A016742 -id:A016742 - cp's OEIS frontend

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

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

A155955 Triangle read by rows: T(n,k) = (k*n)^k, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 16, 1, 3, 36, 729, 1, 4, 64, 1728, 65536, 1, 5, 100, 3375, 160000, 9765625, 1, 6, 144, 5832, 331776, 24300000, 2176782336, 1, 7, 196, 9261, 614656, 52521875, 5489031744, 678223072849, 1, 8, 256, 13824, 1048576, 102400000, 12230590464

Offset: 0

Reinhard Zumkeller, Jan 31 2009

T(n,0)  = 1;
T(n,1)  = n for n > 0;
T(n,2)  = A016742(n) for n > 1;
T(n,3)  = A016767(n) for n > 2;
T(n,4)  = A016804(n) for n > 3;
T(n,5)  = A016853(n) for n > 4;
T(n,6)  = A016914(n) for n > 5;
T(n,7)  = A016987(n) for n > 6;
T(n,8)  = A017072(n) for n > 7;
T(n,9)  = A017169(n) for n > 8;
T(n,10) = A017278(n) for n > 9;
T(n,11) = A017399(n) for n > 10;
T(n,12) = A017532(n) for n > 11;
T(n,n)  = A062206(n).

			Triangle begins:
  1;
  1, 1;
  1, 2,  16;
  1, 3,  36,  729;
  1, 4,  64, 1728,  65536;
  1, 5, 100, 3375, 160000,  9765625;
  1, 6, 144, 5832, 331776, 24300000, 2176782336;
  ...

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A000312.

[[(n*k)^k: k in [0..n]]: n in [0..10]]; // G. C. Greubel, Sep 15 2018

Table[If[n == 0, 1, If[ k == 0, 1, (k*n)^k]], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Sep 15 2018 *)

for(n=0,10, for(k=0,n, print1((k*n)^k, ", "))) \\ G. C. Greubel, Sep 15 2018

A244630 a(n) = 17*n^2.

Original entry on oeis.org

0, 17, 68, 153, 272, 425, 612, 833, 1088, 1377, 1700, 2057, 2448, 2873, 3332, 3825, 4352, 4913, 5508, 6137, 6800, 7497, 8228, 8993, 9792, 10625, 11492, 12393, 13328, 14297, 15300, 16337, 17408, 18513, 19652, 20825, 22032, 23273, 24548, 25857, 27200, 28577, 29988

Offset: 0

Vincenzo Librandi, Jul 03 2014

First bisection of A195047. - Bruno Berselli, Jul 03 2014

Norms of purely imaginary numbers in Z[sqrt(-17)] (for example, 3*sqrt(-17) has norm 153). - Alonso del Arte, Jun 23 2018

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

Cf. A008599, A195047.

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), this sequence (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).

List([0..45],n->17*n^2); # Muniru A Asiru, Jun 29 2018

Magma
```
[17*n^2: n in [0..40]];
```

seq(17*n^2,n=0..45); # Muniru A Asiru, Jun 29 2018

Mathematica
```
Table[17 n^2, {n, 0, 40}]
```

a(n)=17*n^2 \\ Charles R Greathouse IV, Oct 07 2015

for (i <- 0 to 50) yield 17 * (i * i) // Alonso del Arte, Jun 29 2018

G.f.: 17*x*(1 + x)/(1 - x)^3. [corrected by Bruno Berselli, Jul 03 2014]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = 17*A000290(n). - Omar E. Pol, Jul 03 2014
a(n) = a(-n). - Muniru A Asiru, Jun 29 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 17*x*(1 + x)*exp(x).
a(n) = n*A008599(n) = A195047(2*n). (End)

A254963 a(n) = n(11n + 3)/2.

Original entry on oeis.org

0, 7, 25, 54, 94, 145, 207, 280, 364, 459, 565, 682, 810, 949, 1099, 1260, 1432, 1615, 1809, 2014, 2230, 2457, 2695, 2944, 3204, 3475, 3757, 4050, 4354, 4669, 4995, 5332, 5680, 6039, 6409, 6790, 7182, 7585, 7999, 8424, 8860, 9307, 9765, 10234, 10714, 11205, 11707

Offset: 0

Bruno Berselli, Feb 11 2015

This sequence provides the first differences of A254407 and the partial sums of A017473.
Also:
a(n) - n         = A022269(n);
a(n) + n         = n*(11*n+5)/2: 0, 8, 27, 57, 98, 150, 213, 287, ...;
a(n) - 2*n       = A022268(n);
a(n) + 2*n       = n*(11*n+7)/2: 0, 9, 29, 60, 102, 155, 219, 294, ...;
a(n) - 3*n       = n*(11*n-3)/2: 0, 4, 19, 45, 82, 130, 189, 259, ...;
a(n) + 3*n       = A211013(n);
a(n) - 4*n       = A226492(n);
a(n) + 4*n       = A152740(n);
a(n) - 5*n       = A180223(n);
a(n) + 5*n       = n*(11*n+13)/2: 0, 12, 35, 69, 114, 170, 237, 315, ...;
a(n) - 6*n       = A051865(n);
a(n) + 6*n       = n*(11*n+15)/2: 0, 13, 37, 72, 118, 175, 243, 322, ...;
a(n) - 7*n       = A152740(n-1) with A152740(-1) = 0;
a(n) + 7*n       = n*(11*n+17)/2: 0, 14, 39, 75, 122, 180, 249, 329, ...;
a(n) - n*(n-1)/2 = A168668(n);
a(n) + n*(n-1)/2 = A049453(n);
a(n) - n*(n+1)/2 = A202803(n);
a(n) + n*(n+1)/2 = A033580(n).

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

Cf. A008729 and A218530 (seventh column); A017473, A254407.
Cf. similar sequences of the type 4*n^2 + k*n*(n+1)/2: A055999 (k=-7, n>6), A028552 (k=-6, n>2), A095794 (k=-5, n>1), A046092 (k=-4, n>0), A000566 (k=-3), A049450 (k=-2), A022264 (k=-1), A016742 (k=0), A022267 (k=1), A202803 (k=2), this sequence (k=3), A033580 (k=4).
Cf. A069125: (2*n+1)^2 + 3*n*(n+1)/2; A147875: n^2 + 3*n*(n+1)/2.
Cf. A022268, A022269, A049453, A051865, A152740, A168668, A180223, A211013, A226492.

Magma
```
[n*(11*n+3)/2: n in [0..50]];
```

Table[n (11 n + 3)/2, {n, 0, 50}]
LinearRecurrence[{3,-3,1},{0,7,25},50] (* Harvey P. Dale, Mar 25 2018 *)

Maxima
```
makelist(n*(11*n+3)/2, n, 0, 50);
```
PARI
```
vector(50, n, n--; n*(11*n+3)/2)
```
Sage
```
[n*(11*n+3)/2 for n in (0..50)]
```

G.f.: x*(7 + 4*x)/(1 - x)^3.
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: exp(x)*x*(14 + 11*x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A118729 Rectangular array where row r contains the 8 numbers 4r^2 - 3r, 4r^2 - 2r, ..., 4r^2 + 4r.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 64, 68, 72, 76, 80, 85, 90, 95, 100, 105, 110, 115, 120, 126, 132, 138, 144, 150, 156, 162, 168

Offset: 0

Stuart M. Ellerstein (ellerstein(AT)aol.com), May 21 2006

The numbers in row r span the interval ]8*A000217(r-1), 8*A000217(r)].
The first difference between the entries in row r is r.
Partial sums of floor(n/8). - Philippe Deléham, Mar 26 2013
Apart from the initial zeros, the same as A008726. - Philippe Deléham, Mar 28 2013
a(n+7) is the number of key presses required to type a word of n letters, all different, on a keypad with 8 keys where 1 press of a key is some letter, 2 presses is some other letter, etc., and under an optimal mapping of letters to keys and presses (answering LeetCode problem 3014). - Christopher J. Thomas, Feb 16 2024

			The array starts, with row r=0, as
  r=0:   0  0  0  0  0  0  0  0;
  r=1:   1  2  3  4  5  6  7  8;
  r=2:  10 12 14 16 18 20 22 24;
  r=3:  27 30 33 36 39 42 45 48;

G. C. Greubel, Table of n, a(n) for n = 0..1000
LeetCode, 3014. Minimum Number of Pushes to Type Word I.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,1,-2,1).

Cf. similar sequences: A000217, A002620, A130518, A130519, A130520, A174709, A174738, A218470, A131242.

Flatten[Table[4r^2+r(Range[-3,4]),{r,0,6}]] (* or *) LinearRecurrence[ {2,-1,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,1,2},60] (* Harvey P. Dale, Nov 26 2015 *)

From Philippe Deléham, Mar 26 2013: (Start)
a(8k)   = A001107(k).
a(8k+1) = A002939(k).
a(8k+2) = A033991(k).
a(8k+3) = A016742(k).
a(8k+4) = A007742(k).
a(8k+5) = A002943(k).
a(8k+6) = A033954(k).
a(8k+7) = A033996(k). (End)
G.f.: x^8/((1-x)^2*(1-x^8)). - Philippe Deléham, Mar 28 2013
a(n) = floor(n/8)*(n-3-4*floor(n/8)). - Ridouane Oudra, Jun 04 2019
a(n+7) = (1/2)*(n+(n mod 8))*(floor(n/8)+1). - Christopher J. Thomas, Feb 13 2024

Redefined as a rectangular tabf array and description simplified by R. J. Mathar, Oct 20 2010

A129194 a(n) = (n/2)^2*(3 - (-1)^n).

Original entry on oeis.org

0, 1, 2, 9, 8, 25, 18, 49, 32, 81, 50, 121, 72, 169, 98, 225, 128, 289, 162, 361, 200, 441, 242, 529, 288, 625, 338, 729, 392, 841, 450, 961, 512, 1089, 578, 1225, 648, 1369, 722, 1521, 800, 1681, 882, 1849, 968, 2025, 1058, 2209, 1152, 2401, 1250, 2601, 1352

Offset: 0

Paul Barry, Apr 02 2007

The numerator of the integral is 2,1,2,1,2,1,...; the moments of the integral are 2/(n+1)^2. See 2nd formula.
The sequence alternates between twice a square and an odd square, A001105(n) and A016754(n).
Partial sums of the positive elements give the absolute values of A122576. - Omar E. Pol, Aug 22 2011
Partial sums of the positive elements give A212760. - Omar E. Pol, Dec 28 2013
Conjecture: denominator of 4/n - 2/n^2. - Wesley Ivan Hurt, Jul 11 2016
Multiplicative because both A000290 and A040001 are. - Andrew Howroyd, Jul 25 2018

G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 1, Sect. 7, Problem 73.

Vincenzo Librandi, Table of n, a(n) for n = 0..960
Olivier Bordelles, A Multidimensional Cesaro Type Identity and Applications, J.
Int. Seq. 18 (2015) # 15.3.7.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000290, A001105, A007434, A010713, A016742, A016754, A040001.

Cf. A061038, A061040, A061050, A105398, A129204, A152020, A222171, A309337.

[n^2*(3-(-1)^n)/4: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

A129194:=n->n^2*(3-(-1)^n)/4: seq(A129194(n), n=0..80); # Wesley Ivan Hurt, Jul 11 2016

Table[n^2*(3-(-1)^n)/4, {n,0,60}] (* Wesley Ivan Hurt, Jul 11 2016 *)
LinearRecurrence[{0,3,0,-3,0,1},{0,1,2,9,8,25},60] (* Harvey P. Dale, Dec 27 2023 *)

a(n) = lcm(2, n^2)/2; \\ Andrew Howroyd, Jul 25 2018

[n^2*(1+(n%2))/2 for n in range(61)] # G. C. Greubel, Apr 04 2023

G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4)/(1-x^2)^3.
a(n+1) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*n*t)(-((Pi-t)/i)^2)), i=sqrt(-1).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 5. - Paul Curtz, Mar 07 2011
a(n) is the numerator of the coefficient of x^4 in the Maclaurin expansion of exp(-n*x^2). - Francesco Daddi, Aug 04 2011
O.g.f. as a Lambert series: x*Sum_{n >= 1} J_2(n)*x^n/(1 + x^n), where J_2(n) denotes the Jordan totient function A007434(n). See Pólya and Szegő. - Peter Bala, Dec 28 2013
From Ilya Gutkovskiy, Jul 11 2016: (Start)
E.g.f.: x*((2*x + 1)*sinh(x) + (x + 2)*cosh(x))/2.
Sum_{n>=1} 1/a(n) = 5*Pi^2/24. [corrected by Amiram Eldar, Sep 11 2022] (End)
a(n) = A000290(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Sep 11 2022
From Peter Bala, Jan 16 2024: (Start)
a(n) = Sum_{1 <= i, j <= n} (-1)^(1 + gcd(i,j,n)) = Sum_{d | n} (-1)^(d+1) * J_2(n/d), that is, the Dirichlet convolution of the pair of multiplicative functions f(n) = (-1)^(n+1) and the Jordan totient function J_2(n) = A007434(n). Hence this sequence is multiplicative. Cf. A193356 and A309337.
Dirichlet g.f.: (1 - 2/2^s)*zeta(s-2). (End)
a(n) = Sum_{1 <= i, j <= n} (-1)^(n + gcd(i, n)*gcd(j, n)) = Sum_{d|n, e|n} (-1)^(n+e*d) * phi(n/d)*phi(n/e). - Peter Bala, Jan 22 2024

More terms from Michel Marcus, Dec 28 2013

A016826 a(n) = (4n + 2)^2.

Original entry on oeis.org

4, 36, 100, 196, 324, 484, 676, 900, 1156, 1444, 1764, 2116, 2500, 2916, 3364, 3844, 4356, 4900, 5476, 6084, 6724, 7396, 8100, 8836, 9604, 10404, 11236, 12100, 12996, 13924, 14884, 15876, 16900, 17956

Offset: 0

N. J. A. Sloane

A bisection of A016742. Sequence arises from reading the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

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

Equals A001539 + 1.

Cf. A000290, A010503, A016742, A016754, A016802, A016814, A016838.

A016826:=n->(4*n + 2)^2; seq(A016826(n), n=0..40); # Wesley Ivan Hurt, Feb 24 2014

(4*Range[0,40]+2)^2 (* or *) LinearRecurrence[{3,-3,1},{4,36,100},40] (* Harvey P. Dale, Nov 24 2011 *)

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

a(n) = a(n-1) + 32*n (with a(0)=4). - Vincenzo Librandi, Dec 15 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=4, a(1)=36, a(2)=100. - Harvey P. Dale, Nov 24 2011
G.f.: -((4*(x^2+6*x+1))/(x-1)^3). - Harvey P. Dale, Nov 24 2011
a(n) = A000290(A016825(n)). - Wesley Ivan Hurt, Feb 24 2014
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/32.
Sum_{n>=0} (-1)^n/a(n) = G/4, where G is the Catalan constant (A006752). (End)
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/4).
Product_{n>=0} (1 - 1/a(n)) = 1/sqrt(2) (A010503). (End)

A016838 a(n) = (4n + 3)^2.

Original entry on oeis.org

9, 49, 121, 225, 361, 529, 729, 961, 1225, 1521, 1849, 2209, 2601, 3025, 3481, 3969, 4489, 5041, 5625, 6241, 6889, 7569, 8281, 9025, 9801, 10609, 11449, 12321, 13225, 14161, 15129, 16129, 17161, 18225

Offset: 0

N. J. A. Sloane

If Y is a fixed 2-subset of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
A bisection of A016754. Sequence arises from reading the line from 9, in the direction 9, 49, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Using (n,n+1) to generate a Pythagorean triangle with sides of lengths xJ. M. Bergot, Jul 17 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..860
Milan Janjic, Two Enumerative Functions
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001539, A016742, A016754, A016802, A016814, A016826, A068465, A087197.

[(4*n+3)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A016838:=n->(4*n + 3)^2; seq(A016838(n), n=0..50); # Wesley Ivan Hurt, Feb 24 2014

Table[(4*n + 3)^2, {n, 0, 40}] (* Vaclav Kotesovec, Nov 14 2017 *)

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

Denominators of first differences Zeta[2,(4n-1)/4]-Zeta[2,(4(n+1)-1)/4]. - Artur Jasinski, Mar 03 2010
From George F. Johnson, Oct 03 2012: (Start)
G.f.: (9+22*x+x^2)/(1-x)^3.
a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
a(n-1)*a(n+1) = (a(n)-16)^2; a(n+1) - a(n-1) = 16*sqrt(a(n)).
a(n) = A016754(2*n+1) = (A004767(n))^2.
(End)
Sum_{n>=0} 1/a(n) = Pi^2/16 - G/2, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=0} (1 - 1/a(n)) = Gamma(3/4)^2/sqrt(Pi) = A068465^2 * A087197. - Amiram Eldar, Feb 01 2021

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

Original entry on oeis.org

0, 8, 28, 60, 104, 160, 228, 308, 400, 504, 620, 748, 888, 1040, 1204, 1380, 1568, 1768, 1980, 2204, 2440, 2688, 2948, 3220, 3504, 3800, 4108, 4428, 4760, 5104, 5460, 5828, 6208, 6600, 7004, 7420, 7848, 8288, 8740, 9204, 9680, 10168, 10668, 11180, 11704, 12240

Offset: 0

N. J. A. Sloane

Subsequence of A062717: A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
Sequence found by reading the line from 0, in the direction 0, 8,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A139267 in the same spiral - Omar E. Pol, Sep 09 2011
a(n) is the number of edges of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - Emeric Deutsch May 13 2018
The partial sums of this sequence give A035006. - Leo Tavares, Oct 03 2021

Ivan Panchenko, Table of n, a(n) for n = 0..1000
M. K. Siddiqui, M. Naeem, N. A. Rahman, and M. Imran, Computing topological indices of certain networks, J. of Optoelectronics and Advanced Materials, 18, No. 9-10 (2016), pp. 884-892.
Leo Tavares, Illustration: Crossed Stars
Leo Tavares, Illustration: Four Quarter Star Crosses
Leo Tavares, Illustration: Triangulated Star Crosses
Leo Tavares, Illustration: Oblong Star Crosses
Leo Tavares, Illustration: Crossed Diamond Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033579, A049451, A045945, A186423.
Cf. sequences listed in A254963.
Cf. A003154, A016813.
Cf. A035006.
Cf. A005449, A046092, A033996, A002378, A016742, A000290, A016754, A001105.

List([0..50], n-> 2*n*(3*n+1)); # G. C. Greubel, Oct 09 2019

[2*n*(3*n+1): n in [0..50]]; // G. C. Greubel, Oct 09 2019

seq(2*n*(3*n+1), n=0..50); # G. C. Greubel, Oct 09 2019

4*Binomial[3*Range[50]-2, 2]/3 (* G. C. Greubel, Oct 09 2019 *)

a(n)=2*n*(3*n+1) \\ Charles R Greathouse IV, Sep 28 2015

[2*n*(3*n+1) for n in (0..50)] # G. C. Greubel, Oct 09 2019

a(n) = a(n-1) +12*n -4 (with a(0)=0). - Vincenzo Librandi, Aug 05 2010
G.f.: 4*x*(2+x)/(1-x)^3. - Colin Barker, Feb 13 2012
a(-n) = A033579(n). - Michael Somos, Jun 09 2014
E.g.f.: 2*x*(4 + 3*x)*exp(x). - G. C. Greubel, Oct 09 2019
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 3/2 - Pi/(4*sqrt(3)) - 3*log(3)/4.
Sum_{n>=1} (-1)^(n+1)/a(n) =  -3/2 + Pi/(2*sqrt(3)) + log(2). (End)
From Leo Tavares, Oct 12 2021: (Start)
a(n) = A003154(n+1) - A016813(n). See Crossed Stars illustration.
a(n) = 4*A005449(n). See Four Quarter Star Crosses illustration.
a(n) = 2*A049451(n).
a(n) = A046092(n-1) + A033996(n). See Triangulated Star Crosses illustration.
a(n) = 4*A000217(n-1) + 8*A000217(n).
a(n) = 4*A000217(n-1) + 4*A002378. See Oblong Star Crosses illustration.
a(n) = A016754(n) + 4*A000217(n). See Crossed Diamond Stars illustration.
a(n) = 2*A001105(n) + 4*A000217(n).
a(n) = A016742(n) + A046092(n).
a(n) = 4*A000290(n) + 4*A000217(n). (End)

A129370 a(n) = n^2 - (n-1)^2*(1 - (-1)^n)/8.

Original entry on oeis.org

0, 1, 4, 8, 16, 21, 36, 40, 64, 65, 100, 96, 144, 133, 196, 176, 256, 225, 324, 280, 400, 341, 484, 408, 576, 481, 676, 560, 784, 645, 900, 736, 1024, 833, 1156, 936, 1296, 1045, 1444, 1160, 1600, 1281, 1764, 1408

Offset: 0

Paul Barry, Apr 11 2007

Partial sums are A129371.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000567 (odd bisection), A016742 (even bisection), A129371.

[n^2 -(n-1)^2*(n mod 2)/4: n in [0..60]]; // G. C. Greubel, Jan 31 2024

Table[n^2-(n-1)^2 (1-(-1)^n)/8,{n,0,50}] (* Harvey P. Dale, Oct 22 2011 *)

a(n)=n^2-(n-1)^2*(1-(-1)^n)/8 \\ Charles R Greathouse IV, Sep 28 2015

[n^2 -(n-1)^2*(n%2)/4 for n in range(61)] # G. C. Greubel, Jan 31 2024

a(n) = (1/8)*( (7*n^2 + 2*n - 1) + (-1)^n*(n-1)^2 ).
G.f.: x*(1 + 4*x + 5*x^2 + 4*x^3)/(1-x^2)^3.
E.g.f.: (1/4)*( x*(5+4*x)*cosh(x) - (1-4*x-3*x^2)*sinh(x) ). - G. C. Greubel, Jan 31 2024

cp's OEIS Frontend

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

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A155955 Triangle read by rows: T(n,k) = (k*n)^k, 0 <= k <= n.

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A244630 a(n) = 17*n^2.

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A254963 a(n) = n*(11*n + 3)/2.

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A118729 Rectangular array where row r contains the 8 numbers 4*r^2 - 3*r, 4*r^2 - 2*r, ..., 4*r^2 + 4*r.

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A129194 a(n) = (n/2)^2*(3 - (-1)^n).

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A254963 a(n) = n(11n + 3)/2.

A118729 Rectangular array where row r contains the 8 numbers 4r^2 - 3r, 4r^2 - 2r, ..., 4r^2 + 4r.

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).