cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A005900 Octahedral numbers: a(n) = n*(2*n^2 + 1)/3.

Original entry on oeis.org

0, 1, 6, 19, 44, 85, 146, 231, 344, 489, 670, 891, 1156, 1469, 1834, 2255, 2736, 3281, 3894, 4579, 5340, 6181, 7106, 8119, 9224, 10425, 11726, 13131, 14644, 16269, 18010, 19871, 21856, 23969, 26214, 28595, 31116, 33781, 36594, 39559, 42680
Offset: 0

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Author

Keywords

Comments

Series reversion of g.f.: A(x) is Sum_{n>0} - A066357(n)(-x)^n.
Partial sums of centered square numbers A001844. - Paul Barry, Jun 26 2003
Also as a(n) = (1/6)*(4n^3 + 2n), n>0: structured tetragonal diamond numbers (vertex structure 5) (cf. A000447 - structured diamonds); and structured trigonal anti-prism numbers (vertex structure 5) (cf. A100185 - structured anti-prisms). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {3,4}.
If X is an n-set and Y and Z are disjoint 2-subsets of X then a(n-4) is equal to the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Starting with 1 = binomial transform of [1, 5, 8, 4, 0, 0, 0, ...] where (1, 5, 8, 4) = row 3 of the Chebyshev triangle A081277. - Gary W. Adamson, Jul 19 2008
a(n) = largest coefficient of (1 + ... + x^(n-1))^4. - R. H. Hardin, Jul 23 2009
Convolution square root of (1 + 6x + 19x^3 + ...) = (1 + 3x + 5x^2 + 7x^3 + ...) = A005408(x). - Gary W. Adamson, Jul 27 2009
Starting with offset 1 = the triangular series convolved with [1, 3, 4, 4, 4, ...]. - Gary W. Adamson, Jul 28 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral, and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Let b be any product of four different primes. Then the divisor lattice of b^n is of width a(n+1). - Jean Drabbe, Oct 13 2010
Arises in Bezdek's proof on contact numbers for congruent sphere packings (see preprint). - Jonathan Vos Post, Feb 08 2011
Euler transform of length 2 sequence [6, -2]. - Michael Somos, Mar 27 2011
a(n+1) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 2n. - Clark Kimberling, Mar 19 2012
a(n) is the number of semistandard Young tableaux over all partitions of 3 with maximal element <= n. - Alois P. Heinz, Mar 22 2012
Self convolution of the odd numbers. - Reinhard Zumkeller, Apr 04 2012
a(n) is the number of (w,x,y,z) with all terms in {1,...,n} and w+x=y+z; also the number of (w,x,y,z) with all terms in {0,...,n} and |w-x|<=y. - Clark Kimberling, Jun 02 2012
The sequence is the third partial sum of (0, 1, 3, 4, 4, 4, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the number of join-irreducible elements in the Weyl group of type B_n with respect to the strong Bruhat order. - Rafael Mrden, Aug 26 2020
Number of unit octahedra contained in an n-scale octahedron composed of a tetrahedral-octahedral honeycomb. The number of unit tetrahedra in it is 8*A000292(n-1) = 4*(n^3 - n)/3. Also, the number of unit tetrahedra and unit octahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb is respectively A006527(n) = (n^3 + 2*n)/3 and A000292(n-1) = (n^3 - n)/6. - Jianing Song, Feb 24 2025

Examples

			G.f. = x + 6*x^2 + 19*x^3 + 44*x^4 + 85*x^5 + 146*x^6 + 231*x^7 + ...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
  • 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.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sums of 2 consecutive terms give A001845. Cf. A001844.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A022521.
Cf. A081277.
Row n=3 of A210391. - Alois P. Heinz, Mar 22 2012
Cf. A005408.
Cf. A002061.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Similar sequence: A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193(m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

Programs

  • Haskell
    a005900 n = sum $ zipWith (*) odds $ reverse odds
                where odds = take n a005408_list
    a005900_list = scanl (+) 0 a001844_list
    -- Reinhard Zumkeller, Jun 16 2013, Apr 04 2012
    
  • Magma
    [n*(2*n^2+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    I:=[0,1,6,19]; [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, Sep 12 2015
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(3,n), n=0..100)];
    A005900:=(z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation
    with(combinat): seq(fibonacci(4,2*n)/12, n=0..40); # Zerinvary Lajos, Apr 21 2008
  • Mathematica
    Table[(2n^3+n)/3, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,6,19},50] (* Harvey P. Dale, Oct 10 2013 *)
    CoefficientList[Series[x (1 + x)^2/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
  • Maxima
    makelist(n*(2*n^2+1)/3, n, 0, 20); /* Martin Ettl, Jan 07 2013 */
    
  • PARI
    {a(n) = n*(2*n^2+1)/3};
    
  • PARI
    concat([0],Vec(x*(1 + x)^2/(1 - x)^4 + O(x^50))) \\ Indranil Ghosh, Mar 16 2017
    
  • Python
    def a(n): return n*(2*n*n + 1)//3
    print([a(n) for n in range(41)]) # Michael S. Branicky, Sep 03 2021

Formula

a(n) = 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2. - Amarnath Murthy, May 28 2001
G.f.: x * (1 + x)^2 / (1 - x)^4. a(n) = -a(-n) = (2*n^3 + n) / 3.
a(n) = ( ((n+1)^5-n^5) - (n^5-(n-1)^5) )/30. - Xavier Acloque, Oct 17 2003
a(n) is the sum of the products pq, where p and q are both positive and odd and p + q = 2n, e.g., a(4) = 7*1 + 5*3 + 3*5 + 1*7 = 44. - Jon Perry, May 17 2005
a(n) = 4*binomial(n,3) + 4*binomial(n,2) + binomial(n,1). - Mitch Harris, Jul 06 2006
a(n) = binomial(n+2,3) + 2*binomial(n+1,3) + binomial(n,3), (this pair generalizes; see A014820, the 4-cross polytope numbers).
Sum_{n>=1} 1/a(n) = 3*gamma + 3*Psi((I*(1/2))*sqrt(2)) - (1/2)*(3*I)*Pi*coth((1/2)*Pi*sqrt(2)) - (1/2)*(3*I)*sqrt(2) = A175577, where I=sqrt(-1). - Stephen Crowley, Jul 14 2009
a(n) = A035597(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = A000578(n) - 2*A000292(n-1) for n>0. - J. M. Bergot, Apr 05 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, Sep 11 2015
E.g.f.: (1/3)*x*(3 + 6*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Mar 16 2017
a(n) = (A002061(A002061(n+1)) - A002061(A002061(n)))/6. - Daniel Poveda Parrilla, Jun 10 2017
a(n) = 6*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 6*log(2) - 4 = 1/(6 + 2/(6 + 6/(6 + ... + n*(n-1)/(6 + ...)))). See A142983. - Peter Bala, Mar 06 2024

A000447 a(n) = 1^2 + 3^2 + 5^2 + 7^2 + ... + (2*n-1)^2 = n*(4*n^2 - 1)/3.

Original entry on oeis.org

0, 1, 10, 35, 84, 165, 286, 455, 680, 969, 1330, 1771, 2300, 2925, 3654, 4495, 5456, 6545, 7770, 9139, 10660, 12341, 14190, 16215, 18424, 20825, 23426, 26235, 29260, 32509, 35990, 39711, 43680, 47905, 52394, 57155, 62196, 67525, 73150, 79079, 85320, 91881, 98770, 105995, 113564, 121485
Offset: 0

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Keywords

Comments

4 times the variance of the area under an n-step random walk: e.g., with three steps, the area can be 9/2, 7/2, 3/2, 1/2, -1/2, -3/2, -7/2, or -9/2 each with probability 1/8, giving a variance of 35/4 or a(3)/4. - Henry Bottomley, Jul 14 2003
Number of standard tableaux of shape (2n-1,1,1,1) (n>=1). - Emeric Deutsch, May 30 2004
Also a(n) = (1/6)*(8*n^3-2*n), n>0: structured octagonal diamond numbers (vertex structure 9). Cf. A059722 = alternate vertex; A000447 = structured diamonds; and structured tetragonal anti-diamond numbers (vertex structure 9). Cf. A096000 = alternate vertex; A100188 = structured anti-diamonds. Cf. A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The n-th tetrahedral (or pyramidal) number is n(n+1)(n+2)/6. This sequence contains the tetrahedral numbers of A000292 obtained for n= 1,3,5,7,... (see A015219). - Valentin Bakoev, Mar 03 2009
Using three consecutive numbers u, v, w, (u+v+w)^3-(u^3+v^3+w^3) equals 18 times the numbers in this sequence. - J. M. Bergot, Aug 24 2011
This sequence is related to A070893 by A070893(2*n-1) = n*a(n)-sum(i=0..n-1, a(i)). - Bruno Berselli, Aug 26 2011
Number of integer solutions to 1-n <= x <= y <= z <= n-1. - Michael Somos, Dec 27 2011
Partial sums of A016754. - Reinhard Zumkeller, Apr 02 2012
Also the number of cubes in the n-th Haüy square pyramid. - Eric W. Weisstein, Sep 27 2017

Examples

			G.f. = x + 10*x^2 + 35*x^3 + 84*x^4 + 165*x^5 + 286*x^6 + 455*x^7 + 680*x^8 + ...
a(2) = 10 since (-1, -1, -1), (-1, -1, 0), (-1, -1, 1), (-1, 0, 0), (-1, 0, 1), (-1, 1, 1), (0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1) are the 10 solutions (x, y, z) of -1 <= x <= y <= z <= 1.
a(0) = 0, which corresponds to the empty sum.
		

References

  • G. Chrystal, Textbook of Algebra, Vol. 1, A. & C. Black, 1886, Chap. XX, Sect. 10, Example 2.
  • F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • C. V. Durell, Advanced Algebra, Volume 1, G. Bell & Son, 1932, Exercise IIIe, No. 4.
  • L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • 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).

Crossrefs

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 in triangles A008956 and A008958.
A000447 is related to partitions of 2^n into powers of 2, as it is shown in the formula, example and cross-references of A002577. - Valentin Bakoev, Mar 03 2009

Programs

Formula

a(n) = binomial(2*n+1, 3) = A000292(2*n-1).
G.f.: x*(1+6*x+x^2)/(1-x)^4.
a(n) = -a(-n) for all n in Z.
a(n) = A000330(2*n)-4*A000330(n) = A000466(n)*n/3 = A000578(n)+A007290(n-2) = A000583(n)-2*A024196(n-1) = A035328(n)/3. - Henry Bottomley, Jul 14 2003
a(n+1) = (2*n+1)*(2*n+2)(2*n+3)/6. - Valentin Bakoev, Mar 03 2009
a(0)=0, a(1)=1, a(2)=10, a(3)=35, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, May 25 2012
a(n) = v(n,n-1), where v(n,k) is the central factorial numbers of the first kind with odd indices. - Mircea Merca, Jan 25 2014
a(n) = A005917(n+1) - A100157(n+1), where A005917 are the rhombic dodecahedral numbers and A100157 are the structured rhombic dodecahedral numbers (vertex structure 9). - Peter M. Chema, Jan 09 2016
For any nonnegative integers m and n, 8*(n^3)*a(m) + 2*m*a(n) = a(2*m*n). - Ivan N. Ianakiev, Mar 04 2017
E.g.f.: exp(x)*x*(1 + 4*x + (4/3)*x^2). - Wolfdieter Lang, Mar 11 2017
a(n) = A002412(n) + A016061(n-1), for n>0. - Bruce J. Nicholson, Nov 12 2017
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 6*log(2) - 3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3 - 3*log(2). (End)

Extensions

Chrystal and Durell references from R. K. Guy, Apr 02 2004

A002817 Doubly triangular numbers: a(n) = n*(n+1)*(n^2+n+2)/8.

Original entry on oeis.org

0, 1, 6, 21, 55, 120, 231, 406, 666, 1035, 1540, 2211, 3081, 4186, 5565, 7260, 9316, 11781, 14706, 18145, 22155, 26796, 32131, 38226, 45150, 52975, 61776, 71631, 82621, 94830, 108345, 123256, 139656, 157641, 177310, 198765, 222111, 247456, 274911, 304590
Offset: 0

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Keywords

Comments

Number of inequivalent ways to color vertices of a square using <= n colors, allowing rotations and reflections. Group is dihedral group D_8 of order 8 with cycle index (1/8)*(x1^4 + 2*x4 + 3*x2^2 + 2*x1^2*x2); setting all x_i = n gives the formula a(n) = (1/8)*(n^4 + 2*n + 3*n^2 + 2*n^3).
Number of semi-magic 3 X 3 squares with a line sum of n-1. That is, 3 X 3 matrices of nonnegative integers such that row sums and column sums are all equal to n-1. - [Gupta, 1968, page 653; Bell, 1970, page 279]. - Peter Bertok (peter(AT)bertok.com), Jan 12 2002. See A005045 for another version.
Also the coefficient h_2 of x^{n-3} in the shelling polynomial h(x)=h_0*x^n-1 + h_1*x^n-2 + h_2*x^n-3 + ... + h_n-1 for the independence complex of the cycle matroid of the complete graph K_n on n vertices (n>=2) - Woong Kook (andrewk(AT)math.uri.edu), Nov 01 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-4) is equal to the number of 5-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
Starting with offset 1 = binomial transform of [1, 5, 10, 9, 3, 0, 0, 0, ...]. - Gary W. Adamson, Aug 05 2009
Starting with "1" = row sums of triangle A178238. - Gary W. Adamson, May 23 2010
The equation n*(n+1)*(n^2 + n + 2)/8 may be arrived at by solving for x in the following equality: (n^2+n)/2 = (sqrt(8x+1)-1)/2. - William A. Tedeschi, Aug 18 2010
Partial sums of A006003. - Jeremy Gardiner, Jun 23 2013
Doubly triangular numbers are revealed in the sums of row sums of Floyd's triangle.
1, 1+5, 1+5+15, ...
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
- Tony Foster III, Nov 14 2015
From Jaroslav Krizek, Mar 04 2017: (Start)
For n>=1; a(n) = sum of the different sums of elements of all the nonempty subsets of the sets of numbers from 1 to n.
Example: for n = 6; nonempty subsets of the set of numbers from 1 to 3: {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}; sums of elements of these subsets: 1, 2, 3, 3, 4, 5, 6; different sums of elements of these subsets: 1, 2, 3, 4, 5, 6; a(3) = (1+2+3+4+5+6) = 21, ... (End)
a(n) is also the number of 4-cycles in the (n+4)-path complement graph. - Eric W. Weisstein, Apr 11 2018

Examples

			G.f. = x + 6*x^2 + 21*x^3 + 55*x^4 + 120*x^5 + 231*x^6 + 406*x^7 + 666*x^8 + ...
		

References

  • A. Björner, The homology and shellability of matroids and geometric lattices, in Matroid Applications (ed. N. White), Encyclopedia of Mathematics and Its Applications, 40, Cambridge Univ. Press 1992.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 124, #25, Q(3,r).
  • 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).
  • R. P. Stanley, Enumerative Combinatorics I, p. 292.

Crossrefs

Cf. A006003 (first differences), A165211 (mod 2).
Multiple triangular: A000217, A064322, A066370.
Cf. A006528 (square colorings).
Cf. A236770 (see crossrefs).
Row n=3 of A257493 and row n=2 of A331436 and A343097.
Cf. A000332.
Cf. A000292 (3-cycle count of \bar P_{n+4}), A060446 (5-cycle count of \bar P_{n+3}), A302695 (6-cycle count of \bar P_{n+5}).

Programs

  • Maple
    A002817 := n->n*(n+1)*(n^2+n+2)/8;
  • Mathematica
    a[ n_] := n (n + 1) (n^2 + n + 2) / 8; (* Michael Somos, Jul 24 2002 *)
    LinearRecurrence[{5,-10,10,-5,1}, {0,1,6,21,55},40] (* Harvey P. Dale, Jul 18 2011 *)
    nn=50;Join[{0},With[{c=(n(n+1))/2},Flatten[Table[Take[Accumulate[Range[ (nn(nn+1))/2]], {c,c}],{n,nn}]]]] (* Harvey P. Dale, Mar 19 2013 *)
  • PARI
    {a(n) = n * (n+1) * (n^2 + n + 2) / 8}; /* Michael Somos, Jul 24 2002 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(1-x)^5 + O(x^50))) \\ Altug Alkan, Nov 15 2015
    
  • Python
    def A002817(n): return (m:=n*(n+1))*(m+2)>>3 # Chai Wah Wu, Aug 30 2024

Formula

a(n) = 3*binomial(n+2, 4) + binomial(n+1, 2).
G.f.: x*(1 + x + x^2)/(1-x)^5. - Simon Plouffe (in his 1992 dissertation); edited by N. J. A. Sloane, May 13 2008
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Warut Roonguthai, Dec 13 1999
a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + a(n-5) = A000217(A000217(n)). - Ant King, Nov 18 2010
a(n) = Sum(Sum(1 + Sum(3*n))). - Xavier Acloque, Jan 21 2003
a(n) = A000332(n+1) + A000332(n+2) + A000332(n+3), with A000332(n) = binomial(n, 4). - Mitch Harris, Oct 17 2006 and Bruce J. Nicholson, Oct 22 2017
a(n) = Sum_{i=1..C(n,2)} i = C(C(n,2) + 1, 2) = A000217(A000217(n+1)). - Enrique Pérez Herrero, Jun 11 2012
Euler transform of length 3 sequence [6, 0, -1]. - Michael Somos, Nov 19 2015
E.g.f.: x*(8 + 16*x + 8*x^2 + x^3)*exp(x)/8. - Ilya Gutkovskiy, Apr 26 2016
Sum_{n>=1} 1/a(n) = 6 - 4*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) = 1.25269064911978447... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n)*A000124(n)/2.
a(n) = ((n-1)^4 + 3*(n-1)^3 + 2*(n-1)^2 + 2*n))/8. - Bruce J. Nicholson, Apr 05 2017
a(n) = (A016754(n)+ A007204(n)- 2) / 32. - Bruce J. Nicholson, Apr 14 2017
a(n) = a(-1-n) for all n in Z. - Michael Somos, Apr 17 2017
a(n) = T(T(n)) where T are the triangular numbers A000217. - Albert Renshaw, Jan 05 2020
a(n) = 2*n^2 - n + 6*binomial(n, 3) + 3*binomial(n, 4). - Ryan Jean, Mar 20 2021
a(n) = (A008514(n) - 1)/16. - Charlie Marion, Dec 20 2024

Extensions

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999

A004006 a(n) = C(n,1) + C(n,2) + C(n,3), or n*(n^2 + 5)/6.

Original entry on oeis.org

0, 1, 3, 7, 14, 25, 41, 63, 92, 129, 175, 231, 298, 377, 469, 575, 696, 833, 987, 1159, 1350, 1561, 1793, 2047, 2324, 2625, 2951, 3303, 3682, 4089, 4525, 4991, 5488, 6017, 6579, 7175, 7806, 8473, 9177, 9919, 10700, 11521, 12383, 13287, 14234, 15225
Offset: 0

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Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
The Burnside group B(3,n) has order 3^a(n).
Answer to the question: if you have a tall building and 3 plates and you need to find the highest story, a plate thrown from which does not break, what is the number of stories you can handle given n tries? - Leonid Broukhis, Oct 24 2000
Equals row sums of triangle A144329 starting with "1". - Gary W. Adamson, Sep 18 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=4, a(n-1)=-coeff(charpoly(A,x),x^(n-3)). - Milan Janjic, Jan 24 2010
From J. M. Bergot, Aug 03 2011: (Start)
If one formed the 3 X 3 square
| n | n+1 | n+2 |
| n+3 | n+4 | n+5 |
| n+6 | n+7 | n+8 |
and found the sum of the horizontal products n*(n + 1)*(n + 2) + (n + 3)*(n + 4)*(n + 5) + (n + 6)*(n + 7)*(n + 8) and added the sum of the vertical products n*(n + 3)*(n + 6) + (n + 1)*(n + 4)*(n + 7) + (n + 2)*(n + 5)(n + 8) one gets 6*n^3 + 72*n^2 + 318*n + 504. This will give 36 times the values of all the terms in this sequence. (End)
a(n) is divisible by n for n congruent to {1,5} mod 6. (see A007310). - Gary Detlefs, Dec 08 2011
From Beimar Naranjo, Feb 22 2024: (Start)
Number of compositions with at most three parts and sum at most n.
Also the number of compositions with at most one part distinct from 1 and with a sum at most n. (End)
a(n) is the number of strings of length n defined on {0, 1, 2, 3} that contain one 1 and any number of 0's, or two 2's and any number of 0's, or three 3's and any number of 0's. For example, a(6) = 41 since the strings are the 20 permutations of 333000, the 15 permutations of 220000 and the 6 permutations of 100000. - Enrique Navarrete, Jun 18 2025

Examples

			G.f. = x + 3*x^2 + 7*x^3 + 14*x^4 + 25*x^5 + 41*x^6 + 63*x^7 + 92*x^8 + ... - _Michael Somos_, Dec 29 2019
		

References

  • W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.

Crossrefs

Cf. A051576, A055795, A006552. Differences give A000217 + 1 or A000124.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

G.f.: x*(1-x+x^2)/(1-x)^4.
E.g.f.: x*(1 + x/2 + x^2/6) * exp(x).
a(-n) = -a(n).
a(n) = binomial(n+2,n-1) - binomial(n,n-2). - Zerinvary Lajos, May 11 2006
Euler transform of length 6 sequence [3, 1, 1, 0, 0, -1]. - Michael Somos, May 04 2007
Starting (1, 3, 7, 14, ...) = binomial transform of [1, 2, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson, Apr 24 2008
a(0)=0, a(1)=1, a(2)=3, a(3)=7, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 21 2011
a(n+1) = A000292(n) + n + 1. - Reinhard Zumkeller, Mar 31 2012
a(n) = 2*a(n-1) + (n-1) - a(n-2) with a(0) = 0, a(1) = 1. - Richard R. Forberg, Jan 23 2014
a(n) = Sum_{i=1..n} binomial(n-2i,2). - Wesley Ivan Hurt, Nov 18 2017
a(n) = n + Sum_{k=0..n} k*(n-k). - Gionata Neri, May 19 2018
a(n) = Sum_{k=0..n-1} A000124(k). - Torlach Rush, Aug 05 2018
G.f.: ((1 - x^5)/(1 - x)^5 - 1)/5. - Michael Somos, Dec 29 2019
G.f.: g(f(x)), where g is g.f. of A001477 and f is g.f. of A128834. - Oboifeng Dira, Jun 21 2020
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(5)) + polygamma(0, 1+i*sqrt(5)))/5 = 1.6787729555834452106286261834348972248... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

A005917 Rhombic dodecahedral numbers: a(n) = n^4 - (n - 1)^4.

Original entry on oeis.org

1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679, 34481, 39775, 45585, 51935, 58849, 66351, 74465, 83215, 92625, 102719, 113521, 125055, 137345, 150415, 164289, 178991
Offset: 1

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Author

Keywords

Comments

Final digits of a(n), i.e., a(n) mod 10, are repeated periodically with period of length 5 {1,5,5,5,9}. There is a symmetry in this list since the sum of two numbers equally distant from the ends is equal to 10 = 1 + 9 = 5 + 5 = 2*5. Last two digits of a(n), i.e., a(n) mod 100, are repeated periodically with period of length 50. - Alexander Adamchuk, Aug 11 2006
a(n) = VarScheme(n,2) in the scheme displayed in A128195. - Peter Luschny, Feb 26 2007
If Y is a 3-subset of a 2n-set X then, for n >= 2, a(n-2) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
The numbers are the constant number found in magic squares of order n, where n is an odd number, see the comment in A006003. A Magic Square of side 1 is 1; 3 is 15; 5 is 65 and so on. - David Quentin Dauthier, Nov 07 2008
Two times the area of the triangle with vertices at (0,0), ((n - 1)^2, n^2), and (n^2, (n - 1)^2). - J. M. Bergot, Jun 25 2013
Bisection of A006003. - Omar E. Pol, Sep 01 2018
Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals a(n+1). The first five rows of array M are [1, 7, 17, 31, 49, ...]; [3, 5, 15, 29, 47, ...]; [9, 11, 13, 27, 45, ...]; [19, 21, 23, 25, 43, ...]; [33, 35, 37, 39, 41, ...]. - J. M. Bergot, Jul 16 2013 [This contribution was moved here from A047926 by Petros Hadjicostas, Mar 08 2021.]
For n>=2, these are the primitive sides s of squares of type 2 described in A344332. - Bernard Schott, Jun 04 2021
(a(n) + 1) / 2 = A212133(n) is the number of cells in the n-th rhombic-dodecahedral polycube. - George Sicherman, Jan 21 2024

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, p. 53.
  • E. Deza and M. M. Deza, Figurate Numbers, World Scientific Publishing, 2012, pp. 123-124.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

(1/12)*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A063493, A063494, A063495, A063496.
Column k=3 of A047969.

Programs

  • Haskell
    a005917 n = a005917_list !! (n-1)
    a005917_list = map sum $ f 1 [1, 3 ..] where
       f x ws = us : f (x + 2) vs where (us, vs) = splitAt x ws
    -- Reinhard Zumkeller, Nov 13 2014
    
  • Magma
    [n^4 - (n-1)^4: n in [1..50]]; // Vincenzo Librandi, Aug 01 2011
    
  • Mathematica
    Table[n^4-(n-1)^4,{n,40}]  (* Harvey P. Dale, Apr 01 2011 *)
    #[[2]]-#[[1]]&/@Partition[Range[0,40]^4,2,1] (* More efficient than the above Mathematica program because it only has to calculate each 4th power once *) (* Harvey P. Dale, Feb 07 2015 *)
    Differences[Range[0,40]^4] (* Harvey P. Dale, Aug 11 2023 *)
  • PARI
    a(n)=n^4-(n-1)^4 \\ Charles R Greathouse IV, Jul 31 2011
    
  • Python
    A005917_list, m = [], [24, -12, 2, 1]
    for _ in range(10**2):
        A005917_list.append(m[-1])
        for i in range(3):
            m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

Formula

a(n) = (2*n - 1)*(2*n^2 - 2*n + 1).
Sum_{i=1..n} a(i) = n^4 = A000583(n). First differences of A000583.
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
More generally, g.f. for n^m - (n - 1)^m is Euler(m, x)/(1 - x)^m, where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292). E.g.f.: x*(exp(y/(1 - x)) - exp(x*y/(1 - x)))/(exp(x*y/(1 - x))-x*exp(y/(1 - x))). - Vladeta Jovovic, May 08 2002
a(n) = sum of the next (2*n - 1) odd numbers; i.e., group the odd numbers so that the n-th group contains (2*n - 1) elements like this: (1), (3, 5, 7), (9, 11, 13, 15, 17), (19, 21, 23, 25, 27, 29, 31), ... E.g., a(3) = 65 because 9 + 11 + 13 + 15 + 17 = 65. - Xavier Acloque, Oct 11 2003
a(n) = 2*n - 1 + 12*Sum_{i = 1..n} (i - 1)^2. - Xavier Acloque, Oct 16 2003
a(n) = (4*binomial(n,2) + 1)*sqrt(8*binomial(n,2) + 1). - Paul Barry, Mar 14 2004
Binomial transform of [1, 14, 36, 24, 0, 0, 0, ...], if the offset is 0. - Gary W. Adamson, Dec 20 2007
Sum_{i=1..n-1}(a(i) + a(i+1)) = 8*Sum_{i=1..n}(i^3 + i) = 16*A002817(n-1) for n > 1. - Bruno Berselli, Mar 04 2011
a(n+1) = a(n) + 2*(6*n^2 + 1) = a(n) + A005914(n). - Vincenzo Librandi, Mar 16 2011
a(n) = -a(-n+1). a(n) = (1/6)*(A181475(n) - A181475(n-2)). - Bruno Berselli, Sep 26 2011
a(n) = A045975(2*n-1,n) = A204558(2*n-1)/(2*n - 1). - Reinhard Zumkeller, Jan 18 2012
a(n+1) = Sum_{k=0..2*n+1} (A176850(n,k) - A176850(n-1,k))*(2*k + 1), n >= 1. - L. Edson Jeffery, Nov 02 2012
a(n) = A005408(n-1) * A001844(n-1) = (2*(n - 1) + 1) * (2*(n - 1)*n + 1) = A000290(n-1)*12 + 2 + a(n-1). - Bruce J. Nicholson, May 17 2017
a(n) = A007588(n) + A007588(n-1) = A000292(2n-1) + A000292(2n-2) + A000292(2n-3) = A002817(2n-1) - A002817(2n-2). - Bruce J. Nicholson, Oct 22 2017
a(n) = A005898(n-1) + 6*A000330(n-1) (cf. Deza, Deza, 2012, p. 123, Section 2.6.2). - Felix Fröhlich, Oct 01 2018
a(n) = A300758(n-1) + A005408(n-1). - Bruce J. Nicholson, Apr 23 2020
G.f.: polylog(-4, x)*(1-x)/x. See the Simon Plouffe formula above (with expanded numerator), and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 10 2021

A027480 a(n) = n*(n+1)*(n+2)/2.

Original entry on oeis.org

0, 3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, 1092, 1365, 1680, 2040, 2448, 2907, 3420, 3990, 4620, 5313, 6072, 6900, 7800, 8775, 9828, 10962, 12180, 13485, 14880, 16368, 17952, 19635, 21420, 23310, 25308, 27417, 29640, 31980, 34440
Offset: 0

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Author

Olivier Gérard and Ken Knowlton (kcknowlton(AT)aol.com)

Keywords

Comments

Write the integers in groups: 0; 1,2; 3,4,5; 6,7,8,9; ... and add the groups: a(n) = Sum_{j=0..n} (A000217(n)+j), row sums of the triangular view of A001477. - Asher Auel, Jan 06 2000
With offset = 2, a(n) is the number of edges of the line graph of the complete graph of order n, L(K_n). - Roberto E. Martinez II, Jan 07 2002
Also the total number of pips on a set of dominoes of type n. (A "3" domino set would have 0-0, 0-1, 0-2, 0-3, 1-1, 1-2, 1-3, 2-2, 2-3, 3-3.) - Gerard Schildberger, Jun 26 2003. See A129533 for generalization to n-armed "dominoes". - N. J. A. Sloane, Jan 06 2016
Common sum in an (n+1) X (n+1) magic square with entries (0..n^2-1).
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
If Y is a 3-subset of an n-set X then, for n >= 5, a(n-5) is the number of 4-subsets of X which have exactly one element in common with Y. Also, if Y is a 3-subset of an n-set X then, for n >= 5, a(n-5) is the number of (n-5)-subsets of X which have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
These numbers, starting with 3, are the denominators of the power series f(x) = (1-x)^2 * log(1/(1-x)), if the numerators are kept at 1. This sequence of denominators starts at the term x^3/3. - Miklos Bona, Feb 18 2009
a(n) is the number of triples (w,x,y) having all terms in {0..n} and satisfying at least one of the inequalities x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
From Martin Licht, Dec 04 2016: (Start)
Let b(n) = (n+1)(n+2)(n+3)/2 (the same sequence, but with a different offset). Then (see Arnold et al., 2006):
b(n) is the dimension of the Nédélec space of the second kind of polynomials of order n over a tetrahedron.
b(n-1) is the dimension of the curl-conforming Nédélec space of the first kind of polynomials of order n with tangential boundary conditions over a tetrahedron.
b(n) is the dimension of the divergence-conforming Nédélec space of the first kind of polynomials of order n with normal boundary conditions over a tetrahedron. (End)
After a(0), the digital root has period 9: repeat [3, 3, 3, 6, 6, 6, 9, 9, 9]. - Peter M. Chema, Jan 19 2017

Examples

			Row sums of n consecutive integers, starting at 0, seen as a triangle:
.
    0 |  0
    3 |  1  2
   12 |  3  4  5
   30 |  6  7  8  9
   60 | 10 11 12 13 14
  105 | 15 16 17 18 19 20
		

Crossrefs

1/beta(n, 3) in A061928.
A row of array in A129533.
Cf. similar sequences of the type n*(n+1)*(n+k)/2 listed in A267370.
Similar sequences are listed in A316224.
Third column of A003506.
A bisection of A330298.

Programs

  • Magma
    [n*(n+1)*(n+2)/2: n in [0..40]]; // Vincenzo Librandi, Nov 14 2014
    
  • Maple
    [seq(3*binomial(n+2,3),n=0..37)]; # Zerinvary Lajos, Nov 24 2006
    a := n -> add((j+n)*(n+2)/3,j=0..n): seq(a(n),n=0..35); # Zerinvary Lajos, Dec 17 2006
  • Mathematica
    Table[(m^3 - m)/2, {m, 36}] (* Zerinvary Lajos, Mar 21 2007 *)
    LinearRecurrence[{4,-6,4,-1},{0,3,12,30},40] (* Harvey P. Dale, Oct 10 2012 *)
    CoefficientList[Series[3 x / (x - 1)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 14 2014 *)
    With[{nn=50},Total/@TakeList[Range[0,(nn(nn+1))/2-1],Range[nn]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Jun 02 2019 *)
  • PARI
    a(n)=3*binomial(n+2,3) \\ Charles R Greathouse IV, May 23 2011
    
  • Python
    def a(n): return (n**3+3*n**2+2*n)//2 # _Torlach Rush, Jun 16 2024

Formula

a(n) = a(n-1) + A050534(n) = 3*A000292(n-1) = A050534(n) - A050534(n-1).
a(n) = n*binomial(2+n, 2). - Zerinvary Lajos, Jan 10 2006
a(n) = A007531(n+2)/2. - Zerinvary Lajos, Jul 17 2006
Starting with offset 1 = binomial transform of [3, 9, 9, 3, 0, 0, 0]. - Gary W. Adamson, Oct 25 2007
From R. J. Mathar, Apr 07 2009: (Start)
G.f.: 3*x/(x-1)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
a(n) = Sum_{i=0..n} n*(n - i) + 2*i. - Bruno Berselli, Jan 13 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
E.g.f.: x*(6 + 6*x + x^2)*exp(x)/2.
a(n) = Sum_{k=0..n} A045943(k).
Sum_{n>=1} 1/a(n) = 1/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (8*log(2) - 5)/2 = 0.2725887222397812... = A016639/10. (End)
a(n-1) = binomial(n^2,2)/n for n > 0. - Jonathan Sondow, Jan 07 2018
For k > 1, Sum_{i=0..n^2-1} (k+i)^2 = (k*n + a(k-1))^2 + A126275(k). - Charlie Marion, Apr 23 2021

A006527 a(n) = (n^3 + 2*n)/3.

Original entry on oeis.org

0, 1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531, 28424, 30405
Offset: 0

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Author

Keywords

Comments

Number of ways to color vertices (or edges) of a triangle using <= n colors, allowing only rotations.
Also: dot_product (1,2,...,n)*(2,3,...,n,1), n >= 0. - Clark Kimberling
Start from triacid and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002
Starting with offset 1 = row sums of triangle A158822 and binomial transform of (1, 3, 4, 2, 0, 0, 0, ...). - Gary W. Adamson, Mar 28 2009
One-ninth of sum of three consecutive cubes: a(n) = ((n-1)^3 + n^3 + (n+1)^3)/9. - Zak Seidov, Jul 22 2013
For n > 2, number of different cubes, formed after splitting a cube in color C_1, by parallel planes in the colors C_2, C_3, ..., C_n in three spatial dimensions (in the order of the colors from a fixed vertex). Generally, in a large hypercube n^d is f(n,d) = C(n+d-1, d) + C(n, d) different small hypercubes. See below for my formula a(n) = f(n,3). - Thomas Ordowski, Jun 15 2014
a(n) is a square for n = 1, 2 & 24; and for no other values up to 10^7 (see M. Gardner). - Michel Marcus, Sep 06 2015
Number of unit tetrahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb. - Jason Pruski, Aug 23 2017

References

  • M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 of triangle A094414. Row 6 of the array in A107735.
Cf. A000292 (unoriented), A000292(n-2) (chiral), A000290 (achiral) triangle colorings.
Row 2 of A324999 (simplex vertices and facets) and A327083 (simplex edges and ridges).

Programs

  • Haskell
    a006527 n = n * (n ^ 2 + 2) `div` 3  -- Reinhard Zumkeller, Jan 06 2014
  • Magma
    [(n^3 + 2*n)/3: n in [0..50]]; // Vincenzo Librandi, May 15 2011
    
  • Maple
    A006527:=z*(1+z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
    with(combinat):seq(lcm(fibonacci(4,n),fibonacci(2,n))/3,n=0..42); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[ (n^3 + 2*n)/3, {n, 0, 45} ]
    LinearRecurrence[{4,-6,4,-1},{0,1,4,11},46] (* or *) CoefficientList[ Series[(x+x^3)/(x-1)^4,{x,0,49}],x] (* Harvey P. Dale, Jun 13 2011 *)
  • PARI
    a(n)=n*(n^2+2)/3 \\ Charles R Greathouse IV, Jul 25 2011
    

Formula

a(0)=0, a(1)=1, a(2)=4, a(3)=11; for n > 3, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jun 13 2011
From Paul Barry, Mar 13 2003: (Start)
a(n) = 2*binomial(n+1, 3) + binomial(n, 1).
G.f.: x*(1+x^2)/(1-x)^4. (End)
a(n) = A000292(n) + A000292(n-2). - Alexander Adamchuk, May 20 2006
a(n) = n*A059100(n)/3. - Lekraj Beedassy, Feb 06 2007
a(n) = A054602(n)/3. - Zerinvary Lajos, Apr 20 2008
a(n) = ( n + Sum_{i=1..n} A177342(i) )/(n+1), with n > 0. - Bruno Berselli, May 19 2010
a(n) = A002264(A000578(n) + A005843(n)). - Reinhard Zumkeller, Jun 16 2011
a(n) = binomial(n+2, 3) + binomial(n, 3). - Thomas Ordowski, Jun 15 2014
a(n) = A000292(n) - A000292(-n). - Bruno Berselli, Sep 22 2016
E.g.f.: (x/3)*(3 + 3*x + x^2)*exp(x). - G. C. Greubel, Sep 01 2017
From Robert A. Russell, Oct 20 2020: (Start)
a(n) = 1*C(n,1) + 2*C(n,2) + 2*C(n,3), where the coefficient of C(n,k) is the number of oriented triangle colorings using exactly k colors.
a(n) = 2*A000292(n) - A000290(n) = 2*A000292(n-2) + A000290(n). (End)
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/4 = 1.45245201414472469745354677573358867... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

Extensions

More terms from Alexander Adamchuk, May 20 2006
Corrected and replaced 5th formula from Harvey P. Dale, Jun 13 2011
Deleted an erroneous comment. - N. J. A. Sloane, Dec 10 2018

A037270 a(n) = n^2*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 10, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10440, 14365, 19306, 25425, 32896, 41905, 52650, 65341, 80200, 97461, 117370, 140185, 166176, 195625, 228826, 266085, 307720, 354061, 405450, 462241, 524800, 593505, 668746, 750925, 840456, 937765
Offset: 0

Views

Author

Aaron Gulliver (gulliver(AT)elec.canterbury.ac.nz)

Keywords

Comments

Sum of first n^2 positive integers.
Start from xanthene and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002; Amarnath Murthy, Aug 01 2002
Sum of the next n multiples of n. - Amarnath Murthy, Aug 01 2002
The sum of the terms in an n X n spiral. These are also triangular numbers. - William A. Tedeschi, Feb 27 2008
Hypotenuse of Pythagorean triangles with smallest side a cube: A000578(n)^2 + A083374(n)^2 = a(n)^2. - Martin Renner, Nov 12 2011
For n>1, triangular numbers that can be represented as a sum of a square and a triangular number. For example, a(2)=10=4+6=9+1. - Ivan N. Ianakiev, Apr 24 2012
A037270 can be constructed in the following manner: Take A000217 and for every n not in A000290 delete the corresponding A000217(n). - Ivan N. Ianakiev, Apr 26 2012
Starting at a(1)=1 simply take 1*1=1, a(2)= 2*(2+3)=10, a(3)= 3*(4+5+6)=45, a(4)=4*(7+8+9+10) and so on. - J. M. Bergot, May 01 2015
Observation: The digital roots of the terms repeat in the sequence 1, 1, 9; e.g., the digital roots of 1, 10, 45, 136, 325, and 666 are 1, 1, 9, 1, 1, and 9. Verified for the first 10000 terms. - Rob Barton, Mar 28 2018
The above observation is easily explained and proved given that the digital root of a positive number equals the number modulo 9, and a(n + 9k) == a(n) (mod 9). - M. F. Hasler, Apr 05 2018
Number of unoriented rows of length 4 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=10, there are 4 achiral (AAAA, ABBA, BAAB, BBBB) and 6 chiral pairs (AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, BABB-BBAB). - Robert A. Russell, Nov 14 2018
For n > 0, a(2n+1) is the number of non-isomorphic 6C_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k>=2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
Number of achiral colorings of the edges of a tetrahedron with n available colors. - Robert A. Russell, Sep 07 2019

References

  • C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 5.
  • C. Barrientos, Graceful labelings of cyclic snakes, Ars Combin., 60(2001), 85-96.
  • Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 55.
  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.
  • R. A. Wilson, Cosmic Trigger, epilogue of S.-P. Sirag.

Crossrefs

Cf. A000217, A236770 (see crossrefs).
Row 4 of A277504.
Cf. A000583 (oriented), A083374 (chiral), A000290 (achiral).
Cf. A317617.
Row 3 of A327086 (achiral simplex edge colorings).

Programs

  • GAP
    a:=List([0..30],n->n^2*(n^2+1)/2); # Muniru A Asiru, Mar 28 2018
    
  • Magma
    [n^2*(n^2 + 1)/2: n in [0..30]] // Stefano Spezia, Jan 15 2019
  • Maple
    seq(n^2*(n^2+1)/2,n=0..30); # Muniru A Asiru, Mar 28 2018
  • Mathematica
    Table[ n^2*((n^2 + 1)/2), {n, 0, 30} ]
    Table[(1/8) Round[N[Sinh[2 ArcSinh[n]]^2, 100]], {n, 0, 30}] (* Artur Jasinski, Feb 10 2010 *)
    LinearRecurrence[{5,-10,10,-5,1},{0,1,10,45,136},30] (* Harvey P. Dale, Aug 03 2014 *)
  • PARI
    a(n)=binomial(n^2+1,2) \\ Charles R Greathouse IV, Apr 25 2012
    
  • Python
    for n in range(0,30): print(n**2*(n**2+1)/2, end=', ') # Stefano Spezia, Jan 10 2019
    

Formula

a(n) = a(n-1) + n^3 + (n-1)^3.
a(n) = A000537(n)+A000537(n-1), i.e., square of sum of first n integers plus square of sum of first n-1 integers. - Henry Bottomley, Oct 15 2001
a(n) = Sum_{k=0..n^2} k. - William A. Tedeschi, Feb 27 2008
a(n) = (1/8)*sinh(2*arcsinh(n)). - Artur Jasinski, Feb 10 2010
G.f.: x*(1+x)*(1+4*x+x^2)/(1-x)^5. - Colin Barker, Mar 22 2012
a(n) = a(n-1) + A005898(n-1). - Ivan N. Ianakiev, May 13 2012
a(n) = 2 * A000217(n-1) * A000217(n) + A000290(n). - Ivan N. Ianakiev, May 26 2012
a(n) = A000217(n^2). - J. M. Bergot, Jun 07 2012
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5) n>4, a(0)=0, a(1)=1, a(2)=10, a(3)=45, a(4)=136. - Yosu Yurramendi, Sep 02 2013
For n>0, a(n) = A000217(n)^2 + A000217(n-1)^2. - Richard R. Forberg, Dec 25 2013
a(n) = T(T(n)) + T(T(n-1)) + T(T(n)-1) + T(T(n-1)-1), where T(n) = A000217(n). - Charlie Marion, Sep 10 2016
a(n) = t(n-3)*t(n)+t(n-1)*t(n+2), with t(n)=A000217(n). - J. M. Bergot, Apr 07 2018
From Robert A. Russell, Nov 14 2018: (Start)
a(n) = (A000583(n) + A000290(n)) / 2 = (n^4 + n^2) / 2.
a(n) = A000583(n) - A083374(n) = A083374(n) + A000290(n).
G.f.: (Sum_{j=1..4} S2(4,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..2} S2(2,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: Sum_{k=1..4} A145882(4,k) * x^k / (1-x)^5.
E.g.f.: (Sum_{k=1..4} S2(4,k)*x^k + Sum_{k=1..2} S2(2,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>4, a(n) = Sum_{j=1..5} -binomial(j-6,j) * a(n-j). (End)
a(n) = n*A006003(n). - Kritsada Moomuang, Dec 16 2018
For n > 0, a(n) = Sum_{k=1..n} A317617(n,k). - Stefano Spezia, Jan 10 2019
Sum_{n>=1} 1/a(n) = 1 + Pi^2/3 - Pi*coth(Pi) = 1.13652003875929052467672874379... - Vaclav Kotesovec, Jan 21 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi*csch(Pi) + Pi^2/6 - 1. - Amiram Eldar, Nov 02 2021

A007588 Stella octangula numbers: a(n) = n*(2*n^2 - 1).

Original entry on oeis.org

0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, 2651, 3444, 4381, 5474, 6735, 8176, 9809, 11646, 13699, 15980, 18501, 21274, 24311, 27624, 31225, 35126, 39339, 43876, 48749, 53970, 59551, 65504, 71841, 78574, 85715, 93276, 101269, 109706, 118599, 127960
Offset: 0

Views

Author

Keywords

Comments

Also as a(n)=(1/6)*(12*n^3-6*n), n>0: structured hexagonal anti-diamond numbers (vertex structure 13) (Cf. A005915 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The only known square stella octangula number for n>1 is a(169) = 169*(2*169^2 - 1) = 9653449 = 3107^2. - Alexander Adamchuk, Jun 02 2008
Ljunggren proved that 9653449 = (13*239)^2 is the only square stella octangula number for n>1. See A229384 and the Wikipedia link. - Jonathan Sondow, Sep 30 2013
4*A007588 = A144138(ChebyshevU[3,n]). - Vladimir Joseph Stephan Orlovsky, Jun 30 2011
If A016813 is regarded as a regular triangle (with leading terms listed in A001844), a(n) provides the row sums of this triangle: 1, 5+9=14, 13+17+21=51 and so on. - J. M. Bergot, Jul 05 2013
Shares its digital root, A267017, with n*(n^2 + 1)/2 ("sum of the next n natural numbers" see A006003). - Peter M. Chema, Aug 28 2016

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 51.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • W. Ljunggren, Zur Theorie der Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Backwards differences give star numbers A003154: A003154(n)=a(n)-a(n-1).
1/12*t*(n^3-n)+ n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A001653 = Numbers n such that 2*n^2 - 1 is a square.
a(169) = (A229384(3)*A229384(4))^2.

Programs

Formula

G.f.: x*(1+10*x+x^2)/(1-x)^4.
a(n) = n*A056220(n).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Harvey P. Dale, Sep 16 2011
From Ilya Gutkovskiy, Jul 02 2016: (Start)
E.g.f.: x*(1 + 6*x + 2*x^2)*exp(x).
Dirichlet g.f.: 2*zeta(s-3) - zeta(s-1). (End)
a(n) = A004188(n) + A135503(n). - Miquel Cerda, Dec 25 2016
a(n) = A061317(n) - A005843(n) = A062392(n) - A062392(n-1). - J.S. Seneschal, Jul 01 2025

Extensions

In the formula given in the 1995 Encyclopedia of Integer Sequences, the second 2 should be an exponent.

A034262 a(n) = n^3 + n.

Original entry on oeis.org

0, 2, 10, 30, 68, 130, 222, 350, 520, 738, 1010, 1342, 1740, 2210, 2758, 3390, 4112, 4930, 5850, 6878, 8020, 9282, 10670, 12190, 13848, 15650, 17602, 19710, 21980, 24418, 27030, 29822, 32800, 35970, 39338, 42910, 46692, 50690, 54910, 59358, 64040, 68962, 74130
Offset: 0

Views

Author

Stuart M. Ellerstein (ellerstein(AT)aol.com), Apr 21 2000

Keywords

Comments

k such that x^3 + x + k factors over the integers. - James R. Buddenhagen, Apr 19 2005
If a(n)=X [A155977], Y=b(n) [A071253], Z=c(n) [A034262], then X^2+Y^2 = n*Z^3; e.g., if n=3, a(3)=270, b(3)=90, c(3)=30, then 270^2+90^2=3*30^3. - Vincenzo Librandi, Nov 24 2010
From Bruno Berselli, Sep 06 2018: (Start)
After 0, sum of next n even numbers:
... 2, 2
... 4, 6, 10
... 8, 10, 12, 30
.. 14, 16, 18, 20, 68
.. 22, 24, 26, 28, 30, 130
.. 32, 34, 36, 38, 40, 42, 222 etc. (End)
Sequence occurs in the binomial identity Sum_{k = 0..n} a(k)* binomial(n,k)/binomial(n+k,k) = n*(n + 1)/2. Cf. A092181 and A155977. - Peter Bala, Feb 12 2019
For n >= 2, a(n) is the sum of the numbers in the 1st and last columns of an n X n square array whose elements are the numbers from 1..n^2, listed in increasing order by rows. - Wesley Ivan Hurt, May 17 2021

Crossrefs

Programs

Formula

a(n) = 2*A006003(n).
a(n) = A002522(n)*A001477(n). - Zerinvary Lajos, Apr 20 2008
For n>1, a(n) = floor(n^5/(n^2-1)). - Gary Detlefs, Feb 10 2010
Sum_{n>=1} 1/a(n) = 0.6718659855... = gamma + Re psi(1+i) = A001620+A248177. [Borwein et al., J. Math. Anal. Appl. 316 (2006) 328]. - R. J. Mathar, Jul 17 2012
a(n) = -a(-n) for all n in Z. - Michael Somos, Jul 11 2017
G.f.: 2*x*(x^2+x+1)/(x-1)^4. - Alois P. Heinz, Oct 08 2022
E.g.f.: x*(2 + 3*x + x^2)*exp(x). - Stefano Spezia, Jun 20 2024
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