A069480 -id:A069480 - 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

A006003 a(n) = n*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775

Offset: 0

N. J. A. Sloane, Simon Plouffe

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
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
J. D. Bell, A translation of Leonhard Euler's "De Quadratis Magicis", E795, arXiv:math/0408230 [math.CO], 2004-2005.
James Grime and Brady Haran, Magic Hexagon, Numberphile video (2014).
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, Journal of Integer Sequences, 17 (2014), Article 14.3.5. - _Felix Fröhlich_, Oct 11 2016
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Ashish Kumar Pandey and Brajesh Kumar Sharma, A Note On Magic Squares And Magic Constants, Appl. Math. E-Notes (2023) Vol. 23, Art. No. 53, 577-582. See p. 577.
A. J. Turner and J. F. Miller, Recurrent Cartesian Genetic Programming Applied to Famous Mathematical Sequences, preprint, Proceedings of the Companion Publication of the 2015 Annual Conference on Genetic and Evolutionary Computation.
Eric Weisstein's World of Mathematics, Magic Constant.
Wikipedia, Floyd's triangle. - _Paul Muljadi_, Jan 25 2010
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for sequences related to magic squares.
Index to sequences related to polygonal numbers.

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
Cf. A007742, A005449, A005448, A118465, A226449, A034262, A080992, A267017.
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Cf. A000292, A011863, A001620, A248177.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018

a006003 n = n * (n ^ 2 + 1) `div` 2
a006003_list = scanl (+) 0 a005448_list
-- Reinhard Zumkeller, Jun 20 2013

% Also works with FreeMat.
for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
% Stefano Spezia, Aug 12 2018

[n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015

[Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015

Table[ n(n^2 + 1)/2, {n, 0, 45}]
LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)

a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */

{a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */

concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016

def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

A078475 Determinant of rank n matrix of 1..n^2 filled successively back and forth along antidiagonals.

Original entry on oeis.org

1, -2, 15, -594, -5187, 23244, 122475, -279292, -1157143, 1850930, 6642839, -8529278, -27810555, 30741424, 93575187, -92784984, -268191855, 244875462, 679807583, -581798410, -1563707379, 1270245108, 3324627195, -2587197204, -6623079687, 4972012474, 12491212135

Offset: 1

Kit Vongmahadlek (kit119(AT)yahoo.com), Jan 03 2003

The matrix is formed by writing numbers 1 .. n^2 in zig-zag pattern as shown in examples below. Every other antidiagonal reads backwards from A069480.
Whereas each antidiagonal of A069480 begins with one more than a triangular number and ends with the next triangular number, here every other antidiagonal begins with one more than a triangular number and the next antidiagonal begins with a triangular number.
The trace of the matrix is the sequence A006003 (proved). - Stefano Spezia, Aug 07 2018
The matrix is defined by A[i,j] = (2 - i - j)*((i + j - 1) mod 2)+(j^2 + (2*i - 1)*j + i^2 - i)/2 + (j - 1)*(1 - 2*((i + j) mod 2)) if i + j <= n + 1 and A[i,j] = n^2 - ((4*n^2 + (- 4*j - 4*i + 6)*n + j^2 + (2*i - 3)*j + i^2 - 3*i + 2)/2 + (i + j - 2*n)*((2*n - i - j + 1) mod 2)) + 1 - (n - j)*(1 - 2*((i + j) mod 2)) if i + j > n + 1 (proved). - Stefano Spezia, Aug 11 2018

			n=2, det=-2: {1 2 / 3 4 }
n=3, det=15: {1 2 6 / 3 5 7 / 4 8 9 }
n=4, det=-594: { 1 2 6 7 / 3 5 8 13 / 4 9 12 14 / 10 11 15 16 }
n=5, det=-5187: { 1 2 6 7 15 / 3 5 8 14 16 / 4 9 13 17 22 / 10 12 18 21 23 / 11 19 20 24 25 }

Vaclav Kotesovec, Table of n, a(n) for n = 1..1450
Index entries for linear recurrences with constant coefficients, signature (0,-9,0,-36,0,-84,0,-126,0,-126,0,-84,0,-36,0,-9,0,-1).

Cf. A023999, A069480, A079340.

A078475 := function(k)
local i, j, n;
for n in [1 .. k] do
   A:=NullMat(n,n);
   for i in [1 .. n] do
      for j in [1 .. n] do
         if i+j<=n+1 then
            A[i][j] := (2-i-j)*RemInt(i+j-1,2)+(j^2+(2*i-1)*j+i^2-i)/2+(j-1)*(1-2*RemInt(i+j,2));;
         else
            A[i][j] := n^2-((4*n^2+(-4*j-4*i+6)*n+j^2+(2*i-3)*j+i^2-3*i+2)/2+(i+j-2*n)*RemInt(2*n-i-j+1,2))+1-(n-j)*(1-2*RemInt(i+j,2));
         fi;
      od;
   od;
   Print(n," ",Determinant(A),"\n");
od;
end;
A078475(27); # Stefano Spezia, Aug 12 2018

a[i_, j_, n_] := If[i+j<=n+1, (2-i-j)*Mod[i+j-1,2]+(j^2+(2*i-1)*j+i^2-i)/2+(j-1)*(1-2*Mod[i+j,2]),n^2-((4*n^2+(-4*j-4*i+6)*n+j^2+(2*i-3)*j+i^2-3*i+2)/2+(i+j-2*n)*Mod[2*n-i-j+1,2])+1-(n-j)*(1-2*Mod[i+j,2])]; f[n_] := Det[ Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 27] (* Stefano Spezia, Aug 11 2018 *)

A(i,j,n) = if (i + j <= n + 1, (2 - i - j)*((i + j - 1) % 2)+(j^2 + (2*i - 1)*j + i^2 - i)/2 + (j - 1)*(1 - 2*((i + j) % 2)), n^2 - ((4*n^2 + (- 4*j - 4*i + 6)*n + j^2 + (2*i - 3)*j + i^2 - 3*i + 2)/2 + (i + j - 2*n)*((2*n - i - j + 1) % 2)) + 1 - (n - j)*(1 - 2*((i + j) % 2)));
a(n) = matdet(matrix(n, n, i, j, A(i, j, n))); \\ Michel Marcus, Aug 11 2018
(MATLAB, FreeMat and Octave)
for(n=1:27)
   A=zeros(n,n);
   for(i=1:n)
      for(j=1:n)
         if(i+j<=n+1)
            A(i,j)=(2-i-j)*mod(i+j-1,2)+(j^2+(2*i-1)*j+i^2-i)/2+(j-1)*(1-2*mod(i+j,2));
         else
            A(i,j)=n^2-((4*n^2+(-4*j-4*i+6)*n+j^2+(2*i-3)*j+i^2-3*i+2)/2+(i+j-2*n)*mod(2*n-i-j+1,2))+1-(n-j)*(1-2*mod(i+j,2));
         end
      end
   end
   fprintf('%d %0.f\n',n,det(A));
end # Stefano Spezia, Aug 12 2018

From Vaclav Kotesovec, Jan 08 2019: (Start)
Recurrence: (5*n^16 - 176*n^15 + 2888*n^14 - 29332*n^13 + 206454*n^12 - 1068276*n^11 + 4205934*n^10 - 12861022*n^9 + 30891328*n^8 - 58524140*n^7 + 87229074*n^6 - 101275380*n^5 + 89823673*n^4 - 58824210*n^3 + 26795412*n^2 - 7559784*n + 985608)*a(n) = 8*(n^14 - 20*n^13 + 169*n^12 - 754*n^11 + 1630*n^10 + 564*n^9 - 15184*n^8 + 52244*n^7 - 109015*n^6 + 167071*n^5 - 202816*n^4 + 191592*n^3 - 125145*n^2 + 45333*n - 5832)*a(n-1) - (5*n^16 - 96*n^15 + 848*n^14 - 4580*n^13 + 16966*n^12 - 45892*n^11 + 94310*n^10 - 151266*n^9 + 192520*n^8 - 195196*n^7 + 155666*n^6 - 94052*n^5 + 39329*n^4 - 6798*n^3 - 4572*n^2 + 5400*n - 1944)*a(n-2).
a(n) ~ ((-1)^n - 3) * (cos(Pi*n/2) + sin(Pi*n/2)) * n^8 / 72. (End)
a(n) = -9*a(n-2) - 36*a(n-4) - 84*a(n-6) - 126*a(n-8) - 126*a(n-10) - 84*a(n-12) - 36*a(n-14) - 9*a(n-16) - a(n-18) for n > 18. - Stefano Spezia, Apr 25 2021, simplified by Boštjan Gec, Sep 21 2023

Edited and extended by Robert G. Wilson v, May 08 2003

cp's OEIS Frontend

A072332 Largest eigenvalue, rounded to the nearest integer, of a rank n matrix of 1..n^2 filled successively along antidiagonals (A069480).

Views

Author

Keywords

Programs

Mathematica

A342797 Irregular triangle read by rows: T(n, k) is the k-th antidiagonal sum of the n X n matrices defined in A069480 and A078475.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

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

A006003 a(n) = n*(n^2 + 1)/2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

MATLAB

Magma

Magma

Mathematica

Maxima

PARI

PARI

Python

Formula

Extensions

A078475 Determinant of rank n matrix of 1..n^2 filled successively back and forth along antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Mathematica

PARI

Formula

Extensions

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.