A001082 -id:A001082 - cp's OEIS frontend

A270469 Number of ordered ways to write n = x^3 + y(y+1) + z(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.

1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 3, 1, 2, 2, 3, 2, 2, 3, 2, 1, 4, 4, 2, 2, 2, 2, 1, 5, 4, 2, 2, 2, 3, 4, 4, 5, 2, 2, 3, 3, 5, 2, 5, 3, 2, 4, 5, 4, 2, 3, 3, 3, 3, 4, 3, 1, 2, 5, 3, 4, 3, 4, 4, 4, 5, 4, 3, 4, 4, 3, 6, 5, 5, 3, 3, 3, 6, 6, 2, 4

Offset: 1

Zhi-Wei Sun, Mar 17 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 5, 6, 10, 12, 20, 27, 56, 101.
(ii) Each nonnegative integer can be written as x^3 + P(y,z) with x >= 0 and y and z integers, provided that P(y,z) is among y^2+z(3z+1), y(2y+1)+z(3z+1), y(2y+1)+z(3z+2), y(2y+1)+z(5z+2), y(2y+1)+z(5z+3), y(2y+1)+2z(3z+1), y(2y+1)+ 2z(3z+2), y(2y+1)+z(6z+5), y(2y+1)+z(7z+2), y(2y+1)+z(7z+6), y(3y+1)+z(4z+1), y(3y+1)+z(7z+2), y(3y+2)+z(4z+1).
(iii) Every n = 0,1,2,... can be expressed as f(x,y,z) with x >= 0 and y and z integers, provided that f(x,y,z) is among 2x^3+y^2+z(3z+1), 2x^3+y(y+1)+z(3z+2), 3x^3+y(y+1)+z(3z+2), 2x^3+y(2y+1)+z(3z+1), 2x^3+y(2y+1)+z(3z+2), 2x^3+y(2y+1)+z(5z+4), 2x^3+y(3y+1)+z(3z+2), 2x^3+y(3y+2)+z(4z+3).
Note that those y(2y+1) with y integral are just triangular numbers.
See also A262813 for a similar conjecture.

			a(10) = 1 since 10 = 0^3 + 1*2 + (-2)(3*(-2)+2).
a(12) = 1 since 12 = 1^3 + 2*3 + 1*(3*1+2).
a(20) = 1 since 20 = 0^3 + 3*4 + (-2)*(3*(-2)+2).
a(27) = 1 since 27 = 0^3 + 2*3 + (-3)*(3*(-3)+2).
a(56) = 1 since 56 = 0^3 + 0*1 + 4*(3*4+2).
a(101) = 1 since 101 = 2^3 + 8*9 + (-3)*(3*(-3)+2).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000290, A000578, A001082, A002378, A262813, A262815, A262816.

pQ[x_]:=pQ[x]=x>0&&IntegerQ[Sqrt[3x+1]]
Do[r=0;Do[If[pQ[n-x^3-y(y+1)],r=r+1],{x,0,n^(1/3)},{y,0,(Sqrt[4(n-x^3)+1]-1)/2}];Print[n," ",r];Continue,{n,1,80}]

A017569 a(n) = 12*n + 4.

Original entry on oeis.org

4, 16, 28, 40, 52, 64, 76, 88, 100, 112, 124, 136, 148, 160, 172, 184, 196, 208, 220, 232, 244, 256, 268, 280, 292, 304, 316, 328, 340, 352, 364, 376, 388, 400, 412, 424, 436, 448, 460, 472, 484, 496, 508, 520, 532, 544, 556, 568, 580, 592, 604, 616, 628

Offset: 0

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(46).
Number of 6 X n 0-1 matrices avoiding simultaneously the right angled numbered polyomino patterns (ranpp) (00;1), (01;0), (11;0) and (01;1). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1A008574; m=3: A016933; m=4: A022144; m=5: A017293. - Sergey Kitaev, Nov 13 2004
Except for 4, exponents e such that x^e - x^2 + 1 is reducible.
If Y and Z are 2-blocks of a (3n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
Terms are perfect squares iff n is a generalized octagonal number (A001082), then n = k*(3*k-2) and a(n) = (2*(3*k-1))^2. - Bernard Schott, Feb 26 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Milan Janjic, Two Enumerative Functions.
Tanya Khovanova, Recursive Sequences.
Sergey Kitaev, On multi-avoidance of right angled numbered polyomino patterns, Integers: Electronic Journal of Combinatorial Number Theory, Vol. 4 (2004), Article A21, 20pp.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N)).
William A. Stein, The modular forms database.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008594, A017533, A017545, A089911.
Cf. A016777, A016933, A017293, A022144.
Cf. A001082.

a017569 = (+ 4) . (* 12)  -- Reinhard Zumkeller, Jul 05 2013

[12*n+4: n in [0..50]]; // Vincenzo Librandi, May 04 2011

12*Range[0,200]+4 (* Vladimir Joseph Stephan Orlovsky, Feb 19 2011 *)

A089911(a(n)) = 3. - Reinhard Zumkeller, Jul 05 2013
Sum_{n>=0} (-1)^n/a(n) = sqrt(3)*Pi/36 + log(2)/12. - Amiram Eldar, Dec 12 2021
From Stefano Spezia, Feb 25 2023: (Start)
O.g.f.: 4*(1 + 2*x)/(1 - x)^2.
E.g.f.: 4*exp(x)*(1 + 3*x). (End)
From Elmo R. Oliveira, Apr 10 2025: (Start)
a(n) = 2*a(n-1) - a(n-2).
a(n) = 2*A016933(n) = 4*A016777(n) = A016777(4*n+1). (End)

A036666 Numbers k such that 5*k + 1 is a square.

Original entry on oeis.org

0, 3, 7, 16, 24, 39, 51, 72, 88, 115, 135, 168, 192, 231, 259, 304, 336, 387, 423, 480, 520, 583, 627, 696, 744, 819, 871, 952, 1008, 1095, 1155, 1248, 1312, 1411, 1479, 1584, 1656, 1767, 1843, 1960, 2040, 2163, 2247, 2376, 2464, 2599, 2691

Offset: 1

N. J. A. Sloane, Dec 11 1999

Third differences are 4, -6, 8, -10, 12, -14, 16, -18, 20, -22, 24, -26, 28, ...
X values of solutions to the equation 5*X^3 + X^2 = Y^2. - Mohamed Bouhamida, Nov 06 2007
Also, numbers 5*i^2 + 2*i for integer i. The characteristic function is A205633(n). - Jason Kimberley, Nov 15 2012
From Gary W. Adamson, Sep 22 2019: (Start)
Match the values a(n) with the squares 5k + 1 as follows:
   3,....7,....16,....24,...  .a, a, a, a,...
  16,...36,....81,...121,...  (base).
Then 1/5 in the matching base is equal to .a, a, a,...
Example: 1/5 in base 36 is equal to .7, 7, 7, 7...
Check: 7/36 + 7/36^2 = 259/1296 = .199845...; close to 1/5.
(End)

Jason Kimberley, Table of n, a(n) for n = 1..2000
S. Cooper and M. D. Hirschhorn, Results of Hurwitz type for three squares. Discrete Math., Vol. 274, No. 1-3 (2004), pp. 9-24. See D(q).
Ralf Stephan, On the solutions to 'px+1 is square'.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001082, A001622, A002378, A005563, A046092, A047209, A109043, A205633.

List([1..50],n->(10*n*(n-1)+(2*n-1)*(-1)^n+1)/8); # Muniru A Asiru, Nov 28 2018

[(n-1)^2+(n-1)+Ceiling((n-1)/2)^2 : n in [1..50]]; // Wesley Ivan Hurt, Jun 05 2014

seq(n^2+n+ceil(n/2)^2, n=0..46); # Gary Detlefs, Feb 23 2010

(Select[ Range[121], Mod[ #, 5] == 1 || Mod[ #, 5] == 4 &]^2 - 1)/5 (* Robert G. Wilson v, Jun 23 2004 *)
Flatten[Position[5*Range[0,3000]+1,?(IntegerQ[Sqrt[#]]&)]]-1 (* or *) LinearRecurrence[{1,2,-2,-1,1},{0,3,7,16,24},50] (* _Harvey P. Dale, Feb 13 2018 *)
Accumulate[Table[n + LCM[n, 2], {n, 0, 121}]] (* Jon Maiga, Nov 28 2018 *)

PARI
```
a(n)=n^2+n+ceil(n/2)^2
```

G.f.: x*(3 + 4*x + 3*x^2) / ((1 - x)*(1 - x^2)).
a(n) has the form ((5*m + 1)^2 - 1)/5 if n is odd; a(n) has the form ((5*m + 4)^2 - 1)/5 if n is even.
a(2*k) = k*(5*k + 2), a(2*k + 1) = 5*k^2 + 8*k + 3. - Mohamed Bouhamida, Nov 06 2007
a(n+1) = n^2 + n + ceiling(n/2)^2. - Gary Detlefs, Feb 23 2010
From Bruno Berselli, Nov 27 2010: (Start)
a(n) = (10*n*(n - 1)+(2*n - 1)*(-1)^n + 1)/8.
5*a(n) + 1 = A047209(n)^2. (End)
a(n) = Sum_{k=0..n} k + A109043(k). - Jon Maiga, Nov 28 2018
E.g.f.: (exp(x)*(1 + 10*x^2) - exp(-x)*(1 + 2*x))/8. - Franck Maminirina Ramaharo, Nov 29 2018
From Amiram Eldar, Mar 15 2022: (Start)
Sum_{n>=2} 1/a(n) = 5/4 - sqrt(1-2/sqrt(5))*Pi/2.
Sum_{n>=2} (-1)^n/a(n) = 5*(log(5)-1)/4 - sqrt(5)*log(phi)/2, where phi is the golden ratio (A001622). (End)

Better description and additional formula from Santi Spadaro, Jul 12 2001

A270516 Number of ordered ways to write n = x^3(x+1) + y(y+1)/2 + z*(3z+2), where x and y are nonnegative integers, and z is an integer.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 2, 4, 2, 3, 4, 1, 3, 1, 2, 3, 3, 3, 2, 2, 3, 4, 3, 5, 3, 4, 2, 4, 4, 3, 5, 2, 5, 2, 5, 5, 2, 5, 5, 3, 4, 3, 5, 4, 5, 7, 2, 4, 1, 5, 2, 4, 3, 2, 5, 3, 6, 3, 3, 5, 6, 2, 5, 2, 4, 5, 4, 8, 3, 4, 5, 1, 5, 3, 1, 4, 3, 5, 4, 5

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and the only values of n > 1428 with a(n) = 1 are 2205, 2259, 3556, 4107, 4337, 5387, 9331, 16561, 22237, 27569, 63947, 78610.
(ii) Any natural number can be written as x*(x^3+2) + y*(y+1)/2 + z*(3z+1), where x and y are nonnegative integers, and z is an integer.
(iii) Every n = 0,1,2,... can be written as x*(x^3+x^2+6) + y*(y+1)/2 + z*(3z+2) (or x*(x^3+x^2+4x+1) + y*(y+1)/2 + z*(3z+1)), where x and y are nonnegative integers, and z is an integer.
See also A270533 for a similar conjecture.

			a(72) = 1 since 72 = 2^3*3 + 5*6/2 + 3*(3*3+2).
a(75) = 1 since 75 = 0^3*1 + 4*5/2 + (-5)*(3*(-5)+2).
a(5387) = 1 since 5387 = 7^3*8 + 2*3/2 + (-30)*(3*(-30)+2).
a(9331) = 1 since 9331 = 8^3*9 + 2*3/2 + (-40)*(3*(-40)+2).
a(16561) = 1 since 16561 = 1^3*2 + 101*102/2 + (-62)*(3*(-62)+2).
a(22237) = 1 since 22237 = 6^3*7 + 104*105/2 + 71*(3*71+2).
a(27569) = 1 since 27569 = 2^3*3 + 49*50/2 + (-94)*(3*(-94)+2).
a(63947) = 1 since 63947 = 0^3*1 + 173*174/2 + (-128)*(3*(-128)+2).
a(78610) = 1 since 78610 = 16^3*17 + 52*53/2 + 50*(3*50+2).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Cf. A000217, A000578, A001082, A001318, A262813, A262815, A262816, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270533.

OQ[x_]:=OQ[x]=IntegerQ[Sqrt[3x+1]]
Do[r=0;Do[If[OQ[n-y(y+1)/2-x^3*(x+1)],r=r+1],{y,0,(Sqrt[8n+1]-1)/2},{x,0,(n-y(y+1)/2)^(1/4)}];Print[n," ",r];Continue,{n,0,80}]

A032765 a(n) = floor(n(n+1)(n+2) / (n + n+1 + n+2)) = floor(n*(n+2)/3).

Original entry on oeis.org

0, 1, 2, 5, 8, 11, 16, 21, 26, 33, 40, 47, 56, 65, 74, 85, 96, 107, 120, 133, 146, 161, 176, 191, 208, 225, 242, 261, 280, 299, 320, 341, 362, 385, 408, 431, 456, 481, 506, 533, 560, 587, 616, 645, 674, 705, 736, 767, 800, 833, 866, 901, 936, 971, 1008

Offset: 0

Patrick De Geest, May 15 1998

Satisfies a(n+1) - 2*a(n) + a(n-1) = (2/3)*(1 + w^(n+1) + w^(2*n+2)), a(0)=0 & a(1)=1 where w is the imaginary cubic root of unity. - Robert G. Wilson v, Jun 24 2002
First differences have this pattern: (+1) +1 +1 +3 +3 +3 +5 +5 +5 +7 +7 +7 +9 +9 +9. - Alexandre Wajnberg, Dec 19 2005
In Duistermaat (2010) section 11.3 The Planar Four-Bar Link on page 516: "It follows from (4.3.2) that the number of k-periodic fibers of the QRT automorphism, counted with multiplicities, is equal to nu((tau^S)^k) = 3*n^2 - 1 when k = 3*n, 3*n^2 + 2*n when k = 3*n + 1, 3*n^2 + 4*n + 1 when k = 3*n + 2, for every integer n." - Michael Somos, Mar 14 2023
a(n-1) is the maximum number of edges in a graph with vertices labeled from 1 to n, such that there is no triangle whose sum of vertex labels is divisible by n. - Hoang Xuan Thanh, Jun 11 2025

			G.f. = x + 2*x^2 + 5*x^3 + 8*x^4 + 11*x^5 + 16*x^6 + 21*x^7 + 26*x^8 + ... - _Michael Somos_, Mar 14 2023
Let n = 5: a(5-1) = 8. Consider the graph G(5) with vertex set {1, 2, 3, 4, 5} and the edge set: E = {12, 23, 34, 45, 51, 13, 24, 35}, which contains 8 edges. The triangles in this graph are: {123}, {234}, {345}, {135}. Their vertex sums are: 1+2+3 = 6, 2+3+4 = 9, 3+4+5 = 12, 1+3+5=9. Since none of these sums are divisible by 5, the graph satisfies the required condition. If we add the edge 14, a new triangle {145} would appear, whose vertex sum is 1+4+5 = 10, divisible by 5. Similarly, if we add the edge 25, the triangle {235} would form, with vertex sum 2+3+5 = 10, also divisible by 5. - _Hoang Xuan Thanh_, Jun 11 2025

Johannes J. Duistermaat, Discrete Integrable Systems, Springer Science+Business Media, 2010.

Johannes Bader, Kobon Triangles.
Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007.
Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007. [Cached copy, with permission]
Taylor Short, The saturation number of carbon nanocones and nanotubes, arXiv:1807.11355 [math.CO], 2018.
Eric Weisstein's World of Mathematics, Kobon Triangle.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A001082, A032766, A000217, A001399.

A032765:=n->floor(n*(n+2)/3); seq(A032765(n), n=0..100); # Wesley Ivan Hurt, Dec 20 2013

Table[Floor[n (n + 1)(n + 2)/(n + (n + 1) + (n + 2))], {n, 0, 55}]
Table[Floor[n (n + 2)/3], {n, 0, 100}] (* Wesley Ivan Hurt, Dec 20 2013 *)
LinearRecurrence[{2, -1, 1, -2, 1}, {0, 1, 2, 5, 8}, 60] (* Harvey P. Dale, Jun 06 2016 *)
Table[(3 n (n + 2) + 2 Cos[2 n Pi/3] + 2 Sqrt[3] Sin[2 n Pi/3] - 2)/9, {n, 0, 20}] (* Eric W. Weisstein, Jul 27 2025 *)
CoefficientList[Series[x (-1 - 2 x^2 + x^3)/((-1 + x)^3 (1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Jul 27 2025 *)

a(n)=n*(n+2)\3 \\ Charles R Greathouse IV, Jun 11 2015

From Ralf Stephan, May 05 2004: (Start)
a(n) = n^2 - ceiling(n*(n-1)/3).
G.f.: x*(1+2x^2-x^3)/((1+x+x^2)(1-x)^3). (End)
a(n) = floor(n*(n+2)/3). - Saburo Tamura, sent by Alexandre Wajnberg, Dec 19 2005
a(3*n - 1) = 3*n^2 - 1, a(3*n) = 3*n^2 + 2*n, a(3*n + 1) = 3*n^2 + 4*n + 1. - Michael Somos, Mar 14 2023
a(n+5) = A000217(n+5) - A001399(n+3) - A001399(n). - Hoang Xuan Thanh, Jun 11 2025
Sum_{n>=1} (-1)^(n+1)/a(n) = -1/4 + (Pi/(2*sqrt(3)))*(2-cosec(Pi/sqrt(3))). - Amiram Eldar, Jun 18 2025

Name change suggested by Wesley Ivan Hurt, Dec 20 2013

A135453 a(n) = 12*n^2.

Original entry on oeis.org

0, 12, 48, 108, 192, 300, 432, 588, 768, 972, 1200, 1452, 1728, 2028, 2352, 2700, 3072, 3468, 3888, 4332, 4800, 5292, 5808, 6348, 6912, 7500, 8112, 8748, 9408, 10092, 10800, 11532, 12288, 13068, 13872, 14700, 15552, 16428, 17328, 18252, 19200, 20172, 21168, 22188

Offset: 0

Ben Paul Thurston, Dec 14 2007

Areas of perfect 4:3 rectangles (for n > 0).
Sequence found by reading the line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Semi-axis opposite to A069190 in the same spiral. - Omar E. Pol, Sep 16 2011
(x,y,z) = (-a(n), 1 + n*a(n), 1 - n*a(n)) are solutions of the Diophantine equation x^3 + 2*y^3 + 2*z^3 = 4. - XU Pingya, Apr 30 2022

			192 is on the list since 16*12 is a 4:3 rectangle with integer sides and an area of 192.

Ivan Panchenko, Table of n, a(n) for n = 0..10000
John Elias, Illustration: Even Ordered Star Perimeters.
Leo Tavares, Illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001105, A016742, A033428, A033581, A069190, A086729.

Cf. A008594, A195143.

List([0..100],n->12*n^2); # Muniru A Asiru, Jan 29 2018

seq(12*h^2,n=0..100); # Muniru A Asiru, Jan 29 2018

Table[12*n^2, {n, 0, 60}] (* Stefan Steinerberger, Dec 17 2007 *)
LinearRecurrence[{3,-3,1},{0,12,48},50] (* Harvey P. Dale, Jan 19 2020 *)

a(n)=12*n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 12*A000290(n) = 6*A001105(n) = 4*A033428(n) = 3*A016742(n) = 2*A033581(n). - Omar E. Pol, Dec 13 2008
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/72 (A086729).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/144.
Product_{n>=1} (1 + 1/a(n)) = 2*sqrt(3)*sinh(Pi/(2*sqrt(3)))/Pi.
Product_{n>=1} (1 - 1/a(n)) = 2*sqrt(3)*sin(Pi/(2*sqrt(3)))/Pi. (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 12*x*(1 + x)/(1-x)^3.
E.g.f.: 12*x*(1 + x)*exp(x).
a(n) = n*A008594(n) = A195143(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

More terms from Stefan Steinerberger, Dec 17 2007

Minor edits from Omar E. Pol, Dec 15 2008

A195849 Column 5 of array A195825. Also column 1 of triangle A195839. Also 1 together with the row sums of triangle A195839.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 5, 7, 10, 12, 13, 13, 14, 16, 21, 27, 32, 34, 36, 38, 44, 54, 67, 77, 84, 88, 95, 107, 128, 152, 174, 188, 200, 215, 242, 281, 329, 370, 402, 428, 462, 513, 589, 674, 754, 816, 873, 940, 1041, 1176, 1333, 1477, 1600, 1710, 1845

Offset: 0

Omar E. Pol, Oct 07 2011

nonn

Note that this sequence contains three plateaus: [1, 1, 1, 1, 1, 1], [4, 4, 4, 4], [13, 13]. For more information see A210843. See also other columns of A195825. - Omar E. Pol, Jun 29 2012

Number of partitions of n into parts congruent to 0, 1 or 6 (mod 7). - Ludovic Schwob, Aug 05 2021

Ludovic Schwob, Table of n, a(n) for n = 0..10000

Cf. A000041, A001082, A006950, A036820, A057077, A118277, A195825, A195829, A195839, A195848, A195850, A195851, A195852, A196933, A210843, A210964, A211971.

A118277 := proc(n)
        7*n^2/8+7*n/8-3/16+3*(-1)^n*(1/16+n/8) ;
end proc:
A195839 := proc(n, k)
        option remember;
        local ks, a, j ;
        if A118277(k) > n then
                0 ;
        elif n <= 5 then
                return 1;
        elif k = 1 then
                a := 0 ;
                for j from 1 do
                        if A118277(j) <= n-1 then
                                a := a+procname(n-1, j) ;
                        else
                                break;
                        end if;
                end do;
                return a;
        else
                ks := A118277(k) ;
                (-1)^floor((k-1)/2)*procname(n-ks+1, 1) ;
        end if;
end proc:
A195849 := proc(n)
        A195839(n+1,1) ;
end proc:
seq(A195849(n), n=0..60) ; # R. J. Mathar, Oct 08 2011

m = 61;
Product[1/((1 - x^(7k))(1 - x^(7k - 1))(1 - x^(7k - 6))), {k, 1, m}] + O[x]^m // CoefficientList[#, x]& (* Jean-François Alcover, Apr 13 2020, after Ilya Gutkovskiy *)

G.f.: Product_{k>=1} 1/((1 - x^(7*k))*(1 - x^(7*k-1))*(1 - x^(7*k-6))). - Ilya Gutkovskiy, Aug 13 2017

a(n) ~ exp(Pi*sqrt(2*n/7)) / (8*sin(Pi/7)*n). - Vaclav Kotesovec, Aug 14 2017

A014642 Even octagonal numbers: a(n) = 4n(3*n-1).

Original entry on oeis.org

0, 8, 40, 96, 176, 280, 408, 560, 736, 936, 1160, 1408, 1680, 1976, 2296, 2640, 3008, 3400, 3816, 4256, 4720, 5208, 5720, 6256, 6816, 7400, 8008, 8640, 9296, 9976, 10680, 11408, 12160, 12936, 13736, 14560, 15408, 16280, 17176, 18096, 19040, 20008, 21000, 22016

Offset: 0

Mohammad K. Azarian

8 times pentagonal numbers. - Omar E. Pol, Dec 11 2008
Sequence found by reading the line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
The sequence forms the even nesting cube-frames (see illustrations in A000567), which separate and appear according to formula along the axes on the zero-centered and one-centered hexagonal number spirals, as well as the axes of the zero-centered and one-centered square number spirals. See illustrations in links. - John Elias, Jul 20 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000
John Elias, Octagonal Nesting Cubes on the Hexagonal Number Spiral Octagonal Nesting Cubes on the Square Number Spiral
Craig Knecht, Number of positions the remaining tiles can occupy in a 4*n length polyiamond bilayer when one tile is missing.
Yaohui Zhu, Kaiming Sun, Zhengdong Luo, and Lingfeng Wang, Progressive Self-Learning for Domain Adaptation on Symbolic Regression of Integer Sequences, Proc. 39th AAAI Conf. Artif. Intel. (2025) Vol. 39, No. 1, 1692-1699. See p. 1698.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A000326, A001082, A014641, A014793, A014794, A033579, A049450.

List([0..50], n-> 8*Binomial(3*n,2)/3); # G. C. Greubel, Oct 09 2019

[8*Binomial(3*n,2)/3: n in [0..50]]; // G. C. Greubel, Oct 09 2019

seq(8*binomial(3*n,2)/3, n=0..50); # G. C. Greubel, Oct 09 2019

LinearRecurrence[{3,-3,1},{0,8,40}, 50] (* G. C. Greubel, Jun 07 2017 *)
PolygonalNumber[8,Range[0,90,2]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 19 2020 *)

vector(51, n, 8*binomial(3*(n-1),2)/3 ) \\ G. C. Greubel, Jun 07 2017

[8*binomial(3*n,2)/3 for n in (0..50)] # G. C. Greubel, Oct 09 2019

a(n) = A000326(n)*8. - Omar E. Pol, Dec 11 2008
a(n) = A049450(n)*4 = A033579(n)*2. - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 24*n - 16 (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
G.f.: x*(8+16*x)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 06 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - G. C. Greubel, Jun 07 2017
E.g.f.: 4*x*(2 + 3*x)*exp(x). - G. C. Greubel, Oct 09 2019
From Amiram Eldar, Mar 24 2021: (Start)
Sum_{n>=1} 1/a(n) = 3*log(3)/8 - Pi/(8*sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 - Pi/(4*sqrt(3)). (End)

More terms from Patrick De Geest

A152746 Six times hexagonal numbers: 6n(2*n-1).

Original entry on oeis.org

0, 6, 36, 90, 168, 270, 396, 546, 720, 918, 1140, 1386, 1656, 1950, 2268, 2610, 2976, 3366, 3780, 4218, 4680, 5166, 5676, 6210, 6768, 7350, 7956, 8586, 9240, 9918, 10620, 11346, 12096, 12870, 13668, 14490, 15336, 16206, 17100

Offset: 0

Omar E. Pol, Dec 12 2008

Sequence found by reading the line from 0, in the direction 0, 6, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Sep 18 2011

a(n) is the number of walks on a cubic lattice of n dimensions that return to the origin, not necessarily for the first time, after 4 steps. - Shel Kaphan, Mar 20 2023

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A001082, A002939, A094159, A085250, A152745.

Column n=2 of A287318.

[6*n*(2*n-1): n in [0..50]]; // G. C. Greubel, Sep 01 2018

6*PolygonalNumber[6,Range[0,40]] (* The program uses the PolygonalNumber function from Mathematica version 10 *) (* Harvey P. Dale, Mar 04 2016 *)
LinearRecurrence[{3,-3,1}, {0,6,36}, 50] (* or *) Table[6*n*(2*n-1), {n,0,50}] (* G. C. Greubel, Sep 01 2018 *)

a(n)=6*n*(2*n-1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 12*n^2 - 6*n = A000384(n)*6 = A002939(n)*3 = A094159(n)*2.
a(n) = a(n-1) + 24*n - 18 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
From G. C. Greubel, Sep 01 2018: (Start)
G.f.: 6*x*(1+3*x)/(1-x)^3.
E.g.f.: 6*x*(1+2*x)*exp(x). (End)
From Amiram Eldar, Mar 30 2023: (Start)
Sum_{n>=1} 1/a(n) = log(2)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/12 - log(2)/6. (End)

A195143 a(n) = n-th concentric 12-gonal number.

Original entry on oeis.org

0, 1, 12, 25, 48, 73, 108, 145, 192, 241, 300, 361, 432, 505, 588, 673, 768, 865, 972, 1081, 1200, 1321, 1452, 1585, 1728, 1873, 2028, 2185, 2352, 2521, 2700, 2881, 3072, 3265, 3468, 3673, 3888, 4105, 4332, 4561, 4800, 5041, 5292, 5545, 5808, 6073, 6348

Offset: 0

Omar E. Pol, Sep 17 2011

Concentric dodecagonal numbers. [corrected by Ivan Panchenko, Nov 09 2013]
Sequence found by reading the line from 0, in the direction 0, 12,..., and the same line from 1, in the direction 1, 25,..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Main axis, perpendicular to A028896 in the same spiral.
Partial sums of A091998. - Reinhard Zumkeller, Jan 07 2012
Column 12 of A195040. - Omar E. Pol, Sep 28 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A135453 and A069190 interleaved.
Cf. A001082, A032527, A032528, A077221, A195043, A195045, A195040, A195142, A195145, A195146, A195147, A195148, A195149.
Cf. A016921 (6n+1), A016969 (6n+5), A091998 (positive integers of the form 12*k +- 1), A092242 (positive integers of the form 12*k +- 5).

a195143 n = a195143_list !! n
a195143_list = scanl (+) 0 a091998_list
-- Reinhard Zumkeller, Jan 07 2012

[(3*n^2+(-1)^n-1): n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

Table[Sum[2*(-1)^(n - k + 1) + 6*k - 3, {k, n}], {n, 0, 47}] (* L. Edson Jeffery, Sep 14 2014 *)

From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = 3*n^2+(-1)^n-1.
a(n) = -a(n-1) + 6*n^2 - 6*n + 1. (End)
G.f.: -x*(1+10*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = Sum_{k=1..n} (2*(-1)^(n-k+1) + 3*(2*k-1)), n>0, a(0) = 0. - L. Edson Jeffery, Sep 14 2014
Sum_{n>=1} 1/a(n) = Pi^2/72 + tan(Pi/sqrt(6))*Pi/(4*sqrt(6)). - Amiram Eldar, Jan 16 2023

cp's OEIS Frontend

A270469 Number of ordered ways to write n = x^3 + y*(y+1) + z*(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A017569 a(n) = 12*n + 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Formula

A036666 Numbers k such that 5*k + 1 is a square.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Formula

Extensions

A270516 Number of ordered ways to write n = x^3*(x+1) + y*(y+1)/2 + z*(3z+2), where x and y are nonnegative integers, and z is an integer.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A032765 a(n) = floor(n*(n+1)*(n+2) / (n + n+1 + n+2)) = floor(n*(n+2)/3).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A135453 a(n) = 12*n^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Formula

Extensions

A195849 Column 5 of array A195825. Also column 1 of triangle A195839. Also 1 together with the row sums of triangle A195839.

Views

A270469 Number of ordered ways to write n = x^3 + y(y+1) + z(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.

A270516 Number of ordered ways to write n = x^3(x+1) + y(y+1)/2 + z*(3z+2), where x and y are nonnegative integers, and z is an integer.

A032765 a(n) = floor(n(n+1)(n+2) / (n + n+1 + n+2)) = floor(n*(n+2)/3).

A014642 Even octagonal numbers: a(n) = 4n(3*n-1).

A152746 Six times hexagonal numbers: 6n(2*n-1).