A132124 -id:A132124 - cp's OEIS frontend

A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n(n+1)(2*n+1)/6.

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370

Offset: 0

N. J. A. Sloane

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]
Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001
Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025
a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004
Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.
Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004
Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007
Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007
The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007
Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008
Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008
Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008
Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009
The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012
a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010
Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010
From J. M. Bergot, Aug 23 2011: (Start)
Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).
Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)
From Ant King, Oct 17 2012: (Start)
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.
For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)
Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012
Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013
The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013
For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013
A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014
The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014
Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014
Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014
From Bui Quang Tuan, Apr 03 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).
Here is an example with n = 5:
  1
  1  2
  1  2  3
  1  2  3  4
  1  2  3  4  5
  1  2  3  4
  1  2  3
  1  2
  1
(End)
The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016
Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018
The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018
Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019
n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020
For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020
Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020
The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024
Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

			G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
A. H. Beiler, Recreations in the Theory of Numbers, Dover Publications, NY, 1964, p. 194.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 215,223.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 122, see #19 (3(1)), I(n); p. 155.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 47-49.
H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 310, pp. 46-196, Ellipses, Paris, 2004.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, p. 293.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 293.
M. Holt, Math puzzles and games, Walker Publishing Company, 1977, p. 2 and p. 89.
Simon Singh, The Simpsons and Their Mathematical Secrets. London: Bloomsbury Publishing PLC (2013): 188.
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).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 126.

Felix Fröhlich, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
L. Ancora, Quadrature of the Parabola with the Square Pyramidal Number, Mondadori Education, Archimede 66, No. 3, 139-144 (2014).
Jack Anderson, Amy Woodall, and Alexandru Zaharescu, Arithmetic Polygons and Sums of Consecutive Squares, arXiv:2411.08398 [math.NT], 2024.
Ben Babcock and Adam Van Tuyl, Revisiting the spreading and covering numbers, arXiv preprint arXiv:1109.5847 [math.AC], 2011.
Joshua L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
Joshua L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359. [Annotated scanned copy]
Michael A. Bennett, Lucas' square pyramid problem revisited, Acta Arithmetica 105 (2002), 341-347.
Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Fritz Beukers and Jaap Top, On oranges and integral points on certain plane cubic curves, Nieuw Arch. Wiskd., IV (1988), Ser. 6, No. 3, 203-210.
Henry Bottomley, Illustration of initial terms.
Steve Butler and Pavel Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, p=1 in first displayed equation page 4.
Bikash Chakraborty, Proof Without Words: Sums of Powers of Natural numbers, arXiv:2012.11539 [math.HO], 2020.
Robert Dawson, On Some Sequences Related to Sums of Powers, J. Int. Seq., Vol. 21 (2018), Article 18.7.6.
Alexander Diaz-Lopez, Pamela E. Harris, Erik Insko, and Darleen Perez-Lavin, Peaks Sets of Classical Coxeter Groups, arXiv preprint arXiv:1505.04479 [math.GR], 2015.
Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, Cannonball Polygons with Multiplicities, arXiv:2507.18057 [math.NT], 2025. See p. 1.
Michael Dougherty, Christopher French, Benjamin Saderholm, and Wenyang Qian, Hankel Transforms of Linear Combinations of Catalan Numbers, Journal of Integer Sequences, Vol. 14 (2011), Article 11.5.1.
David Galvin and Courtney Sharpe, Independent set sequence of linear hyperpaths, arXiv:2409.15555 [math.CO], 2024. See p. 7.
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G. Xiao, Sigma Server, Operate on"n^2".
Index entries for "core" sequences.
Index to sequences related to pyramidal numbers.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for two-way infinite sequences.

Cf. A000217, A000292, A000537, A005408, A006003, A006331, A033994, A033999, A046092, A050409, A050446, A050447, A060493, A100157, A132124, A132112, A156921, A157702, A258708, A351105, A351830.
Sums of 2 consecutive terms give A005900.
Column 0 of triangle A094414.
Column 1 of triangle A008955.
Right side of triangle A082652.
Row 2 of array A103438.
Partial sums of A000290.
Cf. similar sequences listed in A237616 and A254142.
Cf. |A084930(n, 1)|.
Cf. A253903 (characteristic function).
Cf. A034705 (differences of any two terms).

List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019

a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
a000330_list = scanl1 (+) a000290_list
-- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012

[n*(n+1)*(2*n+1)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 28 2014

[0] cat [((2*n+3)*Binomial(n+2,2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014

A000330 := n -> n*(n+1)*(2*n+1)/6;
a := n->(1/6)*n*(n+1)*(2*n+1): seq(a(n),n=0..53); # Emeric Deutsch
with(combstruct): ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*2), m=1..45) ; # Zerinvary Lajos, Jan 02 2008
nmax := 44; for n from 0 to nmax do fz(n) := product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n) := abs(coeff(fz(n),z,1)); end do: a := n-> c(n): seq(a(n), n=0..nmax); # Johannes W. Meijer, Mar 07 2009

Table[Binomial[w+2, 3] + Binomial[w+1, 3], {w, 0, 30}]
CoefficientList[Series[x(1+x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Accumulate[Range[0,50]^2] (* Harvey P. Dale, Sep 25 2014 *)

A000330(n):=binomial(n+2,3)+binomial(n+1,3)$
makelist(A000330(n),n,0,20); /* Martin Ettl, Nov 12 2012 */

PARI
```
{a(n) = n * (n+1) * (2*n+1) / 6};
```

upto(n) = [x*(x+1)*(2*x+1)/6 | x<-[0..n]] \\ Cino Hilliard, Jun 18 2007, edited by M. F. Hasler, Jan 02 2024

a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017

[n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019

G.f.: x*(1+x)/(1-x)^4. - Simon Plouffe (in his 1992 dissertation: generating function for sequence starting at a(1))
E.g.f.: (x + 3*x^2/2 + x^3/3)*exp(x).
a(n) = n*(n+1)*(2*n+1)/6 = binomial(n+2, 3) + binomial(n+1, 3).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
Can be extended to Z with a(n) = -a(-1-n) for all n in Z.
a(n) = A002492(n)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
From Alexander Adamchuk, Oct 26 2004: (Start)
a(n) = sqrt(A271535(n)).
a(n) = (Sum_{k=1..n} Sum_{j=1..n} Sum_{i=1..n} (i*j*k)^2)^(1/3). (End)
a(n) = Sum_{i=1..n} i*(2*n-2*i+1); sum of squares gives 1 + (1+3) + (1+3+5) + ... - Jon Perry, Dec 08 2004
a(n+1) = A000217(n+1) + 2*A000292(n). - Creighton Dement, Mar 10 2005
Sum_{n>=1} 1/a(n) = 6*(3-4*log(2)); Sum_{n>=1} (-1)^(n+1)*1/a(n) = 6*(Pi-3). - Philippe Deléham, May 31 2005
Sum of two consecutive tetrahedral (or pyramidal) numbers a(n) = A000292(n-1) + A000292(n). - Alexander Adamchuk, May 17 2006
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
a(n) = binomial(n, 2) + 2*binomial(n, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, corrected by M. F. Hasler, Jan 02 2024
a(n) = A168559(n) + 1 for n > 0. - Reinhard Zumkeller, Feb 03 2012
a(n) = Sum_{i=1..n} J_2(i)*floor(n/i), where J_2 is A007434. - Enrique Pérez Herrero, Feb 26 2012
a(n) = s(n+1, n)^2 - 2*s(n+1, n-1), where s(n, k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
a(n) = A001477(n) + A000217(n) + A007290(n+2) + 1. - J. M. Bergot, May 31 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
a(n) = Sum_{i = 1..n} Sum_{j = 1..n} min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A000217(n) + A007290(n+1). - Ivan N. Ianakiev, May 10 2013
a(n) = (A047486(n+2)^3 - A047486(n+2))/24. - Richard R. Forberg, Dec 25 2013
a(n) = Sum_{i=0..n-1} (n-i)*(2*i+1), with a(0) = 0. After 0, row sums of the triangle in A101447. - Bruno Berselli, Feb 10 2014
a(n) = n + 1 + Sum_{i=1..n+1} (i^2 - 2i). - Wesley Ivan Hurt, Feb 25 2014
a(n) = A000578(n+1) - A002412(n+1). - Wesley Ivan Hurt, Jun 28 2014
a(n) = Sum_{i = 1..n} Sum_{j = i..n} max(i,j). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A055112(n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
For n >= 2, a(n) = A028347(n+1) + A101986(n-2). - Bui Quang Tuan, Apr 03 2015
For n > 0: a(n) = A258708(n+3,n-1). - Reinhard Zumkeller, Jun 23 2015
a(n) = A175254(n) + A072481(n), n >= 1. - Omar E. Pol, Aug 12 2015
a(n) = A000332(n+3) - A000332(n+1). - Antal Pinter, Dec 27 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A080851(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = (A005408(n) * A046092(n))/12 = (2*n+1)*(2*n*(n+1))/12. - Bruce J. Nicholson, May 18 2017
12*a(n) = (n+1)*A001105(n) + n*A001105(n+1). - Bruno Berselli, Jul 03 2017
a(n) = binomial(n-1, 1) + binomial(n-1, 2) + binomial(n, 3) + binomial(n+1, 2) + binomial(n+1, 3). - Tony Foster III, Aug 24 2018
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Nathan Fox, Dec 04 2019
Let T(n) = A000217(n), the n-th triangular number.  Then a(n) = (T(n)+1)^2 + (T(n)+2)^2 + ... + (T(n)+n)^2 - (n+2)*T(n)^2. - Charlie Marion, Dec 31 2019
a(n) = 2*n - 1 - a(n-2) + 2*a(n-1). - Boštjan Gec, Nov 09 2023
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j). Cf. A060493. - Peter Bala, Mar 31 2025

Partially edited by Joerg Arndt, Mar 11 2010

A014105 Second hexagonal numbers: a(n) = n(2n + 1).

Original entry on oeis.org

0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528, 595, 666, 741, 820, 903, 990, 1081, 1176, 1275, 1378, 1485, 1596, 1711, 1830, 1953, 2080, 2211, 2346, 2485, 2628, 2775, 2926, 3081, 3240, 3403, 3570, 3741, 3916, 4095, 4278

Offset: 0

N. J. A. Sloane, Jun 14 1998

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015
a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002
Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004
If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011
Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011
a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012
Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]
a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013
For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015
a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015
Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018
This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019
A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021
Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

			For n=6, a(6) = 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2 + 8^2 - 9^2 + 10^2 - 11^2 + 12^2 = 78. - _Bruno Berselli_, Aug 29 2013

Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Matthew Cho, Anton Dochtermann, Ryota Inagaki, Suho Oh, Dylan Snustad, and Bailee Zacovic, Chip-firing and critical groups of signed graphs, arXiv:2306.09315 [math.CO], 2023. See p. 22.
Robert FERREOL, Illustration: triangular numbers of even order
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjic, Two Enumerative Functions, University of Banja Luka (Bosnia and Herzegovina, 2017).
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Kival Ngaokrajang, Illustration of half circle spiral.
Markus Scheuer, show that; strange sum yields triangular numbers, Mathematics StackExchange.
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).
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Leo Tavares, Illustration: Squared Hexagons.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A000384, A002378, A002943, A100040, A100041, A081266, A144312.
Cf. A033567, A033585, A059255, A130757, A132440, A027907.
Second column of array A094416.
Equals A033586(n) divided by 4.
See Comments of A132124.
Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.
Row sums in triangle A253580.
Cf. A016061, A003215, A000290, A188859.

List([0..50],n->n*(2*n+1)); # Muniru A Asiru, Oct 31 2018

a014105 n = n * (2 * n + 1)
a014105_list = scanl (+) 0 a004767_list  -- Reinhard Zumkeller, Oct 03 2012

[ n*(2*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014

seq(binomial(2*n+1,2), n=0..46); # Zerinvary Lajos, Jan 21 2007

Table[n*(2*n+1), {n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
LinearRecurrence[{3,-3,1},{0,3,10},50] (* Harvey P. Dale, Feb 10 2015 *)
CoefficientList[Series[x*(3 + x)/(1 - x)^3,{x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)

PARI
```
a(n)=n*(2*n+1)
```

[n*(2*n+1) for n in range(50)] # G. C. Greubel, Dec 16 2018

a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - Ignacio Larrosa Cañestro, Apr 17 2001
a(n)^2 = n*(a(n) + 1 + a(n) + 2 + ... + a(n) + 2*n); e.g., 10^2 = 2*(11 + 12 + 13 + 14). - Charlie Marion, Jun 15 2003
From N. J. A. Sloane, Sep 13 2003: (Start)
G.f.: x*(3 + x)/(1 - x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = A000217(2*n) = A000384(-n). (End)
a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - Reinhard Zumkeller, Dec 30 2006
a(2*n) = A033585(n); a(3*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - Bruno Berselli, Aug 29 2013
a(n) = A242342(2*n + 1). - Reinhard Zumkeller, May 11 2014
a(n) = Sum_{k=0..2} C(n-2+k, n-2) * C(n+2-k, n), for n > 1. - J. M. Bergot, Jun 14 2014
a(n) = floor(Sum_{j=(n^2 + 1)..((n+1)^2 - 1)} sqrt(j)). Fractional portion of each sum converges to 1/6 as n -> infinity. See A247112 for a similar summation sequence on j^(3/2) and references to other such sequences. - Richard R. Forberg, Dec 02 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10. - Harvey P. Dale, Feb 10 2015
Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... (A188859). - Vaclav Kotesovec, Apr 27 2016
From Wolfdieter Lang, Apr 27 2018: (Start)
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n-1)), for n >= 1, with the trinomial irregular triangle A027907; i.e., trinomial(n,k) = A027907(n,k).
a(n) = (1/Pi) * Integral_{x=0..2} (1/sqrt(4 - x^2)) * (x^2 - 1)^(2*n) * R(4*(n-1), x), for n >= 0, with the R polynomial coefficients given in A127672, and R(-m, x) = R(m, x). [See Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 2, rewritten with x = 2*cos(phi).] (End)
a(n) = A002943(n)/2. - Ralf Steiner, Jul 23 2019
a(n) = A000290(n) + A002378(n). - Torlach Rush, Nov 02 2020
a(n) = A003215(n) - A000290(n+1). See Squared Hexagons illustration. Leo Tavares, Nov 23 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/2 + log(2) - 2. - Amiram Eldar, Nov 28 2021

Link added and minor errors corrected by Johannes W. Meijer, Feb 04 2010

A033994 a(n) = n(n+1)(5*n+1)/6.

Original entry on oeis.org

2, 11, 32, 70, 130, 217, 336, 492, 690, 935, 1232, 1586, 2002, 2485, 3040, 3672, 4386, 5187, 6080, 7070, 8162, 9361, 10672, 12100, 13650, 15327, 17136, 19082, 21170, 23405, 25792, 28336, 31042, 33915, 36960, 40182, 43586, 47177, 50960, 54940

Offset: 1

Barry E. Williams, Dec 16 1999

Partial sums of A005476.

a(n) is the dot product of the vectors of the first n positive integers and the next n integers. - Michel Marcus, Sep 02 2020

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A005476, A016873, A000330, A132124, A132112, A050409.

Cf. A000292, A001622, A002411, A002412.

a:=List([1..40],n->n*(n+1)*(5*n+1)/6);; Print(a); # Muniru A Asiru, Jan 01 2019

[n*(n+1)*(5*n+1)/6 : n in [1..40]]; // Vincenzo Librandi, Jan 01 2019

[n*(n+1)*(5*n+1)/6$n=1..40]; # Muniru A Asiru, Jan 01 2019

Table[Range[x].Range[x+1,2x],{x,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{2,11,32,70},40] (* Harvey P. Dale, Jun 01 2018 *)

PARI
```
a(n) = n*(n+1)*(5*n+1)/6;
```

G.f.: x*(2+3*x)/(1-x)^4.
a(n) = A132121(n,1). - Reinhard Zumkeller, Aug 12 2007
a(n) = A000292(n) + A002412(n) = A000330(n) + A002411(n). - Omar E. Pol, Jan 11 2013
a(n) = Sum_{i=1..n} Sum_{j=1..n} i+min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = Sum_{i=1..n} i*(n+i). - Charlie Marion, Apr 10 2013
Sum_{n>=1} 1/a(n) = 36 - 3*Pi*5^(3/4)*phi^(3/2)/4 - 15*sqrt(5)*log(phi)/4 - 75*log(5)/8 = 0.66131826232008423794478..., where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 01 2018
E.g.f.: exp(x)*x*(12 + 21*x + 5*x^2)/6. - Stefano Spezia, Feb 21 2024

More terms from James Sellers, Jan 19 2000

A050409 Truncated square pyramid numbers: a(n) = Sum_{k = n..2*n} k^2.

Original entry on oeis.org

0, 5, 29, 86, 190, 355, 595, 924, 1356, 1905, 2585, 3410, 4394, 5551, 6895, 8440, 10200, 12189, 14421, 16910, 19670, 22715, 26059, 29716, 33700, 38025, 42705, 47754, 53186, 59015, 65255, 71920, 79024, 86581, 94605, 103110, 112110, 121619

Offset: 0

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 22 1999

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A033994, A129371, A132112, A132121, A132124.
Cf. A225144. - Bruno Berselli, Jun 06 2013
Cf. A045943: Sum_{k = n..2*n} k.
Cf. A304993: Sum_{k = n..2*n} k*(k+1)/2.

List([0..40], n-> n*(n+1)*(14*n+1)/6); # G. C. Greubel, Oct 30 2019

[&+[k^2: k in [n..2*n]]: n in [0..40]]; // Bruno Berselli, Feb 11 2011

I:=[0, 5, 29, 86]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012

seq(add((n+k)^2,k=0..n),n=0..40); # Zerinvary Lajos, Dec 01 2006

LinearRecurrence[{4,-6,4,-1},{0,5,29,86},40] (* Vincenzo Librandi, Jun 22 2012 *)
Table[(n(n+1)(14n+1))/6,{n,0,40}] (* Harvey P. Dale, Mar 08 2020 *)

PARI
```
a(n)=sum(k=n,n+n,k^2)
```

vector(40, n, n*(n-1)*(14*n-13)/6) \\ G. C. Greubel, Oct 30 2019

[n*(n+1)*(14*n+1)/6 for n in (0..40)] # G. C. Greubel, Oct 30 2019

a(n) = n*(n+1)*(14*n+1)/6.
a(n) = A132121(n,4) for n>3. - Reinhard Zumkeller, Aug 12 2007
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5+9*x)/(1-x)^4.
a(n) = A129371(2*n). (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 22 2012
E.g.f.: x*(30 + 57*x + 14*x^2)*exp(x)/6. - G. C. Greubel, Oct 30 2019

A069011 Triangle with T(n,k) = n^2 + k^2.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 13, 18, 16, 17, 20, 25, 32, 25, 26, 29, 34, 41, 50, 36, 37, 40, 45, 52, 61, 72, 49, 50, 53, 58, 65, 74, 85, 98, 64, 65, 68, 73, 80, 89, 100, 113, 128, 81, 82, 85, 90, 97, 106, 117, 130, 145, 162, 100, 101, 104, 109, 116, 125, 136, 149, 164, 181, 200

Offset: 0

Henry Bottomley, Apr 02 2002

For any i,j >=0 a(i)*a(j) is a member of this sequence, since (a^2 + b^2)*(c^2 + d^2) = (a*c + b*d)^2 + (a*d - b*c)^2. - Boris Putievskiy, May 05 2013
A227481(n) = number of squares in row n. - Reinhard Zumkeller, Oct 11 2013
Norm of the complex numbers n +- i*k and k +- i*n, where i denotes the imaginary unit. - Stefano Spezia, Aug 07 2025

			Triangle T(n,k) begins:
    0;
    1,   2;
    4,   5,   8;
    9,  10,  13,  18;
   16,  17,  20,  25,  32;
   25,  26,  29,  34,  41,  50;
   36,  37,  40,  45,  52,  61,  72;
   49,  50,  53,  58,  65,  74,  85,  98;
   64,  65,  68,  73,  80,  89, 100, 113, 128;
   81,  82,  85,  90,  97, 106, 117, 130, 145, 162;
  100, 101, 104, 109, 116, 125, 136, 149, 164, 181, 200;
  ...

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened

Cf. A001481 for terms in this sequence, A000161 for number of times each term appears, A048147 for square array.
Column k=0 gives A000290.
Main diagonal gives A001105.
Row sums give A132124.
T(2n,n) gives A033429.

a069011 n k = a069011_tabl !! n !! k
a069011_row n = a069011_tabl !! n
a069011_tabl = map snd $ iterate f (1, [0]) where
   f (i, xs@(x:_)) = (i + 2, (x + i) : zipWith (+) xs [i + 1, i + 3 ..])
-- Reinhard Zumkeller, Oct 11 2013

Table[n^2 + k^2, {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Aug 07 2025 *)

T(n+1,k+1) = T(n,k) + 2*(n+k+1), k=0..n; T(n+1,0) = T(n,0) + 2*n + 1. - Reinhard Zumkeller, Oct 11 2013

G.f.: x*(1 + 2*y + 5*x^3*y^2 - x^2*y*(2 + 5*y) + x*(1 - 4*y + 2*y^2))/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Aug 04 2025

A132112 a(n) = n(n+1)(11*n+1)/6.

Original entry on oeis.org

0, 4, 23, 68, 150, 280, 469, 728, 1068, 1500, 2035, 2684, 3458, 4368, 5425, 6640, 8024, 9588, 11343, 13300, 15470, 17864, 20493, 23368, 26500, 29900, 33579, 37548, 41818, 46400, 51305, 56544, 62128, 68068, 74375, 81060, 88134, 95608, 103493, 111800

Offset: 0

Reinhard Zumkeller, Aug 10 2007

Sums of rows of the triangle in A132111.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A033994, A050409, A132111, A132121, A132124, A254407.

I:=[0, 4, 23, 68]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 29 2012

CoefficientList[Series[x*(4+7*x)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 29 2012 *)
LinearRecurrence[{4,-6,4,-1},{0,4,23,68},40] (* Harvey P. Dale, Jun 28 2021 *)

a(n)=n*(n+1)*(11*n+1)/6 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A132121(n,3) for n > 2.
G.f.: x*(4+7*x)/(1-x)^4. - Colin Barker, Jun 06 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 29 2012
a(-n) = -A254407(n-1). - Bruno Berselli, Jan 31 2015
E.g.f.: exp(x)*x*(24 + 45*x + 11*x^2)/6. - Stefano Spezia, Feb 21 2024

A132121 Triangle read by rows: T(n,k)=n(n+1)((3k+2)n+1)/6, 0<=k<=n.

Original entry on oeis.org

0, 1, 2, 5, 11, 17, 14, 32, 50, 68, 30, 70, 110, 150, 190, 55, 130, 205, 280, 355, 430, 91, 217, 343, 469, 595, 721, 847, 140, 336, 532, 728, 924, 1120, 1316, 1512, 204, 492, 780, 1068, 1356, 1644, 1932, 2220, 2508, 285, 690, 1095, 1500, 1905, 2310, 2715, 3120, 3525, 3930

Offset: 0

Reinhard Zumkeller, Aug 12 2007

Row sums give A132122; central terms give A132123.
T(n,0) = A000330(n);
T(n,1) = A033994(n) for n>0;
T(n,2) = A132124(n) for n>1;
T(n,3) = A132112(n) for n>2;
T(n,4) = A050409(n) for n>3.

			0;
1, 2;
5, 11, 17;
14, 32, 50, 68;
30, 70, 110, 150, 190;
55, 130, 205, 280, 355, 430;
91, 217, 343, 469, 595, 721, 847;
140, 336, 532, 728, 924, 1120, 1316, 1512;
204, 492, 780, 1068, 1356, 1644, 1932, 2220, 2508;

Cf. A132122, A132123.

Cf. A000330, A033994, A132124, A132112, A050409.

A132121 := proc(n,k)
    n*(n+1)*((3*k+2)*n+1)/6 ;
end proc:
seq(seq(A132121(n,k),k=0..n),n=0..13) ; # R. J. Mathar, Feb 19 2020

Table[n(n+1)((3k+2)n+1)/6,{n,0,9},{k,0,n}]//Flatten (* James C. McMahon, Mar 04 2025 *)

G.f.: Sum_{n>=0} Sum_{k>=0} T(n,k)*x^n*y^k = x*(x*y+1+x)/((1-x)^4*(1-y)^2). - R. J. Mathar, Jul 28 2016. Note that this generates a full array, not just the triangular subspace.

a(53)-a(54) from James C. McMahon, Mar 04 2025

A135713 a(n) = n(n+1)(4*n+1)/2.

Original entry on oeis.org

0, 5, 27, 78, 170, 315, 525, 812, 1188, 1665, 2255, 2970, 3822, 4823, 5985, 7320, 8840, 10557, 12483, 14630, 17010, 19635, 22517, 25668, 29100, 32825, 36855, 41202, 45878, 50895, 56265, 62000, 68112, 74613, 81515, 88830, 96570, 104747, 113373, 122460, 132020

Offset: 0

N. J. A. Sloane, Mar 05 2008

This sequence is related to A045944 by a(n) = n*A045944(n)-Sum_{i=0..n-1} A045944(i); this is the case d=6 in the identity n^2*(d*n+d-2)/2 - sum(k*(d*k+d-2)/2, k=0..n-1) = n*(n+1)*(2*d*n+d-3)/6 . - Bruno Berselli, Nov 19 2010

Bisection (even part) of A002717. See the Conway and Guy reference. - Wolfdieter Lang, Apr 16 2020

J. H. Conway and R. K. Guy, The Book of Numbers, p. 83.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
M. E. Larsen, The eternal triangle - a history of a counting problem, College Math. J., 20 (1989), 370-392.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Bisection of A002717 (even part).

Cf. A045944, A160378; A132124, A006331.

[n*(n+1)*(4*n+1)/2: n in [0..40]];  // Bruno Berselli, Aug 23 2011

LinearRecurrence[{4,-6,4,-1}, {0, 5, 27, 78}, 50] (* Vincenzo Librandi, Mar 01 2012 *)
Table[n*(n+1)*(4*n+1)/2,{n,0,25}] (* G. C. Greubel, Oct 29 2016 *)
Table[PolygonalNumber[n](4n+1),{n,0,40}] (* Harvey P. Dale, Apr 26 2025 *)

O.g.f.: x*(7*x+5)/(x-1)^4. - R. J. Mathar, Apr 22 2008.
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) for n>3. - Bruno Berselli, Nov 19 2010
a(-n-1) = -A051895(n).  - Bruno Berselli, Aug 23 2011
E.g.f.: (1/2)*x*(10 + 17*x + 4*x^2)*exp(x). - G. C. Greubel, Oct 29 2016
Sum_{n>=1} 1/a(n) = 2*(5 - 2*Pi/3 - 4*log(2)) = 0.26603235073404654... - Ilya Gutkovskiy, Oct 29 2016

A218485 Positive numbers differing from next greater square by a square.

Original entry on oeis.org

3, 5, 8, 12, 15, 16, 21, 24, 27, 32, 35, 40, 45, 48, 55, 60, 63, 65, 72, 77, 80, 84, 91, 96, 99, 105, 112, 117, 120, 128, 135, 140, 143, 144, 153, 160, 165, 168, 171, 180, 187, 192, 195, 200, 209, 216, 221, 224, 231, 240, 247, 252, 255, 264, 273, 280, 285

Offset: 1

Michel Marcus, Oct 30 2012

nonn

Square terms in the sequence are 16, 144, 576, 1600, 3600, 7056, ..., that is, A060300 ((2n(n+1))^2) except 0. And their indices, ind(n), are: 6, 34, 100, 220, 410, 686, ..., that is, ind(n) = 2*A132124(.). - Michel Marcus, suggested by Zak Seidov, Nov 26 2013

			8 = 3^2 - 1^2.

Zak Seidov, Table of n, a(n) for n = 1..1515 (all terms up to 20000)
E. J. Barbeau, Numbers differing from consecutive squares by squares, Canad. Math. Bull. 28(1985), pp. 337-342.

Cf. A060300, A218486, A218487, A218488.

Select[Range[300],IntegerQ[Sqrt[(1+Floor[Sqrt[#]])^2-#]]&] (* Zak Seidov, Nov 26 2013 *)

sq1(n) = {for (i=1, n, a = sqrtint(i) + 1; if (issquare(a^2-i), print1(i, ", ")););}

cp's OEIS Frontend

A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

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A014105 Second hexagonal numbers: a(n) = n*(2*n + 1).

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A033994 a(n) = n*(n+1)*(5*n+1)/6.

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A050409 Truncated square pyramid numbers: a(n) = Sum_{k = n..2*n} k^2.

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A069011 Triangle with T(n,k) = n^2 + k^2.

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A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n(n+1)(2*n+1)/6.

A014105 Second hexagonal numbers: a(n) = n(2n + 1).

A033994 a(n) = n(n+1)(5*n+1)/6.

A132112 a(n) = n(n+1)(11*n+1)/6.

A132121 Triangle read by rows: T(n,k)=n(n+1)((3k+2)n+1)/6, 0<=k<=n.

A135713 a(n) = n(n+1)(4*n+1)/2.