A004188 -id:A004188 - cp's OEIS frontend

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

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

A063491 a(n) = (2n - 1)(3n^2 - 3n + 2)/2.

Original entry on oeis.org

1, 12, 50, 133, 279, 506, 832, 1275, 1853, 2584, 3486, 4577, 5875, 7398, 9164, 11191, 13497, 16100, 19018, 22269, 25871, 29842, 34200, 38963, 44149, 49776, 55862, 62425, 69483, 77054, 85156, 93807, 103025, 112828, 123234, 134261, 145927, 158250, 171248, 184939

Offset: 1

N. J. A. Sloane, Aug 01 2001

A triangle has sides of lengths 6*n-3, 6*n^2-6*n+4, and 6*n^2-6*n+7; for n>2 its area is 6*sqrt(a(n)^2 - 1). - J. M. Bergot, Aug 30 2013

[The source of this is using (n,n+1), (n+1,n+2), and (n+2,n+3) as (a,b) in the creation of three Pythagorean triangles with sides b^2-a^2, 2*a*b, and a^2+b^2. Combine the three respective sides to create a new larger triangle, then find its area. It is not simply working backwards from the sequence. As well, the sequence has this as its first comment to show that the numbers are actually doing something to find a solution.]

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Harry J. Smith, Table of n, a(n) for n = 1..1000
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*(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, A005917, A063493, A063494, A063495, A063496.

Cf. A005448, A004188.

[(2*n-1)*(3*n^2 -3*n +2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

LinearRecurrence[{4,-6,4,-1},{1,12,50,133},40] (* Harvey P. Dale, Jun 05 2016 *)
Table[(2*n-1)*(3*n^2 -3*n +2)/2, {n,1,30}] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(3*n^2 - 3*n + 2)/2 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

a <- c(0, 1, 9, 38, 110)
for(n in (length(a)+1):40)
  a[n] <- +4*a[n-1]-6*a[n-2]+4*a[n-3]-a[n-4]
a [Yosu Yurramendi, Sep 04 2013]

G.f.: x*(1+x)*(1+7*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = +4*a(n-1) -6*a(n-2) +4*a(n-3) -1*a(n-4) n > 3, a(1)=1, a(2)=12, a(3)=50, a(4)=133. - Yosu Yurramendi, Sep 04 2013
E.g.f.: (-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017
From Bruce J. Nicholson, Jun 17 2020: (Start)
a(n) = A005448(n) * A005408(n-1).
a(n) = A004188(n) + A004188(n-1). (End)

A063522 a(n) = n(5n^2 - 3)/2.

Original entry on oeis.org

0, 1, 17, 63, 154, 305, 531, 847, 1268, 1809, 2485, 3311, 4302, 5473, 6839, 8415, 10216, 12257, 14553, 17119, 19970, 23121, 26587, 30383, 34524, 39025, 43901, 49167, 54838, 60929, 67455, 74431, 81872, 89793, 98209, 107135, 116586, 126577, 137123, 148239, 159940

Offset: 0

N. J. A. Sloane, Aug 02 2001

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.

Bisections: A160674, A160699.

[n*(5*n^2 -3)/2: n in [0..30]]; // G. C. Greubel, May 02 2018

lst={};Do[AppendTo[lst, LegendreP[3, n]], {n, 10^2}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[x*(1 + 13*x + x^2)/(1-x)^4, {x, 0, 50}], x] (* G. C. Greubel, Sep 01 2017 *)
LinearRecurrence[{4,-6,4,-1},{0,1,17,63},40] (* Harvey P. Dale, Sep 06 2023 *)

a(n) = { n*(5*n^2 - 3)/2 } \\ Harry J. Smith, Aug 25 2009

G.f.: x*(1 + 13*x + x^2)/(1-x)^4. - Colin Barker, Jan 10 2012

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

A004467 a(n) = n(11n^2 - 5)/6.

Original entry on oeis.org

0, 1, 13, 47, 114, 225, 391, 623, 932, 1329, 1825, 2431, 3158, 4017, 5019, 6175, 7496, 8993, 10677, 12559, 14650, 16961, 19503, 22287, 25324, 28625, 32201, 36063, 40222, 44689, 49475, 54591, 60048

Offset: 0

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

3-dimensional analog of centered polygonal numbers, that is: centered hendecagonal pyramidal numbers (see Deza paper in References).

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.

[n*(11*n^2-5)/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011

Table[n(11n^2-5)/6,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,13,47},80] (* Harvey P. Dale, Sep 22 2013 *)

a(n)=n*(11*n^2-5)/6 \\ Charles R Greathouse IV, Sep 28 2011

G.f.: x*(1+9*x+x^2)/(1-x)^4. - Colin Barker, Jan 08 2012
a(0)=0, a(1)=1, a(2)=13, a(3)=47; for n>3, a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Sep 22 2013
E.g.f.: (x/6)*(6 + 33*x + 11*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A062025 a(n) = n(13n^2 - 7)/6.

Original entry on oeis.org

0, 1, 15, 55, 134, 265, 461, 735, 1100, 1569, 2155, 2871, 3730, 4745, 5929, 7295, 8856, 10625, 12615, 14839, 17310, 20041, 23045, 26335, 29924, 33825, 38051, 42615, 47530, 52809, 58465, 64511, 70960, 77825, 85119, 92855, 101046, 109705, 118845, 128479, 138620, 149281

Offset: 0

N. J. A. Sloane, Aug 02 2001

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.

Table[n(13n^2-7)/6,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)

a(n) = n*(13*n^2 - 7)/6 \\ Harry J. Smith, Jul 29 2009

From G. C. Greubel, Sep 01 2017: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (x + 11*x^2 + x^3)/(1 - x)^4.
E.g.f.: (x/6)*(6 + 39*x + 13*x^2)*exp(x). (End)

A063523 a(n) = n(8n^2 - 5)/3.

Original entry on oeis.org

0, 1, 18, 67, 164, 325, 566, 903, 1352, 1929, 2650, 3531, 4588, 5837, 7294, 8975, 10896, 13073, 15522, 18259, 21300, 24661, 28358, 32407, 36824, 41625, 46826, 52443, 58492, 64989, 71950, 79391, 87328, 95777, 104754, 114275, 124356, 135013, 146262, 158119, 170600

Offset: 0

N. J. A. Sloane, Aug 02 2001

Also as a(n)=(1/6)*(16*n^3-10*n), n>0: structured octagonal anti-diamond numbers (vertex structure 17) (Cf. A100187 = 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.

Table[n(8n^2-5)/3,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,18,67},81] (* or *) CoefficientList[ Series[ (x+14 x^2+x^3)/(x-1)^4,{x,0,80}],x] (* Harvey P. Dale, Jul 11 2011 *)

a(n) = n*(8*n^2 - 5)/3 \\ Harry J. Smith, Aug 25 2009

a(0)=0, a(1)=1, a(2)=18, a(3)=67, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 11 2011
G.f.: (x+14*x^2+x^3)/(x-1)^4. - Harvey P. Dale, Jul 11 2011
E.g.f.: (x/3)*(3 + 24*x + 8*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.

Original entry on oeis.org

0, 1, 12, 51, 145, 330, 651, 1162, 1926, 3015, 4510, 6501, 9087, 12376, 16485, 21540, 27676, 35037, 43776, 54055, 66045, 79926, 95887, 114126, 134850, 158275, 184626, 214137, 247051, 283620, 324105, 368776, 417912, 471801, 530740, 595035, 665001, 740962

Offset: 0

Bruno Berselli, Jan 31 2014

After 0, first trisection of A011779 and right border of A177708.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Partial sums of A004188.
Cf. A000217, A000332, A011779, A177708.
Cf. similar sequences on the polygonal numbers: A002817(n) = A000217(A000217(n)); A000537(n) = A000290(A000217(n)); A037270(n) = A000217(A000290(n)); A062392(n) = A000384(A000217(n)).
Cf. sequences of the form A000217(m)+k*A000332(m+2): A062392 (k=12); A264854 (k=11); A264853 (k=10); this sequence (k=9); A006324 (k=8); A006323 (k=7); A000537 (k=6); A006322 (k=5); A006325 (k=4), A002817 (k=3), A006007 (k=2), A006522 (k=1).
Cf. A232713, A260810.

[n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];

Table[n (n + 1) (3 n^2 + 3 n - 2)/8, {n, 0, 40}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,12,51,145},40] (* Harvey P. Dale, Aug 22 2016 *)

for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));

G.f.: x*(1 + 7*x + x^2)/(1 - x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A000326(A000217(n)).
a(n) = A000217(n) + 9*A000332(n+2).
Sum_{n>=1} 1/a(n) = 2 + 4*sqrt(3/11)*Pi*tan(sqrt(11/3)*Pi/2) = 1.11700627139319... . - Vaclav Kotesovec, Apr 27 2016

A135503 a(n) = n*(n^2 - 1)/2.

Original entry on oeis.org

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

Offset: 0

Cino Hilliard, Feb 09 2008

Previous name was: Integer values of sqrt(b) solving sqrt(d) + sqrt(b) = sqrt(c) with d^2 + b = c.
Squaring the first equation and setting the result equal to the second, we need d + b + 2*sqrt(d*b) = d^2+b -> d + 2*sqrt(d*b) = d^2 -> d^2 - d = 2*sqrt(d*b)
-> d^2*(d-1)^2 = 4*d*b -> b = d*(d-1)^2/4 -> sqrt(b) = (d-1)*sqrt(d)/2. Setting d = (n+1)^2 yields sqrt(b) = A027480(n).
This is the case k = 2 for FLTR, Fermat's Last Theorem with rational exponents 1/k: Consider x + y = x + y. Then (x^k)^(1/k) + (y^k)^(1/k) = ((x+y)^k)^(1/k).
For k > 2, there are infinitely many solutions to d^(1/k) + b^(1/k) = c^(1/k). E.g., 8^(1/3) + 27^(1/3) = 125^(1/3) at k = 3. However, in conjunction with d^2 + b = c, I could not find any nontrivial solutions.
A shifted version of A027480. - R. J. Mathar, Apr 07 2009
For n > 2, a(n) is the maximum value of the magic constant in a perimeter-magic n-gon of order n (see A342758). - Stefano Spezia, Mar 21 2021
a(n) is equal to the total number of P_3 edge-disjoint subgraphs of the complete graph on n vertices. - Samuel J. Bevins, May 09 2023

			For d = 9, b = 144, c = 225, 9^(1/2) + 144^(1/2) = 225^(1/2) and 9^2 + 144 = 225. So b^(1/2) = 12 is the 4th entry in the sequence.

G. C. Greubel, Table of n, a(n) for n = 0..1000
C. D. Bennet, A. M. W. Glass and G. J. Székely, Fermat's Last Theorem for Rational Exponents, Am. Math. Monthly 111 (2004), 322-329. - _R. J. Mathar_, Apr 21 2009
Terrel Trotter, Perimeter-Magic Polygons, Journal of Recreational Mathematics Vol. 7, No. 1, 1974, pp. 14-20.

Cf. A000217, A000578, A002411, A004188, A006003, A007531, A007588, A027480, A342758.

Array[# (#^2 - 1)/2 &, 42, 0] (* Michael De Vlieger, Feb 20 2018 *)

flt2(n,p) = { local(a,b); for(a=0,n, b = (a^3-a)/2; print1(b", ") ) }

a(n) = 3*A000292(n-1).
From R. J. Mathar Feb 20 2008: (Start)
O.g.f.: 3*x^2/(-1+x)^4.
a(n) = n*(n^2 - 1)/2 = A007531(n+1)/2. (End)
G.f.: 3*x^2*G(0)/2, where G(k) = 1 + 1/(1 - x/(x + (k+1)/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
a(n) = A006003(n+1) - A000326(n+1). - J. M. Bergot, Dec 04 2014
E.g.f.: (1/2)* x^2 *(3 + x)*exp(x). - G. C. Greubel, Oct 15 2016
From Miquel Cerda, Dec 25 2016: (Start)
a(n) = A000578(n) - A006003(n).
a(n) = A004188(n) - A000578(n).
a(n) = A007588(n) - A004188(n). (End)
a(n) = A002411(n) - A000217(n). - Justin Gaetano, Feb 20 2018
From Amiram Eldar, Jan 09 2021: (Start)
Sum_{n>=2} 1/a(n) = 1/2.
Sum_{n>=2} (-1)^n/a(n) = 4*log(2) - 5/2. (End)

Edited by R. J. Mathar, Apr 21 2009

New name using R. J. Mathar's formula, Joerg Arndt, Dec 05 2014

A116689 Partial sums of dodecahedral numbers (A006566).

Original entry on oeis.org

0, 1, 21, 105, 325, 780, 1596, 2926, 4950, 7875, 11935, 17391, 24531, 33670, 45150, 59340, 76636, 97461, 122265, 151525, 185745, 225456, 271216, 323610, 383250, 450775, 526851, 612171, 707455, 813450, 930930, 1060696, 1203576, 1360425

Offset: 0

Jonathan Vos Post, Mar 15 2006

Geometrically, the partial sums of dodecahedral numbers may be interpreted as 4-dimensional dodecahedral hyperpyramidal numbers. It is somewhat surprising that this is (with proper offset) the triangular number of the "second pentagonal numbers, minus 1."

Also, the sequence is related to A004188 by a(n) = n*A004188(n)-sum(A004188(i), i=0..n-1). [Bruno Berselli, Apr 05 2012]

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A001844, A004188, A005449, A006566, A095794.

A116689:=n->binomial(n*(3*n+1)/2,2); seq(A116689(k), k=0..100); # Wesley Ivan Hurt, Oct 06 2013

Table[Binomial[n(3n+1)/2,2], {n,0,100}] (* Wesley Ivan Hurt, Oct 06 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{0,1,21,105,325},40] (* Harvey P. Dale, Apr 01 2018 *)

a(n) = Sum_{i = 0..n} A006566(i).
a(n) = Sum_{i = 0..n} i*(3*i-1)*(3*i-2)/2.
a(n+1) = A000217(A095794(n)).
a(n+1) = A000217(A005449(n) - 1).
a(n+1) = A000217(n*(3n+1)/2-1).
a(n+1) = A000217(A001844(n) - A000217(n+1) - 1).
a(n) = n*(9*n^3+6*n^2-5*n-2)/8. G.f.: x*(1+16*x+10*x^2)/(1-x)^5. [Colin Barker, Apr 04 2012]
a(n) = binomial(A005449(n), 2). - Wesley Ivan Hurt, Oct 06 2013
From Peter Bala, Sep 03 2023: (Start)
a(n) = n*(n + 1)*(3*n + 1)*(3*n - 2)/8.
a(n) = Sum_{0 <= i <= j <= n-1} (3*i + 1)*(3*j + 1). Cf. A024212 (End)

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

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

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A063491 a(n) = (2*n - 1)*(3*n^2 - 3*n + 2)/2.

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A063522 a(n) = n*(5*n^2 - 3)/2.

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A004467 a(n) = n*(11*n^2 - 5)/6.

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A062025 a(n) = n*(13*n^2 - 7)/6.

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A063523 a(n) = n*(8*n^2 - 5)/3.

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

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

A063491 a(n) = (2n - 1)(3n^2 - 3n + 2)/2.

A063522 a(n) = n(5n^2 - 3)/2.

A004467 a(n) = n(11n^2 - 5)/6.

A062025 a(n) = n(13n^2 - 7)/6.

A063523 a(n) = n(8n^2 - 5)/3.

A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.