A028895 -id:A028895 - cp's OEIS frontend

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

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)

A130520 a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265, 275, 286, 297, 308, 319, 330, 342, 354, 366

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary with A130483 regarding triangular numbers, in that A130483(n) + 5*a(n) = n*(n+1)/2 = A000217(n).

Given a sequence b(n) defined by variables b(0) to b(5) and recursion b(n) = -(b(n-6) * a(n-2) * (b(n-4) * b(n-2)^3 - b(n-3)^3 * b(n-1)) - b(n-5) * b(n-3) * b(n-1) * (b(n-5) * b(n-2)^2 - b(n-4)^2 * b(n-1)))/(b(n-4) * (b(n-5) * b(n-3)^3 - b(n-4)^3 * b(n-2))). The denominator of b(n+1) has a factor of (b(1) * b(3)^3 - b(2)^3 * b(4))^a(n+1). For example, if b(0) = 2, b(1) = b(2) = b(3) = 1, b(4) = 1+x, b(5) = 4, then the denominator of b(n+1) is x^a(n+1). - Michael Somos, Nov 15 2023

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

Cf. A002264, A002265, A004526, A010872, A010873, A010874, A130481, A130482.

List([0..70], n-> Int((n-1)*(n-2)/10)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-3)/10): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

seq(floor((n-1)*(n-2)/10), n=0..70); # G. C. Greubel, Aug 31 2019

Accumulate[Floor[Range[0,70]/5]] (* Harvey P. Dale, May 25 2016 *)

a(n) = sum(k=0, n, k\5); \\ Michel Marcus, May 13 2016

[floor((n-1)*(n-2)/10) for n in (0..70)] # G. C. Greubel, Aug 31 2019

a(n) = floor(n/5)*(2*n - 3 - 5*floor(n/5))/2.
a(n) = A002266(n)*(2*n - 3 - 5*A002266(n))/2.
a(n) = A002266(n)*(n -3 +A010874(n))/2.
G.f.: x^5/((1-x^5)*(1-x)^2) =  x^5/( (1+x+x^2+x^3+x^4)*(1-x)^3 ).
a(n) = floor((n-1)*(n-2)/10). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-3)/10) = ceiling((n+1)*(n-4)/10) = round((n^2 - 3*n - 1)/10). - Mircea Merca, Nov 28 2010
a(n) = A008732(n-5), n > 4. - R. J. Mathar, Nov 22 2008
a(n) = a(n-5) + n - 4, n > 4. - Mircea Merca, Nov 28 2010
a(5n) = A000566(n), a(5n+1) = A005476(n), a(5n+2) = A005475(n), a(5n+3) = A147875(n), a(5n+4) = A028895(n). - Philippe Deléham, Mar 26 2013
From Amiram Eldar, Sep 17 2022: (Start)
Sum_{n>=5} 1/a(n) = 518/45 - 2*sqrt(2*(sqrt(5)+5))*Pi/3.
Sum_{n>=5} (-1)^(n+1)/a(n) = 8*sqrt(5)*arccoth(3/sqrt(5))/3 + 92*log(2)/15 - 418/45. (End)

A022267 a(n) = n(9n + 1)/2.

Original entry on oeis.org

0, 5, 19, 42, 74, 115, 165, 224, 292, 369, 455, 550, 654, 767, 889, 1020, 1160, 1309, 1467, 1634, 1810, 1995, 2189, 2392, 2604, 2825, 3055, 3294, 3542, 3799, 4065, 4340, 4624, 4917, 5219, 5530, 5850, 6179

Offset: 0

N. J. A. Sloane

From Floor van Lamoen, Jul 21 2001: (Start)
Write 0, 1, 2, 3, 4, ... in a triangular spiral; then a(n) is the sequence found by reading the line from 0 in the direction 0, 5, ... . The spiral begins:
.
            15
            / \
          16  14
          /     \
        17   3  13
        /   / \   \
      18   4   2  12
      /   /     \   \
    19   5   0---1  11
    /   /             \
  20   6---7---8---9--10
.
(End)
a(n) is the sum of n consecutive integers starting from 4*n+1: (5), (9+10), (13+14+15), ... - Klaus Purath, Jul 07 2020
a(n) with n>0 are the numbers with the periodic length 3 in the Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022

Lei Zhou, Table of n, a(n) for n = 0..10000
Milan Janjic, Two Enumerative Functions
Leo Tavares, Illustration: X Triangles
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A051682, A110449, A235332.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. similar sequences listed in A254963.
Cf. similar sequences listed in A022289.
Cf. A060544, A016813.

seq(binomial(9*n+1,2)/9, n=0..37); # Zerinvary Lajos, Jan 21 2007

Table[ n (9 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 19}, 40] (* Harvey P. Dale, Jul 01 2013 *)

vector(100,n,(n-1)*(9*n-8)/2) \\ Derek Orr, Feb 06 2015

a(n) = A110449(n, 4) for n>3.
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 4*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A000566(n). (End)
a(n) = 9*n + a(n-1) - 4 with n>0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = A218470(9*n+4). - Philippe Deléham, Mar 27 2013
a(n) = A000217(5*n) - A000217(4*n). - Bruno Berselli, Oct 13 2016
E.g.f.: (1/2)*(9*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
a(n) = A060544(n+1) - A016813(n). - Leo Tavares, Mar 20 2022

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

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)

A195142 Concentric 10-gonal numbers.

Original entry on oeis.org

0, 1, 10, 21, 40, 61, 90, 121, 160, 201, 250, 301, 360, 421, 490, 561, 640, 721, 810, 901, 1000, 1101, 1210, 1321, 1440, 1561, 1690, 1821, 1960, 2101, 2250, 2401, 2560, 2721, 2890, 3061, 3240, 3421, 3610, 3801, 4000, 4201, 4410, 4621, 4840, 5061, 5290

Offset: 0

Omar E. Pol, Sep 17 2011

Also concentric decagonal numbers. Also sequence found by reading the line from 0, in the direction 0, 10, ..., and the same line from 1, in the direction 1, 21, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. Main axis, perpendicular to A028895 in the same spiral.

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

A033583 and A069133 interleaved.
Cf. A028895, A032527, A032528, A077221, A085787, A195042, A195143, A195145, A195146, A195147, A195148, A195149.
Cf. A090771 (first differences).
Column 10 of A195040. - Omar E. Pol, Sep 28 2011

a195142 n = a195142_list !! n
a195142_list = scanl (+) 0 a090771_list
-- Reinhard Zumkeller, Jan 07 2012

[(10*n^2+3*(-1)^n-3)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

RecurrenceTable[{a[0]==0,a[1]==1,a[n]==a[n-2]+10(n-1)},a[n],{n,50}] (* or *) LinearRecurrence[{2,0,-2,1},{0,1,10,21},50] (* Harvey P. Dale, Sep 29 2011 *)

G.f.: -x*(1+8*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = -a(n-1) + 5*n^2 - 5*n + 1, a(0)=0. - Vincenzo Librandi, Sep 27 2011
From Bruno Berselli, Sep 27 2011: (Start)
a(n) = a(-n) = (10*n^2 + 3*(-1)^n - 3)/4.
a(n) = a(n-2) + 10*(n-1). (End)
a(n) = 2*a(n-1) + 0*a(n-2) - 2*a(n-3) + a(n-4); a(0)=0, a(1)=1, a(2)=10, a(3)=21. - Harvey P. Dale, Sep 29 2011
Sum_{n>=1} 1/a(n) = Pi^2/60 + tan(sqrt(3/5)*Pi/2)*Pi/(2*sqrt(15)). - Amiram Eldar, Jan 16 2023

A195014 Vertex number of a square spiral whose edges have length A195013.

Original entry on oeis.org

0, 2, 5, 9, 15, 21, 30, 38, 50, 60, 75, 87, 105, 119, 140, 156, 180, 198, 225, 245, 275, 297, 330, 354, 390, 416, 455, 483, 525, 555, 600, 632, 680, 714, 765, 801, 855, 893, 950, 990, 1050, 1092, 1155, 1199, 1265, 1311, 1380, 1428, 1500, 1550, 1625, 1677

Offset: 0

Omar E. Pol, Sep 09 2011

Zero together with the partial partial sums of A195013.
Second bisection is 2, 9, 21, 38, 60, 87, 119, ...: A005476. - Omar E. Pol, Sep 25 2011
Number of pairs (x,y) with even x in {0,...,n}, odd y in {0,...,3n}, and xClark Kimberling, Jul 02 2012

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

Cf. A005476, A028895, A195013, A195020.

[(10*n^2 + 18*n + 3 + (2*n - 3)*(-1)^n)/16 : n in [0..50]]; // Vincenzo Librandi, Oct 26 2014

LinearRecurrence[{1,2,-2,-1,1},{0,2,5,9,15},60] (* Harvey P. Dale, May 20 2019 *)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
G.f.: f(x)/g(x), where f(x) = 2*x + 3*x^2 and g(x) = (1+x)^2 * (1-x)^3. - Clark Kimberling, Jul 02 2012
a(n) = (10*n^2 + 18*n + 3 + (2*n - 3)*(-1)^n)/16. - Luce ETIENNE, Aug 11 2014

A008732 Molien series for 3-dimensional group [2,n] = *22n.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265

Offset: 0

N. J. A. Sloane

			From _Philippe Deléham_, Apr 05 2013: (Start)
Stored in five columns:
    1   2   3   4   5
    7   9  11  13  15
   18  21  24  27  30
   34  38  42  46  50
   55  60  65  70  75
   81  87  93  99 105
  112 119 126 133 140
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 188
Brian O'Sullivan and Thomas Busch, Spontaneous emission in ultra-cold spin-polarised anisotropic Fermi seas, arXiv 0810.0231v1 [quant-ph], 2008. [Eq 8a, lambda=5]
Index entries for Molien series
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Cf. A130520.

List([0..50], n-> Int((n+3)*(n+4)/10)); # G. C. Greubel, Jul 30 2019

[Floor((n+3)*(n+4)/10): n in [0..50] ]; // Vincenzo Librandi, Aug 21 2011

A092202 := proc(n) op(1+(n mod 5),[0,1,0,-1,0]) ; end proc:
A010891 := proc(n) op(1+(n mod 5),[1,-1,0,0,0]) ; end proc:
A008732 := proc(n) (n+2)*(n+5)/10+(A010891(n-1)+2*A092202(n-1))/5 ; end proc:

LinearRecurrence[{2, -1, 0, 0, 1, -2, 1}, {1, 2, 3, 4, 5, 7, 9}, 50] (* Jean-François Alcover, Jan 18 2018 *)

a(n)=(n+3)*(n+4)\10 \\ Charles R Greathouse IV, Oct 07 2015

[floor((n+3)*(n+4)/10) for n in (0..50)] # G. C. Greubel, Jul 30 2019

a(n) = floor( (n+3)*(n+4)/10 ) = (n+2)*(n+5)/10 + b(n)/5 where b(n) = A010891(n-2) + 2*A092202(n-1) = 0, 1, 1, 0, -2, ... with period length 5.
G.f.: 1/((1-x)^2*(1-x^5)).
a(n) = a(n-5) + n + 1. - Paul Barry, Jul 14 2004
From Mitch Harris, Sep 08 2008: (Start)
a(n) = Sum_{j=0..n+5} floor(j/5).
a(n-5) = (1/2)floor(n/5)*(2*n - 3 - 5*floor(n/5)). (End)
a(n) = A130520(n+5). - Philippe Deléham, Apr 05 2013
a(5n) = A000566(n+1), a(5n+1) = A005476(n+1), a(5n+2) = A005475(n+1), a(5n+3) = A147875(n+2), a(5n+4) = A028895(n+1); these formulas correspond to the 5 columns of the array shown in example. - Philippe Deléham, Apr 05 2013

A152745 5 times hexagonal numbers: 5n(2*n-1).

Original entry on oeis.org

0, 5, 30, 75, 140, 225, 330, 455, 600, 765, 950, 1155, 1380, 1625, 1890, 2175, 2480, 2805, 3150, 3515, 3900, 4305, 4730, 5175, 5640, 6125, 6630, 7155, 7700, 8265, 8850, 9455, 10080, 10725, 11390, 12075, 12780, 13505, 14250, 15015

Offset: 0

Omar E. Pol, Dec 12 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Sep 18 2011

Also sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. This is one of the four semi-diagonals of the spiral. - Omar E. Pol, Oct 14 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. A000384, A085250, A152746.

Bisection of A028895.

[5*n*(2*n-1): n in [0..50]]; // G. C. Greubel, Sep 01 2018

LinearRecurrence[{3,-3,1}, {0, 5, 30}, 50] (* or *) Table[5*n*(2*n-1), {n,0,50}] (* G. C. Greubel, Sep 01 2018 *)

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

a(n) = 10*n^2 - 5*n = A000384(n)*5.
a(n) = a(n-1) + 20*n-15 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
From G. C. Greubel, Sep 01 2018: (Start)
G.f.: 5*x*(1+ 3*x)/(1-x)^3.
E.g.f.: 5*x*(1+2*x)*exp(x). (End)
From Vaclav Kotesovec, Sep 02 2018: (Start)
Sum_{n>=1} 1/a(n) = 2*log(2)/5.
Sum_{n>=1} (-1)^n/a(n) = log(2)/5 - Pi/10. (End)

A194715 15 times triangular numbers.

Original entry on oeis.org

0, 15, 45, 90, 150, 225, 315, 420, 540, 675, 825, 990, 1170, 1365, 1575, 1800, 2040, 2295, 2565, 2850, 3150, 3465, 3795, 4140, 4500, 4875, 5265, 5670, 6090, 6525, 6975, 7440, 7920, 8415, 8925, 9450, 9990, 10545, 11115, 11700, 12300, 12915, 13545, 14190, 14850, 15525

Offset: 0

Omar E. Pol, Oct 03 2011

Sequence found by reading the line from 0, in the direction 0, 15, ... and the same line from 0, in the direction 0, 45, ..., in the square spiral whose vertices are the generalized 17-gonal numbers.
Sum of the numbers from 7*n to 8*n. - Wesley Ivan Hurt, Dec 23 2015
Also the number of 4-cycles in the (n+6)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 28 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Eric Weisstein's World of Mathematics, Graph Cycle.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A028895, A035008, A045943, A069128, A163756.

Cf. A001105 (3-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017

[15*n*(n+1)/2: n in [0..50]]; // Vincenzo Librandi, Oct 04 2011

A194715:=n->15*n*(n+1)/2: seq(A194715(n), n=0..60); # Wesley Ivan Hurt, Dec 23 2015

15*Accumulate[Range[0, 60]] (* Harvey P. Dale, Feb 12 2012 *)
Table[15 n (n + 1)/2, {n, 0, 60}] (* Wesley Ivan Hurt, Dec 23 2015 *)
15 Binomial[Range[20], 2] (* Eric W. Weisstein, Jul 28 2017 *)
15 PolygonalNumber[Range[0, 20]] (* Eric W. Weisstein, Jul 28 2017 *)

a(n)=15*n*(n+1)/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 15*n*(n+1)/2 = 15*A000217(n) = 5*A045943(n) = 3*A028895(n) = A069128(n+1) - 1.
From Wesley Ivan Hurt, Dec 23 2015: (Start)
G.f.: 15*x/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = Sum_{i=7*n..8*n} i. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2/15.
Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/15.
Product_{n>=1} (1 - 1/a(n)) = -(15/(2*Pi))*cos(sqrt(23/15)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (15/(2*Pi))*cos(sqrt(7/15)*Pi/2). (End)
E.g.f.: 15*exp(x)*x*(2 + x)/2. - Elmo R. Oliveira, Dec 25 2024

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A130520 a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A022267 a(n) = n*(9*n + 1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A139273 a(n) = n*(8*n - 3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A195142 Concentric 10-gonal numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Formula

A195014 Vertex number of a square spiral whose edges have length A195013.

Views

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

A022267 a(n) = n(9n + 1)/2.

A139273 a(n) = n(8n - 3).

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

A152745 5 times hexagonal numbers: 5n(2*n-1).