A003154 -id:A003154 - cp's OEIS frontend

A195317 Centered 40-gonal numbers.

1, 41, 121, 241, 401, 601, 841, 1121, 1441, 1801, 2201, 2641, 3121, 3641, 4201, 4801, 5441, 6121, 6841, 7601, 8401, 9241, 10121, 11041, 12001, 13001, 14041, 15121, 16241, 17401, 18601, 19841, 21121, 22441, 23801, 25201, 26641, 28121, 29641, 31201, 32801, 34441, 36121

Offset: 1

Omar E. Pol, Sep 16 2011

Also centered tetracontagonal numbers or centered tetrakaicontagonal numbers. Also sequence found by reading the line from 1, in the direction 1, 41, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Semi-axis opposite to A195322 in the same spiral.

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

Bisection of A195148.
Cf. A003154, A069129, A069133, A069190, A195314, A195315, A195316, A195318.
Cf. A195162, A195322.

[20*n^2 - 20*n + 1: n in [1..50]]; // Vincenzo Librandi, Sep 21 2011

Table[20*n^2 - 20*n + 1, {n, 1, 40}] (* Amiram Eldar, Feb 11 2022 *)

a(n)=20*n^2-20*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 20*n^2 - 20*n + 1.
Sum_{n>=1} 1/a(n) = Pi*tan(Pi/sqrt(5))/(8*sqrt(5)). - Amiram Eldar, Feb 11 2022
G.f.: -x*(1+38*x+x^2)/(x-1)^3. - R. J. Mathar, May 07 2024
From Elmo R. Oliveira, Nov 15 2024: (Start)
E.g.f.: exp(x)*(20*x^2 + 1) - 1.
a(n) = 2*A069133(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3.  (End)

A144390 a(n) = 3*n^2 - n - 1.

Original entry on oeis.org

1, 9, 23, 43, 69, 101, 139, 183, 233, 289, 351, 419, 493, 573, 659, 751, 849, 953, 1063, 1179, 1301, 1429, 1563, 1703, 1849, 2001, 2159, 2323, 2493, 2669, 2851, 3039, 3233, 3433, 3639, 3851, 4069, 4293, 4523, 4759, 5001, 5249, 5503, 5763, 6029, 6301, 6579

Offset: 1

Paul Curtz, Oct 02 2008

Sequence's original Name was "First bisection of A135370."

The partial sums of this sequence give A081437. - Leo Tavares, Dec 26 2021

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
John Elias, Illustration: Belted Hexagrams
Leo Tavares, Illustration: Bounded Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A016933, A049450, A144391.
Cf. A003215, A005408, A016754, A002378.
Cf. A081437 (partial sums).

[3*n^2-n-1: n in [1..60]]; // Vincenzo Librandi, Jun 14 2011

A144390:=n->3*n^2-n-1; seq(A144390(n), n=1..50); # Wesley Ivan Hurt, Mar 26 2014

Table[3*n^2 -n -1 , {n,0,50}] (* G. C. Greubel, Jul 19 2017 *)

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

a(n+1) = a(n) + 6*n + 2; see A016933.
G.f.: x*(1+6*x-x^2)/(1-x)^3. a(n) = A049450(n)-1. - R. J. Mathar, Oct 24 2008
a(-n) = A144391(n). - Michael Somos, Mar 27 2014
E.g.f.: (3*x^2 + 2*x -1)*exp(x) + 1. - G. C. Greubel, Jul 19 2017
From Leo Tavares, Dec 26 2021: (Start)
a(n) = A003215(n) - 2*A005408(n). See Bounded Hexagons illustration.
a(n) = A016754(n-1) - A002378(n-2). (End)
a(n) = A003154(n) - A049451(n-1). - John Elias, Dec 22 2022

Edited by R. J. Mathar, Oct 24 2008

More terms from Vladimir Joseph Stephan Orlovsky, Oct 25 2008

A227776 a(n) = 6*n^2 + 1.

Original entry on oeis.org

1, 7, 25, 55, 97, 151, 217, 295, 385, 487, 601, 727, 865, 1015, 1177, 1351, 1537, 1735, 1945, 2167, 2401, 2647, 2905, 3175, 3457, 3751, 4057, 4375, 4705, 5047, 5401, 5767, 6145, 6535, 6937, 7351, 7777, 8215, 8665, 9127, 9601, 10087, 10585, 11095, 11617, 12151

Offset: 0

Clark Kimberling, Jul 30 2013

Least splitter is defined for x < y at A227631 as the least positive integer d such that x <= c/d < y for some integer c; the number c/d is called the least splitting rational of x and y. Conjecture: a(n) is the least splitter of s(n) and s(n+1), where s(n) = n*sin(1/n).

			The first eight least splitting rationals for {n*sin(1/n), n >=1 } are these fractions: 6/7, 24/25, 54/55, 96/97, 150/151, 216/217, 294/295, 384/385.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Hexagonal Star Rays.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A227631, A287326.

Cf. A003154, A008458, A003215, A000217, A028896, A054554, A046092, A080855, A045943, A172043, A002378, A033581.

z = 40; r[x_, y_] := Module[{c, d}, d = NestWhile[#1 + 1 &, 1, ! (c = Ceiling[#1 x - 1]) < Ceiling[#1 y] - 1 &]; (c + 1)/d]; s[n_] := s[n] = n*Sin[1/n]; t = Table[r[s[n], s[n + 1]], {n, 1, z}] (* least splitting rationals *); fd = Denominator[t] (* Peter J. C. Moses, Jul 15 2013 *)
Array[6 #^2 + 1 &, 45] (* Michael De Vlieger, Nov 08 2017 *)
LinearRecurrence[{3,-3,1},{7,25,55},50] (* Harvey P. Dale, Dec 16 2017 *)

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

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (1 + 4*x + 7*x^2)/(1 - x)^3.
a(n) = A287326(2n, n). - Kolosov Petro, Nov 06 2017
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(6))*coth(Pi/sqrt(6)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(6))*csch(Pi/sqrt(6)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(6))*sinh(Pi/sqrt(3)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(6))*csch(Pi/sqrt(6)).(End)
From Leo Tavares, Nov 20 2021: (Start)
a(n) = A003154(n+1) - A008458(n). See Hexagonal Star Rays illustration.
a(n) = A003215(n) + A028896(n-1).
a(n) = A054554(n+1) + A046092(n).
a(n) = A080855(n) + A045943(n).
a(n) = A172043(n) + A002378(n).
a(n) = A033581(n) + 1. (End)
E.g.f.: exp(x)*(1 + 6*x + 6*x^2). - Stefano Spezia, Sep 14 2024

a(0) = 1 prepended by Robert P. P. McKone, Oct 09 2023

A033571 a(n) = (2n + 1)(5*n + 1).

Original entry on oeis.org

1, 18, 55, 112, 189, 286, 403, 540, 697, 874, 1071, 1288, 1525, 1782, 2059, 2356, 2673, 3010, 3367, 3744, 4141, 4558, 4995, 5452, 5929, 6426, 6943, 7480, 8037, 8614, 9211, 9828, 10465, 11122, 11799, 12496, 13213, 13950, 14707, 15484, 16281, 17098, 17935, 18792, 19669, 20566, 21483

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. This is one of the diagonals in the spiral. - Omar E. Pol, Sep 10 2011

Also sequence found by reading the line from 1, in the direction 1, 18, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. This is a line perpendicular to the main axis A195015 in the same spiral. - Omar E. Pol, Oct 14 2011

G. C. Greubel, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Stellar Layers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A153127. - Reinhard Zumkeller, Dec 20 2008
Cf. A000566, A008596, A010010, A153126, A158186.
Cf. A001622, A003154, A007742, A019952.

List([0..50], n-> (2*n+1)*(5*n+1)); # G. C. Greubel, Oct 12 2019

[(2*n+1)*(5*n+1): n in [0..50]]; // G. C. Greubel, Oct 12 2019

seq((2*n+1)*(5*n+1), n=0..50); # G. C. Greubel, Oct 12 2019

Table[(2*n+1)*(5*n+1), {n,0,50}] (* G. C. Greubel, Oct 12 2019 *)

a(n)=(2*n+1)*(5*n+1) \\ Charles R Greathouse IV, Jun 17 2017

[(2*n+1)*(5*n+1) for n in range(50)] # G. C. Greubel, Oct 12 2019

a(n) = A153126(2*n) = A000566(2*n+1). - Reinhard Zumkeller, Dec 20 2008
From Reinhard Zumkeller, Mar 13 2009: (Start)
a(n) = A008596(n) + A158186(n), for n > 0.
a(n) = A010010(n) - A158186(n). (End)
a(n) = a(n-1) + 20*n - 3 (with a(0)=1). - Vincenzo Librandi, Nov 17 2010
From G. C. Greubel, Oct 12 2019: (Start)
G.f.: (1 + 15*x + 4*x^2)/(1-x)^3.
E.g.f.: (1 + 17*x + 10*x^2)*exp(x). (End)
a(n) = A003154(n+1) + A007742(n). - Leo Tavares, Mar 27 2022
Sum_{n>=0} 1/a(n) = sqrt(1+2/sqrt(5))*Pi/6 + sqrt(5)*log(phi)/6 + 5*log(5)/12 - 2*log(2)/3, where phi is the golden ratio (A001622). - Amiram Eldar, Aug 23 2022

Terms a(36) onward added by G. C. Greubel, Oct 12 2019

A007667 The sum of both two and three consecutive squares.

Original entry on oeis.org

5, 365, 35645, 3492725, 342251285, 33537133085, 3286296790925, 322023548377445, 31555021444198565, 3092070077983081805, 302991312620897818205, 29690056566770003102165

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

			a(2) = 365 = 13^2+14^2 = 10^2+11^2+12^2.

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for sequences related to sums of squares
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A003154, A031138, A006061, A054320.

a:=[5, 365, 35645];; for n in [4..20] do a[n]:=99*a[n-1]-99*a[n-2] + a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( 5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[5*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)),{x,0,20}],x] (* Vincenzo Librandi, Apr 16 2012 *)
LinearRecurrence[{99,-99,1},{5,365,35645},20] (* Harvey P. Dale, Dec 10 2024 *)

my(x='x+O('x^20)); Vec(5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2))) \\ G. C. Greubel, Jul 23 2019

(5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

From Ignacio Larrosa Cañestro, Feb 27 2000: (Start)
a(n) = (b(n)-1)^2 + b(n)^2 + (b(n)+1)^2 = c(n)^2 + (c(n)+1)^2, where b(n) = A054320(n) and c(n) = A031138(n).
a(n) = 3*A006061(n) + 2.
a(n) = 99*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 3*(5 - 2*sqrt(6))/8*(sqrt(3) + sqrt(2))^(4*n) + 3*(5 + 2*sqrt(6))/8*(sqrt(3) - sqrt(2))^(4*n) + 5/4. (End)
G.f.: 5*x*(1-26*x+x^2)/((1-x)*(1-98*x+x^2)). - Colin Barker, Apr 14 2012

Corrected by T. D. Noe, Nov 07 2006

A067726 a(n) = 6n^2 + 12n.

Original entry on oeis.org

18, 48, 90, 144, 210, 288, 378, 480, 594, 720, 858, 1008, 1170, 1344, 1530, 1728, 1938, 2160, 2394, 2640, 2898, 3168, 3450, 3744, 4050, 4368, 4698, 5040, 5394, 5760, 6138, 6528, 6930, 7344, 7770, 8208, 8658, 9120, 9594, 10080, 10578, 11088, 11610

Offset: 1

Robert G. Wilson v, Feb 05 2002

Positive numbers k such that 6*(6 + k) is a perfect square.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Leo Tavares, Illustration: Star Cut Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. numbers k such that k*(k + m) is a perfect square: A028560 (k=9), A067728 (k=8), A067727 (k=7), A067724 (k=5), A028347 (k=4), A067725 (k=3), A054000 (k=2), A005563 (k=1).

Cf. A003154, A003215.

List([1..45], n-> 6*n*(n+2)); # G. C. Greubel, Sep 01 2019

[6*n*(n+2): n in [1..50]]; // Vincenzo Librandi, Jul 08 2012

seq(6*n*(n+2), n=1..45); # G. C. Greubel, Sep 01 2019

Select[ Range[15000], IntegerQ[ Sqrt[ 6(6 + # )]] & ]
CoefficientList[Series[6*(3-x)/(1-x)^3,{x,0,50}],x] (* Vincenzo Librandi, Jul 08 2012 *)
6*(Range[2, 45]^2 -1) (* G. C. Greubel, Sep 01 2019 *)
LinearRecurrence[{3,-3,1},{18,48,90},60] (* Harvey P. Dale, May 10 2022 *)

a(n)=6*n*(n+2) \\ Charles R Greathouse IV, Dec 07 2011

[6*n*(n+2) for n in (1..45)] # G. C. Greubel, Sep 01 2019

G.f.: 6*x*(3 - x)/(1 - x)^3. - Vincenzo Librandi, Jul 08 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 08 2012
E.g.f.: 6*x*(3 + x)*exp(x). - G. C. Greubel, Sep 01 2019
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/24. (End)
a(n) = A003215(2*n) - A003154(n). - Leo Tavares, May 20 2023
a(n) = 6*A005563(n). - Hugo Pfoertner, May 24 2023

A188386 a(n) = numerator(H(n+2)-H(n-1)), where H(n) = Sum_{k=1..n} 1/k is the n-th harmonic number.

Original entry on oeis.org

11, 13, 47, 37, 107, 73, 191, 121, 299, 181, 431, 253, 587, 337, 767, 433, 971, 541, 1199, 661, 1451, 793, 1727, 937, 2027, 1093, 2351, 1261, 2699, 1441, 3071, 1633, 3467, 1837, 3887, 2053, 4331, 2281, 4799, 2521, 5291, 2773, 5807, 3037, 6347, 3313, 6911, 3601

Offset: 1

Gary Detlefs, Mar 29 2011

Denominators are listed in A033931.
A027446 appears to be divisible by a(n).
The sequence lists also the largest odd divisors of 3*m^2-1 (A080663) for m>1. In fact, for m even, the largest odd divisor is 3*m^2-1 itself; for m odd, the largest odd divisor is (3*m^2-1)/2. From this follows the second formula given in Formula field. - Bruno Berselli, Aug 27 2013

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

Cf. A033931 (denominators), A001008, A001711, A027446, A002805, A003154, A080663, A158463, A213998.

import Data.Ratio ((%), numerator)
a188386 n = a188386_list !! (n-1)
a188386_list = map numerator $ zipWith (-) (drop 3 hs) hs
   where hs = 0 : scanl1 (+) (map (1 %) [1..])
-- Reinhard Zumkeller, Jul 03 2012

[Numerator((3*n^2+6*n+2)/((n*(n+1)*(n+2)))): n in [1..50]]; // Vincenzo Librandi, Mar 30 2011

seq((3-(-1)^n)*(3*n^2+6*n+2)/4, n=1..100);

Table[(3 - (-1)^n)*(3*n^2 + 6*n + 2)/4, {n, 40}] (* Wesley Ivan Hurt, Jan 29 2017 *)
Numerator[#[[4]]-#[[1]]]&/@Partition[HarmonicNumber[Range[0,50]],4,1] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{11,13,47,37,107,73},50] (* Harvey P. Dale, Dec 31 2017 *)

a(n) = numerator((3*n^2+6*n+2)/(n*(n+1)*(n+2))).
a(n) = (3-(-1)^n)*(3*n^2+6*n+2)/4.
a(2n+1) = A158463(n+1), a(2n) = A003154(n+1).
G.f.: -x*(11+13*x+14*x^2-2*x^3-x^4+x^5) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Apr 09 2011
a(n) = numerator of coefficient of x^3 in the Maclaurin expansion of sin(x)*exp((n+1)*x). - Francesco Daddi, Aug 04 2011
H(n+3) = 3/2 + 2*f(n)/((n+2)*(n+3)), where f(n) = Sum_{k=0..n}((-1)^k*binomial(-3,k)/(n+1-k)). - Gary Detlefs, Jul 17 2011
a(n) = A213998(n+2,2). - Reinhard Zumkeller, Jul 03 2012
Sum_{n>=1} 1/a(n) = c*(tan(c) - cot(c)/2) - 1/2, where c = Pi/(2*sqrt(3)). - Amiram Eldar, Sep 27 2022

A195314 Centered 28-gonal numbers.

Original entry on oeis.org

1, 29, 85, 169, 281, 421, 589, 785, 1009, 1261, 1541, 1849, 2185, 2549, 2941, 3361, 3809, 4285, 4789, 5321, 5881, 6469, 7085, 7729, 8401, 9101, 9829, 10585, 11369, 12181, 13021, 13889, 14785, 15709, 16661, 17641, 18649, 19685, 20749, 21841, 22961, 24109, 25285, 26489

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 29, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Semi-axis opposite to A144555 in the same spiral.

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

Bisection of A195145.
Cf. A003154, A069129, A069133, A069190, A195315, A195316, A195317, A195318.
Cf. A069127, A118277, A144555.

[(14*n^2-14*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011

Table[14n^2-14n+1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,29,85},50]

a(n)=14*n^2-14*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - 14*n + 1.
G.f.: -x*(1 + 26*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Oct 01 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(5/7)*Pi/2)/(2*sqrt(35)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(14*x^2 + 1) - 1.
a(n) = 2*A069127(n) - 1. (End)

A195315 Centered 32-gonal numbers.

Original entry on oeis.org

1, 33, 97, 193, 321, 481, 673, 897, 1153, 1441, 1761, 2113, 2497, 2913, 3361, 3841, 4353, 4897, 5473, 6081, 6721, 7393, 8097, 8833, 9601, 10401, 11233, 12097, 12993, 13921, 14881, 15873, 16897, 17953, 19041, 20161, 21313, 22497, 23713, 24961, 26241, 27553, 28897, 30273

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 33, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. Semi-axis opposite to A016802 in the same spiral.

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

Bisection of A195146.
Cf. A003154, A069129, A069133, A069190, A195314, A195316, A195317, A195318.
Cf. A016802, A074377.

[(16*n^2-16*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011

Table[16*n^2 - 16*n + 1, {n, 1, 41}] (* Amiram Eldar, Feb 11 2022 *)
LinearRecurrence[{3,-3,1},{1,33,97},50] (* Harvey P. Dale, Feb 11 2024 *)

a(n)=16*n^2-16*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 16*n^2 - 16*n + 1.
G.f.: -x*(1 + 30*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(3)*Pi/4)/(8*sqrt(3)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(16*x^2 + 1) - 1.
a(n) = 2*A069129(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A195316 Centered 36-gonal numbers.

Original entry on oeis.org

1, 37, 109, 217, 361, 541, 757, 1009, 1297, 1621, 1981, 2377, 2809, 3277, 3781, 4321, 4897, 5509, 6157, 6841, 7561, 8317, 9109, 9937, 10801, 11701, 12637, 13609, 14617, 15661, 16741, 17857, 19009, 20197, 21421, 22681, 23977, 25309, 26677, 28081, 29521, 30997, 32509

Offset: 1

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 1, in the direction 1, 37, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Semi-axis opposite to A195321 in the same spiral.

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

Bisection of A195147.
Cf. A003154, A069129, A069133, A069190, A195314, A195315, A195317, A195318.
Cf. A069131, A195160, A195321.

[(18*n^2-18*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011

Table[18*n^2 - 18*n + 1, {n, 1, 40}] (* Amiram Eldar, Feb 11 2022 *)

a(n)=18*n^2-18*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 18*n^2 - 18*n + 1.
G.f.: -x*(1 + 34*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(7)*Pi/6)/(6*sqrt(7)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(18*x^2 + 1) - 1.
a(n) = 2*A069131(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

cp's OEIS Frontend

A195317 Centered 40-gonal numbers.

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A144390 a(n) = 3*n^2 - n - 1.

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A227776 a(n) = 6*n^2 + 1.

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A033571 a(n) = (2*n + 1)*(5*n + 1).

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A007667 The sum of both two and three consecutive squares.

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A067726 a(n) = 6*n^2 + 12*n.

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A033571 a(n) = (2n + 1)(5*n + 1).

A067726 a(n) = 6n^2 + 12n.