A156798 -id:A156798 - cp's OEIS frontend

A002522 a(n) = n^2 + 1.

1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501

Offset: 0

N. J. A. Sloane

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).
a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.
As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001
a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003
Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006
Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.
The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005
Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov
1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006
For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007
Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that 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) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007
{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010
For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010
The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010
n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.
a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011
Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011
a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011
Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011
If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014
The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014
a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016
Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016
For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017
Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018
a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019
a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022
From Klaus Purath, Apr 03 2025: (Start)
The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.
It follows from the above that this sequence is a subsequence of A031396. (End)

			G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 17*x^4 + 26*x^5 + 37*x^6 + 50*x^7 + 65*x^8 + ...

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
E. Gura and M. Maschler, Insights into Game Theory: An Alternative Mathematical Experience, Cambridge, 2008; p. 26.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, New York, 2001.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000. Format corrected by _Peter Kagey_, Jan 25 2016
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, On the Salient Regularities of Strings of Assembly Theory, Preprints (2024). See p. 19.
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints.org (2025). See p. 28.
R. P. Boas and N. J. A. Sloane, Correspondence, 1974
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 44, 56.
Giulio Cerbai and Luca Ferrari, Permutation patterns in genome rearrangement problems: the reversal model, arXiv:1903.08774 [math.CO], 2019. See p. 19.
S. Chaiken et al., Nonattacking Queens in a Rectangular Strip, arXiv:1105.5087 [math.CO], 2011.
Bryan Ek, Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics, arXiv:1804.05933 [math.CO], 2018.
R. M. Green and Tianyuan Xu, 2-roots for simply laced Weyl groups, arXiv:2204.09765 [math.RT], 2022.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv:1410.2657 [math.CO], 2014.
C. Homberger and V. Vatter, On the effective and automatic enumeration of polynomial permutation classes, arXiv:1308.4946 [math.CO], 2013.
L. B. W. Jolley, Summation of Series, Dover, 1961, p. 176.
S. J. Leon, Linear Algebra with Applications: the Perron-Frobenius Theorem [Cached copy at the Wayback Machine]
T. Mansour and J. West, Avoiding 2-letter signed patterns, arXiv:math/0207204 [math.CO], 2002.
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv:1508.07894 [math.NT], 2015.
Eric Weisstein's World of Mathematics, Number Picking
Eric Weisstein's World of Mathematics, Near-Square Prime
Helmut Wielandt, Unzerlegbare, nicht negative Matrizen, Math. Z. 52 (1950), 642-648. volume 52
Reinhard Zumkeller, Enumerations of Divisors
Index to values of cyclotomic polynomials of integer argument
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Left edge of A055096.
Cf. A059100, A117950, A087475, A117951, A114949, A117619 (sequences of form n^2 + K).
a(n+1) = A101220(n, n+1, 3).
Cf. A059592, A124808, A132411, A132414, A028872, A005408, A000124, A016813, A086514, A000125, A058331, A080856, A000127, A161701-A161704, A161706, A161707, A161708, A161710-A161713, A161715, A006261.
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), this sequence (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A002496 (primes).
Cf. A254858.
Cf. A302612, A302644, A302645, A302646.
Subsequence of A031396.

a002522 = (+ 1) . (^ 2)
a002522_list = scanl (+) 1 [1,3..]
-- Reinhard Zumkeller, Apr 06 2012

[n^2 + 1: n in [0..50]]; // Vincenzo Librandi, May 01 2011

A002522 := proc(n)
        numtheory[cyclotomic](4,n) ;
end proc:
seq(A002522(n),n=0..20) ; # R. J. Mathar, Feb 07 2014

Table[n^2 + 1, {n, 0, 50}]; (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)

A002522(n):=n^2+1$ makelist(A002522(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

a(n)=n^2+1 \\ Charles R Greathouse IV, Jun 10 2011

O.g.f.: (1-x+2*x^2)/((1-x)^3). - Eric Werley, Jun 27 2011
Sequences of the form a(n) = n^2 + K with offset 0 have o.g.f. (K - 2*K*x + K*x^2 + x + x^2)/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a*(n-3). - R. J. Mathar, Apr 28 2008
For n > 0: a(n-1) = A143053(A000290(n)) - 1. - Reinhard Zumkeller, Jul 20 2008
A143053(a(n)) = A000290(n+1). - Reinhard Zumkeller, Jul 20 2008
a(n)*a(n-2) = (n-1)^4 + 4. - Reinhard Zumkeller, Feb 12 2009
a(n) = A156798(n)/A087475(n). - Reinhard Zumkeller, Feb 16 2009
From Reinhard Zumkeller, Mar 08 2010: (Start)
a(n) = A170949(A002061(n+1));
A170949(a(n)) = A132411(n+1);
A170950(a(n)) = A002061(n+1). (End)
For n > 1, a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n - 2)^2 + (a(n) + n - 1 + a(n) + n)^2 = (n+1) *(6*n^4 + 18*n^3 + 26*n^2 + 19*n + 6) / 6 = (a(n) + n)^2 + ... + (a(n) + 2*n)^2. - Charlie Marion, Jan 10 2011
From Eric Werley, Jun 27 2011: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2.
a(n) = a(n-1) + 2*n - 1. (End)
a(n) = (n-1)^2 + 2(n-1) + 2 = 122 read in base n-1 (for n > 3). - Jason Kimberley, Oct 20 2011
a(n)*a(n+1) = a(n*(n+1) + 1) so a(1)*a(2) = a(3). More generally, a(n)*a(n+k) = a(n*(n+k) + 1) + k^2 - 1. - Jon Perry, Aug 01 2012
a(n) = (n!)^2* [x^n] BesselI(0, 2*sqrt(x))*(1+x). - Peter Luschny, Aug 25 2012
a(n) = A070216(n,1) for n > 0. - Reinhard Zumkeller, Nov 11 2012
E.g.f.: exp(x)*(1 + x + x^2). - Geoffrey Critzer, Aug 30 2013
a(n) = A254858(n-2,3) for n > 2. - Reinhard Zumkeller, Feb 09 2015
Sum_{n>=0} (-1)^n / a(n) = (1+Pi/sinh(Pi))/2 = 0.636014527491... = A367976 . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} 1/a(n) = (1 + Pi*coth(Pi))/2 = 2.076674... = A113319. - Vaclav Kotesovec, Apr 10 2016
4*a(n) = A001105(n-1) + A001105(n+1). - Bruno Berselli, Jul 03 2017
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi)*sinh(sqrt(2)*Pi).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi). (End)
Sum_{n>=0} a(n)/n! = 3*e. - Davide Rotondo, Feb 16 2025

Partially edited by Joerg Arndt, Mar 11 2010

A059100 a(n) = n^2 + 2.

Original entry on oeis.org

2, 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403, 2502, 2603

Offset: 0

Henry Bottomley, Feb 13 2001

Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. - Benoit Cloitre, Aug 15 2002
Binomial transformation yields A081908, with A081908(0)=1 dropped. - R. J. Mathar, Oct 05 2008
1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). - Wolfdieter Lang, Mar 01 2013
a(n) is the number of election results for an election with n+2 candidates, say C1, C2, ..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. - Dennis P. Walsh, May 08 2013
This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) = A005248; at n=2, b(k) = 2*A001541; at n=3, b(k)= A057076; at n=4, b(k) = 2*A023039. This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) = A005248. The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also see A000032 for a related family of sequences. - Richard R. Forberg, Nov 22 2014
Also the number of maximum matchings in the n-gear graph. - Eric W. Weisstein, Dec 31 2017
Also the Wiener index of the n-dipyramidal graph. - Eric W. Weisstein, Jun 14 2018
Numbers of the form n^2+2 have no factors that are congruent to 7 (mod 8). - Gordon E. Michaels, Sep 12 2019
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {n, 2n}]. - Magus K. Chu, Sep 10 2022

			For n = 2, a(2) = 6 since there are 6 election results in a 4-candidate, 2-voter election that have candidates c1 and c2 tied. Letting <i,j> denote voter 1 voting for candidate i and voter 2 voting for candidate j, the six election results are <1,2>, <2,1>, <3,3>, <3,4>, <4,3>, and <4,4>. - _Dennis P. Walsh_, May 08 2013

Harry J. Smith, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 52, 56.
Hesam Dashti, A New Upper Bound on the Length of Shortest Permutation Strings; An Algorithm for Generating Permutation Strings, arXiv:1009.5053 [math.CO], 2010. - _Jonathan Vos Post_, Sep 28 2010
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Alexander Soifer, Coffee Hour and the Conway-Soifer Cover-Up, In: How Does One Cut a Triangle? (2009), pp. 147-156. See also here
Dennis P. Walsh, Notes on a tied election.
Eric Weisstein's World of Mathematics, Dipyramidal Graph.
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Matching.
Eric Weisstein's World of Mathematics, Maximum Independent Edge Set.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Eric Weisstein's World of Mathematics, Pappus Chain.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000032, A000196, A000290, A001541, A002522, A005248, A008865, A056105, A056109, A056899, A057076, A069987, A081908, A114964, A156798, A166464.
Apart from initial terms, same as A010000.
2nd row/column of A295707.

a059100 = (+ 2) . (^ 2)
a059100_list = scanl (+) (2) [1, 3 ..]
-- Reinhard Zumkeller, Feb 09 2015

with(combinat, fibonacci):seq(fibonacci(3, i)+1, i=0..49); # Zerinvary Lajos, Mar 20 2008

Table[n^2 + 2, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)
LinearRecurrence[{3, -3, 1}, {2, 3, 6}, 50] (* Vincenzo Librandi, Feb 15 2012 *)
Range[0, 20]^2 + 2 (* Eric W. Weisstein, Dec 31 2017 *)
CoefficientList[Series[(-2 + 3 x - 3 x^2)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 31 2017 *)

a(n) = { n^2+2 } \\ Harry J. Smith, Jun 24 2009

[lucas_number1(3,n,-2) for n in range(0, 50)] # Zerinvary Lajos, May 16 2009

G.f.: (2 - 3*x + 3*x^2)/(1 - x)^3. - R. J. Mathar, Oct 05 2008
a(n) = ((n - 2)^2 + 2*(n + 1)^2)/3. - Reinhard Zumkeller, Feb 13 2009
a(n) = A000196(A156798(n) - A000290(n)). - Reinhard Zumkeller, Feb 16 2009
a(n) = 2*n + a(n-1) - 1 with a(0) = 2. - Vincenzo Librandi, Aug 07 2010
a(n+3) = (A166464(n+5) - A166464(n))/20. - Paul Curtz, Nov 07 2012
From Paul Curtz, Nov 07 2012: (Start)
a(3*n) mod 9 = 2.
a(3*n+1) = 3*A056109(n).
a(3*n+2) = 3*A056105(n+1). (End)
Sum_{n >= 1} 1/a(n) = Pi * coth(sqrt(2)*Pi) / 2^(3/2) - 1/4. - Vaclav Kotesovec, May 01 2018
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(2)*Pi*(csch(sqrt(2)*Pi)))/4.
Product_{n>=0} (1 + 1/a(n)) = sqrt(3/2)*csch(sqrt(2)*Pi)*sinh(sqrt(3)*Pi).
Product_{n>=0} (1 - 1/a(n)) = csch(sqrt(2)*Pi)*sinh(Pi)/sqrt(2). (End)
E.g.f.: exp(x)*(2 + x + x^2). - Stefano Spezia, Aug 07 2024

A028552 a(n) = n*(n+3).

Original entry on oeis.org

0, 4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180, 208, 238, 270, 304, 340, 378, 418, 460, 504, 550, 598, 648, 700, 754, 810, 868, 928, 990, 1054, 1120, 1188, 1258, 1330, 1404, 1480, 1558, 1638, 1720, 1804, 1890, 1978, 2068, 2160, 2254, 2350, 2448, 2548, 2650

Offset: 0

Patrick De Geest

n*(n-3), for n >= 3, is the number of [body] diagonals of an n-gonal prism. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr)
a(n) = A028387(n)-1. Half of the difference between n(n+1)(n+2)(n+3) and the largest square less than it. Calling this difference "SquareMod": a(n) = (1/2)*SquareMod(n(n+1)(n+2)(n+3)). - Rainer Rosenthal, Sep 04 2004
n != -2 such that x^4 + x^3 - n*x^2 + x + 1 is reducible over the integers. Starting at 10: n such that x^4 + x^3 - n*x^2 + x + 1 is a product of irreducible quadratic factors over the integers. - James R. Buddenhagen, Apr 19 2005
If a 3-set Y and a 3-set Z, having two element in common, are subsets of an n-set X then a(n-4) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 03 2007
Starting with offset 1 = binomial transform of [4, 6, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jan 09 2009
The sequence provides all nonnegative integers m such that 4*m + 9 is a square. - Vincenzo Librandi, Mar 03 2013
The second-order linear recurrence relations b(n)=3*b(n-1) + a(m-3)*b(n-2), n>=2, b(0)=0, b(1)=1, have closed form solutions involving only powers of m and 3-m where m>=4 is a positive integer; and lim_{n->infinity} b(n+1)/b(n) = 4. - Felix P. Muga II, Mar 18 2014
If a rook is placed at a corner of an n X n chessboard, the expected number of moves for it to reach the opposite corner is a(n-1). (See Mathematics Stack Exchange link.) - Eric M. Schmidt, Oct 29 2014
Partial sums of the even composites (which are A005843 without the 2). - R. J. Mathar, Sep 09 2015
a(n) is the number of segments necessary to represent n squares of area 1, 4, ..., n^2 having the upper and left sides overlapped:
         __      __      __  __
  ||   ||  |   ||  |  |  ||  |  |  |
         |___|   | __|  |  | __|  |  |
                   | __ |  | __ |  |
                               | __  __|
    4       10         18            28       - Stefano Spezia, May 29 2023

			G.f. = 4*x + 10*x^2 + 18*x^3 + 28*x^4 + 40*x^5 + 54*x^6 + 70*x^7 + 88*x^8 + ...

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Patrick De Geest, Palindromic Quasipronics of the form n(n+x).
Milan Janjic, Two Enumerative Functions.
Mathematics Stack Exchange, Expected number of turns for a rook to move to top right-most corner?.
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Aleksandar Petojević, A Note about the Pochhammer Symbol, Mathematica Moravica, Vol. 12-1 (2008), 37-42.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A002522, A005843, A028387, A062145, A156798.

[n*(n+3): n in [0..150]]; // Vincenzo Librandi, Apr 21 2011

A028552 := proc(n) n*(n+3); end proc: # R. J. Mathar, Jan 29 2011

LinearRecurrence[{3,-3,1},{0,4,10},50] (* Harvey P. Dale, Feb 05 2012 *)
Table[ChineseRemainder[{n, n + 1}, {n + 2, n + 3}], {n, -1, 80}] (* Zak Seidov, Oct 25 2014 *)
Table[ChineseRemainder@@TakeDrop[Range[n,n+3],2],{n,-1,50}] (* Harvey P. Dale, Mar 14 2025 *)

makelist(n*(n+3),n,0,20); /* Martin Ettl, Jan 22 2013 */

a(n)=n*(n+3) \\ Charles R Greathouse IV, Mar 16 2012

a(n) = 2*A000096(n).
a(A002522(n)) = A156798(n). - Reinhard Zumkeller, Feb 16 2009
a(n) = a(n-1) + 2*(n+1) for n>0, with a(0)=0. - Vincenzo Librandi, Aug 05 2010
Sum_{n>=1} 1/a(n) = 11/18 via Sum_{n>=0} 1/((n+x)*(n+y)) = (psi(x)-psi(y))/ (x-y). - R. J. Mathar, Mar 22 2011
G.f.: 2*x*(2 - x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 31 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=0, a(1)=4, a(2)=10. - Harvey P. Dale, Feb 05 2012
a(n) = 4*C(n+1,2) - 2*C(n,2) for n>=0. - Felix P. Muga II, Mar 11 2014
a(-3 - n) = a(n) for all n in Z. - Michael Somos, Mar 18 2014
E.g.f.: (x^3 + 4*x)*exp(x). - G. C. Greubel, Jul 20 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - 5/18. - Amiram Eldar, Jan 15 2021
From Amiram Eldar, Feb 12 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = 2*cos(sqrt(13)*Pi/2)/Pi.
Product_{n>=1} (1 + 1/a(n)) = -6*cos(sqrt(5)*Pi/2)/Pi. (End)

A087475 a(n) = n^2 + 4.

Original entry on oeis.org

4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 104, 125, 148, 173, 200, 229, 260, 293, 328, 365, 404, 445, 488, 533, 580, 629, 680, 733, 788, 845, 904, 965, 1028, 1093, 1160, 1229, 1300, 1373, 1448, 1525, 1604, 1685, 1768, 1853, 1940, 2029, 2120, 2213, 2308, 2405, 2504

Offset: 0

Gary W. Adamson, Sep 09 2003

Schroeder, p. 330, states "For positive n, these winding numbers are precisely those whose continued fraction expansion is periodic and has period length 1".
Positive X values of solutions to the equation X^3 - 4*X^2 = Y^2. To find Y values: b(n) = n*(n^2 + 4). - Mohamed Bouhamida, Nov 06 2007
From Artur Jasinski, Oct 03 2008: (Start)
General formula for cotangent recurrences type:
a(n+1) = a(n)^3 + 3*a(n) and a(1)=k is
a(n) = floor(((k + sqrt(k^2 + 4))/2)^(3^(n-1))). (End)
Given sequences of the form S(n) = N*S(n-1) + S(n-2) starting (1, N, ...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p-1)/2) mod p, for n>0, p = odd prime. Example: with N = 2 we have the Pell series (1, 2, 5, 12, 29, 70, 169, ...) with P(7) = 169. Then 169 == 8^3 mod 7, with a(2) = 8. Cf. Schroeder, "Number Theory in Science and Communication", p. 90, for N = 1: F(p) == 5^((p-1)/2) mod p. - Gary W. Adamson, Feb 23 2009
The only two real solutions of the form f(x) = A*x^p with positive p that satisfy f^(n)(x) = f^[-1](x), x >= 0, n >= 1, with f^(n) the n-th derivative and f^[-1] the compositional inverse of f, are obtained for p = p1(n) = (n + sqrt(a(n)))/2 and p = p2(n) = (n - sqrt(a(n)))/2, n >= 1, and A = A(n) = (fallfac(p,n))^(-p/(p+1)), for p = p1(n) and p = p2(n), respectively. Here fallfac(x, k) := product(x - j, j = 0..k-1), the falling factorials. See the T. Koshy reference, pp. 263-264 (there is also a solution for negative p if n is even; see the corresponding comment in A002522). - Wolfdieter Lang, Oct 21 2010, Oct 28 2010
(n + sqrt(a(n)))/2 = [n;n,n,...], with the regular continued fraction with period length 1. For a simple proof see, e.g., the Schroeder reference, pp. 330-331. See also the first comment above.

			a(2) = 8, discriminant of algebraic representation of barover(2) = [2,2,2,...] = sqrt 2 - 1 = 0.41421356... = ((sqrt 8) - 2)/2. a(3) = 13, discriminant of barover(3) = [3,3,3,...] = 0.3027756... = ((sqrt 13) - 3)/2.

Manfred R. Schroeder, "Fractals, Chaos, Power Laws"; W.H. Freeman & Co, 1991, p. 330-331.
Manfred R. Schroeder, "Number Theory in Science and Communication", Springer Verlag, 5th ed., 2009. [From Gary W. Adamson, Feb 23 2009]
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, New York, 2001. [From Wolfdieter Lang, Oct 21 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Near-Square Prime.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001082, A002378, A002522, A005563, A028347, A036666, A046092, A053755, A062717, A155965, A155966, A156701, A156798.

Range[0, 50]^2 + 4 (* Harvey P. Dale, Jan 05 2011 *)

a(n)=n^2+4 \\ Charles R Greathouse IV, Jun 10 2011

(0 to 49).map(n => n * n + 4) // Alonso del Arte, May 29 2019

n^2 + 4 are discriminant terms in the formula for Positive Silver Mean Constants, defined as barover(n), = (sqrt (n^2 + 4) - n)/2. Such constants barover(n) = C have the property: 1/C - C = n.
a(n) = A156701(n) / A053755(n). - Reinhard Zumkeller, Feb 13 2009
a(n) = A156798(n)/A002522(n). - Reinhard Zumkeller, Feb 16 2009
a(n) = a(n-1) + 2*n-1 (with a(0)=4). - Vincenzo Librandi, Nov 22 2010
G.f.: (4 - 7*x + 5*x^2)/(1 - x)^3. - Colin Barker, Jan 06 2012
a(n)^3 = A155965(n)^2 + A155966(n)^2. - Vincenzo Librandi, Feb 22 2012
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 2*Pi*coth(2*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + 2*Pi*cosech(2*Pi))/8 = A371803. (End)
E.g.f.: exp(x)*(4 + x + x^2). - Stefano Spezia, Jul 08 2023
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(3)*sinh(sqrt(3)*Pi)/(2*sinh(2*Pi)).
Product_{n>=0} (1 + 1/a(n)) = sqrt(5)*sinh(sqrt(5)*Pi)/(2*sinh(2*Pi)). (End)

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A028552 a(n) = n*(n+3).

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A087475 a(n) = n^2 + 4.

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