A024966 -id:A024966 - cp's OEIS frontend

A022265 a(n) = n(7n + 1)/2.

0, 4, 15, 33, 58, 90, 129, 175, 228, 288, 355, 429, 510, 598, 693, 795, 904, 1020, 1143, 1273, 1410, 1554, 1705, 1863, 2028, 2200, 2379, 2565, 2758, 2958, 3165, 3379, 3600, 3828, 4063, 4305, 4554, 4810

Offset: 0

N. J. A. Sloane

For n >= 4, a(n) is the sum of the numbers appearing in the 4th row of an n X n square array whose elements are the numbers from 1..n^2, listed in increasing order by rows. - Wesley Ivan Hurt, May 17 2021

			From _Bruno Berselli_, Oct 27 2017: (Start)
After 0:
4  =       -(1)       +             (2 + 3).
15 =     -(1 + 2)     +         (3 + 4 + 5 + 6).
33 =   -(1 + 2 + 3)   +     (4 + 5 + 6 + 7 + 8 + 9).
58 = -(1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + 9 + 10 + 11 + 12). (End)

G. C. Greubel, Table of n, a(n) for n = 0..5000
Leo Tavares, Illustration: Square-sided Triangles.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001106, A022264, A024966, A110449, A174738, A179986, A186029, A218471.
Cf. similar sequences listed in A022289.
Cf. A000217, A000290.

seq(binomial(7*n+1,2)/7, n=0..37); # Zerinvary Lajos, Jan 21 2007
seq(binomial(6*n+1,2)/3-binomial(5*n+1,2)/5, n=0..42); # Zerinvary Lajos, Jan 21 2007

Table[n (7 n + 1)/2, {n, 0, 40}] (* Bruno Berselli, Oct 13 2016 *)
LinearRecurrence[{3,-3,1},{0,4,15},40] (* Harvey P. Dale, Oct 09 2018 *)

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

a(n) = A110449(n, 3) for n>2.
a(n) = A049453(n) - A005475(n). - Zerinvary Lajos, Jan 21 2007
a(n) = 7*n + a(n-1) - 3 for n>0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0)=0, a(1)=4, a(2)=15. - Philippe Deléham, Mar 26 2013
a(n) = A174738(7n+3). - Philippe Deléham, Mar 26 2013
a(n) = A000217(4*n) - A000217(3*n). - Bruno Berselli, Oct 13 2016
G.f.: x*(4 + 3*x)/(1 - x)^3. - Ilya Gutkovskiy, Oct 13 2016
E.g.f.: (x/2)*(7*x + 8)*exp(x). - G. C. Greubel, Aug 23 2017
a(n) = A000217(n) + 3*A000290(n). - Leo Tavares, Mar 15 2025

A195145 Concentric 14-gonal numbers.

Original entry on oeis.org

0, 1, 14, 29, 56, 85, 126, 169, 224, 281, 350, 421, 504, 589, 686, 785, 896, 1009, 1134, 1261, 1400, 1541, 1694, 1849, 2016, 2185, 2366, 2549, 2744, 2941, 3150, 3361, 3584, 3809, 4046, 4285, 4536, 4789, 5054, 5321, 5600, 5881, 6174, 6469, 6776, 7085, 7406

Offset: 0

Omar E. Pol, Sep 17 2011

Also concentric tetradecagonal numbers or concentric tetrakaidecagonal numbers. Also sequence found by reading the line from 0, in the direction 0, 14, ..., and the same line from 1, in the direction 1, 29, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Main axis, perpendicular to A024966 in the same spiral.

Partial sums of A113801. - Reinhard Zumkeller, Jan 07 2012

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

A144555 and A195314 interleaved.
Cf. A024966, A032527, A032528, A077221, A113801, A195045, A195046, A195142, A195143, A195146, A195147, A195148, A195149.
Column 14 of A195040. - Omar E. Pol, Sep 28 2011

a195145 n = a195145_list !! n
a195145_list = scanl (+) 0 a113801_list
-- Reinhard Zumkeller, Jan 07 2012

[(14*n^2+5*(-1)^n-5)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

LinearRecurrence[{2, 0, -2, 1}, {0, 1, 14, 29}, 50] (* Amiram Eldar, Jan 16 2023 *)

G.f.: -x*(1+12*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = (14*n^2 + 5*(-1)^n - 5)/4;
a(n) = a(-n) = -a(n-1) + 7*n^2 - 7*n + 1. (End)
Sum_{n>=1} 1/a(n) = Pi^2/84 + tan(sqrt(5/7)*Pi/2)*Pi/(2*sqrt(35)). - Amiram Eldar, Jan 16 2023
E.g.f.: (7*x*(x + 1)*cosh(x) + (7*x^2 + 7*x - 5)*sinh(x))/2. - Stefano Spezia, Nov 30 2024

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

Original entry on oeis.org

1, 5, 10, 17, 26, 36, 49, 62, 79, 95, 116, 135, 160, 182, 211, 236, 269, 297, 334, 365, 406, 440, 485, 522, 571, 611, 664, 707, 764, 810, 871, 920, 985, 1037, 1106, 1161, 1234, 1292, 1369, 1430, 1511, 1575, 1660, 1727, 1816, 1886, 1979, 2052, 2149, 2225, 2326

Offset: 0

Bruno Berselli, Oct 20 2011 - based on remarks and sequences by Omar E. Pol

For an origin of this sequence, see the numerical spiral illustrated in the Links section.

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

Cf. A195020 (vertices of the numerical spiral in link).

Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A195021, A195023-A195030, A195320, A198017 [incomplete list].

[(14*n*(n+3)+(2*n-5)*(-1)^n+21)/16: n in [0..50]];

Table[(14*n*(n + 3) + (2*n - 5)*(-1)^n + 21)/16, {n, 0, 50}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{1,5,10,17,26},60] (* Harvey P. Dale, Jun 19 2020 *)

for(n=0, 50, print1((14*n*(n+3)+(2*n-5)*(-1)^n+21)/16", "));

O.g.f.: (1 + 4*x + 3*x^2 - x^3)/((1 + x)^2*(1 - x)^3).
E.g.f.: (1/16)*((21 + 56*x + 14*x^2)*exp(x) - (5 + 2*x)*exp(-x)). - G. C. Greubel, Aug 19 2017
a(n) = A195020(n) + n + 1.
a(n)   - a(-n-1)  = A047336(n+1).
a(n+1) - a(-n)    = A113804(n+1).
a(n+2) - a(n)     = A047385(n+3).
a(n+4) - a(n)     = A113803(n+4).
a(2*n) + a(2*n-1) = A069127(n+1).
a(2*n) - a(2*n-1) = A016813(n).
a(2*n+1) - a(2*n) = A016777(n+1).
a(n+2) + 2*a(n+1) + a(n) = A024966(n+2).

A195320 7 times hexagonal numbers: a(n) = 7n(2*n-1).

Original entry on oeis.org

0, 7, 42, 105, 196, 315, 462, 637, 840, 1071, 1330, 1617, 1932, 2275, 2646, 3045, 3472, 3927, 4410, 4921, 5460, 6027, 6622, 7245, 7896, 8575, 9282, 10017, 10780, 11571, 12390, 13237, 14112, 15015, 15946, 16905, 17892, 18907, 19950, 21021, 22120, 23247, 24402, 25585

Offset: 0

Omar E. Pol, Sep 18 2011

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277.

Also sequence found by reading the same line (mentioned above) in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-diagonals of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5]. - Omar E. Pol, Oct 13 2011

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 A024966.

Cf. A000384, A118277, A152746, A152750, A185019, A193053, A195019, A195020, A198017, A316466.

[7*n*(2*n-1): n in [0..50]]; // Vincenzo Librandi, Sep 28 2011

Table[7*n*(2*n - 1), {n, 0, 50}] (* Paolo Xausa, Jan 29 2025 *)
7*PolygonalNumber[6, Range[0, 50]] (* Paolo Xausa, Jan 29 2025 *)

a(n)=7*n*(2*n-1) \\ Charles R Greathouse IV, Sep 28 2015

a(n) = 14*n^2 - 7*n = 7*A000384(n).
G.f.: -7*x*(1+3*x)/(x-1)^3. - R. J. Mathar, Sep 27 2011
From Elmo R. Oliveira, Dec 27 2024: (Start)
E.g.f.: 7*exp(x)*x*(2*x + 1).
a(n) = A316466(n) - n = A024966(2*n+1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A218471 a(n) = n(7n-3)/2.

Original entry on oeis.org

0, 2, 11, 27, 50, 80, 117, 161, 212, 270, 335, 407, 486, 572, 665, 765, 872, 986, 1107, 1235, 1370, 1512, 1661, 1817, 1980, 2150, 2327, 2511, 2702, 2900, 3105, 3317, 3536, 3762, 3995, 4235, 4482, 4736, 4997, 5265, 5540, 5822, 6111, 6407, 6710, 7020, 7337

Offset: 0

Philippe Deléham, Mar 26 2013

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

Cf. A001106, A022264.

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=7). - Bruno Berselli, Jun 10 2013

List([0..50], n-> n*(7*n-3)/2); # G. C. Greubel, Aug 31 2019

[n*(7*n-3)/2: n in [0..50]]; // G. C. Greubel, Aug 31 2019

seq(n*(7*n-3)/2, n=0..50); # G. C. Greubel, Aug 31 2019

Table[n*(7*n-3)/2, {n,0,50}] (* G. C. Greubel, Aug 23 2017 *)

a(n)=n*(7*n-3)/2 \\ Charles R Greathouse IV, Jun 17 2017

[n*(7*n-3)/2 for n in (0..50)] # G. C. Greubel, Aug 31 2019

G.f.: x*(2+5*x)/(1-x)^3.
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) with a(0)=0, a(1)=2, a(2)=11.
a(n) = A001106(n) +  n.
a(n) = A022264(n) -  n.
a(n) = A022265(n) - 2*n.
a(n) = A186029(n) - 3*n.
a(n) = A179986(n) - 4*n.
a(n) = A024966(n) - 5*n.
a(n) = A174738(7*n+1).
E.g.f.: (x/2)*(7*x + 4)*exp(x). - G. C. Greubel, Aug 23 2017

A163756 14 times triangular numbers.

Original entry on oeis.org

0, 14, 42, 84, 140, 210, 294, 392, 504, 630, 770, 924, 1092, 1274, 1470, 1680, 1904, 2142, 2394, 2660, 2940, 3234, 3542, 3864, 4200, 4550, 4914, 5292, 5684, 6090, 6510, 6944, 7392, 7854, 8330, 8820, 9324, 9842, 10374, 10920, 11480, 12054, 12642, 13244, 13860

Offset: 0

Vincenzo Librandi, Aug 03 2009

Sequence found by reading the line from 0, in the direction 0, 14, ... and the same line from 0, in the direction 0, 42, ..., in the square spiral whose vertices are the generalized 16-gonal numbers. - Omar E. Pol, Oct 03 2011

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

Cf. A000217, A002378, A069127.

Cf. A274978 (generalized 16-gonal numbers).

Table[7*n*(n-1),{n,100}] (* Vladimir Joseph Stephan Orlovsky, Jul 06 2011 *)
14*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,14,42},50] (* Harvey P. Dale, May 11 2021 *)

a(n)=7*n*(n+1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 7*n*(n+1) = 14*A000217(n).
G.f.: 14*x/(1-x)^3.
a(n) = 7*A002378(n) = 2*A024966(n) = A069127(n+1) - 1. - Omar E. Pol, Oct 03 2011
E.g.f.: 7*x*(x + 2)*exp(x). - G. C. Greubel, Aug 02 2017
From Amiram Eldar, Feb 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/7.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/7.
Product_{n>=1} (1 - 1/a(n)) = -(7/Pi)*cos(sqrt(11/7)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (7/Pi)*cos(sqrt(3/7)*Pi/2). (End)

Extended by R. J. Mathar, Aug 06 2009

A101447 Triangle read by rows: T(n,k) = (2k+1)(n+1-k), 0 <= k < n.

Original entry on oeis.org

1, 2, 3, 3, 6, 5, 4, 9, 10, 7, 5, 12, 15, 14, 9, 6, 15, 20, 21, 18, 11, 7, 18, 25, 28, 27, 22, 13, 8, 21, 30, 35, 36, 33, 26, 15, 9, 24, 35, 42, 45, 44, 39, 30, 17, 10, 27, 40, 49, 54, 55, 52, 45, 34, 19, 11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21, 12, 33, 50, 63, 72, 77, 78, 75, 68, 57, 42, 23

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 19 2005

The triangle is generated from the product of matrix A and matrix B, i.e., A * B where A = the infinite lower triangular matrix:
  1 0 0 0 0 ...
  1 1 0 0 0 ...
  1 1 1 0 0 ...
  1 1 1 1 0 ...
  1 1 1 1 1 ...
... and B = the infinite lower triangular matrix:
  1 0 0 0 0 ...
  1 3 0 0 0 ...
  1 3 5 0 0 ...
  1 3 5 7 0 ...
  1 3 5 7 9 ...
  ...
Row sums give the square pyramidal numbers A000330.
T(n+0,0)=1*n=A000027(n+1); T(n+1,1)=3*n=A008585(n); T(n+2,2)=5*n=A008587(n); T(n+3,3)=7*n=A008589(n); etc. So T(n,0)*T(n,1)=3*n*(n+1)=A028896(n) (6 times triangular numbers). T(n,1)*T(n,2)/10=3*n*(n+1)/2=A045943(n) for n>0 T(n,2)*T(n,3)/10=7/2*n*(n+1)=A024966(n) for n>1 (7 times triangular numbers), etc.
From Gary W. Adamson, Apr 25 2010: (Start)
Consider the following array, signed as shown:
  ...
  1,   3,  5,   7,  9,  11, ...
  2,  -6, 10, -14, 18, -22, ...
  3,   9, 15,  21, 27,  33, ...
  4, -12, 20, -28, 36, -44, ...
  5,  15, 25,  35, 45,  55, ...
  6, -18, 30, -42, 54, -66, ...
  7,  21, 35,  49, 63,  77, ...
  ...
Let each term (+, -)k = (+, -) phi^(-k).
Consider the inverse terms of the Lucas series (1/1, 1/3, 1/4, 1/7, ...).
By way of example, let q = phi = 1.6180339...; then
...
1/1  = q^(-1) + q^(-3) + q^(-5) + q^(-7) + q^(-9) + ...
1/3  = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
1/4  = q^(-3) + q^(-9) + q^(-15) + q^(-21) + q^(-27) +...
1/7  = q^(-4) - q^(-12) + q^(-20) - q^(-28) + q^(-36) + ...
1/11 = q^(-5) + q^(-15) + q^(-25) + q^(-35) + q^(-45) + ...
...
Relating to the Pell series, the corresponding "Lucas"-like series is (2, 6, 14, 34, 82, 198, ...) such that herein, q = 2.414213... = (1 + sqrt(2)).
Then analogous to the previous set,
...
1/2 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + ...
1/6 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
... (End)

			From _Bruno Berselli_, Feb 10 2014: (Start)
Triangle begins:
   1;
   2,  3;
   3,  6,  5;
   4,  9, 10,  7;
   5, 12, 15, 14,  9;
   6, 15, 20, 21, 18, 11;
   7, 18, 25, 28, 27, 22, 13;
   8, 21, 30, 35, 36, 33, 26, 15;
   9, 24, 35, 42, 45, 44, 39, 30, 17;
  10, 27, 40, 49, 54, 55, 52, 45, 34, 19;
  11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21;
  etc.
(End)

Cf. A094728 (triangle generated by B*A), A000330.

t[n_, k_] := If[n < k, 0, (2*k + 1)*(n - k + 1)]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 20 2005 *)

PARI
```
T(n,k)=if(n
				
```

A124110 Primes of the form A124080 (10 times triangular numbers) +- 1.

Original entry on oeis.org

11, 29, 31, 59, 61, 101, 149, 151, 211, 281, 359, 449, 659, 661, 911, 1049, 1051, 1201, 1361, 1531, 1709, 1901, 2099, 2309, 2311, 2531, 2999, 3001, 3251, 3511, 3779, 4349, 4649, 4651, 5279, 5281, 6299, 6301, 6659, 6661, 7411, 8609, 9029, 9461, 9901, 11279

Offset: 1

Jonathan Vos Post, Nov 26 2006

Numbers j such that A124080(j)-1 is prime or A124080(j)+1 is prime, where repetition means a twin prime, are 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 8, 9, 11, 11, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 21, 22, 24, 24, 25, ..., . - Robert G. Wilson v, Nov 29 2006

			a(1) = A124080(1)+1 = (10*T(1)) - 1 = 10*(1*(1+1)/2) + 1 = 10+1 = 11 is prime.
a(2) = A124080(2)-1 = (10*T(2))-1 = 10*(2*(2+1)/2) - 1 = 30-1 = 29 is prime.
a(3) = A124080(2)+1 = (10*T(2))+1 = 10*(2*(2+1)/2) + 1 = 30+1 = 31 is prime.

Cf. A000040, A000217, A028895, A046092, A045943, A002378, A028896, A024966, A033996, A027468.

s = {}; Do[t = 5n(n + 1); If[PrimeQ[t - 1], AppendTo[s, t - 1]]; If[PrimeQ[t + 1], AppendTo[s, t + 1]], {n, 47}]; s (* Robert G. Wilson v *)

{A124080(j)-1 when prime} U {A124080(j)+1 when prime} = {i = 10*T(j)-1 such that i is prime} U {i = 10*T(j)+1 such that i is prime} where T(j) = A000217(j) = j*(j+1)/2.

More terms from Robert G. Wilson v, Nov 29 2006

A168622 Triangle read by rows: T(n, k) = [x^k]( 7(1+x)^n - 6(1+x^n) ) with T(0, 0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 14, 1, 1, 21, 21, 1, 1, 28, 42, 28, 1, 1, 35, 70, 70, 35, 1, 1, 42, 105, 140, 105, 42, 1, 1, 49, 147, 245, 245, 147, 49, 1, 1, 56, 196, 392, 490, 392, 196, 56, 1, 1, 63, 252, 588, 882, 882, 588, 252, 63, 1, 1, 70, 315, 840, 1470, 1764, 1470, 840, 315, 70, 1

Offset: 0

Roger L. Bagula and Gary W. Adamson, Dec 01 2009

			Triangle begins as:
  1;
  1,  1;
  1, 14,   1;
  1, 21,  21,   1;
  1, 28,  42,  28,    1;
  1, 35,  70,  70,   35,    1;
  1, 42, 105, 140,  105,   42,    1;
  1, 49, 147, 245,  245,  147,   49,   1;
  1, 56, 196, 392,  490,  392,  196,  56,   1;
  1, 63, 252, 588,  882,  882,  588, 252,  63,  1;
  1, 70, 315, 840, 1470, 1764, 1470, 840, 315, 70, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A022090, A131115, A132047, A168620, A168623.

Columns (essentially): A008589 (k=1), A024966 (k=2).

A168622:= func< n,k | k eq 0 or k eq n select 1 else 7*Binomial(n,k) >;
[A168622(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 10 2025

(* First program *)
p[x_, n_]:= With[{m=3}, If[n==0, 1, (2*m+1)(1+x)^n - 2*m*(1+x^n)]];
Table[CoefficientList[p[x,n], x], {n,0,12}]//Flatten
(* Second program *)
A168622[n_, k_]:= If[k==0 || k==n, 1, 7*Binomial[n,k]];
Table[A168622[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 10 2025 *)

def A168622(n,k):
    if k==0 or k==n: return 1
    else: return 7*binomial(n,k)
print(flatten([[A168622(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 10 2025

From G. C. Greubel, Apr 10 2025: (Start)
T(n, k) = 7*binomial(n, k), with T(n, 0) = T(n, n) = 1.
T(n, n-k) = T(n, k).
Sum_{k=0..n} T(n, k) = 2*A048489(n-1) + 6*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = -6*(1 + (-1)^n) + 13*[n=0].
Sum_{k=0..floor(n/2)} T(n-k, k) = A022090(n+1) - 3*(3 + (-1)^n) + 6*[n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (14/sqrt(3))*(-1)^n*cos((4*n+1)*Pi/6) - 6*(1 + (-1)^n*cos(n*Pi/2)) + 6*[n=0]. (End)

A329530 a(n) = n * (7*binomial(n, 2) + 1).

Original entry on oeis.org

0, 1, 16, 66, 172, 355, 636, 1036, 1576, 2277, 3160, 4246, 5556, 7111, 8932, 11040, 13456, 16201, 19296, 22762, 26620, 30891, 35596, 40756, 46392, 52525, 59176, 66366, 74116, 82447, 91380, 100936, 111136, 122001, 133552, 145810, 158796, 172531, 187036, 202332, 218440

Offset: 0

Ilya Gutkovskiy, Nov 15 2019

Centered heptagonal prism numbers.

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), 144.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Centered m-gonal prism numbers: A100175 (m = 3), A059722 (m = 4), A006564 (m = 5), A005915 (m = 6), this sequence (m = 7), A139757 (m = 8), A006566 (m = 9).

Cf. A000217, A002413, A006597, A024966, A069099.

Table[n (7 Binomial[n, 2] + 1), {n, 0, 40}]
nmax = 40; CoefficientList[Series[x (1 + 12 x + 8 x^2)/(1 - x)^4, {x, 0, nmax}], x]
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 16, 66}, 41]

G.f.: x * (1 + 12*x + 8*x^2) / (1 - x)^4.
E.g.f.: exp(x) * x * (2 + 14*x + 7*x^2) / 2.
a(n) = n * (7*n^2 - 7*n + 2) / 2.
a(n) = n * (7*A000217(n-1) + 1).
a(n) = n * A069099(n).

cp's OEIS Frontend

A022265 a(n) = n*(7*n + 1)/2.

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A195145 Concentric 14-gonal numbers.

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Haskell

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A193053 a(n) = (14*n*(n+3) + (2*n-5)*(-1)^n + 21)/16.

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A195320 7 times hexagonal numbers: a(n) = 7*n*(2*n-1).

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Magma

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A218471 a(n) = n*(7*n-3)/2.

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GAP

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A163756 14 times triangular numbers.

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A101447 Triangle read by rows: T(n,k) = (2*k+1)*(n+1-k), 0 <= k < n.

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A022265 a(n) = n(7n + 1)/2.

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

A195320 7 times hexagonal numbers: a(n) = 7n(2*n-1).

A218471 a(n) = n(7n-3)/2.

A101447 Triangle read by rows: T(n,k) = (2k+1)(n+1-k), 0 <= k < n.

A168622 Triangle read by rows: T(n, k) = [x^k]( 7(1+x)^n - 6(1+x^n) ) with T(0, 0) = 1.