A097321 -id:A097321 - cp's OEIS frontend

6, 60, 210, 504, 990, 1716, 2730, 4080, 5814, 7980, 10626, 13800, 17550, 21924, 26970, 32736, 39270, 46620, 54834, 63960, 74046, 85140, 97290, 110544, 124950, 140556, 157410, 175560, 195054, 215940, 238266, 262080, 287430, 314364, 342930

Offset: 0

Benoit Cloitre, Apr 05 2002

Terms are areas of primitive Pythagorean triangles whose odd sides differ by 2; e.g., the triangle with sides 8,15,17 has area 60. - Lekraj Beedassy, Apr 18 2003

Using (n, n+1), (n, n+2), and (n+1, n+2) to generate three unreduced Pythagorean triangles gives a sum of the areas for all three to be (2*n+1)*(2*n+2)*(2*n+3), which are three consecutive numbers. - J. M. Bergot, Aug 22 2011

T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 190.
Jolley, Summation of Series, Oxford (1961).
Konrad Knopp, Theory and application of infinite series, Dover, p. 269.

M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
S. Ramanujan, Notebook entry
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A097321, A069140, A000447, A035328.

Array[Times@@(2#+{1,2,3})&,40,0] (* or *) LinearRecurrence[{4,-6,4,-1},{6,60,210,504},40] (* Harvey P. Dale, Dec 08 2013 *)

a(n)=(2*n+1)*(2*n+2)*(2*n+3) \\ Charles R Greathouse IV, Oct 07 2015

log(2) - 1/2 = Sum_{n>=0} 1/a(n); (1/2)*(1-log(2)) = Sum_{n>=0} (-1)^n/a(n). [Jolley eq 236 and 237]

Sum_{n>=0} x^n/a(n) = ((1+x)/sqrt(x)*log((1+sqrt x)/(1-sqrt x)) + 2*log(1-x)-2)/(4x). [Jolley eq 280 for 0

Sum_{n>=0} (-x)^n/a(n) = (1-log(1+x) -(1-x)/sqrt(x)*arctan(x))/(2x). [Jolley eq 281 for 0

a(n) = 6*A000447(n+1). - Lekraj Beedassy, Apr 18 2003

G.f.: 6*(1 + 6*x + x^2) / (x-1)^4 . - R. J. Mathar, Jun 09 2013

a(0)=6, a(1)=60, a(2)=210, a(3)=504, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 08 2013

a(n) = 2*A035328(n+1). - J. M. Bergot, Jan 02 2015

A054776 a(n) = 3n(3n-1)(3*n-2).

Original entry on oeis.org

0, 6, 120, 504, 1320, 2730, 4896, 7980, 12144, 17550, 24360, 32736, 42840, 54834, 68880, 85140, 103776, 124950, 148824, 175560, 205320, 238266, 274560, 314364, 357840, 405150, 456456, 511920, 571704, 635970, 704880, 778596, 857280, 941094

Offset: 0

Henry Bottomley, May 19 2000

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 46.
Konrad Knopp, Theory and Application of Infinite Series, Dover, p. 268.

Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A006566, A007531, A097321.

A054776:=n->3*n*(3*n-1)*(3*n-2): seq(A054776(n), n=0..50); # Wesley Ivan Hurt, Apr 14 2017

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

a(n) = A007531(3n-2) = 6*A006566(n).
Sum_{n>=1} 1/a(n) = Pi*sqrt(3)/12 - log(3)/4 = 0.178796768891527... [Jolley eq. 250]. - Benoit Cloitre, Apr 05 2002
G.f.: 6*x*(1+16*x+10*x^2)/(1-x)^4.
E.g.f.: 3*exp(x)*x*(2 + 18x + 9x^2). - Indranil Ghosh, Apr 15 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - Pi/(6*sqrt(3)). - Amiram Eldar, Mar 08 2022

A069140 a(n) = (4n-1)4n(4n+1).

Original entry on oeis.org

0, 60, 504, 1716, 4080, 7980, 13800, 21924, 32736, 46620, 63960, 85140, 110544, 140556, 175560, 215940, 262080, 314364, 373176, 438900, 511920, 592620, 681384, 778596, 884640, 999900, 1124760, 1259604, 1404816, 1560780, 1727880, 1906500

Offset: 0

Henry Bottomley, Apr 08 2002

			a(10) = 39*40*41 = 63960.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Henry Bottomley, Medians and area bisectors of a triangle.
S. Ramanujan, Notebook entry.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A007531, A069072, A097321.

Table[64n^3-4n,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,60,504,1716},40] (* Harvey P. Dale, Dec 24 2012 *)

a(n)=(4*n-1)*4*n*(4*n+1) \\ Charles R Greathouse IV, Oct 07 2015

Sum_{i>0} 1/a(i) = log(2)*3/4 - 1/2 = 0.019860..., which is the ratio of the area of the deltoid envelope formed by area bisectors of a triangle to the area of the triangle.
a(n) = 64n^3 - 4n = A007531(4n) = A069072(2n-1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 24 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/2 - log(2)/4 + log(tan(Pi/8))/(2*sqrt(2)). Amiram Eldar, Mar 20 2022

A228889 a(n) = 3n(3n + 1)(3*n + 2).

Original entry on oeis.org

60, 336, 990, 2184, 4080, 6840, 10626, 15600, 21924, 29760, 39270, 50616, 63960, 79464, 97290, 117600, 140556, 166320, 195054, 226920, 262080, 300696, 342930, 388944, 438900, 492960, 551286, 614040, 681384, 753480, 830490, 912576, 999900, 1092624, 1190910

Offset: 1

Peter Bala, Sep 09 2013

Related sequences are A054776 and A097321.

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. A054776, A097321, A228888.

[3*n*(3*n+1)*(3*n+2): n in [1..40]]; // Vincenzo Librandi, Sep 10 2013

Maple
```
seq(3*n*(3*n+1)*(3*n+2), n = 1..35);
```

CoefficientList[Series[6 (10 + 16 x + x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Sep 10 2013 *)
Table[Times@@(3n+{0,1,2}),{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{60,336,990,2184},40] (* Harvey P. Dale, Dec 20 2023 *)

a(n) = 3*n*(3*n + 1)*(3*n + 2) = 6*binomial(3*n + 2,3) = 6*A228888(n).
a(-n) = - A054776(n).
G.f.: 6*x*(10 + 16*x + x^2)/(1 - x)^4 = 60*x + 336*x^2 + 990*x^3 + ....
Sum {n >= 1} 1/a(n) = 3/4 - log(3)/4 - 1/12*sqrt(3)*Pi;
Sum {n >= 1} (-1)^n/a(n) = 3/4 - 2/3*log(2) - 1/18*sqrt(3)*Pi.

A157024 a(0)=1, a(n) = (3n-1)3n(3n+1)/2 for n>0.

Original entry on oeis.org

1, 12, 105, 360, 858, 1680, 2907, 4620, 6900, 9828, 13485, 17952, 23310, 29640, 37023, 45540, 55272, 66300, 78705, 92568, 107970, 124992, 143715, 164220, 186588, 210900, 237237, 265680, 296310, 329208, 364455, 402132, 442320, 485100, 530553, 578760, 629802

Offset: 0

Jaume Oliver Lafont, Feb 21 2009

nonn

Cf. A002391, A027480, A058962, A097321, A154920.

nxt[{n_,a_}]:={n+1,((3n)(3n-1)(3n+1))/2}; NestList[nxt,{0,1},40][[All,2]]/.(0->Nothing) (* Harvey P. Dale, Sep 24 2016 *)

Sum_{n>=0} 1/a(n) = log(3).
G.f.: (1+8x+63x^2+8x^3+x^4)/(1-x)^4.
a(n) = A027480(3n-1), n>0. - R. J. Mathar, Apr 07 2009
Sum_{n>=0} (-1)^n/a(n) = 4*log(2)/3. - Amiram Eldar, Feb 27 2022

Showing 1-5 of 5 results.

cp's OEIS Frontend

A069072 a(n) = (2n+1)*(2n+2)*(2n+3).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A054776 a(n) = 3*n*(3*n-1)*(3*n-2).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Maple

PARI

Formula

A069140 a(n) = (4n-1)*4n*(4n+1).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A228889 a(n) = 3*n*(3*n + 1)*(3*n + 2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A157024 a(0)=1, a(n) = (3n-1)*3n*(3n+1)/2 for n>0.

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Formula

A069072 a(n) = (2n+1)(2n+2)(2n+3).

A054776 a(n) = 3n(3n-1)(3*n-2).

A069140 a(n) = (4n-1)4n(4n+1).

A228889 a(n) = 3n(3n + 1)(3*n + 2).

A157024 a(0)=1, a(n) = (3n-1)3n(3n+1)/2 for n>0.