A028896 -id:A028896 - cp's OEIS frontend

A028895 5 times triangular numbers: a(n) = 5n(n+1)/2.

0, 5, 15, 30, 50, 75, 105, 140, 180, 225, 275, 330, 390, 455, 525, 600, 680, 765, 855, 950, 1050, 1155, 1265, 1380, 1500, 1625, 1755, 1890, 2030, 2175, 2325, 2480, 2640, 2805, 2975, 3150, 3330, 3515, 3705, 3900, 4100, 4305, 4515, 4730, 4950, 5175, 5405, 5640

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

Sequence found by reading the line from 0, in the direction 0, 5, ... and the same line from 0, in the direction 0, 15, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. Axis perpendicular to A195142 in the same spiral. - Omar E. Pol, Sep 18 2011
Bisection of A195014. Sequence found by reading the line from 0, in the direction 0, 5, ..., and the same line from 0, in the direction 0, 15, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. This is the main diagonal of the spiral. - Omar E. Pol, Sep 25 2011
a(n) = the Wiener index of the graph obtained by applying Mycielski's construction to the complete graph K(n) (n>=2). - Emeric Deutsch, Aug 29 2013
Sum of the numbers from 2*n to 3*n for n=0,1,2,... - Wesley Ivan Hurt, Nov 27 2015
Numbers k such that the concatenation k625 is a square, where also 625 is a square. - Bruno Berselli, Nov 07 2018
From Paul Curtz, Nov 29 2019: (Start)
Main column of the pentagonal spiral for n (A001477):
                         50
                      49 30 31
                   48 29 15 16 32
                47 28 14  5  6 17 33
             46 27 13  4  0  1  7 18 34
               45 26 12  3  2  8 19 35
                 44 25 11 10  9 20 36
                   43 24 23 22 21 37
                    42 41 40 39 38
(End)

D. B. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, NJ, 2001, p. 205.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Rangaswami Balakrishnan and S. Francis Raj, The Wiener number of powers of the Mycielskian, Discussiones Math. Graph Theory, Vol. 30, No. 3 (2010), pp. 489-498 (see Theorem 2.1).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001068, A005891, A028896, A046092, A055998, A085787, A130520, A195013, A195014, A195142.
Cf. index to numbers of the form n*(d*n+10-d)/2 in A140090.
Cf. A000566, A005475, A005476, A033583, A085787, A147875, A192136, A326725 (all in the spiral).

[5*n*(n+1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014

[seq(5*binomial(n,2), n=1..45)]; # Zerinvary Lajos, Nov 24 2006

Table[Sum[i + 2*n - 1, {i, 2, n}], {n, 45}] (* Zerinvary Lajos, Jul 11 2009 *)
Table[5 n (n + 1)/2, {n, 0, 50}] (* Bruno Berselli, Sep 23 2016 *)
5 Accumulate[Range[0,60]] (* Harvey P. Dale, Jun 17 2025 *)

a(n)=5*n*(n+1)/2 \\ Charles R Greathouse IV, Sep 24 2015

def A028895(n): return 5*n*(n+1)>>1 # Chai Wah Wu, Aug 07 2025

G.f.: 5*x/(1-x)^3.
a(n) = 5*n*(n+1)/2 = 5*A000217(n).
a(n+1) = 5*n+a(n). - Vincenzo Librandi, Aug 05 2010
a(n) = A005891(n) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A130520(5n+4). - Philippe Deléham, Mar 26 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. - Wesley Ivan Hurt, Nov 27 2015
a(n) = Sum_{i=0..n} A001068(4i). - Wesley Ivan Hurt, May 06 2016
E.g.f.: 5*x*(2 + x)*exp(x)/2. - Ilya Gutkovskiy, May 06 2016
a(n) = A055998(3*n) - A055998(2*n). - Bruno Berselli, Sep 23 2016
From Amiram Eldar, Feb 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/5)*(2*log(2) - 1). (End)
Product_{n>=1} (1 - 1/a(n)) = -(5/(2*Pi))*cos(sqrt(13/5)*Pi/2). - Amiram Eldar, Feb 21 2023

A027468 9 times the triangular numbers A000217.

Original entry on oeis.org

0, 9, 27, 54, 90, 135, 189, 252, 324, 405, 495, 594, 702, 819, 945, 1080, 1224, 1377, 1539, 1710, 1890, 2079, 2277, 2484, 2700, 2925, 3159, 3402, 3654, 3915, 4185, 4464, 4752, 5049, 5355, 5670, 5994, 6327, 6669, 7020, 7380, 7749, 8127, 8514, 8910, 9315

Offset: 0

Olivier Gérard

Staggered diagonal of triangular spiral in A051682, between (0,1,11) spoke and (0,8,25) spoke. - Paul Barry, Mar 15 2003
Number of permutations of n distinct letters (ABCD...) each of which appears thrice with n-2 fixed points. - Zerinvary Lajos, Oct 15 2006
Number of n permutations (n>=2) of 4 objects u, v, z, x with repetition allowed, containing n-2=0 u's. Example: if n=2 then n-2 =zero (0) u, a(1)=9 because we have vv, zz, xx, vx, xv, zx, xz, vz, zv. A027465 formatted as a triangular array: diagonal: 9, 27, 54, 90, 135, 189, 252, 324, ... . - Zerinvary Lajos, Aug 06 2008
a(n) is also the least weight of self-conjugate partitions having n different parts such that each part is a multiple of 3. - Augustine O. Munagi, Dec 18 2008
Also sequence found by reading the line from 0, in the direction 0, 9, ..., and the same line from 0, in the direction 0, 27, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Axis perpendicular to A195147 in the same spiral. - Omar E. Pol, Sep 18 2011
Sum of the numbers from 4*n to 5*n. - Wesley Ivan Hurt, Nov 01 2014

			The first such self-conjugate partitions, corresponding to a(n)=1,2,3,4 are 3+3+3, 6+6+6+3+3+3, 9+9+9+6+6+6+3+3+3, 12+12+12+9+9+9+6+6+6+3+3+3. - _Augustine O. Munagi_, Dec 18 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Augustine O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., Vol. 308, No. 12 (2008), pp. 2492-2501.
Enrique Navarrete and Daniel Orellana, Finding Prime Numbers as Fixed Points of Sequences, arXiv:1907.10023 [math.NT], 2019.
Leo Tavares, Illustration: Centroid Triangles.
D. Zvonkine, Counting ramified coverings and intersection theory on Hurwitz spaces II (local structure of Hurwitz spaces and combinatorial results), Moscow Mathematical Journal, Vol. 7, No. 1 (2007), pp. 135-162.
D. Zvonkine, Home Page.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Index entries for two-way infinite sequences.

Third diagonal of A027465.
Cf. A000459, A002378, A008585, A024966, A028895, A028896, A038764, A033996, A045943, A046092, A049598, A059073, A080855, A134171, A283394.
Cf. A060544.

[9*n*(n+1)/2: n in [0..50]]; // Vincenzo Librandi, Dec 29 2012

[seq(9*binomial(n+1,2), n=0..50)]; # Zerinvary Lajos, Nov 24 2006

Table[(9/2)*n*(n+1), {n,0,50}] (* G. C. Greubel, Aug 22 2017 *)

PARI
```
a(n)=9*n*(n+1)/2
```

[9*binomial(n+1, 2) for n in (0..50)] # G. C. Greubel, May 20 2021

Numerators of sequence a[n, n-2] in (a[i, j])^2 where a[i, j] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i.
a(n) = (9/2)*n*(n+1).
a(n) = 9*C(n, 1) + 9*C(n, 2) (binomial transform of (0, 9, 9, 0, 0, ...)). - Paul Barry, Mar 15 2003
G.f.: 9*x/(1-x)^3.
a(-1-n) = a(n).
a(n) = 9*C(n+1,2), n>=0. - Zerinvary Lajos, Aug 06 2008
a(n) = a(n-1) + 9*n (with a(0)=0). - Vincenzo Librandi, Nov 19 2010
a(n) = A060544(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A218470(9*n+8). - Philippe Deléham, Mar 27 2013
E.g.f.: (9/2)*x*(x+2)*exp(x). - G. C. Greubel, Aug 22 2017
a(n) = A060544(n+1) - 1. See Centroid Triangles illustration. - Leo Tavares, Dec 27 2021
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/9 - 2/9. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(9/(2*Pi))*cos(sqrt(17)*Pi/6).
Product_{n>=1} (1 + 1/a(n)) = 9*sqrt(3)/(4*Pi). (End)

More terms from Patrick De Geest, Oct 15 1999

A017629 a(n) = 12*n + 9.

Original entry on oeis.org

9, 21, 33, 45, 57, 69, 81, 93, 105, 117, 129, 141, 153, 165, 177, 189, 201, 213, 225, 237, 249, 261, 273, 285, 297, 309, 321, 333, 345, 357, 369, 381, 393, 405, 417, 429, 441, 453, 465, 477, 489, 501, 513, 525, 537, 549, 561, 573, 585, 597, 609, 621, 633

Offset: 0

N. J. A. Sloane

Numbers k such that k mod 2 = (k+1) mod 3 = 1 and (k+2) mod 4 != 1. - Klaus Brockhaus, Jun 15 2004
For n > 3, the number of squares on the infinite 3-column chessboard at <= n knight moves from any fixed point. - Ralf Stephan, Sep 15 2004
A016946 is the subsequence of squares (for n = 3*k*(k+1) = A028896(k), then a(n) = (6k+3)^2 = A016946(k)). - Bernard Schott, Apr 05 2021

Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Subsequences include A072065, A016945, A016813, A008585, and A005408.

Cf. A008594, A017533, A017545, A017137, A004767, A001019, A028896, A016946, A089911.

a017629 = (+ 9) . (* 12)  -- Reinhard Zumkeller, Jul 05 2013

12*Range[0,200]+9 (* Vladimir Joseph Stephan Orlovsky, Feb 19 2011 *)
LinearRecurrence[{2,-1},{9,21},60] (* Harvey P. Dale, Apr 14 2019 *)

a(n)=12*n+9 \\ Charles R Greathouse IV, Jul 10 2016

[i+9 for i in range(525) if gcd(i,12) == 12] # Zerinvary Lajos, May 21 2009

a(n) = 6*(4*n+1) - a(n-1) (with a(0)=9). - Vincenzo Librandi, Dec 17 2010
A089911(2*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
G.f.: (9 + 3*x)/(1 - x)^2. - Alejandro J. Becerra Jr., Jul 08 2020
Sum_{n>=0} (-1)^n/a(n) = (Pi + log(3-2*sqrt(2)))/(12*sqrt(2)). - Amiram Eldar, Dec 12 2021
E.g.f.: 3*exp(x)*(3 + 4*x). - Stefano Spezia, Feb 25 2023

A024966 7 times triangular numbers: 7n(n+1)/2.

Original entry on oeis.org

0, 7, 21, 42, 70, 105, 147, 196, 252, 315, 385, 462, 546, 637, 735, 840, 952, 1071, 1197, 1330, 1470, 1617, 1771, 1932, 2100, 2275, 2457, 2646, 2842, 3045, 3255, 3472, 3696, 3927, 4165, 4410, 4662, 4921, 5187, 5460, 5740, 6027, 6321, 6622

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

Sequence found by reading the line from 0, in the direction 0, 7, ... and the same line from 0, in the direction 1, 21, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the main diagonal in the spiral. - Omar E. Pol, Sep 09 2011
Also sequence found by reading the same line mentioned above in the square spiral whose vertices are the generalized enneagonal numbers A118277. Axis perpendicular to A195145 in the same spiral. - Omar E. Pol, Sep 18 2011
Sequence provides all integers m such that 56*m + 49 is a square. - Bruno Berselli, Oct 07 2015
Sum of the numbers from 3*n to 4*n. - Wesley Ivan Hurt, Dec 22 2015

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

Cf. A000217, A001106, A002378, A022264, A022265, A028895, A028896, A033996.
Cf. A045943, A046092, A069099, A118277, A174738, A179986, A186029, A193053.
Cf. A195019, A195020, A195145, A218471.

[ (7*n^2 + 7*n)/2 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

[seq(7*binomial(n,2), n=1..44)]; # Zerinvary Lajos, Nov 24 2006

7 Table[n (n + 1)/2, {n, 0, 43}] (* or *)
Table[Sum[i, {i, 3 n, 4 n}], {n, 0, 43}] (* or *)
Table[SeriesCoefficient[7 x/(1 - x)^3, {x, 0, n}], {n, 0, 43}] (* Michael De Vlieger, Dec 22 2015 *)
7*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,7,21},50] (* Harvey P. Dale, Jul 20 2025 *)

x='x+O('x^100); concat(0, Vec(7*x/(1-x)^3)) \\ Altug Alkan, Dec 23 2015

a(n) = (7/2)*n*(n+1).
G.f.: 7*x/(1-x)^3.
a(n) = (7*n^2 + 7*n)/2 = 7*A000217(n). - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 7*n with n > 0, a(0)=0. - Vincenzo Librandi, Nov 19 2010
a(n) = A069099(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = a(-n-1), a(n+2) = A193053(n+2) + 2*A193053(n+1) + A193053(n). - Bruno Berselli, Oct 21 2011
From Philippe Deléham, Mar 26 2013: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 7, a(2) = 21.
a(n) = A174738(7*n+6).
a(n) = A179986(n) + n = A186029(n) + 2*n = A022265(n) + 3*n = A022264(n) + 4*n = A218471(n) + 5*n = A001106(n) + 6*n. (End)
a(n) = Sum_{i=3*n..4*n} i. - Wesley Ivan Hurt, Dec 22 2015
E.g.f.: (7/2)*x*(x+2)*exp(x). - G. C. Greubel, Aug 19 2017
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/7.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/7)*(2*log(2) - 1). (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(7/(2*Pi))*cos(sqrt(15/7)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (7/(2*Pi))*cosh(Pi/(2*sqrt(7))). (End)

A055112 a(n) = n(n+1)(2*n+1).

Original entry on oeis.org

0, 6, 30, 84, 180, 330, 546, 840, 1224, 1710, 2310, 3036, 3900, 4914, 6090, 7440, 8976, 10710, 12654, 14820, 17220, 19866, 22770, 25944, 29400, 33150, 37206, 41580, 46284, 51330, 56730, 62496, 68640, 75174, 82110, 89460, 97236, 105450

Offset: 0

Henry Bottomley, Jun 15 2000

Original name: Areas of Pythagorean triangles (X, Y, Z = Y + 1) with X^2 + Y^2 = Z^2.
a(n) is the set of possible y values for 4*x^3 + x^2 = y^2 with the x values being A002378(n). - Gary Detlefs, Feb 22 2010
This sequence is related to A028896 by a(n) = n*A028896(n) - Sum_{i = 0..n-1} A028896(i) and this is the case d = 3 in the identity n*(d*(d+1)*n*(n+1)/4) - Sum_{i = 0..n-1} d*(d+1)*i*(i+1)/4 = d*(d+1)*n*(n+1)*(2*n+1)/12. - Bruno Berselli, Mar 31 2012
Also sums of rows of natural numbers (cf. A001477) seen as triangle with an odd numbers of terms per row, see example. - Reinhard Zumkeller, Jan 24 2013
Without mentioning the connection to Pythagorean triangles, Bolker (1967) gives it as an exercise to prove that these numbers are always divisible by 6. This is easy to prove from the formula that he gives, n(n - 1)(2n - 1): obviously either n or (n - 1) must be even; then, if n is congruent to 2 mod 3 it means that (2n - 1) is a multiple of 3, otherwise either n or (n - 1) is a multiple of 3; thus both prime divisors of 6 are accounted for in a(n). - Alonso del Arte, Oct 13 2013
a(n) = n*(n+1)*(n+(n+1)) is the product of two consecutive integers multiplied by the sum of those two consecutive integers. - Charles Kusniec, Sep 04 2022

			.  n   A001477(n) as triangle with row lengths = 2*n+1   Row sums = a(n)
.  0                         0                                  0
.  1                      1  2  3                               6
.  2                   4  5  6  7  8                           30
.  3                9 10 11 12 13 14 15                        84
.  4            16 17 18 19 20 21 22 23 24                    180
.  5         25 26 27 28 29 30 31 32 33 34 35                 330
.  6      36 37 38 39 40 41 42 43 44 45 46 47 48              546
.  7   49 50 51 52 53 54 55 56 57 58 59 60 61 62 63           840 .
- _Reinhard Zumkeller_, Jan 24 2013

Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.5.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. Roy, The discovery of the series formula for π by Leibniz, Gregory and Nilakantha, Mathematics Magazine, 63 (5) 1990, 291-306.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A005408 (X values), A046092 (Y values), A001844 (Z values), A002939 (perimeter), A033581.
Similar sequences are listed in A316224.
Cf. A000217, A000290, A005898, A083374.

Table[n(n + 1)(2n + 1), {n, 0, 39}] (* Vladimir Joseph Stephan Orlovsky, Nov 21 2010 *)
LinearRecurrence[{4,-6,4,-1},{0,6,30,84},50] (* Harvey P. Dale, Oct 02 2024 *)

PARI
```
a(n)=n*(n+1)*(2*n+1);
```

def A055112(n): return n*(n*((n<<1) + 3) + 1) # Chai Wah Wu, Nov 14 2022

a(n) = n*(n+1)*(2*n+1).
G.f.: 6*x*(1+x)/(1-x)^4. - Bruno Berselli, Mar 31 2012
From Benoit Cloitre, Apr 30 2002: (Start)
a(n) = 6*A000330(n) = A007531(2*n)/4 = 3*A000292(2*n-1)/2 = A005408(n)*A046092(n)/2 = A005408(n)*(A001844(n)-1)/2.
Sum_{n > 0} 1/a(n) = 3 - 4*log(2). (End)
a(n) = Sum_{i = 1..n} A033581(i). - Jonathan Vos Post, Mar 15 2006
a(n) = A000217(2*n)*A000217(2*n+1)/(2*n+1). - Charlie Marion, Feb 17 2012
a(n) = Sum_{i = 1..2*n + 1} (n^2 + (i-1)). - Charlie Marion, Sep 14 2012
Sum_{n >= 1} (-1)^(n+1)/a(n) = Pi - 3, due to Nilakantha, circa 1500. See Roy p. 304. - Peter Bala, Feb 19 2015
a(n) = A002378(n) * (2n+1). - Bruce J. Nicholson, Aug 31 2017
a(n) = Sum_{k=n^2..(n+1)^2-1} k. - Darío Clavijo, Jan 31 2025
E.g.f.: exp(x)*x*(6 + 9*x + 2*x^2). - Stefano Spezia, Feb 02 2025
a(n) = A005898(n) - A005408(n) = A083374(n+1) - A083374(n). - J.S. Seneschal, Jul 08 2025

A124080 10 times triangular numbers: a(n) = 5n(n + 1).

Original entry on oeis.org

0, 10, 30, 60, 100, 150, 210, 280, 360, 450, 550, 660, 780, 910, 1050, 1200, 1360, 1530, 1710, 1900, 2100, 2310, 2530, 2760, 3000, 3250, 3510, 3780, 4060, 4350, 4650, 4960, 5280, 5610, 5950, 6300, 6660, 7030, 7410, 7800, 8200, 8610, 9030, 9460, 9900, 10350

Offset: 0

Zerinvary Lajos, Nov 24 2006

If Y is a 5-subset of an n-set X then, for n >= 5, a(n-4) is equal to the number of 5-subsets of X having exactly three elements in common with Y. Y is a 5-subset of an n-set X then, for n >= 6, a(n-6) is the number of (n-5)-subsets of X having exactly two elements in common with Y. - Milan Janjic, Dec 28 2007

Also sequence found by reading the line from 0, in the direction 0, 10, ... and the same line from 0, in the direction 0, 30, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Axis perpendicular to A195148 in the same spiral. - Omar E. Pol, Sep 18 2011

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

Cf. A028895, A046092, A045943, A002378, A028896, A024966, A033996, A027468, A049598, A000217.

[ 5*n*(n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

[seq(10*binomial(n,2),n=1..51)];
seq(n*(n+1)*5, n=0..39); # Zerinvary Lajos, Mar 06 2007

10*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,10,30},50] (* Harvey P. Dale, Jul 21 2011 *)

a(n)=5*n*(n+1) \\ Charles R Greathouse IV, Sep 28 2015

a(n) = 10*C(n,2), n >= 1.
a(n) = A049598(n) - A002378(n). - Zerinvary Lajos, Mar 06 2007
a(n) = 5*n*(n + 1), n >= 0. - Zerinvary Lajos, Mar 06 2007
a(n) = 5*n^2 + 5*n = 10*A000217(n) = 5*A002378(n) = 2*A028895(n). - Omar E. Pol, Dec 12 2008
a(n) = 10*n + a(n-1) (with a(0) = 0). - Vincenzo Librandi, Nov 12 2009
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0) = 0, a(1) = 10, a(2) = 30. - Harvey P. Dale, Jul 21 2011
a(n) = A062786(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A131242(10*n+9). - Philippe Deléham, Mar 27 2013
From G. C. Greubel, Aug 22 2017: (Start)
G.f.: 10*x/(1 - x)^3.
E.g.f.: 5*x*(x + 2)*exp(x). (End)
From Amiram Eldar, Sep 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2)-1)/5. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(5/Pi)*cos(3*Pi/(2*sqrt(5))).
Product_{n>=1} (1 + 1/a(n)) = (5/Pi)*cos(Pi/(2*sqrt(5))). (End)

A152773 3 times heptagonal numbers: a(n) = 3n(5*n-3)/2.

Original entry on oeis.org

0, 3, 21, 54, 102, 165, 243, 336, 444, 567, 705, 858, 1026, 1209, 1407, 1620, 1848, 2091, 2349, 2622, 2910, 3213, 3531, 3864, 4212, 4575, 4953, 5346, 5754, 6177, 6615, 7068, 7536, 8019, 8517, 9030, 9558, 10101, 10659, 11232, 11820, 12423, 13041, 13674, 14322, 14985

Offset: 0

Omar E. Pol, Dec 13 2008

Also the number of 6-cycles in the (n+5)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jun 25 2017

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Graph Cycle.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000566, A001622, A135706, A226489.
3 times n-gonal numbers: A045943, A033428, A062741, A094159, A152751, A152759, A152767, A153783, A153448, A153875.
Cf. numbers of the form n*(n*k - k + 6)/2, this sequence is the case k=15: see Comments lines of A226492.
Cf. A002378 (3-cycles in triangular honeycomb acute knight graph), A045943 (4-cycles), A028896 (5-cycles).

Table[3 n (5 n - 3)/2, {n, 0, 50}] (* Harvey P. Dale, May 08 2012 *)
LinearRecurrence[{3, -3, 1}, {0, 3, 21}, 50] (* Harvey P. Dale, May 08 2012 *)
CoefficientList[Series[-((3 x^5 (1 + 4 x))/(-1 + x)^3), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 25 2017 *)

a(n)=3*n*(5*n-3)/2 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = (15*n^2 - 9*n)/2 = 3*A000566(n).
a(n) = a(n-1) + 15*n - 12 with n > 0, a(0)=0. - Vincenzo Librandi, Nov 26 2010
G.f.: 3*x*(1+4*x)/(1-x)^3. - Bruno Berselli, Jan 21 2011
a(0)=0, a(1)=3, a(2)=21, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 08 2012
a(n) = n + A226489(n). - Bruno Berselli, Jun 11 2013
Sum_{n>=1} 1/a(n) = tan(Pi/10)*Pi/9 - sqrt(5)*log(phi)/9 + 5*log(5)/18, where phi is the golden ratio (A001622). - Amiram Eldar, May 20 2023
E.g.f.: 3*exp(x)*x*(2 + 5*x)/2. - Elmo R. Oliveira, Dec 24 2024

A174709 Partial sums of floor(n/6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 18, 21, 24, 27, 30, 33, 36, 40, 44, 48, 52, 56, 60, 65, 70, 75, 80, 85, 90, 96, 102, 108, 114, 120, 126, 133, 140, 147, 154, 161, 168, 176, 184, 192

Offset: 0

Mircea Merca, Nov 30 2010

Partial sums of A152467.

			a(7) = floor(0/6) + floor(1/6) + floor(2/6) + floor(3/6) + floor(4/6) + floor(5/6) + floor(6/6) + floor(7/6) = 0 + 0 + 0 + 0 + 0 + 0 + 1 + 1 = 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions, J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,1,-2,1).

Cf. A008724, A152467.

[Round(n*(n-4)/12): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Maple
```
a(n):=round(n*(n-4)/12)
```

Table[Sum[Floor[k/6], {k,0,n}], {n,0,25}] (* G. C. Greubel, Dec 13 2016 *)

a(n)=(n-2)^2\12 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = round(n*(n-4)/12) = round((2*n^2 - 8*n - 1)/24).
a(n) = floor((n-2)^2/12).
a(n) = ceiling((n+1)*(n-5)/12).
a(n) = a(n-6) + n - 5, n > 5.
From R. J. Mathar, Nov 30 2010: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-6) - 2*a(n-7) + a(n-8).
G.f.: -x^6 / ( (1+x)*(x^2-x+1)*(1+x+x^2)*(x-1)^3 ).
a(n) = -n/3 + 5/72 + n^2/12 + (-1)^n/24 + A057079(n+5)/6 + A061347(n)/18. (End)
a(6n) = A000567(n), a(6n+1) = 2*A000326(n), a(6n+2) = A033428(n), a(6n+3) = A049451(n), a(6n+4) = A045944(n), a(6n+5) = A028896(n). - Philippe Deléham, Mar 26 2013
a(n) = A008724(n-2). - R. J. Mathar, Jul 10 2015
Sum_{n>=6} 1/a(n) = Pi^2/18 - Pi/(2*sqrt(3)) + 49/12. - Amiram Eldar, Aug 13 2022

A195143 a(n) = n-th concentric 12-gonal number.

Original entry on oeis.org

0, 1, 12, 25, 48, 73, 108, 145, 192, 241, 300, 361, 432, 505, 588, 673, 768, 865, 972, 1081, 1200, 1321, 1452, 1585, 1728, 1873, 2028, 2185, 2352, 2521, 2700, 2881, 3072, 3265, 3468, 3673, 3888, 4105, 4332, 4561, 4800, 5041, 5292, 5545, 5808, 6073, 6348

Offset: 0

Omar E. Pol, Sep 17 2011

Concentric dodecagonal numbers. [corrected by Ivan Panchenko, Nov 09 2013]
Sequence found by reading the line from 0, in the direction 0, 12,..., and the same line from 1, in the direction 1, 25,..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Main axis, perpendicular to A028896 in the same spiral.
Partial sums of A091998. - Reinhard Zumkeller, Jan 07 2012
Column 12 of A195040. - Omar E. Pol, Sep 28 2011

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

A135453 and A069190 interleaved.
Cf. A001082, A032527, A032528, A077221, A195043, A195045, A195040, A195142, A195145, A195146, A195147, A195148, A195149.
Cf. A016921 (6n+1), A016969 (6n+5), A091998 (positive integers of the form 12*k +- 1), A092242 (positive integers of the form 12*k +- 5).

a195143 n = a195143_list !! n
a195143_list = scanl (+) 0 a091998_list
-- Reinhard Zumkeller, Jan 07 2012

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

Table[Sum[2*(-1)^(n - k + 1) + 6*k - 3, {k, n}], {n, 0, 47}] (* L. Edson Jeffery, Sep 14 2014 *)

From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = 3*n^2+(-1)^n-1.
a(n) = -a(n-1) + 6*n^2 - 6*n + 1. (End)
G.f.: -x*(1+10*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = Sum_{k=1..n} (2*(-1)^(n-k+1) + 3*(2*k-1)), n>0, a(0) = 0. - L. Edson Jeffery, Sep 14 2014
Sum_{n>=1} 1/a(n) = Pi^2/72 + tan(Pi/sqrt(6))*Pi/(4*sqrt(6)). - Amiram Eldar, Jan 16 2023

A199530 T(n,k)=Number of -k..k arrays x(0..n-1) of n elements with zero sum and no two consecutive zero elements.

Original entry on oeis.org

1, 1, 2, 1, 4, 6, 1, 6, 18, 12, 1, 8, 36, 72, 32, 1, 10, 60, 212, 320, 80, 1, 12, 90, 464, 1324, 1414, 200, 1, 14, 126, 860, 3734, 8342, 6346, 520, 1, 16, 168, 1432, 8470, 30484, 53302, 28766, 1336, 1, 18, 216, 2212, 16682, 84852, 252154, 343710, 131246, 3472, 1, 20, 270

Offset: 1

R. H. Hardin Nov 07 2011

Table starts
....1......1........1.........1.........1..........1...........1...........1
....2......4........6.........8........10.........12..........14..........16
....6.....18.......36........60........90........126.........168.........216
...12.....72......212.......464.......860.......1432........2212........3232
...32....320.....1324......3734......8470......16682.......29750.......49284
...80...1414.....8342.....30484.....84852.....197962......407946......766664
..200...6346....53302....252154....860854....2378412.....5662636....12071420
..520..28766...343710...2105064...8815392...28844590....79345982...191873280
.1336.131246..2232322..17701326..90927530..352355640..1119873360..3071898666
.3472.602390.14582218.149708146.943302430.4329146404.15897133212.49465959068

			Some solutions for n=6 k=5
..1...-1....0...-4...-1....4....1...-5....1....2...-1....1....3....3....1....5
..0...-1...-1...-3....0....5...-3....0....1...-3...-3...-4....2...-5....1....2
.-4....5....3...-1...-2...-3....5....1...-1....5....0....2...-3....4...-4...-5
..3...-2...-3....2....2...-4...-1....2...-4....4...-2....2....2...-5....5...-1
..1...-3...-2....4...-4...-4....0....5....2...-5....5....4...-4....0...-2....1
.-1....2....3....2....5....2...-2...-3....1...-3....1...-5....0....3...-1...-2

R. H. Hardin, Table of n, a(n) for n = 1..1096

Column 1 is A102881

Row 3 is A028896

Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k
T(3,k) = 3*k^2 + 3*k
T(4,k) = (16/3)*k^3 + 8*k^2 - (4/3)*k
T(5,k) = (115/12)*k^4 + (115/6)*k^3 + (41/12)*k^2 - (1/6)*k
T(6,k) = (88/5)*k^5 + 44*k^4 + (58/3)*k^3 - 3*k^2 + (31/15)*k
T(7,k) = (5887/180)*k^6 + (5887/60)*k^5 + (620/9)*k^4 + (11/12)*k^3 + (433/180)*k^2 - (91/30)*k

cp's OEIS Frontend

A028895 5 times triangular numbers: a(n) = 5*n*(n+1)/2.

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A027468 9 times the triangular numbers A000217.

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A017629 a(n) = 12*n + 9.

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A024966 7 times triangular numbers: 7*n*(n+1)/2.

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A055112 a(n) = n*(n+1)*(2*n+1).

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A124080 10 times triangular numbers: a(n) = 5*n*(n + 1).

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A028895 5 times triangular numbers: a(n) = 5n(n+1)/2.

A024966 7 times triangular numbers: 7n(n+1)/2.

A055112 a(n) = n(n+1)(2*n+1).

A124080 10 times triangular numbers: a(n) = 5n(n + 1).

A152773 3 times heptagonal numbers: a(n) = 3n(5*n-3)/2.