A006003 -id:A006003 - cp's OEIS frontend

A004068 Number of atoms in a decahedron with n shells.

0, 1, 7, 23, 54, 105, 181, 287, 428, 609, 835, 1111, 1442, 1833, 2289, 2815, 3416, 4097, 4863, 5719, 6670, 7721, 8877, 10143, 11524, 13025, 14651, 16407, 18298, 20329, 22505, 24831, 27312, 29953, 32759, 35735, 38886, 42217, 45733, 49439

Offset: 0

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

Also as a(n)=(n/6)*(5*n^2+1), n>0: structured pentagonal diamond numbers (vertex structure 6) (cf. A081436 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of atoms in decahedron with n shells, number = 5/6*(n^3) + 1/6*(n) (T. P. Martin, Shells of atoms, eq.(3)). - Brigitte Stepanov, Jul 02 2011
a(n+1) is the number of triples (w,x,y) having all terms in {0,...,n} and x+y >= w. - Clark Kimberling, Jun 14 2012
a(n) = Sum_{k=1..n} A215630(n,k) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) - a(n-2) = A010001(n-1), for n>1. - K. G. Stier, Dec 21 2012
a(n) is also a figurate number representing a cube of side n with a vertex cut off by a tetrahedron of side n-1. As such, a(n) = A000578(n) - A000292(n-1), n > 0. - Jean M. Morales, Aug 11 2013
The sequence starting with 1 is the third partial sum of (1, 4, 5, 5, 5, ...) and the binomial transform of (1, 6, 10, 5, 0, 0, 0, ...). - Gary W. Adamson, Sep 27 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (3).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-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. A000292, A005891, A006322 (partial sums), A010001, A081436, A100145, A215630.

[5*n^3/6+n/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011

Table[5*n^3/6+n/6,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)

A004068(n):=5*n^3/6+n/6$ makelist(A004068(n),n,0,20); /* Martin Ettl, Jan 07 2013 */

a(n)=5*n^3/6+n/6 \\ Charles R Greathouse IV, Sep 24 2015

def A004068(n): return n*(5*n**2+1)//6 # Chai Wah Wu, Mar 25 2025

a(n) = 5*binomial(n + 1, 3) + binomial(n, 1).
a(n) = 5*n^3/6 + n/6.
a(n) = Sum_{i=0..n-1} A005891(i). - Xavier Acloque, Oct 08 2003
G.f.: x*(1+3*x+x^2) / (1-x)^4. - R. J. Mathar, Jun 05 2011
E.g.f.: (x/6)*(5x^2 + 15x + 6)*exp(x). - G. C. Greubel, Sep 27 2015
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i/sqrt(5)) + polygamma(0, 1+i/sqrt(5))) = 1.233988011257952852492845364799197179252... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

Typo in definition corrected by Jean M. Morales, Aug 11 2013

A002432 Denominators of zeta(2n)/Pi^(2n).

Original entry on oeis.org

2, 6, 90, 945, 9450, 93555, 638512875, 18243225, 325641566250, 38979295480125, 1531329465290625, 13447856940643125, 201919571963756521875, 11094481976030578125, 564653660170076273671875, 5660878804669082674070015625, 62490220571022341207266406250

Offset: 0

N. J. A. Sloane

Also denominators in expansion of Psi(x).
For n >= 1 a(n) is always divisible by 3 (by the von Staudt-Clausen theorem, see A002445).
A comment due to G. Campbell: The original approach taken by Euler was to derive the infinite product for sin(Pi*x)/(Pi*x) equal to (1 - x^2/1^2) (1 - x^2/2^2)(1 - x^2/3^2) ... treating sin(Pi*x)/(Pi*x) as if it were a polynomial. Differentiating the logarithm of both sides and equating coefficients gives all of the zeta function values for 2, 4, 6, 8, .... - M. F. Hasler, Mar 29 2015
Note that 2n+1 divides a(n) for every n. If 2n+1 > 9 is composite, then (2n+1)^2 divides a(n). If 2n+1 is prime, then (2n+1)^2 does not divide a(n). My theorem: for n > 4, (2n+1)^2 divides a(n) if and only if the number 2n+1 is composite. - Thomas Ordowski, Nov 07 2022

			(zeta(2n)/Pi^(2n), n = 0, 1, 2, 3, ...) = (-1/2, 1/6, 1/90, 1/945, 1/9450, 1/93555, 691/638512875, 2/18243225, 3617/325641566250, ...), i.e.: zeta(0) = -1/2, zeta(2) = Pi^2/6, zeta(4) = Pi^4/90, zeta(6) = Pi^6/945, zeta(8) = Pi^8/9450, zeta(10) = Pi^10/93555, zeta(12) = Pi^12*691/638512875, ...
In Maple, series(Psi(x),x,20) gives -1*x^(-1) + (-gamma) + 1/6*Pi^2*x + (-Zeta(3))*x^2 + 1/90*Pi^4*x^3 + (-Zeta(5))*x^4 + 1/945*Pi^6*x^5 + (-Zeta(7))*x^6 + 1/9450*Pi^8*x^7 + (-Zeta(9))*x^8 + 1/93555*Pi^10*x^9 + ...
a(5) = 93555 = 10!/(2^9 * B(10)) = 3628800/(512*5/66). - _Frank Ellermann_, Apr 03 2020

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 84.
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).

T. D. Noe and J.P. Martin-Flatin, Table of n, a(n) for n = 0..250 (first 100 terms were computed by T. D. Noe).
G. Campbell, Some series like ζ(3), ζ(5), ζ(7). Number Theory group on LinkedIn.com, March 2015.
N. D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001-2003.
N. D. Elkies, On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n), Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.
Masato Kobayashi and Shunji Sasaki, Values of zeta-one functions at positive even integers, arXiv:2202.11835 [math.NT], 2022. See p. 4.
J. Sondow and E. W. Weisstein, MathWorld: Riemann Zeta Function
I. Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256.
I. Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256. [Annotated scanned copy]
Index entries for zeta function.

Cf. A046988 (numerators), A006003.

seq(denom(Zeta(2*n)/Pi^(2*n)),n=0..24); # Martin Renner, Sep 07 2016
A002432List := proc(len) series(-x*cot(x)/2, x, 2*len+1):
seq(denom(coeff(%, x, n)), n=0..2*len-1, 2) end:
A002432List(17); # Peter Luschny, Jun 07 2020

Table[Denominator[Zeta[2 n]/Pi^(2 n)], {n, 0, 30}] (* Artur Jasinski, Mar 11 2010 *)
Denominator[Zeta[2*Range[0, 20]]] (* Harvey P. Dale, Sep 05 2013 *)

a(n)=numerator(bestappr(Pi^(2*n)/zeta(2*n))) \\ Requires sufficient realprecision. The standard value of 38 digits yields erroneous values for n>9. \p99 is more than enough to get the 3 lines of displayed data. - M. F. Hasler, Mar 29 2015

a002432(n) = denominator(polcoeff((1-x*cotan(x))/2,n*2))
default(seriesprecision, 33); for(i=0,16,print1(a002432(i),",")) \\ Chris Boyd, Dec 21 2015

Sum_{n>=1} 2/(n^2 + 1) = Pi*coth(Pi)-1. 2*Sum_{k>=1} (-1)^(k + 1)/n^(2*k) = 2/(n^2+1). - Shmuel Spiegel (shmualm(AT)hotmail.com), Aug 13 2001
zeta(2n)/(2i * ( log(1-i)-log(1+i) ))^(2n) = zeta(2n)/(-i*log(-1))^(2n). - Eric Desbiaux, Dec 12 2008
zeta(2n) = Sum_{k >= 1} k^(-2n) = (-1)^(n-1)*B_{2n}*2^(2n-1)*Pi^(2n)/(2n)!.
a(n) = -A046988(n)*A010050(n)*A002445(n)/(A009117(n)*A000367(n))
a(n) = sqrt(denominator(Sum_{i>=1} A000005(i)/i^2n)). - Enrique Pérez Herrero, Jan 19 2012
Sum_{k >= 1} zeta(2k)*x^(2k) = (1-Pi*x*cot(Pi*x))/2. - Chris Boyd, Dec 21 2015
a(n) = denominator([x^(2*n)] -x*cot(x)/2). - Peter Luschny, Jun 07 2020

Formula and link from Henry Bottomley, Mar 10 2000
Formula corrected by Bjoern Boettcher, May 15 2003
Corrected and edited by M. F. Hasler, Mar 29 2015
a(0) = 2 prepended by Peter Luschny, Jun 07 2020

A004188 a(n) = n(3n^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

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

3-dimensional analog of centered polygonal numbers.

(1), (4+7), (10+13+16), (19+22+25+28), ... - Jon Perry, Sep 10 2004

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-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. A002412, A016061, A051682.
Cf. A236770 (partial sums).

[n*(3*n^2-1)/2: n in [0..50]]; //Vincenzo Librandi, May 15 2011

seq(binomial(2*n+1,3)+binomial(n+1,3), n=0..37); # Zerinvary Lajos, Jan 21 2007

Table[n(3n^2-1)/2,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,11,39},40] (* Harvey P. Dale, Jul 19 2019 *)

PARI
```
vector(40, n, n*(3*n^2-1)/2)
```

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

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

Original entry on oeis.org

1, 4, 15, 32, 65, 108, 175, 256, 369, 500, 671, 864, 1105, 1372, 1695, 2048, 2465, 2916, 3439, 4000, 4641, 5324, 6095, 6912, 7825, 8788, 9855, 10976, 12209, 13500, 14911, 16384, 17985, 19652, 21455, 23328, 25345, 27436, 29679, 32000, 34481, 37044, 39775, 42592

Offset: 1

Stefano Spezia, Aug 01 2018

Terms are obtained as partial sums in an algorithm for the generation of the sequence of the fourth powers (A000583). Starting with the sequence of the positive integers (A000027), it is necessary to delete every 4th term and to consider the partial sums of the obtained sequence, then to delete every 3rd term, and lastly to consider again the partial sums (see References).
a(n) is the trace of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern as shown in the examples below. Specifically, M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even, and it has det(M(n)) = 0 for n > 2 (proved).
From Saeed Barari, Oct 31 2021: (Start)
Also the sum of the entries in an n X n matrix whose elements start from 1 and increase as they approach the center. For instance, in case of n=5, the entries of the following matrix sum to 65:
  1 2 3 2 1
  2 3 4 3 2
  3 4 5 4 3
  2 3 4 3 2
  1 2 3 2 1. (End)
The n X n square matrix of the preceding comment is defined as: A[i,j,n] = n - abs((n + 1)/2 - j) - abs((n + 1)/2 - i). - Stefano Spezia, Nov 05 2021

			For n = 1 the matrix M(1) is
  1
with trace Tr(M(1)) = a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with Tr(M(2)) = a(2) = 4.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with Tr(M(3)) = a(3) = 15.

Edward A. Ashcroft, Anthony A. Faustini, Rangaswami Jagannathan, and William W. Wadge, Multidimensional Programming, Oxford University Press 1995, p. 12.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
G. Polya, Mathematics and Plausible Reasoning: Induction and analogy in mathematics, Princeton University Press 1990, p. 118.
Shailesh Shirali, A Primer on Number Sequences, Universities Press (India) 2004, p. 106.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.7.3 on pages 122-123.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, München 1952. Sitzungsberichte: 1951,3.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A000583, A000027, A186424 (first differences).
Cf. A000578, A000035, A193356, A210378.
Cf. A006003, A317297, A001489, A322844.
Cf. related to the M matrices: A074147 (antidiagonals), A130130 (rank), A241016 (row sums), A317617 (column sums), A322277 (permanent), A323723 (subdiagonal sums), A323724 (superdiagonal sums).

a_n:=List([1..nmax], n->(1/2)*(n^3 + n*RemInt(n, 2)));

List([1..50],n->(1/2)*(n^3+n*(n mod 2))); # Muniru A Asiru, Aug 24 2018

[IsEven(n) select n^3/2 else (n^3+n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 07 2018

a:=n->(1/2)*(n^3+n*modp(n,2)): seq(a(n),n=1..50); # Muniru A Asiru, Aug 24 2018

CoefficientList[Series[1/4 E^-x (1 + 3 E^(2 x) + 6 E^(2 x) x + 2 E^(2 x) x^2), {x, 0, 45}], x]*Table[(k + 1)!, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)*(1 + x)^3), {x, 0, 45}], x]*Table[(k + 1)*(-1)^k, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)^3*(1 + x)), {x, 0, 45}], x]*Table[(k + 1), {k, 0, 45}]
From Robert G. Wilson v, Aug 01 2018: (Start)
a[i_, j_, n_] := If[OddQ@ i, j + n (i - 1), n*i - j + 1]; f[n_] := Tr[Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 45]
CoefficientList[Series[(x^4 + 2x^3 + 6x^2 + 2x + 1)/((x - 1)^4 (x + 1)^2), {x, 0, 45}], x]
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 4, 15, 32, 65, 108}, 45]
(End)

a(n):=(1/2)*(n^3 + n*mod(n,2))$ makelist(a(n), n, 1, nmax);

Vec(x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2) + O(x^40)) \\ Colin Barker, Aug 02 2018

M(i, j, n) = if (i % 2, j + n*(i-1), n*i - j + 1);
a(n) = sum(k=1, n, M(k, k, n)); \\ Michel Marcus, Aug 07 2018

for (n in 1:nmax){
   a <- (n^3+n*n%%2)/2
   output <- c(n, a)
   cat(output, "\n")
}
(MATLAB and FreeMat)
for(n=1:nmax); a=(n^3+n*mod(n,2))/2; fprintf('%d\t%0.f\n',n,a); end

a(n) = (1/2)*(A000578(n) + n*A000035(n)).
a(n) = A006003(n) - (n/2)*(1 - (n mod 2)).
a(n) = Sum_{k=1..n} T(n,k), where T(n,k) = ((n + 1)*k - n)*(n mod 2) + ((n - 1)*k + 1)*(1 - (n mod 2)).
E.g.f.: E(x) = (1/4)*exp(-x)*x*(1 + 3*exp(2*x) + 6*exp(2*x)*x + 2*exp(2*x)*x^2).
L.g.f.: L(x) = -x*(1 + x^2)/((-1 + x)*(1 + x)^3).
H.l.g.f.: LH(x) = -x*(1 + x^2)/((-1 + x)^3*(1 + x)).
Dirichlet g.f.: (1/2)*(Zeta(-3 + s) + 2^(-s)*(-2 + 2^s)*Zeta(-1 + s)).
From Colin Barker, Aug 02 2018: (Start)
G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = n^3/2 for n even.
a(n) = (n^3+n)/2 for n odd. (End)
a(2*n) = A317297(n+1) + A001489(n). - Stefano Spezia, Dec 28 2018
Sum_{n>0} 1/a(n) = (1/2)*(-2*polygamma(0, 1/2) + polygamma(0, (1-i)/2)+ polygamma(0, (1+i)/2)) + zeta(3)/4 approximately equal to 1.3959168891658447368440622669882813003351669... - Stefano Spezia, Feb 11 2019
a(n) = (A000578(n) + A193356(n))/2. - Stefano Spezia, Jun 27 2022
a(n) = A210378(n-1)/n. - Stefano Spezia, Jul 15 2024

A004466 a(n) = n(5n^2 - 2)/3.

Original entry on oeis.org

0, 1, 12, 43, 104, 205, 356, 567, 848, 1209, 1660, 2211, 2872, 3653, 4564, 5615, 6816, 8177, 9708, 11419, 13320, 15421, 17732, 20263, 23024, 26025, 29276, 32787, 36568, 40629, 44980, 49631, 54592

Offset: 0

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

3-dimensional analog of centered polygonal numbers.
Also as a(n)=(1/6)*(10*n^3-4*n), n>0: structured pentagonal anti-diamond numbers (vertex structure 11) (Cf. A051673 = alternate vertex A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n+1)-10*a(n) = (n+1)*(5*(n+1)^2-2)/3 - (10n(n+1)(n+2)/6) = n. The unit digits are 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,... . - Eric Desbiaux, Aug 18 2008

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A062786 (first differences), A264853 (partial sums).

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.

[n*(5*n^2-2)/3: n in [0..50]]; // Vincenzo Librandi, May 15 2011

A004466:=n->n*(5*n^2 - 2)/3; seq(A004466(n), n=0..50); # Wesley Ivan Hurt, Mar 10 2014

Table[n(5n^2-2)/3,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)

a(n)=n*(5*n^2-2)/3 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(1+8*x+x^2)/(1-x)^4. - Colin Barker, Jan 08 2012

E.g.f.: (x/3)*(3 + 15*x + 5*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A063521 a(n) = n(7n^2-4)/3.

Original entry on oeis.org

0, 1, 16, 59, 144, 285, 496, 791, 1184, 1689, 2320, 3091, 4016, 5109, 6384, 7855, 9536, 11441, 13584, 15979, 18640, 21581, 24816, 28359, 32224, 36425, 40976, 45891, 51184, 56869, 62960, 69471, 76416, 83809, 91664, 99995, 108816, 118141, 127984, 138359, 149280, 160761

Offset: 0

N. J. A. Sloane, Aug 02 2001

Also as a(n)=(1/6)*(14*n^3-8*n), n>0: structured heptagonal anti-diamond numbers (vertex structure 15) (Cf. A100186 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Harry J. Smith, Table of n, a(n) for n = 0..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -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.

A063521:=n->n*(7*n^2-4)/3; seq(A063521(k), k=0..100); # Wesley Ivan Hurt, Oct 24 2013

lst={};Do[AppendTo[lst, n*(7*n^2-4)/3], {n, 1, 6!}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 02 2008 *)
CoefficientList[Series[x*(1+12*x+x^2)/(1-x)^4, {x, 0, 50}], x] (* G. C. Greubel, Sep 01 2017 *)

a(n) = { n*(7*n^2 - 4)/3 } \\ Harry J. Smith, Aug 25 2009

G.f.: x*(1+12*x+x^2)/(1-x)^4. - Colin Barker, Jan 10 2012

E.g.f.: (x/3)*(3 + 21*x + 7*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A004126 a(n) = n(7n^2 - 1)/6.

Original entry on oeis.org

0, 1, 9, 31, 74, 145, 251, 399, 596, 849, 1165, 1551, 2014, 2561, 3199, 3935, 4776, 5729, 6801, 7999, 9330, 10801, 12419, 14191, 16124, 18225, 20501, 22959, 25606, 28449, 31495, 34751, 38224, 41921, 45849, 50015, 54426, 59089, 64011

Offset: 0

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

3-dimensional analog of centered polygonal numbers.
Sum of n triangular numbers starting from T(n), where T = A000217. E.g., a(4) = T(4) + T(5) + T(6) + T(7) = 10 + 15 + 21 + 28 = 74. - Amarnath Murthy, Jul 16 2004
Also as a(n) = (1/6)*(7*n^3-n), n>0: structured heptagonal diamond numbers (vertex structure 8). Cf. A100179 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Partial sums of A069099, centered heptagonal numbers (A000566). - Jonathan Vos Post, Mar 16 2006
Binomial transform of (0, 1, 7, 7, 0, 0, 0, ...) and third partial sum of (0, 1, 6, 7, 7, 7, ...). - Gary W. Adamson, Oct 05 2015

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-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. A000217, A000566, A016993, A069099.
Cf. A000447, A000292.

[n*(7*n^2-1)/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011

seq(binomial(2*n+1,3)-binomial(n+1,3), n=0..38); # Zerinvary Lajos, Jan 21 2007

Table[n (7 n^2 - 1)/6, {n, 0, 80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)

makelist(n*(7*n^2-1)/6,n,0,30); /* Martin Ettl, Jan 08 2013 */

vector(100, n, n--; n*(7*n^2 - 1)/6) \\ Altug Alkan, Oct 06 2015

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+5*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
E.g.f.: (x/6)*(7*x^2 + 21*x + 6)*exp(x). - G. C. Greubel, Oct 05 2015
a(n) = Sum_{i = n..2*n-1} A000217(i). - Bruno Berselli, Sep 06 2017
a(n) = n^3 + Sum_{k=0..n-1} k*(k+1)/2. Alternately, a(n) = A000578(n) + A000292(n-1) for n>0. - Bruno Berselli, May 23 2018

A063488 a(n) = (2n-1)(n^2 -n +2)/2.

Original entry on oeis.org

1, 6, 20, 49, 99, 176, 286, 435, 629, 874, 1176, 1541, 1975, 2484, 3074, 3751, 4521, 5390, 6364, 7449, 8651, 9976, 11430, 13019, 14749, 16626, 18656, 20845, 23199, 25724, 28426, 31311, 34385, 37654, 41124, 44801, 48691, 52800, 57134, 61699, 66501, 71546, 76840

Offset: 1

N. J. A. Sloane, Aug 01 2001

Sum of two consecutive terms of A006003(n) = n*(n^2+1)/2. a(n) = A006003(n-1) + A006003(n). - Alexander Adamchuk, Jun 03 2006

If a 2-set Y and a 3-set Z are disjoint subsets of an n-set X then a(n-4) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

Harry J. Smith, Table of n, a(n) for n = 1..1000
Milan Janjic, Two Enumerative Functions
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*n*(2*n^2 - 3*n + 1) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Cf. A000217, A006003.
Partial sums of A005918.

[(2*n-1)*(n^2 -n +2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2 n - 1) (n^2 - n + 2)/2, {n, 1, 40}] (* Bruno Berselli, Oct 14 2016 *)
LinearRecurrence[{4,-6,4,-1}, {1,6,20,49}, 50] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(n^2 - n + 2)/2 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2 + 4*x + 3*x^2 + 2*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

G.f.: (1 + x)*(1 + x + x^2)/(1 - x)^4. - Jaume Oliver Lafont, Aug 30 2009
a(n) = A000217(A000217(n)) - A000217(A000217(n-2)). - Bruno Berselli, Oct 14 2016
E.g.f.: (-2 + 4*x + 3*x^2 + 2*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017

A064808 a(n) is the (n+1)st (n+2)-gonal number.

Original entry on oeis.org

1, 3, 9, 22, 45, 81, 133, 204, 297, 415, 561, 738, 949, 1197, 1485, 1816, 2193, 2619, 3097, 3630, 4221, 4873, 5589, 6372, 7225, 8151, 9153, 10234, 11397, 12645, 13981, 15408, 16929, 18547, 20265, 22086, 24013, 26049, 28197, 30460, 32841, 35343, 37969, 40722

Offset: 0

Floor van Lamoen, Oct 22 2001

Sum of n terms of the arithmetic progression with first term 1 and common difference n-1. - Amarnath Murthy, Aug 04 2005
a(n) is the sum of (n+1)-th row terms of triangle A144693. - Gary W. Adamson, Sep 19 2008
See also A131685(k) = smallest positive number m such that c(i) = m*(i^1 + 1)*(i^2 + 2)* ... *(i^k+ k) / k! takes integral values for all i>=0: For k=2, A131685(k)=1, which implies that this is a well-defined integer sequence. - Alexander R. Povolotsky, Apr 24 2015

Harry J. Smith, Table of n, a(n) for n = 0..1000
Justin Crum, Cyrus Cheng, David A. Ham, Lawrence Mitchell, Robert C. Kirby, Joshua A. Levine, and Andrew Gillette, Bringing Trimmed Serendipity Methods to Computational Practice in Firedrake, arXiv:2104.12986 [math.NA], 2021.
Index to divisibility sequences
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Main diagonal of A057145.
Row sums of A076110.
Cf. A144693. - Gary W. Adamson, Sep 19 2008
Cf. A000096, A002378, A006002, A006003.

[(n+1)*(n^2+2)/2 : n in [0..50]]; // Wesley Ivan Hurt, Feb 21 2015

A064808:=n->(n+1)*(n^2+2)/2: seq(A064808(n), n=0..50); # Wesley Ivan Hurt, Feb 21 2015

Table[(n + 1) (n^2 + 2)/2, {n, 0, 50}] (* Wesley Ivan Hurt, Feb 21 2015 *)

a(n) = { (n + 1)*(n^2 + 2)/2 } \\ Harry J. Smith, Sep 26 2009

a(n) = (n+1)*(n^2 + 2)/2.
From Paul Barry, Nov 18 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} (k-(k-1)*C(0, j-k)).
a(n) = A006002(n) - A000096(n-2). (End)
G.f.: (1 - x + 3x^2)/(1 - x)^4. - R. J. Mathar, Jul 07 2009
a(n) = A006003(n+1) - A002378(n). - Rick L. Shepherd, Feb 21 2015
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, Feb 21 2015
a(n) = A057145(n+2,n+1). - R. J. Mathar, Jul 28 2016

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Original entry on oeis.org

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

Offset: 1

Clark Kimberling, Feb 03 2011

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:
  1....2.....4.....7...11...16...22...29...
  3....5.....8....12...17...23...30...38...
  6....9....13....18...24...31...39...48...
  10...14...19....25...32...40...49...59...
  15...20...26....33...41...50...60...71...
  21...27...34....42...51...61...72...84...
  28...35...43....52...62...73...85...98...
Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787.  Deleting the main diagonal and then summing give A185787.  Analogous treatments to the left of the main diagonal give A100182 and A101165.  Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.
Examples:
row 1:  A000124
row 2:  A022856
row 3:  A016028
row 4:  A145018
row 5:  A077169
col 1:  A000217
col 2:  A000096
col 3:  A034856
col 4:  A055998
col 5:  A046691
col 6:  A052905
col 7:  A055999
diag. (1,5,...) ...... A001844
diag. (2,8,...) ...... A001105
diag. (4,12,...)...... A046092
diag. (7,17,...)...... A056220
diag. (11,23,...) .... A132209
diag. (16,30,...) .... A054000
diag. (22,38,...) .... A090288
diag. (3,9,...) ...... A058331
diag. (6,14,...) ..... A051890
diag. (10,20,...) .... A005893
diag. (15,27,...) .... A097080
diag. (21,35,...) .... A093328
antidiagonal sums:  (1,5,15,34,...)=A006003=partial sums of A002817.
Let S(n,k) denote the n-th partial sum of column k.  Then
S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.
S(n,1)=n(n+1)(n+2)/6
S(n,2)=n(n+1)(n+5)/6
S(n,3)=n(n+2)(n+7)/6
S(n,4)=n(n^2+12n+29)/6
S(n,5)=n(n+5)(n+10)/6
S(n,6)=n(n+7)(n+11)/6
S(n,7)=n(n+10)(n+11)/6
Weight array of T: A144112
Accumulation array of T: A185506
Second rectangular sum array of T: A185507
Third rectangular sum array of T: A185508
Fourth rectangular sum array of T: A185509

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

Cf. A000027, A185788, A100182, A101165, A079824.

[n*(7*n^2-6*n+5)/6: n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

f[n_,k_]:=n+(n+k-2)(n+k-1)/2;
s[k_]:=Sum[f[n,k],{n,1,k}];
Factor[s[k]]
Table[s[k],{k,1,70}]  (* A185787 *)
CoefficientList[Series[(3*x^2+3*x+1)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

a(n)=n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

Edited by Clark Kimberling, Feb 25 2023

cp's OEIS Frontend

A004068 Number of atoms in a decahedron with n shells.

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A002432 Denominators of zeta(2*n)/Pi^(2*n).

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

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A317614 a(n) = (1/2)*(n^3 + n*(n mod 2)).

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A004466 a(n) = n*(5*n^2 - 2)/3.

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A002432 Denominators of zeta(2n)/Pi^(2n).

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

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

A004466 a(n) = n(5n^2 - 2)/3.

A063521 a(n) = n(7n^2-4)/3.

A004126 a(n) = n(7n^2 - 1)/6.

A063488 a(n) = (2n-1)(n^2 -n +2)/2.