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A000330
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Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.
(Formerly M3844 N1574)
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462
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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
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OFFSET
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0,3
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COMMENTS
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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]
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 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
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.
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
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
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
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)
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
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
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
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
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REFERENCES
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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.
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.
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).
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LINKS
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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].
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
S. Butler and P. Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, p=1 in first displayed equation page 4.
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
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FORMULA
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G.f.: x*(1+x)/(1-x)^4.
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).
a(n) = -a(-1-n) for all n in Z.
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = binomial(2*(n+1), 3)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^2).
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
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
Starting n (-1, 0, 1, 2, ...), a(n) = binomial(n+2, 2) + 2*binomial(n+2, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
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) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
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) = (2*n^3 + 3*n^2 + n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
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
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EXAMPLE
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G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(PARI) {a(n) = n * (n+1) * (2*n+1) / 6};
(PARI) sumsq(n) = for(x=0, n, y=x*(x+1)*(2*x+1)/6; (print1(y", "))) \\ Cino Hilliard, Jun 18 2007
(Haskell)
a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
a000330_list = scanl1 (+) a000290_list
(Maxima) A000330(n):=binomial(n+2, 3)+binomial(n+1, 3)$
(Magma) [0] cat [((2*n+3)*Binomial(n+2, 2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
(Python) a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017
(Sage) [n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019
(GAP) List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019
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CROSSREFS
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Cf. A000217, A000292, A000537, A005408, A006003, A006331, A033994, A046092, A050409, A050446, A050447, A100157, A132124, A132112, A156921, A157702, A258708.
Sums of 2 consecutive terms give A005900.
Cf. A253903 (characteristic function).
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KEYWORD
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nonn,easy,core,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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