A072481 -id:A072481 - 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.
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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

Partial sums of A024916. - Omar E. Pol, Jul 03 2014
a(n) is also the volume of the stepped pyramid with n levels described in A245092. - Omar E. Pol, Aug 12 2015
Also the alternating row sums of A262612. - Omar E. Pol, Nov 23 2015
From Omar E. Pol, Jan 20 2021: (Start)
Convolution of A000203 and A000027.
Convolution of A340793 and the nonzero terms of A000217.
Antidiagonal sums of A319073.
Row sums of A274824. (End)
Row sums of A345272. - Omar E. Pol, Jun 14 2021
Also the alternating row sums of A353690. - Omar E. Pol, Jun 05 2022

a(n) is also the volume of another version of the pyramid with n levels (starting from the top) described in A245092. Both pyramids have the same top view and the same front view, but in this pyramid the volume in the n-th level is equal to A294016(n) instead of A024916(n).

a(n) is also the volume of a stepped pyramid with n levels which is another version of the stepped pyramid described in A244050. Both pyramids have the same top view and the same front view, that is to say externally both pyramids are equal, but this pyramid with n levels contains a central chamber whose volume is 4*A072481(n). For more information about the central chamber see the diagrams in A294629.

a(n) is the number of unit cubes of the pyramid with n levels.

From Omar E. Pol, Jan 20 2021: (Start)
Conjectures:
1. Convolution of A001065 and A000027.
2. Partial sums of A153485.
3. a(n) is also the difference of volume (the difference of number of cells) between two polycubes: the stepped pyramid described in A245092 which has volume A175254(n) and the stepped pyramid that represents the n-th tetrahedral number which has volume A000292(n).
In the three conjectures assuming that here the offset is 1.
For more information about the first pyramid see A237593. (End)

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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A175254 a(n) = Sum_{k<=n} A000203(k)*(n-k+1), where A000203(m) is the sum of divisors of m.

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A294017 Partial sums of A294016.

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A294630 Partial sums of A294629.

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A123329 Let M be the matrix defined in A111490. Sequence gives M(2,1)-M(1,2), M(2,1)+M(3,1)+M(3,2)-M(1,2)-M(1,3)-M(2,3), etc.

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A367379 a(n) = Sum_{j=1..n} Sum_{i=1..n} (j mod i).

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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.