A300758 -id:A300758 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

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).

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Mario Defranco and Paul E. Gunnells, Hypergraph matrix models and generating functions, arXiv:2204.11361 [math.CO], 2022.
Milan Janjic, Two Enumerative Functions
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (9).
Andy Nicol, Illustration of Rhombic Dodecahedral Numbers
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Rhombic Dodecahedral Number.
Eric Weisstein's World of Mathematics, Nexus Number.
D. Zeitlin, A family of Galileo sequences, Amer. Math. Monthly 82 (1975), 819-822.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(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.
Cf. A128195, A176850, A005408, A176271, A212133.
Cf. A001844, A000583, A000290.
Cf. A007588, A000292, A000332, A002817, A342354.
Cf. A031215, A008292.
Cf. A016754, A344330, A344332.

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

[n^4 - (n-1)^4: n in [1..50]]; // Vincenzo Librandi, Aug 01 2011

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 *)

a(n)=n^4-(n-1)^4 \\ Charles R Greathouse IV, Jul 31 2011

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

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

A217843 Numbers which are the sum of one or more consecutive nonnegative cubes.

Original entry on oeis.org

0, 1, 8, 9, 27, 35, 36, 64, 91, 99, 100, 125, 189, 216, 224, 225, 341, 343, 405, 432, 440, 441, 512, 559, 684, 729, 748, 775, 783, 784, 855, 1000, 1071, 1196, 1241, 1260, 1287, 1295, 1296, 1331, 1584, 1728, 1729, 1800, 1925, 1989, 2016, 2024, 2025, 2197

Offset: 1

T. D. Noe, Oct 23 2012

nonn

Contains A000578 (cubes), A005898 (two consecutive cubes), A027602 (three consecutive cubes), A027603 (four consecutive cubes) etc. - R. J. Mathar, Nov 04 2012
See A265845 for sums of consecutive positive cubes in more than one way. - Reinhard Zumkeller, Dec 17 2015
From Lamine Ngom, Apr 15 2021: (Start)
a(n) can always be expressed as the difference of the squares of two triangular numbers (A000217).
A168566 is the subsequence A000217(n)^2 - 1.
a(n) is also the product of two nonnegative integers whose sum and difference are both promic.
See example and formula sections for details. (End)

			From _Lamine Ngom_, Apr 15 2021: (Start)
Arrange the positive terms in a triangle as follows:
n\k |   1    2    3    4    5    6    7
----+-----------------------------------
  0 |   1;
  1 |   8,   9;
  2 |  27,  35,  36;
  3 |  64,  91,  99, 100;
  4 | 125, 189, 216, 224, 225;
  5 | 216, 341, 405, 432, 440, 441;
  6 | 343, 559, 684, 748, 775, 783, 784;
Column 1: cubes = A000217(n+1)^2 - A000217(n)^2.
The difference of the squares of two consecutive triangular numbers (A000217) is a cube (A000578).
Column 2: sums of 2 consecutive cubes (A027602).
Column 3: sums of 3 consecutive cubes (A027603).
etc.
Column k: sums of k consecutive cubes.
Row n: A000217(n)^2 - A000217(m)^2, m < n.
T(n,n) = A000217(n)^2 (main diagonal).
T(n,n-1) = A000217(n)^2 - 1 (A168566) (2nd diagonal).
Now rectangularize this triangle as follows:
n\k |   1    2     3     4    5     6   ...
----+--------------------------------------
  0 |   1,   9,   36,  100,  225,  441, ...
  1 |   8,  35,   99,  224,  440,  783, ...
  2 |  27,  91,  216,  432,  775, 1287, ...
  3 |  64, 189,  405,  748, 1260, 1989, ...
  4 | 125, 341,  684, 1196, 1925, 2925, ...
  5 | 216, 559, 1071, 1800, 2800, 4131, ...
  6 | 343, 855, 1584, 2584, 3915, 5643, ...
The general form of terms is:
T(n,k) = [n^4 + A016825(k)*n^3 + A003154(k)*n^2 + A300758(k)*n]/4, sum of n consecutive cubes after k^3.
This expression can be factorized into [n*(n + A005408(k))*(n*(n + A005408(k)) + 4*A000217(k))]/4.
For k = 1, the sequence provides all cubes: T(n,1) = A000578(k).
For k = 2, T(n,2) = A005898(k), centered cube numbers, sum of two consecutive cubes.
For k = 3, T(n,3) = A027602(k), sum of three consecutive cubes.
For k = 4, T(n,4) = A027603(k), sum of four consecutive cubes.
For k = 5, T(n,5) = A027604(k), sum of five consecutive cubes.
T(n,n) = A116149(n), sum of n consecutive cubes after n^3 (main diagonal).
For n = 0, we obtain the subsequence T(0,k) = A000217(n)^2, product of two numbers whose difference is 0*1 (promic) and sum is promic too.
For n = 1, we obtain the subsequence T(1,k) = A168566(x), product of two numbers whose difference is 1*2 (promic) and sum is promic too.
For n = 2, we obtain the subsequence T(2,k) = product of two numbers whose difference is 2*3 (promic) and sum is promic too.
etc.
For n = x, we obtain the subsequence formed by products of two numbers whose difference is the promic x*(x+1) and sum is promic too.
Consequently, if m is in the sequence, then m can be expressed as the product of two nonnegative integers whose sum and difference are both promic. (End)

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A034705, A217844-A217850, A062682, A131643, A240137.
Cf. A000578, A005898, A027602, A027603, A027604.
Cf. A265845 (subsequence).
Cf. A000217 (triangular numbers), A046092 (4*A000217).
Cf. A168566 (A000217^2 - 1).
Cf. A002378 (promics), A016825 (singly even numbers), A003154 (stars numbers).
Cf. A000330 (square pyramidal numbers), A300758 (12*A000330).
Cf. A005408 (odd numbers).

import Data.Set (singleton, deleteFindMin, insert, Set)
a217843 n = a217843_list !! (n-1)
a217843_list = f (singleton (0, (0,0))) (-1) where
   f s z = if y /= z then y : f s'' y else f s'' y
              where s'' = (insert (y', (i, j')) $
                           insert (y' - i ^ 3 , (i + 1, j')) s')
                    y' = y + j' ^ 3; j' = j + 1
                    ((y, (i, j)), s') = deleteFindMin s
-- Reinhard Zumkeller, Dec 17 2015, May 12 2015

nMax = 3000; t = {0}; Do[k = n; s = 0; While[s = s + k^3; s <= nMax, AppendTo[t, s]; k++], {n, nMax^(1/3)}]; t = Union[t]

lista(nn) = {my(list = List([0])); for (i=1, nn, my(s = 0); forstep(j=i, 1, -1, s += j^3; if (s > nn^3, break); listput(list, s););); Set(list);} \\ Michel Marcus, Nov 13 2020

a(n) >> n^2. Probably a(n) ~ kn^2 for some k but I cannot prove this. - Charles R Greathouse IV, Aug 07 2013

a(n) is of the form [x*(x+2*k+1)*(x*(x+2*k+1)+2*k*(k+1))]/4, sum of n consecutive cubes starting from (k+1)^3. - Lamine Ngom, Apr 15 2021

Name edited by N. J. A. Sloane, May 24 2021

cp's OEIS Frontend

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

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