A147973 -id:A147973 - cp's OEIS frontend

A001844 Centered square numbers: a(n) = 2n(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513

Offset: 0

N. J. A. Sloane

These are Hogben's central polygonal numbers denoted by
...2...
....P..
...4.n.
Numbers of the form (k^2+1)/2 for k odd.
(y(2x+1))^2 + (y(2x^2+2x))^2 = (y(2x^2+2x+1))^2. E.g., let y = 2, x = 1; (2(2+1))^2 + (2(2+2))^2 = (2(2+2+1))^2, (2(3))^2 + (2(4))^2 = (2(5))^2, 6^2 + 8^2 = 10^2, 36 + 64 = 100. - Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 08 2002
a(n) is also the number of 3 X 3 magic squares with sum 3(n+1). - Sharon Sela (sharonsela(AT)hotmail.com), May 11 2002
For n > 0, a(n) is the smallest k such that zeta(2) - Sum_{i=1..k} 1/i^2 <= zeta(3) - Sum_{i=1..n} 1/i^3. - Benoit Cloitre, May 17 2002
Number of convex polyominoes with a 2 X (n+1) minimal bounding rectangle.
The prime terms are given by A027862. - Lekraj Beedassy, Jul 09 2004
First difference of a(n) is 4n = A008586(n). Any entry k of the sequence is followed by k + 2*(1 + sqrt(2k - 1)). - Lekraj Beedassy, Jun 04 2006
Integers of the form 1 + x + x^2/2 (generating polynomial is Schur's polynomial as in A127876). - Artur Jasinski, Feb 04 2007
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Row sums of triangle A132778. - Gary W. Adamson, Sep 02 2007
Binomial transform of [1, 4, 4, 0, 0, 0, ...]; = inverse binomial transform of A001788: (1, 6, 24, 80, 240, ...). - Gary W. Adamson, Sep 02 2007
Narayana transform (A001263) of [1, 4, 0, 0, 0, ...]. Equals A128064 (unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Dec 29 2007
k such that the Diophantine equation x^3 - y^3 = x*y + k has a solution with y = x-1. If that solution is (x,y) = (m+1,m) then m^2 + (m+1)^2 = k. Note that this Diophantine equation is an elliptic curve and (m+1,m) is an integer point on it. - James R. Buddenhagen, Aug 12 2008
Numbers k such that (k, k, 2*k-2) are the sides of an isosceles triangle with integer area. Also, k such that 2*k-1 is a square. - James R. Buddenhagen, Oct 17 2008
a(n) is also the least weight of self-conjugate partitions having n+1 different odd parts. - Augustine O. Munagi, Dec 18 2008
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) = A153869 * (1, 2, 3, ...). - Gary W. Adamson, Jan 03 2009
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) where a(n) = 2n*(n-1)+1, all tuples of square numbers (X-Y, X, X+Y) are produced by ((m*(a(n)-2n))^2, (m*a(n))^2, (m*(a(n)+2n-2))^2) where m is a whole number. - Doug Bell, Feb 27 2009
Equals (1, 2, 3, ...) convolved with (1, 3, 4, 4, 4, ...). E.g., a(3) = 25 = (1, 2, 3, 4) dot (4, 4, 3, 1) = (4 + 8 + 9 + 4). - Gary W. Adamson, May 01 2009
The running sum of squares taken two at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 18 2009
Equals the odd integers convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 25 2009
Equals the triangular numbers convolved with [1, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
When the positive integers are written in a square array by diagonals as in A038722, a(n) gives the numbers appearing on the main diagonal. - Joshua Zucker, Jul 07 2009
The finite continued fraction [n,1,1,n] = (2n+1)/(2n^2 + 2n + 1) = (2n+1)/a(n); and the squares of the first two denominators of the convergents = a(n). E.g., the convergents and value of [4,1,1,4] = 1/4, 1/5, 2/9, 9/41 where 4^2 + 5^2 = 41. - Gary W. Adamson, Jul 15 2010
From Keith Tyler, Aug 10 2010: (Start)
Running sum of A008574.
Square open pyramidal number; that is, the number of elements in a square pyramid of height (n) with only surface and no bottom nodes. (End)
For k>0, x^4 + x^2 + k factors over the integers iff sqrt(k) is in this sequence. - James R. Buddenhagen, Aug 15 2010
Create the simple continued fraction from Pythagorean triples to get [2n + 1; 2n^2 + 2n, 2n^2 + 2n + 1]; its value equals the rational number 2n + 1 + a(n) / (4n^4 + 8n^3 + 6n^2 + 2n + 1). - J. M. Bergot, Sep 30 2011
a(n), n >= 1, has in its prime number factorization only primes of the form 4*k+1, i.e., congruent to 1 (mod 4) (see A002144). This follows from the fact that a(n) is a primitive sum of two squares and odd. See Theorem 3.20, p. 164, in the given Niven-Zuckerman-Montgomery reference. E.g., a(3) = 25 = 5^2, a(6) = 85 = 5*17. - Wolfdieter Lang, Mar 08 2012
From Ant King, Jun 15 2012: (Start)
a(n) is congruent to 1 (mod 4) for all n.
The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 5, 4, 7, 5, 7, 4, 5, 1.
The units' digits of the a(n) form a purely periodic palindromic 5-cycle 1, 5, 3, 5, 1.
(End)
Number of integer solutions (x,y) of |x| + |y| <= n. Geometrically: number of lattice points inside a square with vertices (n,0), (0,-n), (-n,0), (0,n). - César Eliud Lozada, Sep 18 2012
(a(n)-1)/a(n) = 2*x / (1+x^2) where x = n/(n+1). Note that in this form, this is the velocity-addition formula according to the special theory of relativity (two objects traveling at 1/(n+1) slower than c relative to each other appear to travel at 1/a(n) less than c to a stationary observer). - Christian N. K. Anderson, May 20 2013 [Corrected by Rémi Guillaume, May 22 2025]
A geometric curiosity: the envelope of the circles x^2 + (y-a(n)/2)^2 = ((2n+1)/2)^2 is the parabola y = x^2, the n=0 circle being the osculating circle at the parabola vertex. - Jean-François Alcover, Dec 02 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; a(n-1) gives the number of internal regions into which the plane is divided (cf. A051890, A046092); a(n-1) = A051890(n) - 1 = A046092(n-1) + 1. - Jaroslav Krizek, Dec 27 2013
a(n) is also, of course, the scalar product of the 2-vector (n, n+1) (or (n+1, n)) with itself. The unique inverse of (n, n+1) as vector in the Clifford algebra over the Euclidean 2-space is (1/a(n))(0, n, n+1, 0) (similarly for the other vector). In general the unique inverse of such a nonzero vector v (odd element in Cl_2) is v^(-1) = (1/|v|^2) v. Note that the inverse with respect to the scalar product is not unique for any nonzero vector. See the P. Lounesto reference, sects. 1.7 - 1.12, pp. 7-14. See also the Oct 15 2014 comment in A147973. - Wolfdieter Lang, Nov 06 2014
Subsequence of A004431, for n >= 1. - Bob Selcoe, Mar 23 2016
Numbers k such that 2k - 1 is a perfect square. - Juri-Stepan Gerasimov, Apr 06 2016
The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 574", based on the 5-celled von Neumann neighborhood. - Robert Price, May 13 2016
a(n) is the first integer in a sum of (2*n + 1)^2 consecutive integers that equals (2*n + 1)^4. - Patrick J. McNab, Dec 24 2016
Central elements of odd-length rows of the triangular array of positive integers. a(n) is the mean of the numbers in the (2*n + 1)-th row of this triangle. - David James Sycamore, Aug 01 2018
Intersection of A000982 and A080827. - David James Sycamore, Aug 07 2018
An off-diagonal of the array of Delannoy numbers, A008288, (or a row/column when the array is shown as a square). As such, this is one of the crystal ball sequences. - Jack W Grahl, Feb 15 2021 and Shel Kaphan, Jan 18 2023
a(n) appears as a solution to a "Riddler Express" puzzle on the FiveThirtyEight website. The Jan 21 2022 issue (problem) and the Jan 28 2022 issue (solution) present the following puzzle and include a proof. - Fold a square piece of paper in half, obtaining a rectangle. Fold again to obtain a square with 1/4 the size of the original square. Then make n cuts through the folded paper. a(n) is the greatest number of pieces of the unfolded paper after the cutting. - Manfred Boergens, Feb 22 2022
a(n) is (1/6) times the number of 2 X 2 triangles in the n-th order hexagram with 12*n^2 cells. - Donghwi Park, Feb 06 2024
If k is a centered square number, its index in this sequence is n = (sqrt(2k-1)-1)/2. - Rémi Guillaume, Mar 30 2025.
Row sums of the symmetric triangle of odd numbers [1]; [1, 3, 1]; [1, 3, 5, 3, 1]; [1, 3, 5, 7, 5, 3, 1]; .... - Marco Zárate, Jun 15 2025

			G.f.: 1 + 5*x + 13*x^2 + 25*x^3 + 41*x^4 + 61*x^5 + 85*x^6 + 113*x^7 + 145*x^8 + ...
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first four such partitions, corresponding to n = 0,1,2,3, i.e., to a(n) = 1,5,13,25, are 1, 3+1+1, 5+3+3+1+1, 7+5+5+3+3+1+1. - _Augustine O. Munagi_, Dec 18 2008

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
Pertti Lounesto, Clifford Algebras and Spinors, second edition, Cambridge University Press, 2001.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Travers et al., The Mysterious Lost Proof, Using Advanced Algebra, (1976), pp. 27.

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
U. Alfred, n and n+1 consecutive integers with equal sums of squares, Math. Mag., 35 (1962), 155-164.
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Section 2.3.
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Matthias Beck, Moshe Cohen, Jessica Cuomo, and Paul Gribelyuk, The number of "magic" squares and hypercubes, arXiv:math/0201013 [math.CO], 2002-2005.
Arthur T. Benjamin and Doron Zeilberger, Pythagorean Primes and Palindromic Continued Fractions, Electronic Journal of Combinatorial Number Theory, 5(1) 2005, #A30.
J. A. De Loera, D. C. Haws and M. Koppe, Ehrhart Polynomials of Matroid Polytopes and Polymatroids, arXiv:0710.4346 [math.CO], 2007; Discrete Comput. Geom., 42 (2009), 670-702.
FiveThirtyEight, "Riddler Express" paper cutting problem and solution, Jan 28 2022.
D. C. Haws, Matroids [Broken link, Oct 30 2017]
D. C. Haws, Matroids [Copy on website of Matthias Koeppe]
D. C. Haws, Matroids [Cached copy, pdf file only]
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, pp. 22 and 36.
Milan Janjic, Two Enumerative Functions. [Broken link; WayBackMachine archive.]
Milan Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
Ron Knott, Pythagorean Triples and Online Calculators.
G. Kreweras, Sur les hiérarchies de segments, Cahiers Bureau Universitaire Recherche Opérationnelle, Cahier 20, Inst. Statistiques, Univ. Paris, 1973.
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
A. O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501.
Mitchell Paukner, Lucy Pepin, Manda Riehl, and Jarred Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
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
John A. Jr. Rochowicz, Harmonic Numbers: Insights, Approximations and Applications, Spreadsheets in Education (eJSiE), 2015, Vol. 8: Iss. 2, Article 4.
Amelia Carolina Sparavigna, Groupoid of OEIS A001844 Numbers (centered square numbers), Politecnico di Torino, Italy (2019).
R. G. Stanton and D. D. Cowan, Note on a "square" functional equation, SIAM Rev., 12 (1970), 277-279.
David James Sycamore, Triangular array
Leo Tavares, Illustration: Diamond Rows
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Centered Polygonal Number, Centered Square Number, Diamond, Pythagorean Triple, and von Neumann Neighborhood.
Index entries for sequences related to centered polygonal numbers
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Nicolay Avilov, Graphical representation of the sequence members

X values are A005408; Y values are A046092.
Cf. A008586 (first differences), A005900 (partial sums), A254373 (digital roots).
Cf. A069894, A000217, A000290, A001263, A001788, A001845, A002061, A002144, A003215, A005448, A005891, A005917, A008844 (square terms), A027862 (prime terms), A048395, A051890, A056106, A101096, A127876, A128064, A132778, A147973, A153869, A240876, A251599, A000982, A080827.
Subsequence of A004431.
Right edge of A055096; main diagonal of A069480, A078475, A129312.
Row n=2 (or column k=2) of A008288.
Cf. A016754.

a001844 n = 2 * n * (n + 1) + 1
a001844_list = zipWith (+) a000290_list $ tail a000290_list
-- Reinhard Zumkeller, Dec 04 2012

[2*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Jan 19 2013

[n: n in [0..4400] | IsSquare(2*n-1)]; // Juri-Stepan Gerasimov, Apr 06 2016

A001844:=-(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation

Table[2n(n + 1) + 1, {n, 0, 50}]
FoldList[#1 + #2 &, 1, 4 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
maxn := 47; Flatten[Table[SeriesCoefficient[Series[(n + (n - 1)*x)/(1 - x)^2, {x, 0, maxn}], k], {n, maxn}, {k, n - 1, n - 1}]] (* L. Edson Jeffery, Aug 24 2014 *)
CoefficientList[ Series[-(x^2 + 2x + 1)/(x - 1)^3, {x, 0, 48}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 5, 13}, 48] (* Robert G. Wilson v, Aug 01 2018 *)
Total/@Partition[Range[0,50]^2,2,1] (* Harvey P. Dale, Dec 05 2020 *)
Table[ j! Coefficient[Series[Exp[x]*(1 + 4*x + 2*x^2), {x, 0, 20}], x,
j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 07 2023 *)

PARI
```
{a(n) = 2*n*(n+1) + 1};
```

x='x+O('x^200); Vec((1+x)^2/(1-x)^3) \\ Altug Alkan, Mar 23 2016

print([2*n*(n+1)+1 for n in range(48)]) # Michael S. Branicky, Jan 05 2021

[i**2 + (i + 1)**2 for i in range(46)] # Zerinvary Lajos, Jun 27 2008

a(n) = 2*n^2 + 2*n + 1 = n^2 + (n+1)^2.
a(n) = 1 + 3 + 5 + ... + 2*n-1 + 2*n+1 + 2*n-1 + ... + 3 + 1. - Amarnath Murthy, May 28 2001
a(n) = 1/real(z(n+1)) where z(1)=i, (i^2=-1), z(k+1) = 1/(z(k)+2i). - Benoit Cloitre, Aug 06 2002
Nearest integer to 1/Sum_{k>n} 1/k^3. - Benoit Cloitre, Jun 12 2003
G.f.: (1+x)^2/(1-x)^3.
E.g.f.: exp(x)*(1+4x+2x^2).
a(n) = a(n-1) + 4n.
a(-n) = a(n-1).
a(n) = A064094(n+3, n) (fourth diagonal).
a(n) = 1 + Sum_{j=0..n} 4*j. - Xavier Acloque, Oct 08 2003
a(n) = A046092(n)+1 = (A016754(n)+1)/2. - Lekraj Beedassy, May 25 2004
a(n) = Sum_{k=0..n+1} (-1)^k*binomial(n, k)*Sum_{j=0..n-k+1} binomial(n-k+1, j)*j^2. - Paul Barry, Dec 22 2004
a(n) = ceiling((2n+1)^2/2). - Paul Barry, Jul 16 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=5, a(2)=13. - Jaume Oliver Lafont, Dec 02 2008
a(n)*a(n-1) = 4*n^4 + 1 for n > 0. - Reinhard Zumkeller, Feb 12 2009
Prefaced with a "1" (1, 1, 5, 13, 25, 41, ...): a(n) = 2*n*(n-1)+1. - Doug Bell, Feb 27 2009
a(n) = sqrt((A056220(n)^2 + A056220(n+1)^2) / 2). - Doug Bell, Mar 08 2009
a(n) = floor(2*(n+1)^3/(n+2)). - Gary Detlefs, May 20 2010
a(n) = A000330(n) - A000330(n-2). - Keith Tyler, Aug 10 2010
a(n) = A069894(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Jun 12 2012
Sum_{n>=0} 1/a(n) = (Pi/2)*tanh(Pi/2) = 1.4406595199775... = A228048. - Ant King, Jun 15 2012
a(n) = A209297(2*n+1,n+1). - Reinhard Zumkeller, Jan 19 2013
a(n)^3 = A048395(n)^2 + A048395(-n-1)^2. - Vincenzo Librandi, Jan 19 2013
a(n) = A000217(2n+1) - n. - Ivan N. Ianakiev, Nov 08 2013
a(n) = A251599(3*n+1). - Reinhard Zumkeller, Dec 13 2014
a(n) = A101321(4,n). - R. J. Mathar, Jul 28 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = Sum_{k=0..n} A008574(k).
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = exp(-1) = A068985. (End)
a(n) = 4 * A000217(n) + 1. - Bruce J. Nicholson, Jul 10 2017
a(n) = A002522(n) + A005563(n) = A002522(n+1) + A005563(n-1). - Bruce J. Nicholson, Aug 05 2017
Sum_{n>=0} a(n)/n! = 7*e. Sum_{n>=0} 1/a(n) = A228048. - Amiram Eldar, Jun 20 2020
a(n) = A000326(n+1) + A000217(n-1). - Charlie Marion, Nov 16 2020
a(n) = Integral_{x=0..2n+2} |1-x| dx. - Pedro Caceres, Dec 29 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi)*sinh(Pi/2). (End)
a(n) = A001651(n+1) + 1 - A028242(n). - Charlie Marion, Apr 05 2022
a(n) = A016754(n) - A046092(n). - Leo Tavares, Sep 16 2022
For n>0, a(n) = A101096(n+2) / 30. - Andy Nicol, Feb 06 2025
From Rémi Guillaume, Apr 21 2025: (Start)
a(n) = (2*A003215(n)+1)/3.
a(n) = (4*A005448(n+1)-1)/3.
a(n) + a(n-1) = A001845(n) - A001845(n-1), for n >= 1.
a(n) = (A005917(n+1))/(2n+1). (End)

Partially edited by Joerg Arndt, Mar 11 2010

A001105 a(n) = 2*n^2.

Original entry on oeis.org

0, 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418

Offset: 0

Bernd.Walter(AT)frankfurt.netsurf.de

Number of edges of the complete bipartite graph of order 3n, K_{n,2n}. - Roberto E. Martinez II, Jan 07 2002
"If each period in the periodic system ends in a rare gas ..., the number of elements in a period can be found from the ordinal number n of the period by the formula: L = ((2n+3+(-1)^n)^2)/8..." - Nature, Jun 09 1951; Nature 411 (Jun 07 2001), p. 648. This produces the present sequence doubled up.
Let z(1) = i = sqrt(-1), z(k+1) = 1/(z(k)+2i); then a(n) = (-1)*Imag(z(n+1))/Real(z(n+1)). - Benoit Cloitre, Aug 06 2002
Maximum number of electrons in an atomic shell with total quantum number n. Partial sums of A016825. - Jeremy Gardiner, Dec 19 2004
Arithmetic mean of triangular numbers in pairs: (1+3)/2, (6+10)/2, (15+21)/2, ... . - Amarnath Murthy, Aug 05 2005
These numbers form a pattern on the Ulam spiral similar to that of the triangular numbers. - G. Roda, Oct 20 2010
Integral areas of isosceles right triangles with rational legs (legs are 2n and triangles are nondegenerate for n > 0). - Rick L. Shepherd, Sep 29 2009
Even squares divided by 2. - Omar E. Pol, Aug 18 2011
Number of stars when distributed as in the U.S.A. flag: n rows with n+1 stars and, between each pair of these, one row with n stars (i.e., n-1 of these), i.e., n*(n+1)+(n-1)*n = 2*n^2 = A001105(n). - César Eliud Lozada, Sep 17 2012
Apparently the number of Dyck paths with semilength n+3 and an odd number of peaks and the central peak having height n-3. - David Scambler, Apr 29 2013
Sum of the partition parts of 2n into exactly two parts. - Wesley Ivan Hurt, Jun 01 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective odd leg a (A180620); sequence gives values c-a, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
Number of roots in the root systems of type B_n and C_n (for n > 1). - Tom Edgar, Nov 05 2013
Area of a square with diagonal 2n. - Wesley Ivan Hurt, Jun 18 2014
This sequence appears also as the first and second member of the quartet [a(n), a(n), p(n), p(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p(n) = A046092(n). See an Oct 15 2014 comment on A147973 where also a reference is given. - Wolfdieter Lang, Oct 16 2014
a(n) are the only integers m where (A000005(m) + A000203(m)) = (number of divisors of m + sum of divisors of m) is an odd number. - Richard R. Forberg, Jan 09 2015
a(n) represents the first term in a sum of consecutive integers running to a(n+1)-1 that equals (2n+1)^3. - Patrick J. McNab, Dec 24 2016
Also the number of 3-cycles in the (n+4)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 29 2017
Also the Wiener index of the n-cocktail party graph for n > 1. - Eric W. Weisstein, Sep 07 2017
Numbers represented as the palindrome 242 in number base B including B=2 (binary), 3 (ternary) and 4: 242(2)=18, 242(3)=32, 242(4)=50, ... 242(9)=200, 242(10)=242, ... - Ron Knott, Nov 14 2017
a(n) is the square of the hypotenuse of an isosceles right triangle whose sides are equal to n. - Thomas M. Green, Aug 20 2019
The sequence contains all odd powers of 2 (A004171) but no even power of 2 (A000302). - Torlach Rush, Oct 10 2019
From Bernard Schott, Aug 31 2021 and Sep 16 2021: (Start)
Apart from 0, integers such that the number of even divisors (A183063) is odd.
Proof: every n = 2^q * (2k+1), q, k >= 0, then 2*n^2 = 2^(2q+1) * (2k+1)^2; now, gcd(2, 2k+1) = 1, tau(2^(2q+1)) = 2q+2 and tau((2k+1)^2) = 2u+1 because (2k+1)^2 is square, so, tau(2*n^2) = (2q+2) * (2u+1).
The 2q+2 divisors of 2^(2q+1) are {1, 2, 2^2, 2^3, ..., 2^(2q+1)}, so 2^(2q+1) has 2q+1 even divisors {2^1, 2^2, 2^3, ..., 2^(2q+1)}.
Conclusion: these 2q+1 even divisors create with the 2u+1 odd divisors of (2k+1)^2 exactly (2q+1)*(2u+1) even divisors of 2*n^2, and (2q+1)*(2u+1) is odd. (End)
a(n) with n>0 are the numbers with period length 2 for Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022
Number of points at L1 distance = 2 from any given point in Z^n. - Shel Kaphan, Feb 25 2023
Integer that multiplies (h^2)/(m*L^2) to give the energy of a 1-D quantum mechanical particle in a box whenever it is an integer multiple of (h^2)/(m*L^2), where h = Planck's constant, m = mass of particle, and L = length of box. - A. Timothy Royappa, Mar 14 2025

			a(3) = 18; since 2(3) = 6 has 3 partitions with exactly two parts: (5,1), (4,2), (3,3).  Adding all the parts, we get: 1 + 2 + 3 + 3 + 4 + 5 = 18. - _Wesley Ivan Hurt_, Jun 01 2013

Peter Atkins, Julio De Paula, and James Keeler, "Atkins' Physical Chemistry," Oxford University Press, 2023, p. 31.
Arthur Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
Martin Gardner, The Colossal Book of Mathematics, Classic Puzzles, Paradoxes and Problems, Chapter 2 entitled "The Calculus of Finite Differences," W. W. Norton and Company, New York, 2001, pages 12-13.
L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 44.
Alain M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000, p. 213.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Lancelot Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013. - _N. J. A. Sloane_, Feb 13 2013
Vladimir Ladma, Magic Numbers.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv:1406.3081 [math.CO], 2014.
Eric Weisstein's World of Mathematics, Cocktail Party Graph.
Eric Weisstein's World of Mathematics, Graph Cycle.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A006331 (partial sums), A016742, A056106, A116471, A245508, A251599, A002061, A165900.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Cf. A058331 and A247375. - Bruno Berselli, Sep 16 2014
Cf. A194715 (4-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017
Cf. A139098, A077591.
Cf. A000217, A002266.
Integers such that: this sequence (the number of even divisors is odd), A028982 (the number of odd divisors is odd), A028983 (the number of odd divisors is even), A183300 (the number of even divisors is even).

List([0..50],n->2*n^2); # Muniru A Asiru, Feb 24 2019

a001105 = a005843 . a000290  -- Reinhard Zumkeller, Dec 12 2012

[2*n^2: n in [0..50] ]; // Vincenzo Librandi, Apr 30 2011

A001105:=n->2*n^2; seq(A001105(k), k=0..100); # Wesley Ivan Hurt, Oct 29 2013

2 Range[0, 50]^2 (* Harvey P. Dale, Jan 23 2011 *)
LinearRecurrence[{3, -3, 1}, {2, 8, 18}, {0, 20}] (* Eric W. Weisstein, Jul 28 2017 *)
2 PolygonalNumber[4, Range[0, 20]] (* Eric W. Weisstein, Jul 28 2017 *)

a(n) = 2*n^2; \\ Charles R Greathouse IV, Jun 16 2011

def A001105(n): return n**2<<1 # Chai Wah Wu, Aug 04 2025

[2*n^2 for n in (0..20)] # G. C. Greubel, Feb 22 2019

a(n) = (-1)^(n+1) * A053120(2*n, 2).
G.f.: 2*x*(1+x)/(1-x)^3.
a(n) = A100345(n, n).
Sum_{n>=1} 1/a(n) = Pi^2/12 =A072691. [Jolley eq. 319]. - Gary W. Adamson, Dec 21 2006
a(n) = A049452(n) - A033991(n). - Zerinvary Lajos, Jun 12 2007
a(n) = A016742(n)/2. - Zerinvary Lajos, Jun 20 2008
a(n) = 2 * A000290(n). - Omar E. Pol, May 14 2008
a(n) = 4*n + a(n-1) - 2, n > 0. - Vincenzo Librandi
a(n) = A002378(n-1) + A002378(n). - Joerg M. Schuetze (joerg(AT)cyberheim.de), Mar 08 2010 [Corrected by Klaus Purath, Jun 18 2020]
a(n) = A176271(n,k) + A176271(n,n-k+1), 1 <= k <= n. - Reinhard Zumkeller, Apr 13 2010
a(n) = A007607(A000290(n)). - Reinhard Zumkeller, Feb 12 2011
For n > 0, a(n) = 1/coefficient of x^2 in the Maclaurin expansion of 1/(cos(x)+n-1). - Francesco Daddi, Aug 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Artur Jasinski, Nov 24 2011
a(n) = A070216(n,n) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) = A014132(2*n-1,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n) = A000217(n) + A000326(n). - Omar E. Pol, Jan 11 2013
(a(n) - A000217(k))^2  = A000217(2*n-1-k)*A000217(2*n+k) + n^2, for all k. - Charlie Marion, May 04 2013
a(n) = floor(1/(1-cos(1/n))), n > 0. - Clark Kimberling, Oct 08 2014
a(n) = A251599(3*n-1) for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+4)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = A002061(n+1) + A165900(n). - Torlach Rush, Feb 21 2019
E.g.f.: 2*exp(x)*x*(1 + x). - Stefano Spezia, Oct 12 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Jul 03 2020
From Amiram Eldar, Feb 03 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sqrt(2)*sinh(Pi/sqrt(2))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(2)*sin(Pi/sqrt(2))/Pi. (End)

A046092 4 times triangular numbers: a(n) = 2n(n+1).

Original entry on oeis.org

0, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480, 544, 612, 684, 760, 840, 924, 1012, 1104, 1200, 1300, 1404, 1512, 1624, 1740, 1860, 1984, 2112, 2244, 2380, 2520, 2664, 2812, 2964, 3120, 3280, 3444, 3612, 3784, 3960, 4140, 4324

Offset: 0

N. J. A. Sloane and Eric W. Weisstein

Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; sequence gives Y values. X values are 1, 3, 5, 7, 9, ... (A005408), Z values are A001844.
In the triple (X, Y, Z) we have X^2=Y+Z. Actually, the triple is given by {x, (x^2 -+ 1)/2}, where x runs over the odd numbers (A005408) and x^2 over the odd squares (A016754). - Lekraj Beedassy, Jun 11 2004
a(n) is the number of edges in n X n square grid with all horizontal and vertical segments filled in. - Asher Auel, Jan 12 2000 [Corrected by Felix Huber, Apr 09 2024]
a(n) is the only number satisfying an inequality related to zeta(2) and zeta(3): Sum_{i>a(n)+1} 1/i^2 < Sum_{i>n} 1/i^3 < Sum_{i>a(n)} 1/i^2. - Benoit Cloitre, Nov 02 2001
Number of right triangles made from vertices of a regular n-gon when n is even. - Sen-Peng Eu, Apr 05 2001
Number of ways to change two non-identical letters in the word aabbccdd..., where there are n type of letters. - Zerinvary Lajos, Feb 15 2005
a(n) is the number of (n-1)-dimensional sides of an (n+1)-dimensional hypercube (e.g., squares have 4 corners, cubes have 12 edges, etc.). - Freek van Walderveen (freek_is(AT)vanwal.nl), Nov 11 2005
From Nikolaos Diamantis (nikos7am(AT)yahoo.com), May 23 2006: (Start)
Consider a triangle, a pentagon, a heptagon, ..., a k-gon where k is odd. We label a triangle with n=1, a pentagon with n=2, ..., a k-gon with n = floor(k/2). Imagine a player standing at each vertex of the k-gon.
Initially there are 2 frisbees, one held by each of two neighboring players. Every time they throw the frisbee to one of their two nearest neighbors with equal probability. Then a(n) gives the average number of steps needed so that the frisbees meet.
I verified this by simulating the processes with a computer program. For example, a(2) = 12 because in a pentagon that's the expected number of trials we need to perform. That is an exercise in Concrete Mathematics and it can be done using generating functions. (End)
A diagonal of A059056. - Zerinvary Lajos, Jun 18 2007
If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n-1) is equal to the number of 2-subsets of X containing none of X_i, (i=1,...,n). - Milan Janjic, Jul 16 2007
X values of solutions to the equation 2*X^3 + X^2 = Y^2. To find Y values: b(n) = 2n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007
Number of (n+1)-permutations of 3 objects u,v,w, with repetition allowed, containing n-1 u's. Example: a(1)=4 because we have vv, vw, wv and ww; a(2)=12 because we can place u in each of the previous four 2-permutations either in front, or in the middle, or at the end. - Zerinvary Lajos, Dec 27 2007
Sequence found by reading the line from 0, in the direction 0, 4, ... and the same line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, May 03 2008
a(n) is also the least weight of self-conjugate partitions having n different even parts. - Augustine O. Munagi, Dec 18 2008
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers of even integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,1)-(-1)^k P(n,2k+1))/2. Here n=2, thus
a(k) = |(P(2,1) - (-1)^k*P(2,2k+1))/2|. (End)
The sum of squares of n+1 consecutive numbers between a(n)-n and a(n) inclusive equals the sum of squares of n consecutive numbers following a(n). For example, for n = 2, a(2) = 12, and the corresponding equation is 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Tanya Khovanova, Jul 20 2009
Number of roots in the root system of type D_{n+1} (for n>2). - Tom Edgar, Nov 05 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of intersections of these ellipses (cf. A051890, A001844); a(n) = A051890(n+1) - 2 = A001844(n) - 1. - Jaroslav Krizek, Dec 27 2013
a(n) appears also as the second member of the quartet [p0(n), a(n), p2(n), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p2(n) = A054000(n+1) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014
a(n) appears also as the third and fourth member of the quartet [p0(n), p0(n), a(n), a(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A001105(n). - Wolfdieter Lang, Oct 16 2014
Consider two equal rectangles composed of unit squares. Then surround the 1st rectangle with 1-unit-wide layers to build larger rectangles, and surround the 2nd rectangle just to hide the previous layers. If r(n) and h(n) are the number of unit squares needed for n layers in the 1st case and the 2nd case, then for all rectangles, we have a(n) = r(n) - h(n) for n>=1. - Michel Marcus, Sep 28 2015
When greater than 4, a(n) is the perimeter of a Pythagorean triangle with an even short leg 2*n. - Agola Kisira Odero, Apr 26 2016
Also the number of minimum connected dominating sets in the (n+1)-cocktail party graph. - Eric W. Weisstein, Jun 29 2017
a(n+1) is the harmonic mean of A000384(n+2) and A014105(n+1). - Bob Andriesse, Apr 27 2019
Consider a circular cake from which wedges of equal center angle c are cut out in clockwise succession and turned around so that the bottom comes to the top. This goes on until the cake shows its initial surface again. An interesting case occurs if 360°/c is not an integer. Then, with n = floor(360°/c), the number of wedges which have to be cut out and turned equals a(n). (For the number of cutting line segments see A005408.) - According to Peter Winkler's book "Mathematical Mind-Benders", which presents the problem and its solution (see Winkler, pp. 111, 115) the problem seems to be of French origin but little is known about its history. - Manfred Boergens, Apr 05 2022
a(n-3) is the maximum irregularity over all maximal 2-degenerate graphs with n vertices.  The extremal graphs are 2-stars (K_2 joined to n-2 independent vertices).  (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
Number of ways of placing a domino on a (n+1)X(n+1) board of squares. - R. J. Mathar, Apr 24 2024
The sequence terms are the exponents in the expansion of (1/(1 + x)) * Sum_{n >= 0} x^n * Product_{k = 1..n} (1 - x^(2*k-1))/(1 + x^(2*k+1)) = 1 - x^4 + x^12 - x^24 + x^40 - x^60 + - ... (Andrews and Berndt, Entry 9.3.3, p. 229). Cf. A153140. - Peter Bala, Feb 15 2025
Number of edges in an (n+1)-dimensional orthoplex. 2D orthoplexes (diamonds) have 4 edges, 3D orthoplexes (octahedrons) have 12 edges, 4D orthoplexes (16-cell) have 24 edges, and so on. - Aaron Franke, Mar 23 2025

			a(7)=112 because 112 = 2*7*(7+1).
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first such partitions, corresponding to a(n)=1,2,3,4, are 2+2, 4+4+2+2, 6+6+4+4+2+2, 8+8+6+6+4+4+2+2. - _Augustine O. Munagi_, Dec 18 2008

George E. Andrews and Bruce C. Berndt, Ramanujan's Lost Notebook, Part I, Springer, 2005.
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
Albert H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
Ronald L. Graham, D. E. Knuth and Oren Patashnik, Concrete Mathematics, Reading, Massachusetts: Addison-Wesley, 1994.
Peter Winkler, Mathematical Mind-Benders, Wellesley, Massachusetts: A K Peters, 2007.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
H. J. Brothers, Pascal's Prism: Supplementary Material.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi 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, Svante Janson, and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, Vol. 13, No. 4 (2017), Article #47.
Z. Janelidze, F. van Niekerk, and J. Viljoen, What is the maximal connected partial symmetry index of a connected graph of a given size?, arXiv:2502.00704 [math.CO], 2025. See p. 3.
Milan Janjic, Two Enumerative Functions
Ron Knott, Pythagorean Triples and Online Calculators
Tanya Khovanova, A Miracle Equation.
Augustine O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501. [From _Augustine O. Munagi_, Dec 18 2008]
Enrique Navarrete and Daniel Orellana, Finding Prime Numbers as Fixed Points of Sequences, arXiv:1907.10023 [math.NT], 2019.
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Amelia Carolina Sparavigna, Some Groupoids and their Representations by Means of Integer Sequences, International Journal of Sciences (2019) Vol. 8, No. 10.
Rusliansyah D. Suprijanto, Observation on Sums of Powers of Integers Divisible by Four, Applied Mathematical Sciences, Vol. 8, 2014, no. 45, 2219 - 2226.
Leo Tavares, Illustration: Diamond Rows
Herman Tulleken, Polyominoes 2.2: How they fit together, (2019).
Eric Weisstein's World of Mathematics, Aztec Diamond.
Eric Weisstein's World of Mathematics, Cocktail Party Graph.
Eric Weisstein's World of Mathematics, Connected Dominating Set.
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Hamiltonian Path.
Eric Weisstein's World of Mathematics, Pythagorean Triple.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A045943, A028895, A002943, A054000, A000330, A007290, A002378, A033996, A124080, A028896, A049598, A005563, A000217, A033586, A085250.
Main diagonal of array in A001477.
Equals A033996/2. Cf. A001844. - Augustine O. Munagi, Dec 18 2008
Cf. A078371, A141530 (see Librandi's comment in A078371).
Cf. A097080, A001845.
Cf. similar sequences listed in A299645.
Cf. A005408.
Cf. A016754.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

a046092 = (* 2) . a002378  -- Reinhard Zumkeller, Dec 15 2013

[2*n*(n+1): n in [0..50]]; // Vincenzo Librandi, Oct 04 2011

Table[2 n (n + 1), {n, 0, 50}] (* Stefan Steinerberger, Apr 03 2006 *)
LinearRecurrence[{3, -3, 1}, {0, 4, 12}, 50] (* Harvey P. Dale, Jul 25 2011 *)
4*Binomial[Range[50], 2] (* Harvey P. Dale, Jul 25 2011 *)

A046092(n):=2*n*(n+1)$
makelist(A046092(n),n,0,30); /* Martin Ettl, Nov 08 2012 */

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

def A046092(n): return n*(n+1)<<1 # Chai Wah Wu, Mar 11 2025

a(n) = A100345(n+1, n-1) for n>0.
a(n) = 2*A002378(n) = 4*A000217(n). - Lekraj Beedassy, May 25 2004
a(n) = C(2n, 2) - n = 4*C(n, 2). - Zerinvary Lajos, Feb 15 2005
From Lekraj Beedassy, Jun 04 2006: (Start)
a(n) - a(n-1)=4*n.
Let k=a(n). Then a(n+1) = k + 2*(1 + sqrt(2k + 1)). (End)
Array read by rows: row n gives A033586(n), A085250(n+1). - Omar E. Pol, May 03 2008
O.g.f.:4*x/(1-x)^3; e.g.f.: exp(x)*(2*x^2+4*x). - Geoffrey Critzer, May 17 2009
From Stephen Crowley, Jul 26 2009: (Start)
a(n) = 1/int(-(x*n+x-1)*(step((-1+x*n)/n)-1)*n*step((x*n+x-1)/(n+1)),x=0..1) where step(x)=piecewise(x<0,0,0<=x,1) is the Heaviside step function.
Sum_{n>=1} 1/a(n) = 1/2. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=12. - Harvey P. Dale, Jul 25 2011
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} (sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A001844(n) - 1. - Omar E. Pol, Oct 03 2011
(a(n) - A000217(k))^2  = A000217(2n-k)*A000217(2n+1+k) - (A002378(n) - A000217(k)), for all k. See also A001105. - Charlie Marion, May 09 2013
From Ivan N. Ianakiev, Aug 30 2013: (Start)
a(n)*(2m+1)^2 + a(m) = a(n*(2m+1)+m), for any nonnegative integers n and m.
t(k)*a(n) + t(k-1)*a(n+1) = a((n+1)*(t(k)-t(k-1)-1)), where k>=2, n>=1, t(k)=A000217(k). (End)
a(n) = A245300(n,n). - Reinhard Zumkeller, Jul 17 2014
2*a(n)+1 = A016754(n) = A005408(n)^2, the odd squares. - M. F. Hasler, Oct 02 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) - 1/2 = A187832. - Ilya Gutkovskiy, Mar 16 2017
a(n) = lcm(2*n,2*n+2). - Enrique Navarrete, Aug 30 2017
a(n)*a(n+k) + k^2 = m^2 (a perfect square), n >= 1, k >= 0. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/2)/(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -2*cos(sqrt(3)*Pi/2)/Pi. (End)
a(n) = A016754(n) - A001844(n). - Leo Tavares, Sep 20 2022

A054000 a(n) = 2*n^2 - 2.

Original entry on oeis.org

0, 6, 16, 30, 48, 70, 96, 126, 160, 198, 240, 286, 336, 390, 448, 510, 576, 646, 720, 798, 880, 966, 1056, 1150, 1248, 1350, 1456, 1566, 1680, 1798, 1920, 2046, 2176, 2310, 2448, 2590, 2736, 2886, 3040, 3198, 3360, 3526, 3696, 3870, 4048, 4230, 4416

Offset: 1

Asher Auel, Jan 12 2000

a(n) is the number of edges in (n+1) X (n+1) square grid with all horizontal, vertical and great diagonal segments filled in.
Nonnegative X values of integer solutions to the equation 2*X^3 + 4*X^2 = Y^2. To find Y values: b(n) = 2*n*(2*n^2 - 2). - Mohamed Bouhamida, Nov 06 2007
Second term of an arithmetic progression of 5 numbers with common difference 2n+1. The sum of squares of such 5 terms equals the sum of squares of 5 consecutive numbers starting a(n) + 2n + 1. - Carmine Suriano, Oct 16 2013
For m > 2, a(m-1) = 2*m*(m-2) is the number of Hamiltonian circuits on an m-gonal bipyramid with labeled vertices. - Stanislav Sykora, Jul 22 2014
a(n+1), n >= 0, appears also as the third member of the quartet [p0(n), p1(n), a(n+1), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1(n) = A046092(n) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014
From Bui Quang Tuan, Mar 31 2015: (Start)
For n >= 2, a(n) is the total sum of all numbers on the perimeter of a square consisting of n columns, each of which contains n numbers 1, 2, 3, ..., n.
Here is an example with n = 5:
  1 1 1 1 1
  2 2 2 2 2
  3 3 3 3 3
  4 4 4 4 4
  5 5 5 5 5
where 1+1+1+1+1 + 2+2 + 3+3 + 4+4 + 5+5+5+5+5 = 48 = a(5).
(End)
Nonnegative k such that k/2+1 is a square. - Bruno Berselli, Apr 10 2018

			For n=5, a(5)=48 and 37^2 + 48^2 + 59^2 + 70^2 + 81^2 = 59^2 + 60^2 + 61^2 + 62^2 + 63^2. - _Carmine Suriano_, Oct 16 2013

G. C. Greubel, Table of n, a(n) for n = 1..5000
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

a(n) = A100345(n+1, n-4) for n>2.

Cf. A000217, A001082, A002378, A002943, A005563, A028347, A036666, A046092, A056220, A062717, A067725, A087475.

Maple
```
[ seq(2*n^2 - 2, n=1..60) ];
```

2 Range[50]^2 - 2 (* or *) LinearRecurrence[{3, -3, 1}, {0, 6, 16}, 50] (* Harvey P. Dale, Feb 03 2012 *)
CoefficientList[Series[2 x (3 - x) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Apr 01 2015 *)

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

a(n) = 4*n + a(n-1) - 2, with n>1, a(1)=0. - Vincenzo Librandi, Aug 06 2010
a(1)=0, a(2)=6, a(3)=16; for n>3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 03 2012
a(n) = (n+i)^2 + (n-i)^2, where i=sqrt(-1). - Bruno Berselli, Jan 23 2014
a(n) = 1*A000290(n-1) + 2*A000217(n-1) + 3*A001477(n-1). - J. M. Bergot, Apr 23 2014
G.f.: 2*x^2*(3 - x)/(1 - x)^3. - Vincenzo Librandi, Apr 01 2015
E.g.f.: 2*(x^2 + x -1)*exp(x) + 2. - G. C. Greubel, Jul 13 2017
a(n) + a(n+2) = A005843(n+1)^2. - Ezhilarasu Velayutham, May 30 2019
From Amiram Eldar, Dec 09 2021: (Start)
Sum_{n>=2} 1/a(n) = 3/8.
Sum_{n>=2} (-1)^n/a(n) = 1/8. (End)

A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

Original entry on oeis.org

0, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 20737, 54288, 142130, 372099, 974170, 2550408, 6677057, 17480760, 45765226, 119814915, 313679522, 821223648, 2149991425, 5628750624, 14736260450, 38580030723, 101003831722, 264431464440, 692290561601, 1812440220360

Offset: 0

Henry Bottomley, Feb 09 2001

Expansion of golden ratio (1+sqrt(5))/2 as an infinite product: phi = Product_{i>=0} (1+1/(Fibonacci(2*i+1) * Fibonacci(2*i+3)-1)) * (1-1/(Fibonacci(2*i+2) * Fibonacci(2i+4)+1)). - Thomas Baruchel, Nov 11 2003
Each of these is one short of or one over the square of a Fibonacci number (A007598). This means that a rectangle sized F(n) by F(n + 2) units can't be converted into a square with sides of length F(n + 1) units unless one square unit of material is added or removed. - Alonso del Arte, May 03 2011
These are the integer parts of the numerators of the numbers with continued fraction representations [1, 2, 2, 2, ...], [1, 1, 2, 2, 2, ...], [1, 1, 1, 2, 2, 2, ...], etc., that is, sqrt(2), (2+sqrt(2))/2, 3-sqrt(2), (10+sqrt(2))/7, (24-sqrt(2))/14, etc. - Geoffrey Caveney, May 03 2014
a(n) appears also as the third component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), A079472(n+1), a(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
Numbers with a continued fraction expansion with the repeating sequence of length n [1, 1, ..., 1, 2], n-1 ones followed by a single two, for n > = 1, appear to be equal to (F(n) + sqrt(a(n)))/F(n+1), where F(n) = A000045(n). - R. James Evans, Nov 21 2018
The preceding conjecture is true. Proof: For n >= 1 let c(n) := confrac(repeat(1^{n-1}, 2)) where 1^{k} denotes 1 taken k times. This can be computed, e.g. from [Perron, third and fourth eq. on p. 62], as c(n) = (F(n) + sqrt(F(n+1)^2 - (-1)^n)) / F(n+1), which is the conjectured formula because F(n+1)^2 - (-1)^n = a(n). - Wolfdieter Lang, Jan 05 2019

			G.f. = 2*x + 3*x^2 + 10*x^3 + 24*x^4 + 65*x^5 + 168*x^6 + ... - _Michael Somos_, Mar 18 2022

Oskar Perron, Die Lehre von den Kettenbrüchen, Band I, 3. Auflage, B. G. Teubner, Stuttgart, 1954, pp. 61-61.

Muniru A Asiru, Table of n, a(n) for n = 0..2374 (first 501 terms from Harry J. Smith)
Tim Jones (producer), Engineering Bits & Bytes: The Fibonacci Puzzle, Wayne State University College of Engineering (2011).
Eric H. Kuo, Applications of graphical condensation for enumerating matchings and tilings, arXiv:math/0304090 [math.CO], 2003.
Marc Renault, Properties of the Fibonacci Sequence Under Various Moduli, Master's Thesis, Wake Forest University, 1996.
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Bisection of A070550.
First differences of A059840.
Cf. A000045, A001519, A001654, A007598, A033999, A035513, A079472, A121801, A126358, A147973, A248161.

a:=List([0..30],n->Fibonacci(n)*Fibonacci(n+2));; Print(a); # Muniru A Asiru, Jan 05 2019

[Fibonacci(n)*Fibonacci(n+2): n in [0..30]]; // Vincenzo Librandi, Jul 02 2014

with(combinat): a:=n->fibonacci(n)*fibonacci(n+2): seq(a(n), n=0..26); # Zerinvary Lajos, Oct 07 2007

Table[Fibonacci[n]*Fibonacci[n+2],{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)

a(n) = fibonacci(n)*fibonacci(n + 2) \\ Harry J. Smith, Jun 30 2009

from sympy import fibonacci
[fibonacci(n)*fibonacci(n+2) for n in range(30)] # Stefano Spezia, Jan 05 2019

[fibonacci(n)*fibonacci(n+2) for n in range(30)] # G. C. Greubel, Nov 21 2018

a(n) = Fibonacci(n+1)^2 - (-1)^n = A007598(n+1) + A033999(n+1) = A000045(n+1)^2 - A033999(n).
G.f.: (2*x-x^2) / ((1+x)*(1-3*x+x^2)).
Sum_{n>=1} 1/a(n) = 1.
Sum_{n>=1} (-1)^n/a(n) = 2 - sqrt(5).
Sum_{n>=1} 1/a(2n-1) = 1/phi = (sqrt(5) - 1)/2. - Franz Vrabec, Sep 15 2005
Sum_{n>=1} 1/a(2n) = (3 - sqrt(5))/2. - Franz Vrabec, Nov 30 2009
a(n) = ((7+3*sqrt(5))/10)*((3+sqrt(5))/2)^(n-1) + ((7-3*sqrt(5))/10)*((3-sqrt(5))/2)^(n-1) + (3/5)*(-1)^(n-1). - Tim Monahan, Aug 09 2011
a(n) = (Lucas(n+1)^2 - Fibonacci(n+1)^2)/4. - Vincenzo Librandi, Aug 02 2014
a(n) = F(n-2)*F(n) + F(n-1)*F(n) + F(n-2)*F(n+1) + F(n-1)*F(n+1), where F=A000045, F(-2)=-1, F(-1)=1. - Bruno Berselli, Nov 03 2015
a(n) = A035513(1,n-1)*A035513(3,n-1)/2 = A035513(1,n-1)*A035513(4,n-1)/3. - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A001519(n+2). - R. J. Mathar, Oct 19 2021
a(n) = 2*A001654(n) - A001654(n-1). - R. J. Mathar, Oct 19 2021
a(n)+a(n+3) = 2*F(2n+5) = A126358(n+2). - Andrés Ventas, Oct 25 2021
Sum_{n>=1} Fibonacci(n+1)/a(n) = 2. - Amiram Eldar, Jan 11 2022
a(n) = a(-2-n) and a(n) + a(n+3) = 2*(a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Mar 18 2022

A139570 a(n) = 2n(n+3).

Original entry on oeis.org

0, 8, 20, 36, 56, 80, 108, 140, 176, 216, 260, 308, 360, 416, 476, 540, 608, 680, 756, 836, 920, 1008, 1100, 1196, 1296, 1400, 1508, 1620, 1736, 1856, 1980, 2108, 2240, 2376, 2516, 2660, 2808, 2960, 3116, 3276, 3440, 3608, 3780, 3956, 4136, 4320, 4508, 4700, 4896

Offset: 0

Omar E. Pol, May 19 2008

Numbers n such that 2*n + 9 is a square. - Vincenzo Librandi, Nov 24 2010

a(n) appears also as the fourth member of the quartet [p0(n), p1(n), p2(n), a(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1(n) = A046092(n), and p2(n) = A054000(n+1). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014

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

Cf. A000096, A001105, A020739, A022998, A028552, A046092, A054000, A067728, A147973.

CoefficientList[Series[4 x (2 - x)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, May 23 2014 *)

a(n)=2*n*(n+3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 2*A028552(n) = 2*n^2 + 6*n = n*(2*n+6).
a(n) = a(n-1) + 4*n + 4 (with a(0)=0). - Vincenzo Librandi, Nov 24 2010
From Paul Curtz, Mar 27 2011: (Start)
a(n) = A022998(n)*A022998(n+3).
a(n) = 4*A000096(n). (End)
G.f.: 4*x*(2 - x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 31 2011
From Amiram Eldar, Dec 23 2022: (Start)
Sum_{n>=1} 1/a(n) = 11/36.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 5/36. (End)
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: 2*exp(x)*x*(4 + x).
a(n) = n*A020739(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A255843 a(n) = 2*n^2 + 4.

Original entry on oeis.org

4, 6, 12, 22, 36, 54, 76, 102, 132, 166, 204, 246, 292, 342, 396, 454, 516, 582, 652, 726, 804, 886, 972, 1062, 1156, 1254, 1356, 1462, 1572, 1686, 1804, 1926, 2052, 2182, 2316, 2454, 2596, 2742, 2892, 3046, 3204, 3366, 3532, 3702, 3876, 4054, 4236, 4422

Offset: 0

Avi Friedlich, Mar 08 2015

This is the case k=2 of the form (n + sqrt(k))^2 + (n - sqrt(k))^2.

Equivalently, numbers m such that 2*m - 8 is a square.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A059100.
Cf. unsigned A147973: numbers of the form 2*m^2-4.
Cf. sequences of the form 2*m^2+2*k: A005893 (k=1), this sequence (k=2), A255844 (k=3), A155966 (k=4), A255845 (k=5), A255842 (k=6), A255846 (k=7), A255847 (k=8), A255848 (k=9).

Magma
```
[2*n^2+4: n in [0..50]];
```
Mathematica
```
Table[2 n^2 + 4, {n, 0, 50}]
```
PARI
```
vector(50, n, n--; 2*n^2+4)
```
Sage
```
[2*n^2+4 for n in (0..50)]
```

G.f.: 2*(2 - 3*x + 3*x^2)/(1 - x)^3.
a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*A059100(n).
a(n) = a(n-1) + 4n - 2. - Bob Selcoe, Mar 25 2020
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(2)*Pi*coth(sqrt(2)*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(2)*Pi*cosech(sqrt(2)*Pi))/8. (End)
E.g.f.: 2*exp(x)*(2 + x + x^2). - Stefano Spezia, Aug 07 2024

Edited by Bruno Berselli, Mar 13 2015

A079472 Number of perfect matchings on an n X n L-shaped graph.

Original entry on oeis.org

0, 2, 4, 12, 30, 80, 208, 546, 1428, 3740, 9790, 25632, 67104, 175682, 459940, 1204140, 3152478, 8253296, 21607408, 56568930, 148099380, 387729212, 1015088254, 2657535552, 6957518400, 18215019650, 47687540548, 124847601996, 326855265438, 855718194320

Offset: 1

Helen King (h.king(AT)uea.ac.uk), Jan 15 2003

a(n+1) = 2*F(n)*F(n+1) appears as the second component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, with F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), a(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
a(n+1) is the numerator of the continued fraction [1,...,1,2,1,...,1] with n 1's to the left of the central 2, and n 1's to the right of the central 2. For the denominators, see A061646. - Greg Dresden and Max Liu, Jun 25 2023
For n >= 3, a(n) equals the sum of the sides of the right triangle with side lengths [F(n)*F(n-3), 2*F(n-1)*F(n-2), F(2*n-3)] (n = 4 corresponds to the 3-4-5 right triangle). - Peter Bala, Nov 03 2023

			a(7) = 2*13*8 = 208 = number of matchings. F(7) = 13 F(6) = 8
a(3) = 4 because in the graph with vertex set {(0,0), (1,0), (2,0), (0,1), (1,1), (2,1), (0,2), (1,2)} and edge set {h(0,0), h(1,0), h(0,1), h(1,1), h(0,2), v(0,0), v(0,1), v(1,0), v(1,1), v(2,0)}, where h(i,j) (v(i,j)) is a horizontal (vertical) edge of unit length starting from vertex (i,j), we have the following four perfect matchings: {h(0,0), h(0,1), h(0,2), v(2,0)}, {h(0,0), v(0,1), v(1,1), v(2,0)}, {v(0,0), v(1,0), v(2,0), h(0,2)} and {v(0,0), h(1,0), h(1,1), h(0,2)}. - _Emeric Deutsch_, Dec 30 2004
G.f. = 2*x^2 + 4*x^3 + 12*x^4 + 30*x^5 + 80*x^6 + 208*x^7 + 546*x^8 + ...

Daniele Corradetti, La Metafisica del Numero, 2008.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 178, 255.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
P. F. F. Espinosa, J. F. González, J. P. Herrán, A. M. Cañadas, and J. L. Ramírez, On some relationships between snake graphs and Brauer configuration algebras, Algebra Disc. Math. (2022) Vol. 33, No. 2, 29-59.
I. Gutman and S. J. Cyvin, A result on 1-factors related to Fibonacci numbers, The Fibonacci Quarterly, 28 (1990), pp. 81-84.
C.-A. Laisant, Observations sur les triangles rectangles en nombres entiers et les suites de Fibonacci, Nouvelles Annales de Math. (1919, in French) Series 4, Vol. 19, 391-397.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001654, A121646.

List([1..30], n -> 2*Fibonacci(n-1)*Fibonacci(n)); # G. C. Greubel, Jan 07 2019

[2*Fibonacci(n)*Fibonacci(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 29 2014

with(combinat,fibonacci):seq(2*fibonacci(n)*fibonacci(n-1),n=1..30);

LinearRecurrence[{2,2,-1}, {0,2,4}, 30] (* Arkadiusz Wesolowski, Sep 15 2012 *)
Table[(2*Fibonacci[n]*Fibonacci[n-1]), {n,30}] (* Vincenzo Librandi, Jun 29 2014 *)

{a(n) = 2 * fibonacci(n) * fibonacci(n-1)}; \\ Michael Somos, Jun 28 2014

concat(0, Vec(2*x^2/((x+1)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Sep 27 2016

[2*fibonacci(n-1)*fibonacci(n) for n in (1..30)] # G. C. Greubel, Jan 07 2019

a(n) = 2*F(n)*F(n-1) where F(n) are the Fibonacci numbers (A000045).
From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003: (Start)
a(n) = 2*A001654(n) = F(2*n) - F(n)^2 = A001906(n) - A007598(n).
a(n) = (F(n+1)^2 - F(n-2)^2)/2 = (A007598(n+1) - A007598(n-2))/2.
a(n) = 2*(L(2*n-1) + (-1)^n)/5 = (2/5)*(A002878(n-1) + A033999(n)), where L(n) = A000032(n).
a(n+1) = a(n) + 2*F(n)^2.
G.f.: 2*x^2/((1+x)*(1-3*x+x^2)). (End)
a(n) = Im( (F(n) + i*F(n+1))^2 ) (cf. A121646). - Daniele Corradetti (d.corradetti(AT)gmail.com), May 02 2008
From Michael Somos, Jun 28 2014: (Start)
a(n) = F(n+1)^2 - F(n)^2 - F(n-1)^2.
a(1 - n) = -a(n). (End)
a(n) = ( 2*(-1)^n - (1+sqrt(5))*((3-sqrt(5))/2)^n - (1-sqrt(5))*((3+sqrt(5))/2)^n )/5. - Colin Barker, Sep 27 2016
From Rigoberto Florez, May 06 2020: (Start)
a(n) = F(2n-2) + F(n-1)^2, where F(n) is the n-th Fibonacci number.
a(n) = M^(n+1)[2,1], for n>0 where M=[0,0,1;0,1,2;1,1,1]. (End)
a(n) = F(n)^2 + F(n-1)^2 - F(n-2)^2. - Michael Somos, Mar 02 2023

More terms from Benoit Cloitre and Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003

cp's OEIS Frontend

A001844 Centered square numbers: a(n) = 2*n*(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Magma

Maple

Mathematica

PARI

PARI

Python

Sage

Formula

Extensions

A001105 a(n) = 2*n^2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

A046092 4 times triangular numbers: a(n) = 2*n*(n+1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Maxima

PARI

Python

Formula

A054000 a(n) = 2*n^2 - 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

A001844 Centered square numbers: a(n) = 2n(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

A046092 4 times triangular numbers: a(n) = 2n(n+1).

A139570 a(n) = 2n(n+3).

A121801 Expansion of 2x^2(3-x)/((1+x)(1-3x+x^2)).