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.

Showing 1-10 of 10 results.

A000537 Sum of first n cubes; or n-th triangular number squared.

Original entry on oeis.org

0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, 4356, 6084, 8281, 11025, 14400, 18496, 23409, 29241, 36100, 44100, 53361, 64009, 76176, 90000, 105625, 123201, 142884, 164836, 189225, 216225, 246016, 278784, 314721, 354025, 396900, 443556, 494209, 549081
Offset: 0

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Author

Keywords

Comments

Number of parallelograms in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Or, number of orthogonal rectangles in an n X n checkerboard, or rectangles in an n X n array of squares. - Jud McCranie, Feb 28 2003. Compare A085582.
Also number of 2-dimensional cage assemblies (cf. A059827, A059860).
The n-th triangular number T(n) = Sum_{r=1..n} r = n(n+1)/2 satisfies the relations: (i) T(n) + T(n-1) = n^2 and (ii) T(n) - T(n-1) = n by definition, so that n^2*n = n^3 = {T(n)}^2 - {T(n-1)}^2 and by summing on n we have Sum_{ r = 1..n } r^3 = {T(n)}^2 = (1+2+3+...+n)^2 = (n*(n+1)/2)^2. - Lekraj Beedassy, May 14 2004
Number of 4-tuples of integers from {0,1,...,n}, without repetition, whose last component is strictly bigger than the others. Number of 4-tuples of integers from {1,...,n}, with repetition, whose last component is greater than or equal to the others.
Number of ordered pairs of two-element subsets of {0,1,...,n} without repetition.
Number of ordered pairs of 2-element multisubsets of {1,...,n} with repetition.
1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
a(n) is the number of parameters needed in general to know the Riemannian metric g of an n-dimensional Riemannian manifold (M,g), by knowing all its second derivatives; even though to know the curvature tensor R requires (due to symmetries) (n^2)*(n^2-1)/12 parameters, a smaller number (and a 4-dimensional pyramidal number). - Jonathan Vos Post, May 05 2006
Also number of hexagons with vertices in an hexagonal grid with n points in each side. - Ignacio Larrosa Cañestro, Oct 15 2006
Number of permutations of n distinct letters (ABCD...) each of which appears twice with 4 and n-4 fixed points. - Zerinvary Lajos, Nov 09 2006
With offset 1 = binomial transform of [1, 8, 19, 18, 6, ...]. - Gary W. Adamson, Dec 03 2008
The sequence is related to A000330 by a(n) = n*A000330(n) - Sum_{i=0..n-1} A000330(i): this is the case d=1 in the identity n*(n*(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. - Bruno Berselli, Apr 26 2010, Mar 01 2012
From Wolfdieter Lang, Jan 11 2013: (Start)
For sums of powers of positive integers S(k,n) := Sum_{j=1..n}j^k one has the recurrence S(k,n) = (n+1)*S(k-1,n) - Sum_{l=1..n} S(k-1,l), n >= 1, k >= 1.
This was used for k=4 by Ibn al-Haytham in an attempt to compute the volume of the interior of a paraboloid. See the Strick reference where the trick he used is shown, and the W. Lang link.
This trick generalizes immediately to arbitrary powers k. For k=3: a(n) = (n+1)*A000330(n) - Sum_{l=1..n} A000330(l), which coincides with the formula given in the previous comment by Berselli. (End)
Regarding to the previous contribution, see also Matem@ticamente in Links field and comments on this recurrences in similar sequences (partial sums of n-th powers). - Bruno Berselli, Jun 24 2013
A rectangular prism with sides A000217(n), A000217(n+1), and A000217(n+2) has surface area 6*a(n+1). - J. M. Bergot, Aug 07 2013, edited with corrected indices by Antti Karttunen, Aug 09 2013
A formula for the r-th successive summation of k^3, for k = 1 to n, is (6*n^2+r*(6*n+r-1)*(n+r)!)/((r+3)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
Note that this sequence and its formula were known to (and possibly discovered by) Nicomachus, predating Ibn al-Haytham by 800 years. - Charles R Greathouse IV, Apr 23 2014
a(n) is the number of ways to paint the sides of a nonsquare rectangle using at most n colors. Cf. A039623. - Geoffrey Critzer, Jun 18 2014
For n > 0: A256188(a(n)) = A000217(n) and A256188(m) != A000217(n) for m < a(n), i.e., positions of first occurrences of triangular numbers in A256188. - Reinhard Zumkeller, Mar 26 2015
There is no cube in this sequence except 0 and 1. - Altug Alkan, Jul 02 2016
Also the number of chordless cycles in the complete bipartite graph K_{n+1,n+1}. - Eric W. Weisstein, Jan 02 2018
a(n) is the sum of the elements in the multiplication table [0..n] X [0..n]. - Michel Marcus, May 06 2021

Examples

			G.f. = x + 9*x^2 + 36*x^3 + 100*x^4 + 225*x^5 + 441*x^6 + ... - _Michael Somos_, Aug 29 2022
		

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.
  • Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 62, eq. (6.3) for k=3.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 110ff.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, pp. 36, 58.
  • Clifford Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Oxford University Press, 2001, p. 325.
  • 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).
  • H. K. Strick, Geschichten aus der Mathematik II, Spektrum Spezial 3/11, p. 13.
  • D. Wells, You Are A Mathematician, "Counting rectangles in a rectangle", Problem 8H, pp. 240; 254, Penguin Books 1995.

Crossrefs

Convolution of A000217 and A008458.
Row sums of triangles A094414 and A094415.
Second column of triangle A008459.
Row 3 of array A103438.
Cf. A236770 (see crossrefs).

Programs

  • GAP
    List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018
    
  • Haskell
    a000537 = a000290 . a000217  -- Reinhard Zumkeller, Mar 26 2015
    
  • Magma
    [(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
    
  • Maple
    a:= n-> (n*(n+1)/2)^2:
    seq(a(n), n=0..40);
  • Mathematica
    Accumulate[Range[0, 50]^3] (* Harvey P. Dale, Mar 01 2011 *)
    f[n_] := n^2 (n + 1)^2/4; Array[f, 39, 0] (* Robert G. Wilson v, Nov 16 2012 *)
    Table[CycleIndex[{{1, 2, 3, 4}, {3, 2, 1, 4}, {1, 4, 3, 2}, {3, 4, 1, 2}}, s] /. Table[s[i] -> n, {i, 1, 2}], {n, 0, 30}] (* Geoffrey Critzer, Jun 18 2014 *)
    Accumulate @ Range[0, 50]^2 (* Waldemar Puszkarz, Jan 24 2015 *)
    Binomial[Range[20], 2]^2 (* Eric W. Weisstein, Jan 02 2018 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 9, 36, 100}, 20] (* Eric W. Weisstein, Jan 02 2018 *)
    CoefficientList[Series[-((x (1 + 4 x + x^2))/(-1 + x)^5), {x, 0, 20}], x] (* Eric W. Weisstein, Jan 02 2018 *)
  • PARI
    a(n)=(n*(n+1)/2)^2
    
  • Python
    def A000537(n): return (n*(n+1)>>1)**2 # Chai Wah Wu, Oct 20 2023

Formula

a(n) = (n*(n+1)/2)^2 = A000217(n)^2 = Sum_{k=1..n} A000578(k), that is, 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
G.f.: (x+4*x^2+x^3)/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = Sum ( Sum ( 1 + Sum (6*n) ) ), rephrasing the formula in A000578. - Xavier Acloque, Jan 21 2003
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j, row sums of A127777. - Alexander Adamchuk, Oct 24 2004
a(n) = A035287(n)/4. - Zerinvary Lajos, May 09 2007
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=1. - Alexander R. Povolotsky, May 17 2008
G.f.: x*F(3,3;1;x). - Paul Barry, Sep 18 2008
Sum_{k > 0} 1/a(k) = (4/3)*(Pi^2-9). - Jaume Oliver Lafont, Sep 20 2009
a(n) = Sum_{1 <= k <= m <= n} A176271(m,k). - Reinhard Zumkeller, Apr 13 2010
a(n) = Sum_{i=1..n} J_3(i)*floor(n/i), where J_ 3 is A059376. - Enrique Pérez Herrero, Feb 26 2012
a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} min(i,j,k). - Enrique Pérez Herrero, Feb 26 2013 [corrected by Ridouane Oudra, Mar 05 2025]
a(n) = 6*C(n+2,4) + C(n+1,2) = 6*A000332(n+2) + A000217(n), (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..3} j*Stirling1(n+1,n+1-j)*Stirling2(n+3-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*(3-4*log(2)). - Vaclav Kotesovec, Feb 13 2015
a(n)*((s-2)*(s-3)/2) = P(3, P(s, n+1)) - P(s, P(3, n+1)), where P(s, m) = ((s-2)*m^2-(s-4)*m)/2 is the m-th s-gonal number. For s=7, 10*a(n) = A000217(A000566(n+1)) - A000566(A000217(n+1)). - Bruno Berselli, Aug 04 2015
From Ilya Gutkovskiy, Jul 03 2016: (Start)
E.g.f.: x*(4 + 14*x + 8*x^2 + x^3)*exp(x)/4.
Dirichlet g.f.: (zeta(s-4) + 2*zeta(s-3) + zeta(s-2))/4. (End)
a(n) = (Bernoulli(4, n+1) - Bernoulli(4, 1))/4, n >= 0, with the Bernoulli polynomial B(4, x) from row n=4 of A053382/A053383. See, e.g., the Ash-Gross reference, p. 62, eq. (6.3) for k=3. - Wolfdieter Lang, Mar 12 2017
a(n) = A000217((n+1)^2) - A000217(n+1)^2. - Bruno Berselli, Aug 31 2017
a(n) = n*binomial(n+2, 3) + binomial(n+2, 4) + binomial(n+1, 4). - Tony Foster III, Nov 14 2017
Another identity: ..., a(3) = (1/2)*(1*(2+4+6)+3*(4+6)+5*6) = 36, a(4) = (1/2)*(1*(2+4+6+8)+3*(4+6+8)+5*(6+8)+7*(8)) = 100, a(5) = (1/2)*(1*(2+4+6+8+10)+3*(4+6+8+10)+5*(6+8+10)+7*(8+10)+9*(10)) = 225, ... - J. M. Bergot, Aug 27 2022
Comment from Michael Somos, Aug 28 2022: (Start)
The previous comment expresses a(n) as the sum of all of the n X n multiplication table array entries.
For example, for n = 4:
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
This array sum can be split up as follows:
+---+---------------+
| 0 | 1 2 3 4 | (0+1)*(1+2+3+4)
| +---+-----------+
| 0 | 2 | 4 6 8 | (1+2)*(2+3+4)
| | +---+-------+
| 0 | 3 | 6 | 9 12 | (2+3)*(3+4)
| | | +---+---+
| 0 | 4 | 8 |12 |16 | (3+4)*(4)
+---+---+---+---+---+
This kind of row+column sums was used by Ramanujan and others for summing Lambert series. (End)
a(n) = 6*A000332(n+4) - 12*A000292(n+1) + 7*A000217(n+1) - n - 1. - Adam Mohamed, Sep 05 2024

Extensions

Edited by M. F. Hasler, May 02 2015

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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Author

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

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

A004188 a(n) = n*(3*n^2 - 1)/2.

Original entry on oeis.org

0, 1, 11, 39, 94, 185, 321, 511, 764, 1089, 1495, 1991, 2586, 3289, 4109, 5055, 6136, 7361, 8739, 10279, 11990, 13881, 15961, 18239, 20724, 23425, 26351, 29511, 32914, 36569, 40485, 44671, 49136, 53889, 58939, 64295, 69966, 75961
Offset: 0

Views

Author

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

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
(1), (4+7), (10+13+16), (19+22+25+28), ... - Jon Perry, Sep 10 2004

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).

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.
Cf. A236770 (partial sums).

Programs

Formula

Partial sums of n-1 3-spaced triangular numbers, e.g., a(4) = t(1) + t(4) + t(7) = 1 + 10 + 28 = 39. - Jon Perry, Jul 23 2003
a(n) = C(2*n+1,3) + C(n+1,3), n >= 0. - Zerinvary Lajos, Jan 21 2007
a(n) = A000447(n) + A000292(n). - Zerinvary Lajos, Jan 21 2007
G.f.: x*(1+7*x+x^2) / (x-1)^4. - R. J. Mathar, Oct 08 2011
From Miquel Cerda, Dec 25 2016: (Start)
a(n) = A000578(n) + A135503(n).
a(n) = A007588(n) - A135503(n). (End)
E.g.f.: (x/2)*(2 + 9*x + 3*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A062392 a(n) = n^4 - (n-1)^4 + (n-2)^4 - ... 0^4.

Original entry on oeis.org

0, 1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945, 28680, 36856, 46665, 58311, 72010, 87990, 106491, 127765, 152076, 179700, 210925, 246051, 285390, 329266, 378015, 431985, 491536, 557040, 628881, 707455, 793170, 886446, 987715
Offset: 0

Views

Author

Henry Bottomley, Jun 21 2001

Keywords

Comments

Number of edges in the join of two complete graphs of order n^2 and n, K_n^2 * K_n. - Roberto E. Martinez II, Jan 07 2002
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-5)(P(4,1)-(-1)^k P(4,2k+1))|. - Peter Luschny, Jul 12 2009
Define an infinite symmetric array by T(n,m) = n*(n-1) + m for 0 <= m <= n and T(n,m) = T(m,n), n >= 0. Then a(n) is the sum of terms in the top left (n+1) X (n+1) subarray: a(n) = Sum_{r=0..n} Sum_{c=0..n} T(r,c). - J. M. Bergot, Jul 05 2013
a(n) is the sum of all positive numbers less than A002378(n). - J. M. Bergot, Aug 30 2013
Except the first term, these are triangular numbers that remain triangular when divided by their index, e.g., 66 divided by 11 gives 6. - Waldemar Puszkarz, Sep 14 2017
a(n) is the semiperimeter of the unique primitive Pythagorean triple such that (a-b+c)/2 = T(n) = A000217(n). Its long leg and hypotenuse are consecutive natural numbers and the triple is (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

Examples

			From _Bruno Berselli_, Oct 30 2017: (Start)
After 0:
1   =                 -(1) + (2);
15  =             -(1 + 2) + (3 + 4 + 5 + 2*3);
66  =         -(1 + 2 + 3) + (4 + 5 + 6 + 7 + ... + 11 + 3*4);
190 =     -(1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + ... + 19 + 4*5);
435 = -(1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + ... + 29 + 5*6), etc. (End)
		

References

  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Crossrefs

Cf. A000538, A000583. A062393 provides the result for 5th powers, A011934 for cubes, A000217 for squares, A001057 (unsigned) for nonnegative integers, A000035 (offset) for 0th powers.
Cf. A236770 (see crossrefs).

Programs

  • Maple
    a := n -> (2*n^2+n^3-1)*n/2; # Peter Luschny, Jul 12 2009
  • Mathematica
    Table[n (n + 1) (n^2 + n - 1)/2, {n, 0, 40}] (* Harvey P. Dale, Oct 19 2011 *)
  • PARI
    { a=0; for (n=0, 1000, write("b062392.txt", n, " ", a=n^4 - a) ) } \\ Harry J. Smith, Aug 07 2009

Formula

a(n) = n*(n+1)*(n^2 + n - 1)/2 = n^4 - a(n-1) = A000583(n) - a(n-1) = A000217(A028387(n-1)) = A000217(n)*A028387(n-1).
a(n) = Sum_{i=0..n} A007588(i) for n > 0. - Jonathan Vos Post, Mar 15 2006
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Harvey P. Dale, Oct 19 2011
G.f.: x*(x*(x + 10) + 1)/(1 - x)^5. - Harvey P. Dale, Oct 19 2011
a(n) = A000384(A000217(n)). - Bruno Berselli, Jan 31 2014
a(n) = A110450(n) - A002378(n). - Gionata Neri, May 13 2015
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*2*Pi/sqrt(5). - Amiram Eldar, Jan 22 2024
a(n) = sqrt(144*A288876(n-2) + 72*A006542(n+2) + A000537(n)). - Yasser Arath Chavez Reyes, Jul 22 2024
E.g.f.: exp(x)*x*(2 + 13*x + 8*x^2 + x^3)/2. - Stefano Spezia, Apr 27 2025
a(n) = A000217(n)*(2*A000217(n)-1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

A232713 Doubly pentagonal numbers: a(n) = n*(3*n-2)*(3*n-1)*(3*n+1)/8.

Original entry on oeis.org

0, 1, 35, 210, 715, 1820, 3876, 7315, 12650, 20475, 31465, 46376, 66045, 91390, 123410, 163185, 211876, 270725, 341055, 424270, 521855, 635376, 766480, 916895, 1088430, 1282975, 1502501, 1749060, 2024785, 2331890, 2672670, 3049501, 3464840, 3921225, 4421275
Offset: 0

Views

Author

Bruno Berselli, Nov 28 2013

Keywords

Crossrefs

Cf. similar sequences: A000583 for A000290(A000290(n)); A002817 for A000217(A000217(n)); A063249 for A000384(A000384(n)).

Programs

  • Magma
    [n*(3*n-2)*(3*n-1)*(3*n+1)/8: n in [0..40]];
    
  • Mathematica
    Table[n (3 n - 2) (3 n - 1) (3 n + 1)/8, {n, 0, 40}]
  • PARI
    a(n)=n*(3*n-2)*(3*n-1)*(3*n+1)/8 \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f.: x*(1 + 30*x + 45*x^2 + 5*x^3) / (1 - x)^5.
a(n) = A000326(A000326(n)) = A000332(3n+1).
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 4 + 2*Pi/sqrt(3) - 6*log(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 32*log(2)/3 - 4*Pi/(3*sqrt(3)) - 4. (End)

A260810 a(n) = n^2*(3*n^2 - 1)/2.

Original entry on oeis.org

0, 1, 22, 117, 376, 925, 1926, 3577, 6112, 9801, 14950, 21901, 31032, 42757, 57526, 75825, 98176, 125137, 157302, 195301, 239800, 291501, 351142, 419497, 497376, 585625, 685126, 796797, 921592, 1060501, 1214550, 1384801, 1572352, 1778337, 2003926, 2250325, 2518776
Offset: 0

Views

Author

Bruno Berselli, Jul 31 2015

Keywords

Comments

Pentagonal numbers with square indices.
After 0, a(k) is a square if k is in A072256.

Crossrefs

Subsequence of A001318 and A245288 (see Formula field).
Cf. A000326, A193218 (first differences).
Cf. A000583 (squares with square indices), A002593 (hexagonal numbers with square indices).
Cf. A232713 (pentagonal numbers with pentagonal indices), A236770 (pentagonal numbers with triangular indices).

Programs

  • Magma
    [n^2*(3*n^2-1)/2: n in [0..40]];
    
  • Magma
    I:=[0,1,22,117,376]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Aug 23 2015
  • Maple
    A260810:=n->n^2*(3*n^2 - 1)/2: seq(A260810(n), n=0..50); # Wesley Ivan Hurt, Apr 25 2017
  • Mathematica
    Table[n^2 (3 n^2 - 1)/2, {n, 0, 40}]
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 22, 117, 376}, 40] (* Vincenzo Librandi, Aug 23 2015 *)
  • PARI
    vector(40, n, n--; n^2*(3*n^2-1)/2)
    
  • Sage
    [n^2*(3*n^2-1)/2 for n in (0..40)]
    

Formula

G.f.: x*(1 + x)*(1 + 16*x + x^2)/(1 - x)^5.
a(n) = a(-n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A245288(2*n^2).
a(n) = A001318(2*n^2-1) with A001318(-1) = 0.
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi^2/3 - sqrt(3)*Pi*cot(Pi/sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi*cosec(Pi/sqrt(3)) - Pi^2/6 - 3. (End)

A011779 Expansion of 1/((1-x)^3*(1-x^3)^2).

Original entry on oeis.org

1, 3, 6, 12, 21, 33, 51, 75, 105, 145, 195, 255, 330, 420, 525, 651, 798, 966, 1162, 1386, 1638, 1926, 2250, 2610, 3015, 3465, 3960, 4510, 5115, 5775, 6501, 7293, 8151, 9087, 10101, 11193, 12376, 13650
Offset: 0

Views

Author

Keywords

Comments

The Ca2 and Ze4 triangle sums of A139600 are related to the sequence given above, e.g., Ze4(n) = A011779(n-1) - A011779(n-2) - A011779(n-4) + 3*A011779(n-5), with A011779(n) = 0 for n <= -1. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 29 2011

Crossrefs

Cf. A011779, A049347, A099254, A139600, A236770 (first trisection, except 0).

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 60);
    Coefficients(R!( 1/((1-x)^3*(1-x^3)^2) )); // G. C. Greubel, Oct 22 2024
    
  • Mathematica
    CoefficientList[Series[1 / ((1 - x)^3 (1 - x^3)^2), {x, 0, 100}], x] (* Vincenzo Librandi, Jun 23 2013 *)
  • PARI
    Vec(1/((1-x)^3*(1-x^3)^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012
    
  • PARI
    a(n)=1/216 * n^4 + 1/12 * n^3 + 37/72 * n^2 + [5/4, 139/108, 131/108][1+n%3] * n + [1, 10/9, 7/9][1+n%3] \\ Yurii Ivanov, Jul 06 2021
    
  • SageMath
    def A011779_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( 1/((1-x)^3*(1-x^3)^2) ).list()
    A011779_list(60) # G. C. Greubel, Oct 22 2024

Formula

a(n) = (1/216)*((208 + 270*n + 111*n^2 + 18*n^3 + n^4) - 8*(-1)^n*(A099254(n) + A099254(n-1)) + 16*(A049347(n) + 2*A049347(n-1)) ). - G. C. Greubel, Oct 22 2024

A177708 Pentagonal triangle.

Original entry on oeis.org

1, 6, 12, 18, 57, 51, 40, 156, 209, 145, 75, 330, 531, 534, 330, 126, 600, 1074, 1278, 1122, 651, 196, 987, 1895, 2488, 2559, 2081, 1162, 288, 1512, 3051, 4275, 4824, 4563, 3537, 1926, 405, 2196, 4599, 6750, 8100, 8370, 7506, 5634, 3015
Offset: 1

Views

Author

Jonathan Vos Post, Dec 11 2010

Keywords

Comments

This is to A093445 as pentagonal numbers A000326 are to triangular numbers A000217. The n-th row of the triangular table begins by considering A000217(n) pentagonal numbers (starting with 1) in order. Now segregate them into n chunks beginning with n members in the first chunk, n-1 members in the second chunk, and so forth. Now sum each chunk. Thus the first term is the sum of first n numbers = n*(3n-1)/2, the second term is the sum of the next n-1 terms (from n+1 to 2n-1), the third term is the sum of the next n-2 terms (2n to 3n-3)... This triangle can be called the pentagonal triangle. The sequence contains the triangle by rows. The first column is A002411 (Pentagonal pyramidal numbers: n^2*(n+1)/2).

Examples

			The row for n = 4 is (1+5+12+22), (35+51+70), (92+117), 145 => 40, 156, 209, 145.
    1;
    6,   12;
   18,   57,   51;
   40,  156,  209,   145;
   75,  330,  531,   534,   330;
  126,  600, 1074,  1278,  1122,   651;
  196,  987, 1895,  2488,  2559,  2081,  1162;
  288, 1512, 3051,  4275,  4824,  4563,  3537,  1926;
  405, 2196, 4599,  6750,  8100,  8370,  7506,  5634, 3015;
  550, 3060, 6596, 10024, 12570, 13775, 13450, 11631, 8534, 4510;
		

Crossrefs

Cf. A000217, A000326, A002411, A093445, A236770 (right border).

Programs

  • Maple
    A000326 :=proc(n) n*(3*n-1)/2 ; end proc:
    A177708 := proc(n,k) kc := 1 ; nsk := n ; ns := 1 ; while kc < k do ns := ns+nsk ; kc := kc+1 ; nsk := nsk-1 ; end do: add(A000326(i),i=ns..ns+nsk-1) ; end proc: # R. J. Mathar, Dec 14 2010
  • Mathematica
    Table[Total/@TakeList[PolygonalNumber[5,Range[60]],Range[n,1,-1]],{n,10}]//Flatten (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Feb 17 2018 *)

Formula

T(n,1) = A002411(n).
T(n,2) = n*(n-1)*(7*n-2)/2.
T(n,3) = (n-2)*(19*n^2-26*n+9)/2 = Sum_{i=2n..3(n-1)} A000326(i).

A264854 a(n) = n*(n + 1)*(11*n^2 + 11*n - 10)/24.

Original entry on oeis.org

0, 1, 14, 61, 175, 400, 791, 1414, 2346, 3675, 5500, 7931, 11089, 15106, 20125, 26300, 33796, 42789, 53466, 66025, 80675, 97636, 117139, 139426, 164750, 193375, 225576, 261639, 301861, 346550, 396025, 450616, 510664, 576521, 648550, 727125, 812631, 905464, 1006031
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 26 2015

Keywords

Comments

Partial sums of centered 11-gonal (or hendecagonal) pyramidal numbers.

Crossrefs

Cf. A004467.
Cf. similar sequences provided by the partial sums of centered k-gonal pyramidal numbers: A006522 (k=1), A006007 (k=2), A002817 (k=3), A006325 (k=4), A006322 (k=5), A000537 (k=6), A006323 (k=7), A006324 (k=8), A236770 (k=9), A264853 (k=10), this sequence (k=11), A062392 (k=12), A264888 (k=13).

Programs

  • Magma
    [n*(n+1)*(11*n^2+11*n-10)/24: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015
    
  • Mathematica
    Table[n (n + 1) (11 n^2 + 11 n - 10)/24, {n, 0, 50}]
  • PARI
    a(n)=n*(n+1)*(11*n^2+11*n-10)/24 \\ Charles R Greathouse IV, Jul 26 2016

Formula

G.f.: x*(1 + 9*x + x^2)/(1 - x)^5.
a(n) = Sum_{k = 0..n} A004467(k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015
Showing 1-10 of 10 results.