A163274 -id:A163274 - cp's OEIS frontend

A006002 a(n) = n*(n+1)^2/2.

0, 2, 9, 24, 50, 90, 147, 224, 324, 450, 605, 792, 1014, 1274, 1575, 1920, 2312, 2754, 3249, 3800, 4410, 5082, 5819, 6624, 7500, 8450, 9477, 10584, 11774, 13050, 14415, 15872, 17424, 19074, 20825, 22680, 24642, 26714, 28899, 31200, 33620, 36162, 38829, 41624

Offset: 0

N. J. A. Sloane

a(n) is the largest number that is not the sum of distinct numbers of form kn+1, k >= 0. - David W. Wilson, Dec 11 1999
Sum of the nontriangular numbers between successive triangular numbers. 1, (2), 3, (4, 5), 6, (7, 8, 9), 10, (11, 12, 13, 14), 15, ... Sum of the terms in brackets. Or sum of n consecutive integers beginning with T(n) + 1, where T(n) = n(n+1)/2. - Amarnath Murthy, Aug 27 2005
Apparently this is also the splittance (as defined by Hammer & Simeone, 1977) of the Kneser graphs of the form K(n+3,2). - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 13 2009
Row sums of triangle A159797. - Omar E. Pol, Jul 24 2009
The same results occur when one plots the points (1,3), (3,6), (6,10), (10,15), and so on, for all the triangular numbers and finds the area beneath. Take three consecutive triangular numbers and label them a, b, c; the area created is simply (b-a)*(b+c)/2. Thus for 6,10,15 the area beneath the line defined by the points (6,10) and (10,15) is (10-6)*(10+15)/2 = 50. - J. M. Bergot, Jun 28 2011
Let P = ab where a and b are nonequal prime numbers > 1. Let Q be the product of all divisors of P^n. Q can be expressed as P^k, where k = n*(n+1)^2/2. This follows from the fact that all divisors are of the form a^i*b^j, for i,j from 0 to n. An example is given below. In the more general case, where P is the product of m nonequal prime numbers, k = n*(n+1)^m/2. When m=3, the sequence is the same as A092364. - James A. Raymond & Douglas Raymond, Dec 04 2011
For n > 0: sum of n-th row in A014132, seen as a triangle read by rows. - Reinhard Zumkeller, Dec 12 2012
Partial sums of A005449. - Omar E. Pol, Jan 12 2013
a(n) is the sum of x (or y) coordinates for an n X n square lattice in the upper right quadrant of Z^2 whose corner points are (0, 0), (0, n), (n, 0), and (n, n). - Joseph Wheat, Feb 03 2018
a(n) is the number of permutations of [n+2] that contain exactly 2 elements which are not left-to-right minimal. E.g., the 9 permutations comprising a(2) are 2134, 2143, 3124, 3142, 4123, 4132, 2314, 2413, 3412. - Andy Niedermaier, May 07 2022

			Let P^n=6^2. The product of the divisors of 36 = 10077796 = 6^9, i.e., for n=2, k=9. - _James A. Raymond_ & Douglas Raymond, Dec 04 2011

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Quang T. Bach, Roshil Paudyal and Jeffrey B. Remmel, A Fibonacci analogue of Stirling numbers, arXiv preprint arXiv:1510.04310 [math.CO], 2015.
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
Dexter Jane L. Indong and Gilbert R. Peralta, Inversions of permutations in Symmetric, Alternating, and Dihedral Gropus, JIS, Vol. 11 (2008), Article 08.4.3.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for two-way infinite sequences.

Cf. A002411: -a(-1-n).
Cf. A000217, A000292, A000330, A014132, A035006, A045991, A092364, A159797, A163274.
Cf. A000914 (partial sums), A005449 (first differences).
Cf. similar sequences of the type n*(n+1)*(n+k)/2 listed in A267370.
Cf. A002378, A027480.
A bisection of A330298.

List([0..10^3],n->n*(n+1)^2/2); # Muniru A Asiru, Feb 04 2018

a006002 n = n * (n + 1) ^ 2 `div` 2  -- Reinhard Zumkeller, Dec 12 2012

[n*(n+1)^2/2 : n in [0..50]]; // Wesley Ivan Hurt, Nov 17 2014

seq(binomial(n+1,2)*(n+1), n=0..36); # Zerinvary Lajos, Apr 25 2007

Table[(n^3-n^2)/2, {n, 41}] (* Zerinvary Lajos, Mar 21 2007 *)
LinearRecurrence[{4,-6,4,-1}, {0,2,9,24}, 40] (* Harvey P. Dale, Aug 14 2012 *)
Accumulate @ # (# + 1) & [Range[0,50]] (* Waldemar Puszkarz, Jan 24 2015 *)

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

G.f.: x*(x + 2)/(1 - x)^4. - Michael Somos, Jan 30 2004
a(n) = (n + 1) * binomial(n+1, 2). - Zerinvary Lajos, Jan 10 2006
a(n) = A035006(n+1)/4. - Johannes W. Meijer, Feb 04 2010
a(n) = 2*binomial(n+1, 2) + 3*binomial(n+1, 3). - Gary Detlefs, Jun 06 2010
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 14 2012
a(n) = A000292(n) + A000330(n). - Omar E. Pol, Jan 11 2013
a(n) = A045991(n+1)/2. - J. M. Bergot, Aug 10 2013
a(n) = Sum_{j=1..n} Sum_{i=1..j} (2*j - i + 1). - Wesley Ivan Hurt, Nov 17 2014
a(n) = Sum_{i=0..n} n*(n - i) + i. - Bruno Berselli, Jan 13 2016
a(n) = t(n, A000217(n)), where t(h,k) = A000217(h) + h*k. - Bruno Berselli, Feb 28 2017
Sum_{n>0} 1/a(n) = 4 - Pi^2/3. - Jaume Oliver Lafont, Jul 11 2017 [corrected by Amiram Eldar, Jan 28 2022]
E.g.f.: exp(x)*x*(4 + 5*x + x^2)/2. - Stefano Spezia, May 21 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/6 + 4*log(2) - 4. - Amiram Eldar, Jan 28 2022
From J.S. Seneschal, Jun 27 2024: (Start)
a(n) = (A002378(n)^2/2)/n = (n+1)/2 * A002378(n).
a(n) = A027480(n) - A000217(n). (End)

A163102 a(n) = n^2*(n+1)^2/2.

Original entry on oeis.org

0, 2, 18, 72, 200, 450, 882, 1568, 2592, 4050, 6050, 8712, 12168, 16562, 22050, 28800, 36992, 46818, 58482, 72200, 88200, 106722, 128018, 152352, 180000, 211250, 246402, 285768, 329672, 378450, 432450, 492032, 557568, 629442, 708050, 793800, 887112, 988418

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163282.
Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - Thomas Zaslavsky, Jun 26 2010
If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - Bruno Berselli, Sep 05 2014
Subsequence of A000982, see formula. - David James Sycamore, Jul 31 2018
Number of edges in the (n+1) X (n+1) rook complement graph. - Freddy Barrera, May 02 2019
Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - Greg Dresden and Snezhana Tuneska, Aug 24 2023

Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - Thomas Zaslavsky, Jun 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
A. Umar, Combinatorial Results for Semigroups of Orientation-Preserving Partial Transformations, J. Int. Seq., Vol. 14 (2011), Article 11.7.5.
Eric Weisstein's World of Mathematics, Rook Complement Graph.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Column k=2 of A144084.
Cf. A000217, A000537, A000982, A035287.
Cf. A006002, A099903, A163274, A163275, A163276, A163277, A163282.
Cf. A060300, A254371.

List([0..40],n->(n*(n+1))^2/2); # Muniru A Asiru, Aug 02 2018

[n^2*(n+1)^2/2: n in [0..40]]; // Vincenzo Librandi, Mar 26 2012

CoefficientList[Series[2*x*(1+4*x+x^2)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Mar 26 2012 *)

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

a(n) = 2*A000537(n) = A035287(n+1)/2. - Omar E. Pol, Nov 29 2011
G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - R. J. Mathar, Nov 30 2011
Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - J. M. Bergot, Jun 21 2012
a(n) = A000982(n*(n+1)). - David James Sycamore, Jul 31 2018
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.
Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)
Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - J. M. Bergot, Aug 25 2022
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A254371(n)/4 = A060300(n)/8. (End)

A163275 a(n) = n^5*(n+1)^2/2.

Original entry on oeis.org

0, 2, 144, 1944, 12800, 56250, 190512, 537824, 1327104, 2952450, 6050000, 11595672, 21026304, 36386714, 60505200, 97200000, 151519232, 230016834, 341067024, 495219800, 705600000, 988352442, 1363135664, 1853666784

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163285.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A006002, A099903, A163102, A163274, A163276, A163277, A163285.

A163275 := proc(n) n^5*(n+1)^2/2 ; end proc: seq(A163275(n),n=0..60) ; # R. J. Mathar, Feb 05 2010

Table[(1/2)*n^5*(n + 1)^2, {n,0,50}] (* or *) LinearRecurrence[{8,-28,56, -70,56,-28,8,-1}, {0,2,144,1944,12800,56250,190512,537824}, 50] (* G. C. Greubel, Dec 12 2016 *)

concat([0], Vec(2*x*(1+64*x+424*x^2+584*x^3+179*x^4+8*x^5)/(x-1)^8 + O(x^50))) \\ G. C. Greubel, Dec 12 2016

From R. J. Mathar, Feb 05 2010: (Start)
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).
G.f.: 2*x*(1 + 64*x + 424*x^2 + 584*x^3 + 179*x^4  +8*x^5)/(x-1)^8. (End)
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 12 -5*Pi^2/3 - 2*Pi^4/45 + 6*zeta(3) + 2*zeta(5).
Sum_{n>=1} (-1)^(n+1)/a(n) = 20*log(2) + 9*zeta(3)/2 + 15*zeta(5)/8 - 12 - Pi^2/2 - 7*Pi^4/180. (End)

Extended by R. J. Mathar, Feb 05 2010

A163276 a(n) = n^6*(n+1)^2/2.

Original entry on oeis.org

0, 2, 288, 5832, 51200, 281250, 1143072, 3764768, 10616832, 26572050, 60500000, 127552392, 252315648, 473027282, 847072800, 1458000000, 2424307712, 3910286178, 6139206432, 9409176200, 14112000000, 20755401282, 29988984608

Offset: 0

Omar E. Pol, Jul 24 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A006002, A099903, A163102, A163274, A163275, A163277.

[n^6*(n+1)^2/2: n in [0..30]]; // Vincenzo Librandi, Dec 13 2016

seq((1/2)*n^6*(n+1)^2, n = 0 .. 25); # Emeric Deutsch, Aug 01 2009

Table[(1/2)*n^6*(n + 1)^2, {n,0,50}] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1}, {0, 2, 288, 5832, 51200, 281250, 1143072, 3764768, 10616832}, 50] (* G. C. Greubel, Dec 12 2016 *)

concat([0], Vec(2*x*(1 + 135*x +1656*x^2 +4456*x^3 +3231*x^4 +585*x^5 +16*x^6)/(1-x)^9 + O(x^50))) \\ G. C. Greubel, Dec 12 2016

G.f.: 2*x*(1+135*x+1656*x^2+4456*x^3+3231*x^4+585*x^5+16*x^6)/(1-x)^9. - Colin Barker, May 05 2012
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2 + Pi^4/15 + 2*Pi^6/945 - 14 - 8*zeta(3) - 4*zeta(5).
Sum_{n>=1} (-1)^(n+1)/a(n) = 14 + 2*Pi^2/3 + 7*Pi^4/120 + 31*Pi^6/15120 - 24*log(2) - 6*zeta(3) - 15*zeta(5)/4. (End)

Extended by Emeric Deutsch, Aug 01 2009

A163277 a(n) = n^7*(n+1)^2/2.

Original entry on oeis.org

0, 2, 576, 17496, 204800, 1406250, 6858432, 26353376, 84934656, 239148450, 605000000, 1403076312, 3027787776, 6149354666, 11859019200, 21870000000, 38788923392, 66474865026, 110505715776, 178774347800, 282240000000

Offset: 0

Omar E. Pol, Jul 24 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A006002, A099903, A163102, A163274, A163275, A163276.

[n^7*(n+1)^2/2: n in [0..30]]; // Vincenzo Librandi, Dec 13 2016

A163277 := proc(n) n^7*(n+1)^2/2 ; end proc: seq(A163277(n),n=0..60) ; \\ R. J. Mathar, Feb 05 2010

Table[(1/2)*n^7*(n + 1)^2, {n,0,50}] (* G. C. Greubel, Dec 12 2016 *)

concat([0], Vec(2*x*(1 +278*x +5913*x^2 +27760*x^3 +38435*x^4 +16434*x^5 +1867*x^6 +32*x^7)/(x-1)^10 + O(x^50))) \\ G. C. Greubel, Dec 12 2016

From R. J. Mathar, Feb 05 2010: (Start)
a(n) = n^2*A163275(n).
G.f.: 2*x*(1 +278*x +5913*x^2 +27760*x^3 +38435*x^4 +16434*x^5 +1867*x^6 +32*x^7)/(x-1)^10. (End)
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 16 - 7*Pi^2/3 - 4*Pi^4/45 - 4*Pi^6/945 + 10*zeta(3) + 6*zeta(5) + 2*zeta(7).
Sum_{n>=1} (-1)^(n+1)/a(n) = 28*log(2) + 15*zeta(3)/2 + 45*zeta(5)/8 + 63*zeta(7)/32 - 16 - 5*Pi^2/6 - 7*Pi^4/90 - 31*Pi^6/7560. (End)

Extended by R. J. Mathar, Feb 05 2010

cp's OEIS Frontend

A006002 a(n) = n*(n+1)^2/2.

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A163102 a(n) = n^2*(n+1)^2/2.

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A163275 a(n) = n^5*(n+1)^2/2.

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A163276 a(n) = n^6*(n+1)^2/2.

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A163277 a(n) = n^7*(n+1)^2/2.

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