A035008 -id:A035008 - cp's OEIS frontend

A139271 a(n) = 2n(4*n-3).

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.
Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Table[8n^2-6n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,2,20},50] (* Harvey P. Dale, Sep 26 2016 *)

a(n)=2*n*(4*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 - 6*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Corrected by Harvey P. Dale, Sep 26 2016

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A060626 Number of right triangles of a given area required to form successively larger squares.

Original entry on oeis.org

2, 14, 34, 62, 98, 142, 194, 254, 322, 398, 482, 574, 674, 782, 898, 1022, 1154, 1294, 1442, 1598, 1762, 1934, 2114, 2302, 2498, 2702, 2914, 3134, 3362, 3598, 3842, 4094, 4354, 4622, 4898, 5182, 5474, 5774, 6082, 6398, 6722, 7054, 7394, 7742, 8098, 8462, 8834, 9214

Offset: 0

Jason Earls, Apr 13 2001

a(n) = number of row of Pascal's triangle in which three consecutive entries appear in the ratio n : n+1 : n+2 (valid for n = 0 if you consider a position of -1 to have value 0). E.g., entries in the ratio 1:2:3 appear in row 14 (1001, 2002, 3003); entries in the ratio 2:3:4 appear in row 34 (927983760, 1391975640, 1855967520); and so on. (The position within the row is given by A091823). - Howard A. Landman, Mar 08 2004
a(n)*(a(n)+1) is an oblong number (Cf. A002378) with the property that the product with the oblong numbers n*(n+1) or (n+1)*(n+2) both are again oblong numbers. Example: For n=3 we have (62*63)*(3*4) = 216*217 and (62*63)*(4*5) = 279*280. - Herbert Kociemba, Apr 13 2008
For n > 0, Hermite polynomial H_2(n) = 4*n^2 - 2. - Vincenzo Librandi, Aug 07 2010
The identity (4*n^2-2)^2 - (n^2-1)*(4*n)^2 = 4 can be written as a(n+1)^2 - A132411(n+2)*A008586(n+2)^2 = 4. - Vincenzo Librandi, Jun 16 2014
Equivalently: positive integers k congruent to 2 mod 4 (A016825) such that k$ / (k/2+1)! is a square when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692, A349496 and A349766 for further information). Integers k multiple of 4 such that that k$ / (k/2+1)! is a square are in A035008. - Bernard Schott, Dec 05 2021

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

Cf. A000178, A002378, A007318, A008586, A016777, A035008, A056220, A091823.
Cf. A132411, A348692.
Twice Column 2 of array A188644.
Subsequence of A016825.
Equals disjoint union of A349496 and A349766.

for n from 0 to 80 do printf(`%d,`,4*n^2+8*n+2) od:

Table[4*n*(n + 2) + 2, {n, 0, 100}] (* Paolo Xausa, Aug 08 2024 *)

a(n) = { 4*n^2 + 8*n + 2 } \\ Harry J. Smith, Jul 08 2009

a(n) = 4*n^2 + 8*n + 2.
a(n) = 8*n + a(n-1) + 4 with n > 0, a(0)=2. - Vincenzo Librandi, Aug 07 2010
G.f.: 2*(1 + 4*x - x^2)/(1-x)^3. - Colin Barker, Jun 28 2012
a(n) = 4*(n+1)^2 - 2 = 2*A056220(n+1). - Bruce J. Nicholson, Aug 31 2017
a(n) + a(n-1) + (n-1)^2 = (3*n + 1)^2 = A016777(n)^2. - Ezhilarasu Velayutham, May 23 2019
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: 2*exp(x)*(1 + 6*x + 2*x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

More terms from James Sellers, Apr 14 2001

A139278 a(n) = n(8n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., in the square spiral whose vertices are the triangular numbers A000217.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139276, A139277, A139279, A139280, A139281, A139282.

Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)

a(n)=n*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 7*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022

A139272 a(n) = n(8n-5).

Original entry on oeis.org

0, 3, 22, 57, 108, 175, 258, 357, 472, 603, 750, 913, 1092, 1287, 1498, 1725, 1968, 2227, 2502, 2793, 3100, 3423, 3762, 4117, 4488, 4875, 5278, 5697, 6132, 6583, 7050, 7533, 8032, 8547, 9078, 9625, 10188, 10767, 11362, 11973, 12600

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 3, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139276 in the same spiral.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(n*k-k+6)/2, this sequence is the case k=16: see Comments lines of A226492.

s=0;lst={s};Do[s+=n++ +3;AppendTo[lst, s], {n, 0, 7!, 16}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
Table[n*(8*n -5), {n,0,50}] (* G. C. Greubel, Jul 18 2017 *)
LinearRecurrence[{3,-3,1},{0,3,22},50] (* Harvey P. Dale, Jan 13 2024 *)

a(n)=n*(8*n-5) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 8*n^2 - 5*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 13 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(13*x + 3)/(1-x)^3.
E.g.f.: (8*x^2 + 3*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = ((sqrt(2)-1)*Pi + 8*log(2) - 2*sqrt(2)*log(sqrt(2)+1))/10. - Amiram Eldar, Mar 17 2022

A035006 Number of possible rook moves on an n X n chessboard.

Original entry on oeis.org

0, 8, 36, 96, 200, 360, 588, 896, 1296, 1800, 2420, 3168, 4056, 5096, 6300, 7680, 9248, 11016, 12996, 15200, 17640, 20328, 23276, 26496, 30000, 33800, 37908, 42336, 47096, 52200, 57660, 63488, 69696, 76296, 83300, 90720, 98568, 106856

Offset: 1

Ulrich Schimke (ulrschimke(AT)aol.com)

Obviously A035005(n) = A002492(n-1) + a(n) since Queen = Bishop + Rook. - Johannes W. Meijer, Feb 04 2010

X values of solutions of the equation: (X-Y)^3-2*X*Y=0. Y values are b(n)=2*n*(n-1)^2 (see A181617). - Mohamed Bouhamida, Jul 06 2023

			On a 3 X 3-board, rook has 9*4 moves, so a(3)=36.

E. Bonsdorff, K. Fabel and O. Riihimaa, Schach und Zahl (Chess and numbers), Walter Rau Verlag, Dusseldorf, 1966.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Alexander M. Haupt, Bijective enumeration of rook walks, arXiv:2007.01018 [math.CO], 2020.
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
M. Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Richard P. Stanley, Bijective Proof Problems, Problem 540 p. 63, (2015).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A033586 (King), A035005 (Queen), A035008 (Knight), A002492 (Bishop) and A049450 (Pawn).

Cf. A006002, A006564, A006566.

[(n-1)*2*n^2: n in [1..40]]; // Vincenzo Librandi, Jun 16 2011

Table[(n-1) 2 n^2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,8,36,96},40] (* Harvey P. Dale, May 12 2012 *)

a(n) = (n-1)*2*n^2.
a(n) = Sum_{j=1..n} ((n+j-1)^2 - (n-j+1)^2). - Zerinvary Lajos, Sep 13 2006
1/a(n+1) = Integral_{x=1/(n+1)..1/n} x*h(x) = Integral_{x=1/(n+1)..1/n} x*(1/x - floor(1/x)) = 1/((2*(n^2+2*n+1))*n) and Sum_{n>=1} 1/((2*(n^2+2*n+1))*n) = 1-Zeta(2)/2 where h(x) is the Gauss (continued fraction) map h(x)={x^-1} and {x} is the fractional part of x. - Stephen Crowley, Jul 24 2009
a(n) = 4 * A006002(n-1). - Johannes W. Meijer, Feb 04 2010
G.f.: 4*x^2*(2+x)/(1-x)^4. - Colin Barker, Mar 11 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=0, a(2)=8, a(3)=36, a(4)=96. - Harvey P. Dale, May 12 2012
a(n) = A006566(n) - A006564(n). - Peter M. Chema, Feb 10 2016
E.g.f.: 2*exp(x)*x^2*(2 + x). - Stefano Spezia, May 10 2022
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=2} 1/a(n) = 1 - Pi^2/12.
Sum_{n>=2} (-1)^n/a(n) = Pi^2/24 + log(2) - 1. (End)

A139274 a(n) = n(8n-1).

Original entry on oeis.org

0, 7, 30, 69, 124, 195, 282, 385, 504, 639, 790, 957, 1140, 1339, 1554, 1785, 2032, 2295, 2574, 2869, 3180, 3507, 3850, 4209, 4584, 4975, 5382, 5805, 6244, 6699, 7170, 7657, 8160, 8679, 9214, 9765, 10332, 10915, 11514, 12129, 12760, 13407, 14070, 14749

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the triangular numbers A000217.

Polygonal number connection: 2*P_n + 5*S_n where P_n is the n-th pentagonal number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010

			a(1) = 16*1 + 0 - 9 = 7; a(2) = 16*2 + 7 - 9 = 30; a(3) = 16*3 + 30 - 9 = 69. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

[n*(8*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

A139274:=n->n*(8*n-1): seq(A139274(n), n=0..100); # Wesley Ivan Hurt, Dec 04 2016

CoefficientList[Series[x (9 x + 7)/(1 - x)^3, {x, 0, 43}], x] (* Michael De Vlieger, Jan 11 2020 *)
Table[n(8n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,30},50] (* Harvey P. Dale, Apr 01 2024 *)

a(n)=n*(8*n-1) \\ Charles R Greathouse IV, Jun 17 2017

Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = (1/3) * Sum_{i=n..(7*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(9*x+7)/(1-x)^3.
E.g.f.: (8*x^2 + 7*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 4*log(2) + sqrt(2)*log(sqrt(2)+1) - (sqrt(2)+1)*Pi/2. - Amiram Eldar, Mar 18 2022

A035005 Number of possible queen moves on an n X n chessboard.

Original entry on oeis.org

0, 12, 56, 152, 320, 580, 952, 1456, 2112, 2940, 3960, 5192, 6656, 8372, 10360, 12640, 15232, 18156, 21432, 25080, 29120, 33572, 38456, 43792, 49600, 55900, 62712, 70056, 77952, 86420, 95480, 105152, 115456, 126412, 138040, 150360

Offset: 1

Ulrich Schimke (ulrschimke(AT)aol.com)

The number of (2 to n) digit sequences that can be found reading in any orientation, including diagonals, in an (n X n) grid. - Paul Cleary, Aug 12 2005

			3 X 3 board: queen has 8*6 moves and 1*8 moves, so a(3)=56.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A033586 (King), A035006 (Rook), A035008 (Knight), A002492 (Bishop) and A049450 (Pawn).

Cf. A162147.

[(n-1)*2*n^2 + (4*n^3-6*n^2+2*n)/3: n in [1..40]]; // Vincenzo Librandi, Jun 16 2011

Table[(n-1)2n^2+(4n^3-6n^2+2n)/3,{n,40}] (* or *) LinearRecurrence[ {4,-6,4,-1},{0,12,56,152},40] (* Harvey P. Dale, Aug 24 2011 *)

a(n) = (n-1)*2*n^2 + (4*n^3-6*n^2+2*n)/3.
From Johannes W. Meijer, Feb 04 2010: (Start)
a(n) = A002492(n-1) + A035006(n) since Queen = Bishop + Rook.
a(n) = 4 * A162147(n-1). (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=12, a(2)=56, a(3)=152. - Harvey P. Dale, Aug 24 2011
From Colin Barker, Mar 11 2012: (Start)
a(n) = 2*n*(1-6*n+5*n^2)/3.
G.f.: 4*x^2*(3+2*x)/(1-x)^4. (End)
E.g.f.: 2*exp(x)*x^2*(9 + 5*x)/3. - Stefano Spezia, Jul 31 2022

More terms from Erich Friedman

A139276 a(n) = n(8n+3).

Original entry on oeis.org

0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.

			a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139277, A139278, A139279, A139280, A139281, A139282.

a[n_]:=n*(8*n+3); a[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3,-3,1},{0,11,38},50] (* Harvey P. Dale, Jul 02 2022 *)

a(n)=n*(8*n+3) \\ Charles R Greathouse IV, Jun 16 2017

a(n) = 8*n^2 + 3*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022

A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!2!...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

Original entry on oeis.org

1, 2, 0, 2, 0, 0, 0, 3, 4, 0, 0, 0, 6, 0, 8, 9, 0, 8, 9, 0, 7, 0, 10, 0, 0, 0, 12, 0, 0, 0, 14, 0, 0, 0, 15, 16, 0, 18, 0, 18, 0, 0, 0, 20, 0, 0, 0, 22, 0, 0, 0, 24, 25, 0, 0, 0, 26, 0, 0, 0, 28, 0, 0, 0, 30, 0, 32, 0, 32, 0, 0, 0, 34, 0, 0, 0, 35, 36, 0, 0, 0, 38, 0, 0, 0, 40

Offset: 1

Bernard Schott, Oct 30 2021

This sequence is the generalization of a problem proposed during the 17th Tournament of Towns (Spring 1996) and also during the first stage of the Moscow Mathematical Olympiad (1995-1996); the problem asked the question for n = 100 (see Andreescu-Gelca reference, Norman Do link, and Examples section).
Exhaustive results coming from Mabry-McCormick's link and adapted for OEIS:
-> n$ (A000178) is never a square if n > 1.
-> There is no solution if n is odd > 1, hence row(2q+1) = 0 when q > 0.
-> When n is even and there is a solution, then m belongs to {n/2 - 2, n/2 - 1, n/2, n/2 + 1, n/2 + 2}.
-> If 4 divides n (A008536), m = n/2 is always a solution because
    (n$) / (n/2)! = ( 2^(n/4) * Product_{j=1..n/2} ((2j-1)!) )^2.
-> For other cases, see Formula section.
-> When n is even, there are 0, 1 or 2 solutions, so, the maximal length of a row is 2.
-> It is not possible to get more than three consecutive 0 terms, and three consecutive 0 terms correspond to three consecutive rows such that (n, n+1, n+2) = (4u+1, 4u+2, 4u+3) for some u >= 1.

			For n = 4, 4$ / 3! = 48, 4$ / 4! = 12 but 4$ / 2! = 12^2, hence, m = 2.
For n = 8, 8$ / 2! is not a square, but m_1 = 3 because 8$ / 3! = 29030400^2 and m_2 = 4 because 8$ / 4! = 14515200^2.
For n = 14, m_1 = 8 because 14$ / 8! = 1309248519599593818685440000000^2 and m_2 = 9 because 14$ / 9! = 436416173199864606228480000000^2.
For n = 16, m_1 = 8 because 16$ / 8! = 6848282921689337839624757371207680000000000^2 and m_2 = 9 because 16$ / 9! = 2282760973896445946541585790402560000000000^2.
For n = 18, m = 7 because 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2 and there is no other solution.
For n = 100, m = 50, unique solution to the Olympiad problems.
Triangle begins:
    1;
    2;
    0;
    2;
    0;
    0;
    0;
    8,  9;
    0;
    ...

Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 725, pp. 253 and 686.
Peter J. Taylor and A. M. Storozhev, Tournament of Towns 1993-1997, Book 4, Tournament 17, Spring 1996, O Level, Senior questions, Australian Mathematics Trust, 1998, problem 3, p. 96.

Diophante, A1963 - Le vilain petit canard (in French).
Norman Do, Factorial fun, Puzzle Corner 13, Gaz. Aust. Math. Soc. 36, 2009, 176-179, page 178.
Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.
Tournament of Towns, Tournament 17, 1995-1996, Spring 1996, O Level, Senior questions, question 3 (in Russian).
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A000178, A001541, A008536, A035008, A060626, A139098.

sf(n)=prod(k=2, n, k!); \\ A000178
row(n) = my(s=sf(n)); Vec(select(issquare, vector(n, k, s/k!), 1));
lista(nn) = {my(list = List()); for (n=1, nn, my(r=row(n)); if (#r, for (k=1, #r, listput(list, r[k])), listput(list, 0));); Vec(list);} \\ Michel Marcus, Oct 30 2021

When there are two such integers m, then m_1 < m_2.
If n = 8*q^2 (A139098), then m_1 = n/2 - 1 = 4q^2-1 (see example for n=8).
If n = 8q*(q+1) (A035008), then m_2 = n/2 + 1 = (2q+1)^2 (see example for n=16).
if n = 4q^2 - 2 (A060626), then m_1 = n/2 + 1 = 2q^2 (see example for n=14).
If n = 2q^2, q>1 in A001541, then m = n/2 - 2 = q^2-2 (see example for n=18).
If n = 2q^2-4, q>1 in A001541, then m_2 = n/2 + 2 = q^2 (see example for n=14).

cp's OEIS Frontend

A139271 a(n) = 2*n*(4*n-3).

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A139273 a(n) = n*(8*n - 3).

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Magma

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A060626 Number of right triangles of a given area required to form successively larger squares.

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Maple

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A139278 a(n) = n*(8*n+7).

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A139272 a(n) = n*(8*n-5).

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A035006 Number of possible rook moves on an n X n chessboard.

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A139274 a(n) = n*(8*n-1).

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A139271 a(n) = 2n(4*n-3).

A139273 a(n) = n(8n - 3).

A139278 a(n) = n(8n+7).

A139272 a(n) = n(8n-5).

A139274 a(n) = n(8n-1).

A139276 a(n) = n(8n+3).

A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!2!...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.