|
| |
|
|
A000330
|
|
Square pyramidal numbers: 0^2 + 1^2 + 2^2 +...+ n^2 = n*(n+1)*(2*n+1)/6.
(Formerly M3844 N1574)
|
|
264
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007
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 You (giawgwan(AT)single.url.com.tw), 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, R.H.S. of relation (i) telescopes to f(n+1)+f(n) = T(n)*{(n+2)+(n-1)}, whence result sum_(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. 7, 2004.
Number of triples of integers from {1,2,...,n} whose last component is greater than or equal to the others.
Kekule numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Euler transform of length 2 sequence [ 5, -1]. - Michael Somos, Sep 04 2006
Sum of the first n squares, or square pyramidal numbers. - Cino Hilliard (hillcino368(AT)hotmail.com), 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 (sfalcon(AT)dma.ulpgc.es), Feb 05 2008
Hankel transform of C(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,C(n+2,i+2)*b(i)}, where b(i)=1,2,0,0,0,... [From Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009]
Convolution of A001477 with A005408: a(n)=SUM((2*k+1)*(n-k):0<=k<=n). [From 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(A000217(i), i=0..n-1) and this is the case d=1 in the identity n^2*(d*n-d+2)/2-sum(i*(d*i-d+2)/2, i=0..n-1) = 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*(i+2*d+1)/2, i=0..n-1) = 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. number k = 195 is quadratic mean (root mean square) of first 337 positive integers. There are other such numbers - see A084231 and A084232. [From Jaroslav Krizek, May 23 2010]
Contribution from Carmine Suriano, Sep 10 2010: (Start)
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. (End)
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 for n is 60 times the numbers in this sequence. Using three consecutive odd numbers a, b, c, (a+b+c)^3 - (a^3 + b^3 + c^3) equals 576=24^2 times the numbers in this sequence. [J. M. Bergot, Aug 23 2011]
Contribution from Ant King, Oct 17 2012: (Start)
For n>0, the digital roots of this sequence A010888(A000330(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(A000330(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
|
|
|
REFERENCES
|
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.
B. Babcock and A. van Tuyl, Revisiting the spreading and covering numbers, Arxiv preprint arXiv:1109.5847, 2011
A. H. Beiler, Recreations in the Theory of Numbers, Dover, 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, Kekule structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
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, pg 293.
Manfred Goebel, Rewriting Techniques and Degree Bounds for Higher Order Symmetric Polynomials, Applicable Algebra in Engineering, Communication and Computing (AAECC), Volume 9, Issue 6 (1999), 559-573.
M. Merca, A Special Case of the Generalized Girard-Waring Formula, Journal of Integer Sequences, Vol. 15 (2012), #12.5.7. - From N. J. A. Sloane, Nov 25 2012
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).
|
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n = 0..1000
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).
H. Bottomley, Illustration of initial terms
T. Aaron Gulliver, Sequences from hexagonal pyramid of integers, International Mathematical Forum, Vol. 6, 2011, no. 17, p. 821 - 827.
Milan Janjic, Two Enumerative Functions
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550, 2013. - From N. J. A. Sloane, Feb 13 2013
R. Jovanovic, First 2500 Pyramidal numbers [Broken link?]
T. Mansour, Restricted permutations by patterns of type 2-1.
Mircea Merca, A Special Case of the Generalized Girard-Waring Formula J. Integer Sequences, Vol. 15 (2012), Article 12.5.7.
_Simon Plouffe_, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
T. Sillke, Square Counting
Eric Weisstein's World of Mathematics, Faulhaber's Formula
Wikipedia, Faulhaber's formula
Eric Weisstein's World of Mathematics, Square Pyramidal Number
G. Xiao, Sigma Server, Operate on"n^2"
Index entries for "core" sequences
Index entries for two-way infinite sequences
Index to sequences with linear recurrences with constant coefficients, signature (4,-6,4,-1).
|
|
|
FORMULA
|
G.f.: x*(1+x)/(1-x)^4.
E.g.f.: (x+3/2*x^2+1/3*x^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).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
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(sum[(i*j)^2, {i, 1, n}), {j, 1, n})). a(n) = sum(sum(sum((i*j*k)^2, {i, 1, n}), {j, 1, n}), {k, 1, n})^(1/3). - Alexander Adamchuk, Oct 26 2004
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-1) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), 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 A000292: C(n+3,3) = (n+1)*(n+2)*(n+3)/6: a(n) = A000292(n-1) + A000292(n-2). - Alexander Adamchuk, May 17 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
Starting n (-1,0,1,2,...), a(n) = C(n+2,2)+2*C(n+2,3). [From Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009]
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. [From 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) = (A000292(n) + A002411(n))/2. - Omar E. Pol, Jan 11 2013
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
|
|
|
MAPLE
|
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]
A000330:=(1+z)/(z-1)^4; [Simon Plouffe in his 1992 dissertation, sequence starting at a(1).]
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 (zerinvarylajos(AT)yahoo.com), Jan 02 2008
a:=n->sum(k^2, k=1..n):seq(a(n), n=0...44); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 15 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
|
|
|
MATHEMATICA
|
Table[Binomial[w+2, 3]+Binomial[w+1, 3], {w, 1, 30}]
|
|
|
PROG
|
(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 (hillcino368(AT)hotmail.com), Jun 18 2007
(PARI) a(n)=sum(m=1, n, sum(i=1, m, (2*i-1))) - Alexander R. Povolotsky, Nov 04 2007
(Haskell)
a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
a000330_list = scanl1 (+) a000290_list
-- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012
(Maxima) A000330(n):=binomial(n+2, 3)+binomial(n+1, 3)$
makelist(A000330(n), n, 0, 20); /* Martin Ettl, Nov 12 2012 */
|
|
|
CROSSREFS
|
Cf. A000217, A050446, A050447, A000537, A006003, A006331.
Cf. A000292, A033994, A132124, A132112, A050409.
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. [From Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009]
Cf. A156921, A157702.
Sequence in context: A096893 A074784 A109678 * A211804 A166068 A070129
Adjacent sequences: A000327 A000328 A000329 * A000331 A000332 A000333
|
|
|
KEYWORD
|
nonn,easy,core,nice,changed
|
|
|
AUTHOR
|
N. J. A. Sloane.
|
|
|
EXTENSIONS
|
Partially edited by Joerg Arndt, Mar 11 2010
|
|
|
STATUS
|
approved
|
| |
|
|
