A078371 -id:A078371 - cp's OEIS frontend

A171621 Numerator of 1/4 - 1/n^2, each fourth term multiplied by 4.

0, 5, 3, 21, 8, 45, 15, 77, 24, 117, 35, 165, 48, 221, 63, 285, 80, 357, 99, 437, 120, 525, 143, 621, 168, 725, 195, 837, 224, 957, 255, 1085, 288, 1221, 323, 1365, 360, 1517, 399, 1677, 440, 1845, 483, 2021, 528

Offset: 2

Paul Curtz, Dec 13 2009

These are the square roots of the fifth column of the array of denominators mentioned in A171522.

Colin Barker, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A005563, A010121, A061037, A078371, A171522.

[-(-5+3*(-1)^n)*(-4+n^2)/8: n in [0..100]]; // G. C. Greubel, Sep 19 2018

A061037 := proc(n) 1/4-1/n^2 ; numer(%) ; end proc:
A171621 := proc(n) if n mod 4 = 2 then 4*A061037(n) ; else A061037(n) ; end if; end proc:
seq(A171621(n),n=2..90) ; # R. J. Mathar, Apr 02 2011

Table[-(-5+3*(-1)^n)*(-4+n^2)/8, {n,0,100}] (* G. C. Greubel, Sep 19 2018 *)
LinearRecurrence[{0,3,0,-3,0,1},{0,5,3,21,8,45},50] (* Harvey P. Dale, Nov 01 2019 *)

concat(0, Vec(x^3*(-5-3*x-6*x^2+x^3+3*x^4)/((x-1)^3*(1+x)^3) + O(x^100))) \\ Colin Barker, Nov 03 2014

a(n) = A061037(n) * A010121(n+2).
a(2n+2) = A005563(n). a(2n+3) = A078371(n).
G.f.: x^3*(-5-3*x-6*x^2+x^3+3*x^4) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Apr 02 2011
a(n) = -(-5+3*(-1)^n)*(-4+n^2)/8. - Colin Barker, Nov 03 2014
Sum_{n>=3} 1/a(n) = 13/12. - Amiram Eldar, Aug 11 2022

A003185 a(n) = (4n+1)(4*n+5).

Original entry on oeis.org

5, 45, 117, 221, 357, 525, 725, 957, 1221, 1517, 1845, 2205, 2597, 3021, 3477, 3965, 4485, 5037, 5621, 6237, 6885, 7565, 8277, 9021, 9797, 10605, 11445, 12317, 13221, 14157, 15125, 16125, 17157, 18221

Offset: 0

N. J. A. Sloane

Bisection of A078371. - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 19 2004
a(n) is the smallest number not in the sequence such that Sum_{k=0..n} 1/a(k) has a denominator 4*n+5. - Derek Orr, Jun 21 2015
a(n) is the number of 2 X 2 matrices with all elements in {-n,..,0,..,n} with permanent = determinant^n except for a(0), where a(0)=0, but A003185(0) = 5. - Indranil Ghosh, Jan 04 2017

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

Cf. A001513, A003185, A063164, A078371.

Table[(4n+1)(4n+5),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{5,45,117},40] (* Harvey P. Dale, Jan 27 2013 *)

a(n) = (4*n+1)*(4*n+5); \\ Michel Marcus, Jan 17 2023

a= lambda n: (4*n+1)*(4*n+5) # Indranil Ghosh, Jan 04 2017

1 = Sum_{n>=0} 4/a(n). Sum_{k=0..n} 4/a(k) = 4(n+1)/[4(n+1)+1]. Integral_{x=0..1} 1/(1 + x^4) = Sum_{n>=0} 4/a(2*n) = Sum_{n>=0} (-1)^n/(4n+1). - Gary W. Adamson, Jun 18 2003
1 = 1/5 + Sum_{n>=1} 16/a(n); with partial sums (4n+1)/(4n+5). - Gary W. Adamson, Jun 18 2003
From R. J. Mathar, Apr 04 2008: (Start)
O.g.f.: (-5-30*x+3*x^2)/(-1+x)^3.
a(3*n) = A001513(2*n).
Conjecture: a(n+1)-a(n) = A063164(n+2). (End)
a(n) = 32*n + a(n-1) + 8 (with a(0)=5). - Vincenzo Librandi, Nov 12 2010
a(0)=5, a(1)=45, a(2)=117, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jan 27 2013
Sum_{n>=0} (-1)^n/a(n) = (log(2*sqrt(2)+3) + Pi)/(8*sqrt(2)) - 1/4. - Amiram Eldar, Oct 08 2023

A142717 First (leftmost) odd term in the n-th row of triangle A120070.

Original entry on oeis.org

3, 5, 15, 21, 35, 45, 63, 77, 99, 117, 143, 165, 195, 221, 255, 285, 323, 357, 399, 437, 483, 525, 575, 621, 675, 725, 783, 837, 899, 957, 1023, 1085, 1155, 1221, 1295, 1365, 1443, 1517, 1599, 1677, 1763, 1845, 1935, 2021, 2115, 2205, 2303, 2397, 2499, 2597

Offset: 1

Paul Curtz, Sep 26 2008

nonn

Also: Records sequence of A100181.

The last (rightmost) term in the n-th row of triangle A120070 is A005408(n).

			The odd terms of A120070 build the irregular triangle
  3;
  5;
  15,7;
  21,9;
  35,27,11;
  45,33,13;
  63,55,39,15;
The leftmost column defines this sequence.

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

Cf. A000466, A078371, A100181.

A142717[n_]:=(n+1)^2-If[OddQ[n],1,4];Array[A142717,100] (* or *)
LinearRecurrence[{2,0,-2,1},{3,5,15,21},100] (* Paolo Xausa, Dec 05 2023 *)

First differences: a(n+1)-a(n) = A142954(n).
From R. J. Mathar, Oct 24 2008: (Start)
a(n) = (n+1)^2-1 = A000466((n+1)/2) if n odd.
a(n) = (n+1)^2-4 = A078371(n/2-1) if n even.
a(n) = 2*a(n-1) -2*a(n-3) +a(n-4).
G.f.: x(3-x+5x^2-3x^3)/((1+x)(1-x)^3). (End)

Edited and extended by R. J. Mathar, Oct 24 2008

A085027 a(n) = (4n+3)(4*n+7).

Original entry on oeis.org

21, 77, 165, 285, 437, 621, 837, 1085, 1365, 1677, 2021, 2397, 2805, 3245, 3717, 4221, 4757, 5325, 5925, 6557, 7221, 7917, 8645, 9405, 10197, 11021, 11877, 12765, 13685, 14637, 15621, 16637, 17685, 18765, 19877, 21021, 22197, 23405, 24645, 25917, 27221, 28557, 29925, 31325, 32757, 34221

Offset: 0

Gary W. Adamson, Jun 19 2003

1 = 3/7 + Sum_{n>=1} 16/a(n) = 3/7 + 16/77 + 16/165 + 16/285...+...; with partial sums: 3/7, 7/11, 11/15, 15/19, 19/23, ...(4n+3)/(4n+7), ... ==> 1.
With A003185(n) = (4*n+1)*(4*n+5), a bisection of A078371(n) which is a bisection of A061037(n+2).
A quadrisection of A061037(n+2). After A002378(n), A003185(n) and A000466(n+1). - Paul Curtz, Mar 30 2011

			21 = (3)(7), 77 = (7)(11), 165 = (11)(15), 285 = (15)(19), 437 = (19)(23)...

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000466, A002378, A003185, A061037, A078371.

[16*n^2+40*n+21: n in [0..35]]; // Vincenzo Librandi, Aug 13 2011

Mathematica
```
Table[(4*n + 3) (4*n + 7), {n, 0, 45}]
```

a(n)=(4*n+3)*(4*n+7) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 16*n^2+40*n+21. - Vincenzo Librandi, Aug 13 2011
From Colin Barker, Jul 11 2012: (Start)
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
G.f.: (21+14*x-3*x^2)/(1-x)^3. (End)
E.g.f.: (21 +56*x +16*x^2)*exp(x). - G. C. Greubel, Sep 20 2018
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=0} 1/a(n) = 1/12.
Sum_{n>=0} (-1)^n/a(n) = Pi/(8*sqrt(2)) + log(sqrt(2)-1)/(4*sqrt(2)) - 1/12. (End)

A247829 a(3k) = k(k+1), a(3k+1) = (2k-1)(2k+1), a(3k+2) = (2k-1)(2k+3).

Original entry on oeis.org

0, -1, -3, 2, 3, 5, 6, 15, 21, 12, 35, 45, 20, 63, 77, 30, 99, 117, 42, 143, 165, 56, 195, 221, 72, 255, 285, 90, 323, 357, 110, 399, 437, 132, 483, 525, 156, 575, 621, 182, 675, 725, 210, 783, 837, 240, 899, 957, 272, 1023, 1085, 306, 1155, 1221, 342, 1295

Offset: 0

Paul Curtz, Dec 01 2014

A permutation of A061037(n) = -1, -3, 0, 5, 3, 21, 2, 45, 15, 77, 6, ... and of A214297(n) = -1, 0, -3, 2, 3, 6, 5, 12, ... .
Among consequences: b(n) = 4*a(n) + (sequence of period 3:repeat 1, 4, 16) = 1, 0, 4, 9, 16, 36, 25, 64, 100, ... , is a permutation of the squares of the nonnegative integers A000290(n).
And a(n)/b(n) = 0/1, -1/0, -3/4, 2/9, 3/16, 5/36, ... is a permutation of the Balmer series A061037(n)/A061038(n) = -1/0, -3/4, 0/1, 5/36, 3/16, ... .
a(5n) is divisible by 5.

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

Cf. A142717 (2 terms out of 3), A000290, A000466, A002378, A061037/A061038, A078371, A214297.

m:=50; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3*x^7-3*x^6-14*x^4-6*x^3-2*x^2+3*x+1)/((x-1)^3*(x^2 +x+1)^3))); // G. C. Greubel, Sep 20 2018

seq(op([k*(k+1),(2*k-1)*(2*k+1),(2*k-1)*(2*k+3)]), k=0..100); # Robert Israel, Dec 01 2014

Table[Sequence @@ {n*(n+1), (2*n-1)*(2*n+1), (2*n-1)*(2*n+3)}, {n, 0, 18}] (* Jean-François Alcover, Dec 16 2014 *)

concat(0, Vec(x*(3*x^7-3*x^6-14*x^4-6*x^3-2*x^2+3*x+1)/((x-1)^3*(x^2+x+1)^3) + O(x^100))) \\ Colin Barker, Dec 02 2014

a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9).
a(3*k) + a(3*k+1) + a(3*k+2) = 9*k^2 + 5*k - 4.
G.f.: x*(3*x^7 - 3*x^6 - 14*x^4 - 6*x^3 - 2*x^2 + 3*x + 1)/((x-1)^3*(x^2 +x+1)^3). - Robert Israel, Dec 01 2014
a(n) = -(n^2 + n + floor(n/3)*(27*floor(n/3)^3 - 18*(n+1)*floor(n/3)^2 + (3*n^2 + 21*n - 14)*floor(n/3) - (5*n^2 - n + 5)))/2. - Luce ETIENNE, Mar 13 2017
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/2.
Sum_{n>=1} (-1)^n/a(n) = Pi/4 + 1 - 2*log(2). (End)

A265056 Partial sums of A234275.

Original entry on oeis.org

1, 5, 21, 45, 77, 117, 165, 221, 285, 357, 437, 525, 621, 725, 837, 957, 1085, 1221, 1365, 1517, 1677, 1845, 2021, 2205, 2397, 2597, 2805, 3021, 3245, 3477, 3717, 3965, 4221, 4485, 4757, 5037, 5325, 5621, 5925, 6237, 6557, 6885, 7221, 7565, 7917, 8277, 8645, 9021, 9405, 9797, 10197, 10605, 11021

Offset: 0

N. J. A. Sloane, Dec 28 2015

The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 899", based on the 5-celled von Neumann neighborhood, initiated with a single black (ON) cell. - Robert Price, May 28 2016

For n > 0, a(n) is the number of unit squares visible when two (2n-1) X (2n+1) grids are placed perpendicular to each other with their centers aligned. - Kiran Ananthpur Bacche, Sep 04 2025

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Colin Barker, Table of n, a(n) for n = 0..1000
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A234275.

Apart from initial term, same as A078371.

Accumulate[LinearRecurrence[{2,-1},{1,4,16,24},60]] (* or *) LinearRecurrence[{3,-3,1},{1,5,21,45},60] (* Harvey P. Dale, Sep 22 2024 *)

Vec((1+2*x+9*x^2-4*x^3)/(1-x)^3 + O(x^100)) \\ Colin Barker, Jan 01 2016

From Colin Barker, Jan 01 2016: (Start)
a(n) = 4*n^2+4*n-3 for n>0.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>3.
G.f.: (1+2*x+9*x^2-4*x^3) / (1-x)^3.
(End)

A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).

Original entry on oeis.org

0, -15, -3, 2, 3, 6, -7, 5, 12, 15, 20, 9, 21, 30, 35, 42, 33, 45, 56, 63, 72, 65, 77, 90, 99, 110, 105, 117, 132, 143, 156, 153, 165, 182, 195, 210, 209, 221, 240, 255, 272, 273, 285, 306, 323, 342, 345, 357, 380, 399, 420, 425, 437

Offset: 0

Paul Curtz, Jun 05 2013

The sequence is the fifth row of the following array:
0,   6,  20, 42, 72, 110, 156, 210, 272, ...   A002943
0,   3,   6, 15, 20,  35,  42,  63,  72, ...   bisections A002943, A000466
0,   2,   3,  6, 12,  15,  20,  30,  35, ...   A226023 (trisections A002943, A000466, A002439)
0,  -3,   2,  3,  6,   5,  12,  15,  20, ...   A214297 (quadrisections A078371)
0, -15,  -3,  2,  3,   6,  -7,   5,  12, ...   a(n)
0, -63, -15, -3,  2,   3,   6, -55,  -7, ...
The principle of construction is that (i) the lower left triangular portion has constant values down the diagonals (6, 3, 2, -3, -15, ...), defined from row 4 on by the negated values of A024036. (ii) The extension along the rows is defined by maintaining bisections, trisections, quadrisections etc of the form (2*n+x)*(2*n+y) with some constants x and y. In the fifth line this needs the quintisections shown in the NAME.
Each row in the array has the subsequences of the previous row plus another subsequence of the format (2*n+1)*(2*n+y) shuffled in; the first A002943, the second also A000466, the third also A002439, the fourth also A078371, and the fifth (2*n+3)*(2*n-5).
Only the first three rows are monotonically increasing everywhere.
a(n) is divisible by A226203(n).
Numerators of: 0, -15/4, -3/4, 2/9, 3/16, 6/25, -7/36, 5/36, 12/49, 15/64, 20/81, ... = a(n)/A226096(n). A permutation of A225948(n+1)/A226008(n+1).
Is the sequence increasing monotonically from 221 on?

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2,-2,0,0,0,-1,1).

Cf. A000466, A002439, A002939, A002943, A024036, A078371, A145923.

Cf. A214297, A225948, A226008, A226023, A226096, A226203.

R:=PowerSeriesRing(Integers(), 60); [0] cat Coefficients(R!( -x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9)/((1-x^5)^2*(1-x)) )); // G. C. Greubel, Mar 23 2024

CoefficientList[Series[x*(15 - 12*x - 5*x^2 - x^3 - 3*x^4 - 17*x^5 + 12*x^6 + 3*x^7 - x^8 + x^9)/((x^4 + x^3 + x^2 + x + 1)^2*(x - 1)^3), {x, 0, 80}], x] (* Wesley Ivan Hurt, Oct 03 2017 *)

def A226379_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( -x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9)/((1-x^5)^2*(1-x)) ).list()
A226379_list(50) # G. C. Greubel, Mar 23 2024

4*a(n) = A226096(n) - period 5: repeat [1, 64, 16, 1, 4].
G.f.: x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9) / ( (x^4+x^3+x^2+x+1)^2 *(x-1)^3 ). - R. J. Mathar, Jun 13 2013
a(n) = a(n-1)+2*a(n-5)-2*a(n-6)-a(n-10)+a(n-11) for n > 10. - Wesley Ivan Hurt, Oct 03 2017

A273408 Partial sums of the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 675", based on the 5-celled von Neumann neighborhood.

Original entry on oeis.org

1, 6, 27, 72, 149, 266, 431, 652, 937, 1294, 1731, 2256, 2877, 3602, 4439, 5396, 6481, 7702, 9067, 10584, 12261, 14106, 16127, 18332, 20729, 23326, 26131, 29152, 32397, 35874, 39591, 43556, 47777, 52262, 57019, 62056, 67381, 73002, 78927, 85164, 91721, 98606

Offset: 0

Robert Price, May 21 2016

Initialized with a single black (ON) cell at stage zero.

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Robert Price, Table of n, a(n) for n = 0..128
Robert Price, Diagrams of the first 20 stages
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata

Cf. A078371.

CAStep[rule_,a_]:=Map[rule[[10-#]]&,ListConvolve[{{0,2,0},{2,1,2},{0,2,0}},a,2],{2}];
code=675; stages=128;
rule=IntegerDigits[code,2,10];
g=2*stages+1; (* Maximum size of grid *)
a=PadLeft[{{1}},{g,g},0,Floor[{g,g}/2]]; (* Initial ON cell on grid *)
ca=a;
ca=Table[ca=CAStep[rule,ca],{n,1,stages+1}];
PrependTo[ca,a];
(* Trim full grid to reflect growth by one cell at each stage *)
k=(Length[ca[[1]]]+1)/2;
ca=Table[Table[Part[ca[[n]][[j]],Range[k+1-n,k-1+n]],{j,k+1-n,k-1+n}],{n,1,k}];
on=Map[Function[Apply[Plus,Flatten[#1]]],ca] (* Count ON cells at each stage *)
Table[Total[Part[on,Range[1,i]]],{i,1,Length[on]}] (* Sum at each stage *)

Conjectures from Colin Barker, May 22 2016: (Start)
a(n) = (4*n^3+12*n^2-n+3)/3.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.
G.f.: (1+2*x+9*x^2-4*x^3) / (1-x)^4. (End)

A069079 a(n) = (2n+1)(2n+2)(2n+4)(2*n+5).

Original entry on oeis.org

40, 504, 2160, 6160, 14040, 27720, 49504, 82080, 128520, 192280, 277200, 387504, 527800, 703080, 918720, 1180480, 1494504, 1867320, 2305840, 2817360, 3409560, 4090504, 4868640, 5752800, 6752200, 7876440, 9135504, 10539760, 12099960, 13827240, 15733120, 17829504

Offset: 0

Benoit Cloitre, Apr 05 2002

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 (5,-10,10,-5,1).

Cf. A033996, A078371.

Table[16n^4+96n^3+196n^2+156n+40,{n,0,40}]

my(x='x+O('x^32)); Vec(-8*(5+38*x+5*x^2)/(x-1)^5) \\ Elmo R. Oliveira, Aug 28 2025

Sum_{n>=1} 1/a(n) = 1/36 Sum_{n>=1} (-1)^n/a(n) = 5/36 - log(2)/6.
From Elmo R. Oliveira, Aug 28 2025: (Start)
G.f.: 8*(5 + 38*x + 5*x^2)/(1 - x)^5.
E.g.f.: 4*exp(x)*(10 + 116*x + 149*x^2 + 48*x^3 + 4*x^4).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A078371(n)*A033996(n+1). (End)

More terms from Elmo R. Oliveira, Aug 28 2025

A144483 A144481(4n-3).

Original entry on oeis.org

5, 0, 6, 5, 6, 0, 5, 3, 3, 5, 0, 6, 5, 6, 0, 5, 3, 3, 5, 0, 6, 5, 6, 0, 5, 3, 3, 5, 0, 6, 5, 6, 0, 5, 3, 3, 5, 0, 6, 5, 6, 0, 5, 3, 3, 5, 0, 6, 5, 6, 0, 5, 3, 3

Offset: 1

Paul Curtz, Oct 12 2008

nonn

Cf. A078371.

Period 9: a(n+9)=a(n).

Edited by R. J. Mathar, Dec 08 2008

cp's OEIS Frontend

A171621 Numerator of 1/4 - 1/n^2, each fourth term multiplied by 4.

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A003185 a(n) = (4*n+1)*(4*n+5).

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A142717 First (leftmost) odd term in the n-th row of triangle A120070.

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A085027 a(n) = (4*n+3)*(4*n+7).

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A247829 a(3*k) = k*(k+1), a(3*k+1) = (2*k-1)*(2*k+1), a(3*k+2) = (2*k-1)*(2*k+3).

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A265056 Partial sums of A234275.

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A226379 a(5n) = 2*n*(2*n+1), a(5n+1) = (2*n-3)*(2*n+5), a(5n+2) = (2*n-1)*(2*n+3), a(5n+3) = (2*n+2)*(2*n+1), a(5n+4) = (2*n+1)*(2*n+3).

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A003185 a(n) = (4n+1)(4*n+5).

A085027 a(n) = (4n+3)(4*n+7).

A247829 a(3k) = k(k+1), a(3k+1) = (2k-1)(2k+1), a(3k+2) = (2k-1)(2k+3).

A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).

A069079 a(n) = (2n+1)(2n+2)(2n+4)(2*n+5).