A139098 -id:A139098 - cp's OEIS frontend

A195323 a(n) = 22*n^2.

0, 22, 88, 198, 352, 550, 792, 1078, 1408, 1782, 2200, 2662, 3168, 3718, 4312, 4950, 5632, 6358, 7128, 7942, 8800, 9702, 10648, 11638, 12672, 13750, 14872, 16038, 17248, 18502, 19800, 21142, 22528, 23958, 25432, 26950, 28512, 30118, 31768, 33462, 35200, 36982, 38808

Offset: 0

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 0, in the direction 0, 22, ..., in the square spiral whose vertices are the generalized tridecagonal numbers A195313. Semi-axis opposite to A195318 in the same spiral.

Surface area of a rectangular prism with dimensions n, 2n and 3n. - Wesley Ivan Hurt, Apr 10 2015

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

Bisection of A195149.
Cf. A016802, A033581, A033583, A135453, A139098, A144555, A195321, A195322.
Cf. A000290, A001105, A008604, A033584, A195313, A195318.

[22*n^2 : n in [0..40]]; // Vincenzo Librandi, Sep 20 2011

A195323:=n->22*n^2: seq(A195323(n), n=0..80); # Wesley Ivan Hurt, Apr 10 2015

22Range[0,50]^2 (* or *) LinearRecurrence[{3,-3,1},{0,22,88},50] (* Harvey P. Dale, Sep 19 2011 *)

vector(50,n,22*(n-1)^2) \\ Derek Orr, Apr 10 2015

a(n) = 22*A000290(n) = 11*A001105(n) = 2*A033584(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Sep 19 2011
G.f.: 22*x*(1+x)/(1-x)^3. - Wesley Ivan Hurt, Apr 10 2015
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 22*x*(1 + x)*exp(x).
a(n) = n*A008604(n) = A195149(2*n). (End)

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).

A016910 a(n) = (6*n)^2.

Original entry on oeis.org

0, 36, 144, 324, 576, 900, 1296, 1764, 2304, 2916, 3600, 4356, 5184, 6084, 7056, 8100, 9216, 10404, 11664, 12996, 14400, 15876, 17424, 19044, 20736, 22500, 24336, 26244, 28224, 30276, 32400, 34596, 36864, 39204, 41616, 44100, 46656, 49284, 51984, 54756, 57600, 60516, 63504, 66564, 69696, 72900

Offset: 0

N. J. A. Sloane

Areas A of two classes of triangles with integer sides (a,b,c) where a = 9k, b=10k and c = 17k, or a = 3k, b = 25k and c = 26k for k=0,1,2,... These areas are given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) = (6k)^2, with the semiperimeter s = (a+b+c)/2. This sequence is a subsequence of A188158. - Michel Lagneau, Oct 11 2013

Sequence found by reading the line from 0, in the direction 0, 36, ..., in the square spiral whose vertices are the generalized 20-gonal numbers A218864. - Omar E. Pol, May 13 2018.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A089491, A188158, A243941.

Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30).

[(6*n)^2: n in [0..40]]; // Vincenzo Librandi, May 03 2011

(6*Range[0,40])^2 (* or *) LinearRecurrence[{3,-3,1},{0,36,144},40] (* Harvey P. Dale, Dec 25 2017 *)
Table[36 n^2, {n, 0, 45}] (* Omar E. Pol, Jun 07 2018 *)

a(n)=36*n^2 \\ Charles R Greathouse IV, Jun 10 2016

From Ilya Gutkovskiy, Jun 09 2016: (Start)
O.g.f.: 36*x*(1 + x)/(1 - x)^3.
E.g.f.: 36*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
Sum_{n>=1} 1/a(n) = Pi^2/216 = A086726. (End)
Product_{n>=1} a(n)/A136017(n) = Pi/3. - Fred Daniel Kline, Jun 09 2016
a(n) = t(9*n) - 9*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(9*n) - 9*A000217(n). - Bruno Berselli, Aug 31 2017
a(n) = 36*A000290(n) = 18*A001105(n) = 12*A033428 = 9*A016742(n) = 6*A033581(n) = 4*A016766(n) = 3*A135453(n) = 2*A195321(n). - Omar E. Pol, Jun 07 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/432. - Amiram Eldar, Jun 27 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/6)/(Pi/6).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/6)/(Pi/6) = 3/Pi (A089491). (End)

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

Original entry on oeis.org

2, 7, 18, 31, 50, 71, 98, 127, 162, 199, 242, 287, 338, 391, 450, 511, 578, 647, 722, 799, 882, 967, 1058, 1151, 1250, 1351, 1458, 1567, 1682, 1799, 1922, 2047, 2178, 2311, 2450, 2591, 2738, 2887, 3042, 3199, 3362, 3527, 3698, 3871, 4050, 4231, 4418, 4607, 4802

Offset: 0

Bruno Berselli, Sep 21 2011 - based on remarks and sequences by Omar E. Pol

Sequence found by reading the numbers in increasing order on the vertical line containing 2 of the square spiral whose vertices are the triangular numbers (A000217) - see Pol's comments in other sequences visible in this numerical spiral.

Also A077591 (without first term) and A157914 interleaved.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A195605.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000217, A077591, A157914.
Cf. A047621 (contains first differences), A016754 (contains the sum of any two consecutive terms).
Cf. A033585, A069129, A077221, A102083, A139098, A139271-A139277, A139592, A139593, A188135, A194268, A194431, A195241 [incomplete list].

[(4*n*(n+2)+(-1)^n+3)/2: n in [0..48]];

CoefficientList[Series[(2 + 3 x + 4 x^2 - x^3) / ((1 + x) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{2,0,-2,1},{2,7,18,31},50] (* Harvey P. Dale, Jan 21 2017 *)

for(n=0, 48, print1((4*n*(n+2)+(-1)^n+3)/2", "));

G.f.: (2+3*x+4*x^2-x^3)/((1+x)*(1-x)^3).
a(n) = a(-n-2) = 2*a(n-1)-2*a(n-3)+a(n-4).
a(n) = A047524(A000982(n+1)).
Sum_{n>=0} 1/a(n) = 1/2 + Pi^2/16 - cot(Pi/(2*sqrt(2)))*Pi/(4*sqrt(2)). - Amiram Eldar, Mar 06 2023

A118057 a(n) = 8n^2 - 4n - 3.

Original entry on oeis.org

1, 21, 57, 109, 177, 261, 361, 477, 609, 757, 921, 1101, 1297, 1509, 1737, 1981, 2241, 2517, 2809, 3117, 3441, 3781, 4137, 4509, 4897, 5301, 5721, 6157, 6609, 7077, 7561, 8061, 8577, 9109, 9657, 10221, 10801, 11397, 12009, 12637, 13281, 13941, 14617

Offset: 1

Charlie Marion, Apr 26 2006

In general, all sequences of equations which contain every positive integer in order exactly once (a pairwise equal summed, ordered partition of the positive integers) may be defined as follows: For all k, let x(k)=A001652(k) and z(k)=A001653(k). Then if we define a(n) to be (x(k)+z(k))n^2-(z(k)-1)n-x(k), the following equation is true: a(n)+(a(n)+1)+...+(a(n)+(x(k)+z(k))n+(2x(k)+z(k)-1)/2)=(a(n)+ (x(k)+z(k))n+(2x(k)+z(k)+1)/2)+...+(a(n)+2(x(k)+z(k))n+x(k)); a(n)+2(x(k)+z(k))n+x(k))=a(n+1)-1; e.g., in this sequence, x(1)=A001652(1)=3 and z(1)=A001653(1)=5; cf. A000290, A118058-A118061.

Sequence found by reading the segment (1, 21) together with the line from 21, in the direction 21, 57, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 04 2011

			a(3)=8*3^2-4*3-3=57, a(4)=8*4^2-4*4-3=109 and 57+58+...+86=87+...+108.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A004767, A139098.

Cf. A000290, A001652, A001653, A118058-A118061.

[8*n^2-4*n-3 : n in [1..60]]; // Wesley Ivan Hurt, Jan 28 2021

Table[8n^2-4n-3,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,21,57},50] (* Harvey P. Dale, Sep 18 2012 *)

a(n)=8*n^2-4*n-3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: x*(1+18*x-3*x^2)/(1-x)^3. - Colin Barker, Jul 01 2012
a(n)+(a(n)+1)+...+(a(n)+8n+5)=(a(n)+8n+6)+...+a(n+1)-1; a(n+1)-1=a(n)+16n+3.
a(n)+(a(n)+1)+...+(a(n)+8n+5)=(4n-1)(4n+1)(4n+3); e.g., 21+22+...+56=693=7*9*11.
a(n) = 16*n+a(n-1)-12 (with a(1)=1). - Vincenzo Librandi, Nov 13 2010
a(n) = A139098(n) - A004767(n). - Omar E. Pol, Sep 18 2012

A195824 a(n) = 24*n^2.

Original entry on oeis.org

0, 24, 96, 216, 384, 600, 864, 1176, 1536, 1944, 2400, 2904, 3456, 4056, 4704, 5400, 6144, 6936, 7776, 8664, 9600, 10584, 11616, 12696, 13824, 15000, 16224, 17496, 18816, 20184, 21600, 23064, 24576, 26136, 27744, 29400, 31104, 32856, 34656, 36504, 38400, 40344

Offset: 0

Omar E. Pol, Sep 28 2011

Sequence found by reading the line from 0, in the direction 0, 24, ..., in the square spiral whose vertices are the generalized tetradecagonal numbers A195818.

Surface area of a cube with side 2n. - Wesley Ivan Hurt, Aug 05 2014

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

Bisection of A195158.

Cf. A000290, A001105, A008606, A016742, A033428, A033581, A135453, A139098, A195818.

[24*n^2 : n in [0..50]]; // Wesley Ivan Hurt, Aug 05 2014

I:=[0,24,96]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 06 2014

A195824:=n->24*n^2: seq(A195824(n), n=0..50); # Wesley Ivan Hurt, Aug 05 2014

24 Range[0, 30]^2 (* or *) Table[24 n^2, {n, 0, 30}] (* or *) CoefficientList[Series[24 x (1 + x)/(1 - x)^3, {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 05 2014 *)
LinearRecurrence[{3,-3,1},{0,24,96},40] (* Harvey P. Dale, Nov 11 2017 *)

a(n) = 24*n^2; \\ Michel Marcus, Aug 05 2014

a(n) = 24*A000290(n) = 12*A001105(n) = 8*A033428(n) = 6*A016742(n) = 4*A033581(n) = 3*A139098(n) = 2*A135453(n).
From Wesley Ivan Hurt, Aug 05 2014: (Start)
G.f.: 24*x*(1+x)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 24*x*(1 + x)*exp(x).
a(n) = n*A008606(n) = A195158(2*n). (End)

A244082 a(n) = 32*n^2.

Original entry on oeis.org

0, 32, 128, 288, 512, 800, 1152, 1568, 2048, 2592, 3200, 3872, 4608, 5408, 6272, 7200, 8192, 9248, 10368, 11552, 12800, 14112, 15488, 16928, 18432, 20000, 21632, 23328, 25088, 26912, 28800, 30752, 32768, 34848, 36992, 39200, 41472, 43808, 46208, 48672, 51200

Offset: 0

Wesley Ivan Hurt, Jun 19 2014

Geometric connections of a(n) to the area and perimeter of a square.
Area:
. half the area of a square with side 8n (cf. A008590);
. area of a square with diagonal 8n (cf. A008590);
. twice the area of a square with side 4n (cf. A008586);
. four times the area of a square with diagonal 4n (cf. A008586);
. eight times the area of a square with side 2n (cf. A005843);
. sixteen times the area of a square with diagonal 2n (cf. A005843);
. thirty two times the area of a square with side n (cf. A001477);
. sixty four times the area of a square with diagonal n (cf. A001477).
Perimeter:
. perimeter of a square with side 8n^2 (cf. A139098);
. twice the perimeter of a square with side 4n^2 (cf. A016742);
. four times the perimeter of a square with side 2n^2 (cf. A001105);
. eight times the perimeter of a square with side n^2 (cf. A000290).
Sequence found by reading the line from 0, in the direction 0, 32, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, May 10 2018

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Pentasection of A077221, A181900.
Cf. A000290, A001105, A010021, A016742, A016802, A139098.
Cf. A001477, A005843, A008586, A008590, A139098, A174312.

Magma
```
[32*n^2 : n in [0..50]];
```

A244082:=n->32*n^2; seq(A244082(n), n=0..50);

32 Range[0, 50]^2 (* or *)
Table[32 n^2, {n, 0, 50}] (* or *)
CoefficientList[Series[32 x (1 + x)/(1 - x)^3, {x, 0, 30}], x]

a(n)=32*n^2 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: 32*x*(1+x)/(1-x)^3.
a(n) =  2 * A016802(n).
a(n) =  4 * A139098(n).
a(n) =  8 * A016742(n).
a(n) = 16 * A001105(n).
a(n) = 32 * A000290(n).
a(n) = A010021(n) - 2 for n > 0. - Bruno Berselli, Jun 24 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Nov 19 2021
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 32*x*(1 + x)*exp(x).
a(n) = n*A174312(n) = A139098(2*n). (End)

A181900 a(n) = n * A022998(n).

Original entry on oeis.org

0, 1, 8, 9, 32, 25, 72, 49, 128, 81, 200, 121, 288, 169, 392, 225, 512, 289, 648, 361, 800, 441, 968, 529, 1152, 625, 1352, 729, 1568, 841, 1800, 961, 2048, 1089, 2312, 1225, 2592, 1369, 2888, 1521, 3200, 1681, 3528, 1849, 3872, 2025, 4232, 2209, 4608, 2401

Offset: 0

Reinhard Zumkeller, Mar 31 2012

Multiplicative because A022998 is. - Andrew Howroyd, Jul 26 2018

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

Cf. A016754, A022998, A086500 (partial sums), A139098.

Haskell
```
a181900 n = a022998 n * n
```

[Numerator(2*n^2/(n^2+1)): n in [0..50]]; // Vincenzo Librandi, Aug 19 2014

LinearRecurrence[{0,3,0,-3,0,1},{0,1,8,9,32,25},50] (* Harvey P. Dale, Dec 01 2018 *)

a(n)=if(n%2,n^2,2*n^2) \\ Charles R Greathouse IV, Aug 07 2012

[n^2*(1 + ((n+1)%2)) for n in (0..60)] # G. C. Greubel, Aug 01 2022

a(2*n) = 8*n^2 = A139098(n).
a(2*n+1) = 1 + 4*n + 4*n^2 = A016754(n).
From Colin Barker, Apr 01 2012: (Start)
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x*(1+8*x+6*x^2+8*x^3+x^4)/((1-x)^3*(1+x)^3). (End)
a(n) = numerator(2*n^2/(n^2+1)). - Vincenzo Librandi, Aug 19 2014
From Amiram Eldar, Feb 22 2022: (Start)
Sum_{n>=1} 1/a(n) = 7*Pi^2/48.
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*Pi^2/48. (End)
From G. C. Greubel, Aug 01 2022: (Start)
a(n) = n^2*(3 + (-1)^n)/2.
E.g.f.: (1/2)*x*(-1 + x + 3*(1 + x)*exp(2*x)). (End)

A194268 a(n) = 8n^2 + 7n + 1.

Original entry on oeis.org

1, 16, 47, 94, 157, 236, 331, 442, 569, 712, 871, 1046, 1237, 1444, 1667, 1906, 2161, 2432, 2719, 3022, 3341, 3676, 4027, 4394, 4777, 5176, 5591, 6022, 6469, 6932, 7411, 7906, 8417, 8944, 9487, 10046, 10621, 11212, 11819, 12442, 13081, 13736, 14407, 15094, 15797

Offset: 0

Omar E. Pol, Sep 05 2011

Sequence found by reading the line from 1, in the direction 1, 16,..., in the square spiral whose vertices are the triangular numbers A000217.

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

Cf. A000217, A014634, A051870, A069129, A139098, A194431.

[8*n^2 +7*n + 1: n in [0..50]]; // Vincenzo Librandi, Sep 07 2011

A194268:=n->8*n^2+7*n+1: seq(A194268(n), n=0..50); # Wesley Ivan Hurt, Jul 15 2014

Table[8n^2+7n+1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,16,47},50] (* Harvey P. Dale, Apr 06 2014 *)

a(n)=8*n^2+7*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(0)=1, a(1)=16, a(2)=47, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Apr 06 2014
From Elmo R. Oliveira, Oct 22 2024: (Start)
G.f.: (1 + 13*x + 2*x^2)/(1 - x)^3.
E.g.f.: (1 + 15*x + 8*x^2)*exp(x). (End)

A194431 a(n) = 8n^2 - 6n - 1.

Original entry on oeis.org

1, 19, 53, 103, 169, 251, 349, 463, 593, 739, 901, 1079, 1273, 1483, 1709, 1951, 2209, 2483, 2773, 3079, 3401, 3739, 4093, 4463, 4849, 5251, 5669, 6103, 6553, 7019, 7501, 7999, 8513, 9043, 9589, 10151, 10729, 11323, 11933, 12559, 13201, 13859, 14533, 15223, 15929

Offset: 1

Omar E. Pol, Sep 05 2011

Sequence found by reading the line from 1, in the direction 1, 19, ..., in the square spiral whose vertices are the triangular numbers A000217.

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

Cf. A000217, A014634, A051870, A069129, A139098, A194268.

[8*n^2 - 6*n - 1: n in [1..50]]; // Vincenzo Librandi, Sep 07 2011

Table[8n^2-6n-1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,19,53},50] (* Harvey P. Dale, May 29 2021 *)

a(n)=8*n^2-6*n-1 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(-1 - 16*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 06 2011
From Elmo R. Oliveira, Jun 04 2025: (Start)
E.g.f.: 1 + (-1 + 2*x + 8*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

cp's OEIS Frontend

A195323 a(n) = 22*n^2.

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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.

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A016910 a(n) = (6*n)^2.

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A195605 a(n) = (4*n*(n+2)+(-1)^n+1)/2 + 1.

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A118057 a(n) = 8*n^2 - 4*n - 3.

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A195824 a(n) = 24*n^2.

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A244082 a(n) = 32*n^2.

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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.

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

A118057 a(n) = 8n^2 - 4n - 3.

A194268 a(n) = 8n^2 + 7n + 1.

A194431 a(n) = 8n^2 - 6n - 1.