A240137 -id:A240137 - cp's OEIS frontend

A000326 Pentagonal numbers: a(n) = n(3n-1)/2.

0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, 376, 425, 477, 532, 590, 651, 715, 782, 852, 925, 1001, 1080, 1162, 1247, 1335, 1426, 1520, 1617, 1717, 1820, 1926, 2035, 2147, 2262, 2380, 2501, 2625, 2752, 2882, 3015, 3151

Offset: 0

N. J. A. Sloane

The average of the first n (n > 0) pentagonal numbers is the n-th triangular number. - Mario Catalani (mario.catalani(AT)unito.it), Apr 10 2003
a(n) is the sum of n integers starting from n, i.e., 1, 2 + 3, 3 + 4 + 5, 4 + 5 + 6 + 7, etc. - Jon Perry, Jan 15 2004
Partial sums of 1, 4, 7, 10, 13, 16, ... (1 mod 3), a(2k) = k(6k-1), a(2k-1) = (2k-1)(3k-2). - Jon Perry, Sep 10 2004
Starting with offset 1 = binomial transform of [1, 4, 3, 0, 0, 0, ...]. Also, A004736 * [1, 3, 3, 3, ...]. - Gary W. Adamson, Oct 25 2007
If Y is a 3-subset of an n-set X then, for n >= 4, a(n-3) is the number of 4-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
Solutions to the duplication formula 2*a(n) = a(k) are given by the index pairs (n, k) = (5,7), (5577, 7887), (6435661, 9101399), etc. The indices are integer solutions to the pair of equations 2(6n-1)^2 = 1 + y^2, k = (1+y)/6, so these n can be generated from the subset of numbers [1+A001653(i)]/6, any i, where these are integers, confined to the cases where the associated k=[1+A002315(i)]/6 are also integers. - R. J. Mathar, Feb 01 2008
a(n) is a binomial coefficient C(n,4) (A000332) if and only if n is a generalized pentagonal number (A001318). Also see A145920. - Matthew Vandermast, Oct 28 2008
Even octagonal numbers divided by 8. - Omar E. Pol, Aug 18 2011
Sequence found by reading the line from 0, in the direction 0, 5, ... and the line from 1, in the direction 1, 12, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011
The hyper-Wiener index of the star-tree with n edges (see A196060, example). - Emeric Deutsch, Sep 30 2011
More generally the n-th k-gonal number is equal to n + (k-2)*A000217(n-1), n >= 1, k >= 3. In this case k = 5. - Omar E. Pol, Apr 06 2013
Note that both Euler's pentagonal theorem for the partition numbers and Euler's pentagonal theorem for the sum of divisors refer more exactly to the generalized pentagonal numbers, not this sequence. For more information see A001318, A175003, A238442. - Omar E. Pol, Mar 01 2014
The Fuss-Catalan numbers are Cat(d,k)= [1/(k*(d-1)+1)]*binomial(k*d,k) and enumerate the number of (d+1)-gon partitions of a (k*(d-1)+2)-gon (cf. Schuetz and Whieldon link). a(n)= Cat(n,3), so enumerates the number of (n+1)-gon partitions of a (3*(n-1)+2)-gon. Analogous sequences are A100157 (k=4) and A234043 (k=5). - Tom Copeland, Oct 05 2014
Binomial transform of (0, 1, 3, 0, 0, 0, ...) (A169585 with offset 1) and second partial sum of (0, 1, 3, 3, 3, ...). - Gary W. Adamson, Oct 05 2015
For n > 0, a(n) is the number of compositions of n+8 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
a(n) is also the number of edges in the Mycielskian of the complete graph K[n]. Indeed, K[n] has n vertices and n(n-1)/2 edges. Then its Mycielskian has n + 3n(n-1)/2 = n(3n-1)/2. See p. 205 of the West reference. - Emeric Deutsch, Nov 04 2016
Sum of the numbers from n to 2n-1. - Wesley Ivan Hurt, Dec 03 2016
Also the number of maximal cliques in the n-Andrásfai graph. - Eric W. Weisstein, Dec 01 2017
Coefficients in the hypergeometric series identity 1 - 5*(x - 1)/(2*x + 1) + 12*(x - 1)*(x - 2)/((2*x + 1)*(2*x + 2)) - 22*(x - 1)*(x - 2)*(x - 3)/((2*x + 1)*(2*x + 2)*(2*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A002412 and A002418. Column 2 of A103450. - Peter Bala, Mar 14 2019
A generalization of the Comment dated Apr 10 2003 follows. (k-3)*A000292(n-2) plus the average of the first n (2k-1)-gonal numbers is the n-th k-gonal number. - Charlie Marion, Nov 01 2020
a(n+1) is the number of Dyck paths of size (3,3n+1); i.e., the number of NE lattice paths from (0,0) to (3,3n+1) which stay above the line connecting these points. - Harry Richman, Jul 13 2021
a(n) is the largest sum of n positive integers x_1, ..., x_n such that x_i | x_(i+1)+1 for each 1 <= i <= n, where x_(n+1) = x_1. - Yifan Xie, Feb 21 2025

			Illustration of initial terms:
.
.                                       o
.                                     o o
.                          o        o o o
.                        o o      o o o o
.                o     o o o    o o o o o
.              o o   o o o o    o o o o o
.        o   o o o   o o o o    o o o o o
.      o o   o o o   o o o o    o o o o o
.  o   o o   o o o   o o o o    o o o o o
.
.  1    5     12       22           35
- _Philippe Deléham_, Mar 30 2013

Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pages 2 and 311.
Raymond Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; p. 129.
Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 40.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 291.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 284.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 64.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 52-53, 129-130, 132.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 7-10.
André Weil, Number theory: an approach through history; from Hammurapi to Legendre, Birkhäuser, Boston, 1984; see p. 186.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 98-100.
Douglas B. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, NJ, 2001.

Daniel Mondot, Table of n, a(n) for n = 0..10000 (first 1000 terms by T. D. Noe)
George E. Andrews, Euler's "De Partitio Numerorum", Bull. Amer. Math. Soc., 44 (No. 4, 2007), 561-573.
S. Barbero, U. Cerruti and N. Murru, Transforming Recurrent Sequences by Using the Binomial and Invert Operators, J. Int. Seq. 13 (2010) # 10.7.7, section 4.4.
Jordan Bell, Euler and the pentagonal number theorem, arXiv:math/0510054 [math.HO], 2005-2006.
Anicius Manlius Severinus Boethius, De institutione arithmetica libri duo, Book 2, sections 13-14.
Charles K. Cook and Michael R. Bacon, Some polygonal number summation formulas, Fib. Q., 52 (2014), 336-343.
Olivier Danvy, Summa Summarum: Moessner's Theorem without Dynamic Programming, arXiv:2412.03127 [cs.DM], 2024. See p. 33.
Stephan Eberhart, Letter to N. J. A. Sloane, Jan 06 1978, also scanned copy of Mathematical-Physical Correspondence, No. 22, Christmas 1977.
Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, par. 1
Leonhard Euler, Observatio de summis divisorum p. 8.
Leonhard Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004, 2009. See p. 8.
Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
Shyam Sunder Gupta, Beauty of Number 153, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 15, 399-410.
Rodney T. Hansen, Arithmetic of pentagonal numbers, Fib. Quart., 8 (1970), 83-87.
Alfred Hoehn, Illustration of initial terms of A000326, A005449, A045943, A115067
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 339
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Jangwon Ju and Daejun Kim, The pentagonal theorem of sixty-three and generalizations of Cauchy's lemma, arXiv:2010.16123 [math.NT], 2020.
Bir Kafle, Florian Luca and Alain Togbé, Pentagonal and heptagonal repdigits, Annales Mathematicae et Informaticae. pp. 137-145.
Sameen Ahmed Khan, Sums of the powers of reciprocals of polygonal numbers, Int'l J. of Appl. Math. (2020) Vol. 33, No. 2, 265-282.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
R. P. Loh, A. G. Shannon and A. F. Horadam, Divisibility Criteria and Sequence Generators Associated with Fermat Coefficients, Preprint, 1980.
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
Omar E. Pol, Illustration of initial terms of A000217, A000290, A000326, A000384, A000566, A000567
Cliff Reiter, Polygonal Numbers and Fifty One Stars, Lafayette College, Easton, PA (2019).
Alison Schuetz and Gwyneth Whieldon, Polygonal Dissections and Reversions of Series, arXiv:1401.7194 [math.CO], 2014.
W. Sierpiński, Sur les nombres pentagonaux, Bull. Soc. Royale Sciences Liège, 33 (No. 9-10, 1964), 513-517. [Annotated scanned copy]
N. J. A. Sloane, Illustration of initial terms of A000217, A000290, A000326.
Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, Andrásfai Graph.
Eric Weisstein's World of Mathematics, Maximal Clique.
Eric Weisstein's World of Mathematics, Pentagonal Number.
Wikipedia, Mycielskian.
Wikipedia, Pentagonal number
Index entries for "core" sequences
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Index entries for two-way infinite sequences

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.
Cf. A001318 (generalized pentagonal numbers), A049452, A033570, A010815, A034856, A051340, A004736, A033568, A049453, A002411 (partial sums), A033579.
See A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.
Cf. A240137: sum of n consecutive cubes starting from n^3.
Cf. similar sequences listed in A022288.
Partial sums of A016777.

List([0..50],n->n*(3*n-1)/2); # Muniru A Asiru, Mar 18 2019

a000326 n = n * (3 * n - 1) `div` 2  -- Reinhard Zumkeller, Jul 07 2012

[n*(3*n-1)/2 : n in [0..100]]; // Wesley Ivan Hurt, Oct 15 2015

A000326 := n->n*(3*n-1)/2: seq(A000326(n), n=0..100);
A000326:=-(1+2*z)/(z-1)**3; # Simon Plouffe in his 1992 dissertation
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]-a[n-2]+3 od: seq(a[n], n=0..50); # Miklos Kristof, Zerinvary Lajos, Feb 18 2008

Table[n (3 n - 1)/2, {n, 0, 60}] (* Stefan Steinerberger, Apr 01 2006 *)
Array[# (3 # - 1)/2 &, 47, 0] (* Zerinvary Lajos, Jul 10 2009 *)
LinearRecurrence[{3, -3, 1}, {0, 1, 5}, 61] (* Harvey P. Dale, Dec 27 2011 *)
pentQ[n_] := IntegerQ[(1 + Sqrt[24 n + 1])/6]; pentQ[0] = True; Select[Range[0, 3200], pentQ@# &] (* Robert G. Wilson v, Mar 31 2014 *)
Join[{0}, Accumulate[Range[1, 312, 3]]] (* Harvey P. Dale, Mar 26 2016 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[5], n], {n, 0, 46}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
CoefficientList[Series[x (-1 - 2 x)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
PolygonalNumber[5, Range[0, 20]] (* Eric W. Weisstein, Dec 01 2017 *)

PARI
```
a(n)=n*(3*n-1)/2
```

vector(100, n, n--; binomial(3*n, 2)/3) \\ Altug Alkan, Oct 06 2015

is_a000326(n) = my(s); n==0 || (issquare (24*n+1, &s) && s%6==5); \\ Hugo Pfoertner, Aug 03 2023

# Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
     x, y = 1, 1
     yield 0
     while True:
         yield x
         x, y = x + y + 3, y + 3
A000326 = aList()
print([next(A000326) for i in range(47)]) # Peter Luschny, Aug 04 2019

Product_{m > 0} (1 - q^m) = Sum_{k} (-1)^k*x^a(k). - Paul Barry, Jul 20 2003
G.f.: x*(1+2*x)/(1-x)^3.
E.g.f.: exp(x)*(x+3*x^2/2).
a(n) = n*(3*n-1)/2.
a(-n) = A005449(n).
a(n) = binomial(3*n, 2)/3. - Paul Barry, Jul 20 2003
a(n) = A000290(n) + A000217(n-1). - Lekraj Beedassy, Jun 07 2004
a(0) = 0, a(1) = 1; for n >= 2, a(n) = 2*a(n-1) - a(n-2) + 3. - Miklos Kristof, Mar 09 2005
a(n) = Sum_{k=1..n} (2*n - k). - Paul Barry, Aug 19 2005
a(n) = 3*A000217(n) - 2*n. - Lekraj Beedassy, Sep 26 2006
a(n) = A126890(n, n-1) for n > 0. - Reinhard Zumkeller, Dec 30 2006
a(n) = A049452(n) - A022266(n) = A033991(n) - A005476(n). - Zerinvary Lajos, Jun 12 2007
Equals A034856(n) + (n - 1)^2. Also equals A051340 * [1,2,3,...]. - Gary W. Adamson, Jul 27 2007
a(n) = binomial(n+1, 2) + 2*binomial(n, 2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0) = 0, a(1) = 1, a(2) = 5. - Jaume Oliver Lafont, Dec 02 2008
a(n) = a(n-1) + 3*n-2 with n > 0, a(0)=0. - Vincenzo Librandi, Nov 20 2010
a(n) = A000217(n) + 2*A000217(n-1). - Vincenzo Librandi, Nov 20 2010
a(n) = A014642(n)/8. - Omar E. Pol, Aug 18 2011
a(n) = A142150(n) + A191967(n). - Reinhard Zumkeller, Jul 07 2012
a(n) = (A000290(n) + A000384(n))/2 = (A000217(n) + A000566(n))/2 = A049450(n)/2. - Omar E. Pol, Jan 11 2013
a(n) = n*A000217(n) - (n-1)*A000217(n-1). - Bruno Berselli, Jan 18 2013
a(n) = A005449(n) - n. - Philippe Deléham, Mar 30 2013
From Oskar Wieland, Apr 10 2013: (Start)
a(n) = a(n+1) - A016777(n),
a(n) = a(n+2) - A016969(n),
a(n) = a(n+3) - A016777(n)*3 = a(n+3) - A017197(n),
a(n) = a(n+4) - A016969(n)*2 = a(n+4) - A017641(n),
a(n) = a(n+5) - A016777(n)*5,
a(n) = a(n+6) - A016969(n)*3,
a(n) = a(n+7) - A016777(n)*7,
a(n) = a(n+8) - A016969(n)*4,
a(n) = a(n+9) - A016777(n)*9. (End)
a(n) = A000217(2n-1) - A000217(n-1), for n > 0. - Ivan N. Ianakiev, Apr 17 2013
a(n) = A002411(n) - A002411(n-1). - J. M. Bergot, Jun 12 2013
Sum_{n>=1} a(n)/n! = 2.5*exp(1). - Richard R. Forberg, Jul 15 2013
a(n) = floor(n/(exp(2/(3*n)) - 1)), for n > 0. - Richard R. Forberg, Jul 27 2013
From Vladimir Shevelev, Jan 24 2014: (Start)
a(3*a(n) + 4*n + 1) = a(3*a(n) + 4*n) + a(3*n+1).
A generalization. Let {G_k(n)}_(n >= 0) be sequence of k-gonal numbers (k >= 3). Then the following identity holds: G_k((k-2)*G_k(n) + c(k-3)*n + 1) = G_k((k-2)*G_k(n) + c(k-3)*n) + G_k((k-2)*n + 1), where c = A000124. (End)
A242357(a(n)) = 1 for n > 0. - Reinhard Zumkeller, May 11 2014
Sum_{n>=1} 1/a(n)= (1/3)*(9*log(3) - sqrt(3)*Pi). - Enrique Pérez Herrero, Dec 02 2014. See the decimal expansion A244641.
a(n) = (A000292(6*n+k-1)-A000292(k))/(6*n-1)-A000217(3*n+k), for any k >= 0. - Manfred Arens, Apr 26 2015 [minor edits from Wolfdieter Lang, May 10 2015]
a(n) = A258708(3*n-1,1) for n > 0. - Reinhard Zumkeller, Jun 23 2015
a(n) = A007584(n) - A245301(n-1), for n > 0. - Manfred Arens, Jan 31 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(sqrt(3)*Pi - 6*log(2))/3 = 0.85501000622865446... - Ilya Gutkovskiy, Jul 28 2016
a(m+n) = a(m) + a(n) + 3*m*n. - Etienne Dupuis, Feb 16 2017
In general, let P(k,n) be the n-th k-gonal number. Then P(k,m+n) = P(k,m) + (k-2)mn + P(k,n). - Charlie Marion, Apr 16 2017
a(n) = A023855(2*n-1) - A023855(2*n-2). - Luc Rousseau, Feb 24 2018
a(n) = binomial(n,2) + n^2. - Pedro Caceres, Jul 28 2019
Product_{n>=2} (1 - 1/a(n)) = 3/5. - Amiram Eldar, Jan 21 2021
(n+1)*(a(n^2) + a(n^2+1) + ... + a(n^2+n)) = n*(a(n^2+n+1) + ... + a(n^2+2n)). - Charlie Marion, Apr 28 2024
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * binomial(k, 2)*binomial(3*n+k-1, 2*k). - Peter Bala, Nov 04 2024

Incorrect example removed by Joerg Arndt, Mar 11 2010

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

A050410 Truncated square pyramid numbers: a(n) = Sum_{k = n..2*n-1} k^2.

Original entry on oeis.org

0, 1, 13, 50, 126, 255, 451, 728, 1100, 1581, 2185, 2926, 3818, 4875, 6111, 7540, 9176, 11033, 13125, 15466, 18070, 20951, 24123, 27600, 31396, 35525, 40001, 44838, 50050, 55651, 61655, 68076, 74928, 82225, 89981, 98210, 106926, 116143

Offset: 0

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 22 1999

Starting with offset 1 = binomial transform of [1, 12, 25, 14, 0, 0, 0, ...]. - Gary W. Adamson, Jan 09 2009

			1^2 + 1;
2^2 + 3^2 = 13;
3^2 + 4^2 + 5^2 = 50; ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A072474, A240137.

List([0..40], n-> n*(7*n-1)*(2*n-1)/6); # G. C. Greubel, Oct 30 2019

[n*(7*n-1)*(2*n-1)/6: n in [0..40]]; // Vincenzo Librandi, Apr 27 2012

seq(n*(7*n-1)*(2*n-1)/6, n=0..36); # Zerinvary Lajos, Dec 01 2006

Table[Sum[k^2,{k,n,2n-1}],{n,0,40}] (* or *) LinearRecurrence[{4,-6,4, -1}, {0,1,13,50},40] (* Harvey P. Dale, Feb 29 2012 *)

for(n=1,100,print1(sum(i=0,n-1,(n+i)^2),","))

vector(40, n, (n-1)*(7*n-8)*(2*n-3)/6) \\ G. C. Greubel, Oct 30 2019

[n*(7*n-1)*(2*n-1)/6 for n in (0..40)] # G. C. Greubel, Oct 30 2019

a(n) = n*(7*n-1)*(2*n-1)/6.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=13, a(3)=50. - Harvey P. Dale, Feb 29 2012
G.f.: x*(1 + 9*x + 4*x^2)/(1-x)^4. - Colin Barker, Mar 23 2012
E.g.f.: x*(6 + 33*x + 14*x^2)*exp(x)/6. - G. C. Greubel, Oct 30 2019

A116149 a(n) = sum of n consecutive cubes after n^3.

Original entry on oeis.org

8, 91, 405, 1196, 2800, 5643, 10241, 17200, 27216, 41075, 59653, 83916, 114920, 153811, 201825, 260288, 330616, 414315, 512981, 628300, 762048, 916091, 1092385, 1292976, 1520000, 1775683, 2062341, 2382380, 2738296, 3132675, 3568193

Offset: 1

Zak Seidov, Apr 14 2007

			a(1) = sum of 1 cube  after 1^3 = 2^3 = 8,
a(2) = sum of 2 cubes after 2^3 = 3^3+4^3 = 91,
a(3) = sum of 3 cubes after 3^3 = 4^3+5^3+6^3 = 405,
a(4) = sum of 4 cubes after 4^3 = 5^3+6^3+7^3+8^3 = 1196.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A240137.

List([1..40], n-> n^2*(1+3*n)*(3+5*n)/4); # G. C. Greubel, May 10 2019

[n^2*(1+3*n)*(3+5*n)/4: n in [1..40]]; // G. C. Greubel, May 10 2019

With[{cbs=Range[100]^3},Table[Total[Take[cbs,{n+1,2n}]],{n,35}]]  (* Harvey P. Dale, Feb 13 2011 *)

{a(n) = n^2*(1+3*n)*(3+5*n)/4}; \\ G. C. Greubel, May 10 2019

[n^2*(1+3*n)*(3+5*n)/4 for n in (1..40)] # G. C. Greubel, May 10 2019

a(n) = n^2*(1 + 3*n)*(3 + 5*n)/4.
G.f.: x*(8 +51*x + 30*x^2 + x^3)/(1-x)^5. - Colin Barker, Dec 17 2012
a(n) = A000217(2*n)^2 - A000217(n)^2. - Bruno Berselli, Aug 31 2017
From G. C. Greubel, May 10 2019: (Start)
a(n) = Sum_{k=(n+1)..2*n} k^3.
E.g.f.: x*(32 + 150*x + 104*x^2 + 15*x^3)*exp(x)/4. (End)

A180920 Numbers k such that the sum of the cubes of the k consecutive integers starting from k is a square.

Original entry on oeis.org

1, 33, 2017, 124993, 7747521, 480221281, 29765971873, 1845010034817, 114360856186753, 7088528073543841, 439374379703531361, 27234123013545400513, 1688076252460111300417, 104633493529513355225313, 6485588522577367912668961, 402001854906267297230250241

Offset: 1

Vladimir Pletser, Sep 24 2010

Numbers k such that A240137(k) is a square.

A240137(k) is a square iff (3n-1)*(5n-3) is a square. 15*n^2-14*n+3 = k^2 can be solved as a Pell equation resulting in formula below confirming linear recurrence from Colin Barker. - Ray Chandler, Jan 12 2024

Colin Barker, Table of n, a(n) for n = 1..558
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Cf. A180921, A240137.

a[1] = 1; a[n_] := a[n] = 31 a[n - 1] - 14 + 8 Sqrt[(3 a[n - 1] - 1)*(5 a[n - 1] - 3)]; Array[a, 14] (* Robert G. Wilson v, Sep 27 2010 *)

default(realprecision, 1000)
vector(20, n, if(n==1, t=1, t=round(31*t-14+8*((3*t-1)*(5*t-3))^(1/2)))) \\ Colin Barker, Feb 19 2015

a(n) = 31*a(n-1) - 14 + 8*sqrt((3*a(n-1) - 1)*(5*a(n-1) - 3)).
From Colin Barker, Feb 18 2015: (Start)
a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3).
G.f.: -x*(x^2-30*x+1) / ((x-1)*(x^2-62*x+1)). (End)
a(n) = (14+2*(4*(31-8*Sqrt(15))^n+Sqrt(15)*(31-8*Sqrt(15))^n+4*(31+8*Sqrt(15))^n-Sqrt(15)*(31+8*Sqrt(15))^n))/30. - Ray Chandler, Jan 12 2024

a(8) onwards from Robert G. Wilson v, Sep 27 2010

Name clarified by Jon E. Schoenfield, Mar 11 2022

A180921 a(n) is the square root of the sum of the cubes of the b(n) consecutive integers starting from b(n), where b(n) = A180920.

Original entry on oeis.org

1, 2079, 7876385, 30254180671, 116236127290689, 446579144331338591, 1715756954644453458529, 6591937773063166150358655, 25326223208345427203876398721, 97303342974524967600723097592479, 373839418381901692962342398114034081

Offset: 1

Vladimir Pletser, Sep 24 2010

Colin Barker's linear recurrence conjecture confirmed, see A180920. - Ray Chandler, Jan 12 2024

			a(3) = 2017*(31*(2079/33) + 8*sqrt(15*((2079/33)^2) + 1)).

Colin Barker, Table of n, a(n) for n = 1..279
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (3904,-238206,3904,-1).

Cf. A180920, A240137.

LinearRecurrence[{3904,-238206,3904,-1},{1,2079,7876385,30254180671},20] (* Harvey P. Dale, Apr 27 2025 *)

default(realprecision, 1000);
b=vector(20, n, if(n==1, t=1, t=round(31*t-14+8*((3*t-1)*(5*t-3))^(1/2))));
vector(#b, n, if(n==1, t=1, t=round(b[n]*(31*(t/b[n-1])+8*(15*((t/b[n-1])^2)+1)^(1/2))))) \\ Colin Barker, Feb 19 2015

a(n) = b(n)*(31*(a(n-1)/b(n-1)) + 8*sqrt(15*((a(n-1)/b(n-1))^2) + 1)) where b(n) = A180920(n).
From Colin Barker, Feb 19 2015: (Start)
a(n) = 3904*a(n-1) - 238206*a(n-2) + 3904*a(n-3) - a(4).
G.f.: x*(x+1)*(x^2-1826*x+1) / ((x^2-3842*x+1)*(x^2-62*x+1)). (End)
a(n) = Sqrt(A240137(A180920(n))). - Ray Chandler, Jan 12 2024

Name clarified by Jon E. Schoenfield, Mar 11 2022

A262925 Sum of n consecutive fourth powers starting with n^4.

Original entry on oeis.org

0, 1, 97, 962, 4578, 14979, 38995, 86996, 173636, 318597, 547333, 891814, 1391270, 2092935, 3052791, 4336312, 6019208, 8188169, 10941609, 14390410, 18658666, 23884427, 30220443, 37834908, 46912204, 57653645, 70278221, 85023342, 102145582, 121921423

Offset: 0

Colin Barker, Oct 04 2015

			a(3) = 3^4 + 4^4 + 5^4 = 962.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A050410, A240137, A262926.

LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,97,962,4578,14979},30] (* Harvey P. Dale, Jul 30 2017 *)

vector(40, n, n--; sum(k=n, 2*n-1, k^4))

concat(0, Vec(x*(16*x^4+241*x^3+395*x^2+91*x+1)/(x-1)^6 + O(x^40)))

a(n) = n*(-1+70*n^2-225*n^3+186*n^4)/30.

G.f.: x*(16*x^4+241*x^3+395*x^2+91*x+1) / (x-1)^6.

A262926 Sum of n consecutive n-th powers starting with n^n.

Original entry on oeis.org

0, 1, 31, 3408, 873580, 405071029, 295716738515, 312086923732368, 449317984129326216, 846136323944158864793, 2018612200059553898143707, 5949463230586042065279268128, 21227845340442717633531647231668, 90172805592203250075964230466892813

Offset: 0

Colin Barker, Oct 04 2015

			a(3) = 3^3 + 4^4 + 5^5 = 3408.

Cf. A050410, A240137, A262925.

Table[Sum[k^k, {k, n, 2 n - 1}], {n, 0, 13}] (* Michael De Vlieger, Oct 05 2015 *)

vector(20, n, n--; sum(k=n, 2*n-1, k^k))

[sum(k^k for k in (n..2*n-1)) for n in (0..20)] # Bruno Berselli, Oct 05 2015

cp's OEIS Frontend

A000326 Pentagonal numbers: a(n) = n*(3*n-1)/2.

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

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A050410 Truncated square pyramid numbers: a(n) = Sum_{k = n..2*n-1} k^2.

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A116149 a(n) = sum of n consecutive cubes after n^3.

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A180920 Numbers k such that the sum of the cubes of the k consecutive integers starting from k is a square.

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A180921 a(n) is the square root of the sum of the cubes of the b(n) consecutive integers starting from b(n), where b(n) = A180920.

A000326 Pentagonal numbers: a(n) = n(3n-1)/2.