, 1767, 1852, 1939, 2028, 2119, 2212, 2307, 2404, 2503

Offset: 0

Eric W. Weisstein, Apr 04 2006

Nonnegative X values of solutions to the equation X^3 - (X + 3)^2 + X + 6 = Y^2. To prove that X = n^2 + 3: Y^2 = X^3 - (X + 3)^2 + X + 6 = X^3 - X^2 - 5X - 3 = (X - 3)(X^2 + 2X + 1) = (X - 3)*(X + 1)^2 it means: (X - 3) must be a perfect square, so X = n^2 + 3 and Y = n(n^2 + 4). - Mohamed Bouhamida, Nov 12 2007
An equivalent technique of integer factorization would work, for example, for the equation X^3 - 3*X^2 - 9*X - 5 = (X-5)(X+1)^2 = Y^2, looking for perfect squares of the form X - 5 = n^2. - R. J. Mathar, Nov 20 2007
Take a square array of (n+1) X (n+1) dots (which correspond to the vertices of a grid of n X n squares). Connect the dots with vertical and horizontal line segments of any length so that each dot is connected to each of its orthogonal neighbors, and so that no line segment crosses any previously drawn line segment. Then the minimum number of line segments is a(n), for n >= 1. - Leroy Quet, Apr 12 2009
a(n) is also the Wiener index of the double fan graph F(n). The double fan graph F(n) is defined as the graph obtained by joining each node of an n-node path graph with two additional nodes. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of the graph F(n) is (3n-1)t + (1/2)(n^2 - 3n + 4)t^2. Example: a(3)=12 because the corresponding double fan graph is the wheel graph on 5 nodes OABCD, O being the center of the wheel. Its Wiener index = number of edges + |AC| +|BD| = 8 + 2 + 2 = 12. - Emeric Deutsch, Sep 24 2010

Ivan Panchenko, Table of n, a(n) for n = 0..1000
B. E. Sagan, Y-N. Yeh and P. Zhang, The Wiener Polynomial of a Graph, Internat. J. of Quantum Chem., 60, 1996, 959-969. - _Emeric Deutsch_, Sep 24 2010
Eric Weisstein's World of Mathematics, Near-Square Prime
Eric Weisstein's World of Mathematics, Fan Graph. - _Emeric Deutsch_, Sep 24 2010
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A027688, A028560, A005563, A132111.

For primes in this sequence see A049422.

Table[n^2 + 3, {n, 0, 49}] (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)

a(n)=n^2+3 \\ Charles R Greathouse IV, Sep 24 2015

[lucas_number1(3,n,-3) for n in range(0, 51)] # Zerinvary Lajos, May 16 2009

G.f.: (3 - 5*x + 4*x^2)/(1-x)^3. - R. J. Mathar, Nov 20 2007
a(n) = A000290(n) + 3. - Omar E. Pol, Dec 20 2008
a(n) = ((n-3)^2 + 3*(n+1)^2)/4. - Reinhard Zumkeller, Feb 13 2009
a(n) = A132111(n-1,2) for n>1. - Reinhard Zumkeller, Aug 10 2007
a(n) = ceiling((n+1/n)^2), n>0. - Vincenzo Librandi, Oct 19 2011
a(n) = 2*n + a(n-1) - 1 (with a(0)=3). - Vincenzo Librandi, Nov 13 2010
a(n)*a(n-1) - 3 = (a(n)-n)^2 = A027688(n-1)^2. - Bruno Berselli, Dec 08 2011
From Amiram Eldar, Jul 21 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(3)*Pi*coth(sqrt(3)*Pi))/6.
Sum_{n>=0} (-1)^n/a(n) = (1 + (sqrt(3)*Pi)*csch(sqrt(3)*Pi))/6. (End)
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = 2*csch(sqrt(3)*Pi)*sinh(2*Pi)/sqrt(3).
Product_{n>=0} (1 - 1/a(n)) = sqrt(2/3)*csch(sqrt(3)*Pi)*sinh(sqrt(2)*Pi). (End)
E.g.f.: exp(x)*(3 + x + x^2). - Stefano Spezia, Aug 26 2024

Edited by N. J. A. Sloane Apr 15 2009 at the suggestion of Leroy Quet

A046301 Product of 3 successive primes.

Original entry on oeis.org

30, 105, 385, 1001, 2431, 4199, 7429, 12673, 20677, 33263, 47027, 65231, 82861, 107113, 146969, 190747, 241133, 290177, 347261, 409457, 478661, 583573, 716539, 871933, 1009091, 1113121, 1201289, 1317919, 1564259, 1879981, 2279269, 2494633, 2837407, 3127361, 3532343

Offset: 1

Patrick De Geest, Jun 15 1998

			From _K. D. Bajpai_, Aug 27 2014: (Start)
a(2) = 105 is in the sequence because 105 = 3* 5 * 7, product of three successive primes.
a(3) = 385 is in the sequence because 385 = 5 * 7 * 11, product of three successive primes.
(End)

Vincenzo Librandi and K. D. Bajpai, Table of n, a(n) for n = 1..10000 (first 1000 terms from Vincenzo Librandi)

Cf. A002110, A006094, A046302, A046303, A046324, A046325, A046326, A046327.

Cf. A000040, A242187, A257891.

a046301 n = a046301_list !! (n-1)
a046301_list = zipWith3 (((*) .) . (*))
               a000040_list (tail a000040_list) (drop 2 a000040_list)
-- Reinhard Zumkeller, May 12 2015

[NthPrime(n)*NthPrime(n+1)*NthPrime(n+2): n in [1..31]]; /* Or: */ [&*[ NthPrime(n+k): k in [0..2] ]: n in [1..31] ]; // Bruno Berselli, Feb 25 2011

A046301:=n->ithprime(n)*ithprime(n+1)*ithprime(n+2): seq(A028560(n), n=1..100);  # K. D. Bajpai, Aug 27 2014

Table[Prime[n] Prime[n+1] Prime[n+2],{n,50}]   (* K. D. Bajpai, Aug 27 2014 *)
Times@@@Partition[Prime[Range[40]],3,1] (* Harvey P. Dale, Mar 25 2019 *)

a(n)=prime(n)*prime(n+1)*prime(n+2); \\ Joerg Arndt, Aug 30 2014

Sum_{n>=1} 1/a(n) = A242187. - Amiram Eldar, Nov 19 2020

Offset changed from 0 to 1 by Vincenzo Librandi, Jan 16 2012

A067725 a(n) = 3n^2 + 6n.

Original entry on oeis.org

0, 9, 24, 45, 72, 105, 144, 189, 240, 297, 360, 429, 504, 585, 672, 765, 864, 969, 1080, 1197, 1320, 1449, 1584, 1725, 1872, 2025, 2184, 2349, 2520, 2697, 2880, 3069, 3264, 3465, 3672, 3885, 4104, 4329, 4560, 4797, 5040, 5289, 5544, 5805, 6072, 6345, 6624

Offset: 0

Robert G. Wilson v, Feb 05 2002

Numbers h such that 3*(3 + h) is a perfect square. - Alexander D. Healy, Tj Tullo, Avery Pickford, Sep 20 2004

Equivalently, numbers k such that k/3+1 is a square. - Bruno Berselli, Apr 10 2018

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005563.

Cf. numbers k such that k*(k + m) is a perfect square: A028560 (k=9), A067728 (k=8), A067727 (k=7), A067726 (k=6), A067724 (k=5), A028347 (k=4), A054000 (k=2), A005563 (k=1).

List([0..50], n-> 3*n*(n+2)); # G. C. Greubel, Sep 01 2019

[3*n*(n+2): n in [0..50]]; // G. C. Greubel, Sep 01 2019

seq(3*n*(n+2), n=0..50); # G. C. Greubel, Sep 01 2019

Select[ Range[10000], IntegerQ[ Sqrt[ 3(3 + # )]] & ]
3*(Range[50]^2 -1) (* G. C. Greubel, Sep 01 2019 *)

a(n)=3*n*(n+2) \\ Charles R Greathouse IV, Dec 07 2011

[3*n*(n+2) for n in (0..50)] # G. C. Greubel, Sep 01 2019

a(n) = 3*A005563(n). - Zerinvary Lajos, Mar 06 2007
a(n) = a(n-1) + 6*n + 3, with n>0, a(0)=0. - Vincenzo Librandi, Aug 08 2010
From Colin Barker, Apr 11 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 3*x*(3-x)/(1-x)^3. (End)
E.g.f.: 3*x*(x + 3)*exp(x). - G. C. Greubel, Jul 20 2017
From Amiram Eldar, Feb 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/12. (End)

Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar

A028884 a(n) = (n + 3)^2 - 8.

Original entry on oeis.org

1, 8, 17, 28, 41, 56, 73, 92, 113, 136, 161, 188, 217, 248, 281, 316, 353, 392, 433, 476, 521, 568, 617, 668, 721, 776, 833, 892, 953, 1016, 1081, 1148, 1217, 1288, 1361, 1436, 1513, 1592, 1673, 1756, 1841, 1928, 2017, 2108, 2201, 2296, 2393

Offset: 0

Patrick De Geest

From Klaus Purath, Jan 04 2023: (Start)
The product of two consecutive terms belongs to the sequence: a(n)*a(n+1) = a(a(n)+n) = (a(n)+n)*(a(n+1)-n-1) + 1.
a(n) is never divisible by primes given in A003629.
Each odd prime factor p divides exactly 2 out of any p consecutive terms. If a(i) and a(k) form such a pair that are divisible by p, then i + k == -6 (mod p).
The prime factors are listed in A038873 and the primes in A028886.
For n > 0, this is a proper subsequence of A079896.
Conjecture: a(n) = A079896(A265284(n-1)). -
(End)

			From _Stefano Spezia_, Nov 08 2022: (Start)
Illustrations for n = 0..4:
          *       * * *     * * * * *
      a(0) = 1    *   *     *       *
                  * * *     *   *   *
                a(1) = 8    *       *
                            * * * * *
                            a(2) = 17
.
   * * * * * * *    * * * * * * * * *
   *           *    *               *
   *   *   *   *    *   *   *   *   *
   *           *    *               *
   *   *   *   *    *   *   *   *   *
   *           *    *               *
   * * * * * * *    *   *   *   *   *
     a(3) = 28      *               *
                    * * * * * * * * *
                        a(4) = 41
(End)

Altug Alkan, Table of n, a(n) for n = 0..10000
Patrick De Geest, Palindromic Quasipronics of the form n(n+x).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005563, A014616, A028560, A082111, A190576.

a014616 n = (n * (n + 6) + 1) `div` 4 -- Reinhard Zumkeller, Apr 07 2013

Range[3, 50]^2 - 8 (* Alonso del Arte, Aug 15 2016 *)

a(n)=(n+3)^2-8 \\ Charles R Greathouse IV, Oct 07 2015

(3 to 49).map(n => n * n - 8) // Alonso del Arte, May 07 2020

a(n) = a(n-1) + 2*n + 5 (with a(0) = 1). - Vincenzo Librandi, Aug 05 2010
a(n) = A028560(n) + 1; A014616(n) = floor(a(n+1)/4). - Reinhard Zumkeller, Apr 07 2013
G.f.: (-1 - 5*x + 4*x^2)/(x - 1)^3. - R. J. Mathar, Mar 24 2013
Sum_{n >= 0} 1/a(n) = 51/112 - Pi*cot(2*Pi*sqrt(2))/(4*sqrt(2)) = 1.3839174974448... . - Vaclav Kotesovec, Apr 10 2016
E.g.f.: (1 + 7*x + x^2)*exp(x). - G. C. Greubel, Aug 19 2017
Sum_{n >= 0} (-1)^n/a(n) = (-19 + 14*sqrt(2)*Pi*cosec(2*sqrt(2)*Pi))/112. - Amiram Eldar, Nov 04 2020
From Klaus Purath, Jan 04 2023: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2, n >= 2.
a(n) = A082111(n) + n.
a(n) = A190576(n+1) - n. (End)
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = 7*Pi/(45*sqrt(2)*sin(2*sqrt(2)*Pi)).
Product_{n>=0} (1 + 1/a(n)) = (4*sqrt(14)/9)*sin(sqrt(7)*Pi)/sin(2*sqrt(2)*Pi). (End)

Definition corrected by Omar E. Pol, Jul 27 2009

A132411 a(0) = 0, a(1) = 1 and a(n) = n^2 - 1 with n >= 2.

Original entry on oeis.org

0, 1, 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

Offset: 0

Mohamed Bouhamida, Nov 12 2007

X values of solutions to the equation X^3 - (X + 1)^2 + X + 2 = Y^2.
To prove that X = 1 or X = n^2 - 1: Y^2 = X^3 - (X + 1)^2 + X + 2 = X^3 - X^2 - X + 1 = (X + 1)(X^2 - 2X + 1) = (X + 1)*(X - 1)^2 it means: X = 1 or (X + 1) must be a perfect square, so X = 1 or X = n^2 - 1 with n >= 1. Which gives: (X, Y) = (0, 1) or (X, Y) = (1, 0) or (X, Y) = (n^2 - 1, n*(n^2 - 2)) with n >= 2.
An equivalent technique of integer factorization would work for example for the equation X^3 + 3*X^2 - 9*X + 5 = (X+5)(X-1)^2 = Y^2, looking for perfect squares of the form X + 5 = n^2. Another example is X^3 + X^2 - 5*X + 3 = (X+3)*(X-1)^2 = Y^2 with solutions generated from perfect squares of the form X + 3 = n^2. - R. J. Mathar, Nov 20 2007
Sum of possible divisors of a prime number up to its square root, with duplicate entries removed. - Odimar Fabeny, Aug 25 2010
a(0) = 0, a(1) = 1 and a(n) is the smallest k different from n such that n divides k and n+1 divides k+1. - Michel Lagneau, Apr 27 2013
The identity (4*n^2-2)^2 - (n^2-1)*(4*n)^2 = 4 can be written as A060626(n+1)^2 - a(n+2)*A008586(n+2)^2 = 4. - Vincenzo Librandi, Jun 16 2014
Conjecture: the sequence terms are the exponents in the expansion of the q-series Sum_{n >= 0} q^(2*n) * Product_{k >= 2*n+2} 1 - (-q)^k = 1 + q^3 + q^8 + q^15 + q^24 + .... - Peter Bala, May 10 2025

			0^3 - 1^2 + 2 = 1^2, 1^3 - 2^2 + 3 = 0^2, 3^3 - 4^2 + 5 = 4^2.
For P(n) = 29 we have sqrt(29) = 5.3851... so possible divisors are 3 and 5; for P(n) = 53 we have sqrt(53) = 7.2801... so possible divisors are 3, 5 and 7. - _Odimar Fabeny_, Aug 25 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A005563, A008586, A028560, A060626, A170949.

[0,1] cat [n^2 - 1: n in [2..60]]; // Vincenzo Librandi, May 01 2014

a:= n-> `if`(n<2, n, n^2-1):
seq(a(n), n=0..55);  # Alois P. Heinz, Jan 24 2021

Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {3, 8, 15}, 80]] (* and *) Table[If[n < 2, n, n^2 - 1], {n, 0, 80}] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2012 *)
Join[{0,1},Range[2,50]^2-1] (* Harvey P. Dale, Feb 27 2013 *)
CoefficientList[Series[x + x^2 (-3 + x)/(-1 + x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, May 01 2014 *)

concat(0, Vec(x+x^2*(-3+x)/(-1+x)^3 + O(x^100))) \\ Altug Alkan, Dec 18 2015

a(n)=if(n>1,n^2-1,n) \\ Charles R Greathouse IV, Dec 18 2015

a(n) = A005563(n-1), n > 1.
G.f.: x + x^2*(-3+x)/(-1+x)^3. - R. J. Mathar, Nov 20 2007
Starting (1, 3, 8, 15, 24, ...) = binomial transform of [1, 2, 3, -1, 1, -1, ...]. - Gary W. Adamson, May 12 2008
a(n) = A170949(A002522(n-1)) for n > 0. - Reinhard Zumkeller, Mar 08 2010
Sum_{n>0} 1/a(n) = 7/4. - Enrique Pérez Herrero, Dec 18 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/4. - Amiram Eldar, Sep 27 2022
From Elmo R. Oliveira, May 29 2025: (Start)
E.g.f.: exp(x)*(x^2 + x - 1) + x + 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 5. (End)

Definition simplified by N. J. A. Sloane, Sep 05 2010

A158405 Triangle T(n,m) = 1+2*m of odd numbers read along rows, 0<=m

Original entry on oeis.org

1, 1, 3, 1, 3, 5, 1, 3, 5, 7, 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 11, 1, 3, 5, 7, 9, 11, 13, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23

Offset: 1

Paul Curtz, Mar 18 2009

Row sums are n^2 = A000290(n).
The triangle sums, see A180662 for their definitions, link this triangle of odd numbers with seventeen different sequences, see the crossrefs. The knight sums Kn14 - Kn110 have been added. - Johannes W. Meijer, Sep 22 2010
A208057 is the eigentriangle of A158405 such that as infinite lower triangular matrices, A158405 * A208057 shifts the latter, deleting the right border of 1's. - Gary W. Adamson, Feb 22 2012
T(n,k) = A099375(n-1,n-k), 1<=k<=n. [Reinhard Zumkeller, Mar 31 2012]

			The triangle contains the first n odd numbers in row n:
  1;
  1,3;
  1,3,5;
  1,3,5,7;
From _Seiichi Manyama_, Dec 02 2017: (Start)
    |       a(n)        |                               | A000290(n)
   -----------------------------------------------------------------
   0|                                                      (=  0)
   1|                 1 = 1/3 * ( 3)                       (=  1)
   2|             1 + 3 = 1/3 * ( 5 +  7)                  (=  4)
   3|         1 + 3 + 5 = 1/3 * ( 7 +  9 + 11)             (=  9)
   4|     1 + 3 + 5 + 7 = 1/3 * ( 9 + 11 + 13 + 15)        (= 16)
   5| 1 + 3 + 5 + 7 + 9 = 1/3 * (11 + 13 + 15 + 17 + 19)   (= 25)
(End)

Seiichi Manyama, Rows n = 1..140, flattened
Daniel Erman, The Josephus Problem, Numberphile video (2016)
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Index entries for sequences related to the Josephus Problem

Cf. A129326, A099375, A005408.
Triangle sums (see the comments): A000290 (Row1; Kn11 & Kn4 & Ca1 & Ca4 & Gi1 & Gi4); A000027 (Row2); A005563 (Kn12); A028347 (Kn13); A028560 (Kn14); A028566 (Kn15); A098603 (Kn16); A098847 (Kn17); A098848 (Kn18); A098849 (Kn19); A098850 (Kn110); A000217 (Kn21. Kn22, Kn23, Fi2, Ze2); A000384 (Kn3, Fi1, Ze3); A000212 (Ca2 & Ze4); A000567 (Ca3, Ze1); A011848 (Gi2); A001107 (Gi3). - Johannes W. Meijer, Sep 22 2010
Cf. A063656, A063657, A208057, A292610, A292611.

a158405 n k = a158405_row n !! (k-1)
a158405_row n = a158405_tabl !! (n-1)
a158405_tabl = map reverse a099375_tabl
-- Reinhard Zumkeller, Mar 31 2012

Table[2 Range[1, n] - 1, {n, 12}] // Flatten (* Michael De Vlieger, Oct 01 2015 *)

a(n) = 2*(n-floor((-1+sqrt(8*n-7))/2)*(floor((-1+sqrt(8*n-7))/2)+1)/2)-1;
vector(100, n, a(n)) \\ Altug Alkan, Oct 01 2015

a(n) = 2*i-1, where i = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Feb 03 2013

a(n) = 2*A002262(n-1) + 1. - Eric Werley, Sep 30 2015

Edited by R. J. Mathar, Oct 06 2009

cp's OEIS Frontend

A147651 First trisection of A028560.

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

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A117950 a(n) = n^2 + 3.

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A046301 Product of 3 successive primes.

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A067725 a(n) = 3*n^2 + 6*n.

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A028884 a(n) = (n + 3)^2 - 8.

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A067725 a(n) = 3n^2 + 6n.