A002411 -id:A002411 - cp's OEIS frontend

A093375 Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 18, 8, 1, 6, 25, 40, 30, 10, 1, 7, 36, 75, 80, 45, 12, 1, 8, 49, 126, 175, 140, 63, 14, 1, 9, 64, 196, 336, 350, 224, 84, 16, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 11, 100, 405, 960, 1470, 1512, 1050, 480, 135, 20, 1, 12

Offset: 1

Ralf Stephan, Apr 28 2004

Number of n-long m-ary words avoiding the pattern 1-1'2'.
T(n,n+1) = Sum_{i=1..n} T(n,i).
Exponential Riordan array [(1+x)e^x, x] as a number triangle. - Paul Barry, Feb 17 2009
From Peter Bala, Jul 22 2014: (Start)
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A059298. (End)

			Array T(m,n) (with rows m >= 1 and columns n >= 1) begins as follows:
   1   1   1   1   1   1 ...
   2   4   6   8  10  12 ...
   3   9  18  30  45  63 ...
   4  16  40  80 140 224 ...
   5  25  75 175 350 630 ...
   ...
Triangle S(n,k) = T(n-k+1, k+1) begins
.n\k.|....0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....2....1
..2..|....3....4....1
..3..|....4....9....6....1
..4..|....5...16...18....8....1
..5..|....6...25...40...30...10....1
..6..|....7...36...75...80...45...12....1
...

Muniru A Asiru, Antidiagonals, n=1..100 flattened
Yasemin Alp and E. Gokcen Kocer, Exponential Almost-Riordan Arrays, Results Math 79, 173 (2024). See page 7.
Sergey Kitaev and Toufik Mansour, Partially ordered generalized patterns and k-ary words, arXiv:math/0210023 [math.CO], 2002.
Wikipedia, Sheffer sequence.

Rows include A045943. Columns include A002411, A027810.
Main diagonal is A037965. Subdiagonals include A002457.
Antidiagonal sums are A001792.
See A103283 for a signed version.
Cf. A103406, A059298, A073107 (unsigned inverse).

nmax:=14;; T:=List([1..nmax],n->List([1..nmax],k->k*Binomial(n+k-2,n-1)));;
b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][1]][b[i][j][2]]))); # Muniru A Asiru, Aug 07 2018

nmax = 10;
T = Transpose[CoefficientList[# + O[z]^(nmax+1), z]& /@ CoefficientList[(1 - x z)/(1 - z - x z)^2 + O[x]^(nmax+1), x]];
row[n_] := T[[n+1, 1 ;; n+1]];
Table[row[n], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Aug 07 2018 *)

# uses[riordan_array from A256893]
riordan_array((1+x)*exp(x), x, 8, exp=true) # Peter Luschny, Nov 02 2019

Triangle = P*M, the binomial transform of the infinite bidiagonal matrix M with (1,1,1,...) in the main diagonal and (1,2,3,...) in the subdiagonal, and zeros elsewhere. P = Pascal's triangle as an infinite lower triangular matrix. - Gary W. Adamson, Nov 05 2006
From Peter Bala, Sep 20 2012: (Start)
E.g.f. for triangle: (1 + z)*exp((1 + x)*z) = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2/2! + ....
O.g.f. for triangle: (1 - x*z)/(1 - z - x*z)^2 = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2 + ....
The n-th row polynomial R(n,x) of the triangle equals (1+x)^n + n*(1+x)^(n-1) for n >= 0 and satisfies d/dx(R(n,x)) = n*R(n-1,x), as well as R(n,x+y) = Sum_{k = 0..n} binomial(n,k)*R(k,x)*y^(n-k). The row polynomials are a Sheffer sequence of Appell type.
Matrix inverse of the triangle is a signed version of A073107. (End)
From Tom Copeland, Oct 20 2015: (Start)
With offset 0 and D = d/dx, the raising operator for the signed row polynomials P(n,x) is RP = x - d{log[e^D/(1-D)]}/dD = x - 1 - 1/(1-D) =  x - 2 - D - D^2 + ..., i.e., RP P(n,x) = P(n+1,x).
The e.g.f. for the signed array is (1-t) * e^(-t) * e^(x*t).
From the Appell formalism, the row polynomials PI(n,x) of A073107 are the umbral inverse of this entry's row polynomials; that is, P(n,PI(.,x)) = x^n = PI(n,P(.,x)) under umbral composition. (End)
From Petros Hadjicostas, Nov 01 2019: (Start)
As a triangle, we let S(n,k) = T(n-k+1, k+1) = (n-k+1)*binomial(n, k) for n >= 0 and 0 <= k <= n. See the example below.
As stated above by Peter Bala, Sum_{n,k >= 0} S(n,k)*z^n*x^k = (1 - x*z)/(1 - z -x*z)^2.
Also, Sum_{n, k >= 0} S(n,k)*z^n*x^k/n! = (1+z)*exp((1+x)*z).
As he also states, the n-th row polynomial is R(n,x) = Sum_{k = 0..n} S(n, k)*x^k = (1 + x)^n + n*(1 + x)^(n-1).
If we define the signed triangle S*(n,k) = (-1)^(n+k) * S(n,k) = (-1)^(n+k) * T(n-k+1, k+1), as Tom Copeland states, Sum_{n,k >= 0} S^*(n,k)*t^n*x^k/n! = (1-t)*exp((1-x)*(-t)) = (1-t) * e^(-t) * e^(x*t).
Apparently, S*(n,k) = A103283(n,k).
As he says above, the signed n-th row polynomial is P(n,x) =  (-1)^n*R(n,-x) = (x - 1)^n - n*(x - 1)^(n-1).
According to Gary W. Adamson, P(n,x) is "the monic characteristic polynomial of the n X n matrix with 2's on the diagonal and 1's elsewhere." (End)

A125232 Triangle T(n,k) read by rows: the (n-k)-th term of the k-fold iterated partial sum of the pentagonal numbers.

Original entry on oeis.org

1, 5, 1, 12, 6, 1, 22, 18, 7, 1, 35, 40, 25, 8, 1, 51, 75, 65, 33, 9, 1, 70, 126, 140, 98, 42, 10, 1, 92, 196, 266, 238, 140, 52, 11, 1, 117, 288, 462, 504, 378, 192, 63, 12, 1, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1, 176, 550, 1155, 1716, 1848, 1452, 825, 330, 88, 14, 1

Offset: 1

Gary W. Adamson, Nov 24 2006

			First few rows of the triangle are:
   1;
   5,   1;
  12,   6,   1;
  22,  18,   7,   1;
  35,  40,  25,   8,   1;
  51,  75,  65,  33,   9,   1;
  70, 126, 140,  98,  42,  10,   1;
  ...
Example: (5,3) = 65 = 25 + 40 = (4,3) + (4,2).

Albert H. Beiler, "Recreations in the Theory of Numbers", Dover, 1966, p 189.

Robert Israel, Table of n, a(n) for n = 1..10011(rows 0 to 140, flattened)

Columns: A000326 (pentagonal numbers), A002411, A001296, A051836, A051923.

Cf. A095264 (row sums).

A125232 := proc(n,k) option remember ; if k = 0 then A000326(n) ; elif k = n-1 then 1 ; else procname(n-1,k)+procname(n-1,k-1) ; fi : end: # R. J. Mathar, Jun 09 2008

nmax = 11; col[1] = Table[n(3n-1)/2, {n, 1, nmax}]; col[k_] := col[k] = Prepend[Accumulate[col[k-1]], 0]; Table[col[k][[n]], {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 25 2019 *)

T(n,0)=A000326(n). T(n,k)=T(n-1,k) + T(n-1,k-1), k>0. - R. J. Mathar, Jun 09 2008

G.f. as triangle: (1+2*x)/((1-x)^2*(1-x-x*y)). - Robert Israel, Nov 07 2016

Edited and extended by R. J. Mathar, Jun 09 2008

A181617 Molecular topological indices of the complete graph K_n.

Original entry on oeis.org

0, 4, 24, 72, 160, 300, 504, 784, 1152, 1620, 2200, 2904, 3744, 4732, 5880, 7200, 8704, 10404, 12312, 14440, 16800, 19404, 22264, 25392, 28800, 32500, 36504, 40824, 45472, 50460, 55800, 61504, 67584, 74052, 80920, 88200, 95904, 104044, 112632, 121680, 131200

Offset: 1

Eric W. Weisstein, Jul 10 2011

a(n) = the area of a trapezoid with vertices at (n-1,n), (n,n-1), ((n-1)^2,n^2), and (n^2,(n-1)^2). - J. M. Bergot, Mar 23 2014
For n > 3, also the detour index of the (n-1)-helm graph. - Eric W. Weisstein, Dec 16 2017
a(n-3) is the maximum sigma irregularity over all maximal 2-degenerate graphs with n vertices.  The extremal graphs are 2-stars (K_2 joined to n-2 independent vertices).  (The sigma irregularity of a graph is the sum of the squares of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 14 2023

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Eric Weisstein's World of Mathematics, Complete Graph.
Eric Weisstein's World of Mathematics, Detour Index.
Eric Weisstein's World of Mathematics, Helm Graph.
Eric Weisstein's World of Mathematics, Molecular Topological Index.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002411.

Cf. A011379, A181617, A270205 (sigma irregularities of maximal k-degenerate graphs).

[2*n*(n-1)^2: n in [1..50]]; // Vincenzo Librandi, Mar 24 2014

CoefficientList[Series[4 x (1 + 2 x)/(1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 24 2014 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 4, 24, 72}, 50] (* Harvey P. Dale, Jun 16 2016 *)
Table[2 n (n - 1)^2, {n, 20}] (* Eric W. Weisstein, Dec 16 2017 *)

a(n) = 2*n*(n-1)^2; \\ Joerg Arndt, Mar 24 2014

a(n) = 2*n*(n-1)^2.
a(n) = 4*A002411(n).
G.f.: 4*x^2*(1+2*x)/(1-x)^4. - Colin Barker, Nov 04 2012
From Amiram Eldar, Jan 22 2023: (Start)
Sum_{n>=2} 1/a(n) = Pi^2/12 - 1/2.
Sum_{n>=2} (-1)^n/a(n) = Pi^2/24 - log(2) + 1/2. (End)

More terms from Joerg Arndt, Mar 24 2014

A220084 a(n) = (n + 1)(20n^2 + 19*n + 6)/6.

Original entry on oeis.org

1, 15, 62, 162, 335, 601, 980, 1492, 2157, 2995, 4026, 5270, 6747, 8477, 10480, 12776, 15385, 18327, 21622, 25290, 29351, 33825, 38732, 44092, 49925, 56251, 63090, 70462, 78387, 86885, 95976, 105680, 116017, 127007, 138670, 151026, 164095, 177897, 192452

Offset: 0

Bruno Berselli, Dec 11 2012

Sequence related to heptagonal pyramidal numbers (A002413) by a(n) = n*A002413(n) - (n-1)*A002413(n-1).
Other sequences of numbers of the form m*P(k,m)-(m-1)*P(k,m-1), where P(k,m) is the m-th k-gonal pyramidal number:
k=3, A002412(m) = m*A000292(m)-(m-1)*A000292(m-1);
k=4, A051662(m) = (m+1)*A000330(m+1)-m*A000330(m);
k=5, A213772(m) = m*A002411(m)-(m-1)*A002411(m-1);
k=6, A213837(m) = m*A002412(m)-(m-1)*A002412(m-1);
k=7, this sequence;
k=8, A130748(m) = m*A002414(m)-(m-1)*A002414(m-1).
Also, first bisection of A212983.
Binomial transform of (1, 14, 33, 20, 0, 0, 0, ...). - Gary W. Adamson, Aug 26 2015

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A000330, A000566, A002411, A002412, A002413, A002414, A051662, A130748, A212983, A213772, A213837.

[(n+1)*(20*n^2+19*n+6)/6: n in [0..40]]; // Bruno Berselli, Jun 28 2016

/* By first comment: */  A002413:=func; [n*A002413(n)-(n-1)*A002413(n-1): n in [1..40]];

I:=[1,15,62,162]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

Table[(n + 1) (20 n^2 + 19 n + 6)/6, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{1,15,62,162},40] (* Harvey P. Dale, Dec 23 2012 *)
CoefficientList[Series[(1 + 11 x + 8 x^2) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist((n+1)*(20*n^2+19*n+6)/6, n, 0, 20); /* Martin Ettl, Dec 12 2012 */

a(n)=(n+1)*(20*n^2+19*n+6)/6 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1+11*x+8*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3, a(0)=1, a(1)=15, a(2)=62, a(3)=162. - Harvey P. Dale, Dec 23 2012
a(n) = (n+1)*A000566(n+1) + Sum_{i=0..n} A000566(i). - Bruno Berselli, Dec 18 2013
E.g.f.: exp(x)*(6 + 84*x + 99*x^2 + 20*x^3)/6. - Elmo R. Oliveira, Aug 06 2025

A270205 Number of 2 X 2 planar subsets in an n X n X n cube.

Original entry on oeis.org

0, 0, 6, 36, 108, 240, 450, 756, 1176, 1728, 2430, 3300, 4356, 5616, 7098, 8820, 10800, 13056, 15606, 18468, 21660, 25200, 29106, 33396, 38088, 43200, 48750, 54756, 61236, 68208, 75690, 83700, 92256

Offset: 0

Craig Knecht, Mar 13 2016

William H. Press looked at the hybrid structure of a most-perfect magic square and the Hilbert space filling curve and thought it might be the "most uniform" way of putting the consecutive integers in a 2-d square. He thought a definition of "most uniform" would be useful.
Al Zimmermann suggested this: Start by defining the "non-uniformity of a distribution of integers among the cells of a square [or cube or hypercube]" to be the standard deviation of the sums of the 2 X 2 planar subsets. Then define a "most uniform distribution of integers" to be a distribution with the smallest non-uniformity. For both the most-perfect square and most-perfect cube the non-uniformity is 0 and so each is a most uniform distribution. (Of course, you'd want a better word for "non-uniformity". Skewness?) Perhaps use "2 X 2 planar subset" instead of "2 X 2 partition"?
Comment from Dwane Campbell: For cubes, the definition of compact is that all 2 X 2 X 2 subcubes add to the same sum. That definition also includes wrap around. Your most perfect space cube is compact. It has the additional constraint that each orthogonal plane is also compact. There are 64 2 X 2 X 2 subcubes that add to 260 and 192 2 X 2 subsquares that add to 130 in your cube. I did not think either result was possible. Congratulations!
The most-perfect order 4 cube and the reversible order 4 cube are the new findings to look at in the link section.
Most-perfect magic squares require every 2 X 2 cell block to have the same sum. This sequence looks at that same subset in the cube.
Most-perfect space is defined as a structure where all these 2 X 2 subsets have the same sum.
What structure provides the most uniform distribution of integers in a cube?
a(n+1) is the number of unit faces required to make an n X n X n cubic lattice. Number of unit edges required for the same is A059986(n). - Mohammed Yaseen, Aug 22 2021
a(n-3) is the maximum sigma irregularity over all maximal 3-degenerate graphs with n vertices.  The extremal graphs are 3-stars (K_3 joined to n-3 independent vertices).  (The sigma irregularity of a graph is the sum of the squares of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 14 2023

			The 2 X 2 X 2 cube labeled with the integers 1 to 8 has the following six 2 X 2 planar subsets each containing 4 cells: 1,2,3,4; 5,6,7,8; 1,2,5,6; 3,4,7,8; 1,4,5,8; 2,3,6,7.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Craig Knecht, Cube assembly from different 2x2 planar criteria.
Craig Knecht, F1 code most-perfect magic cube 960 examples.
Craig Knecht, F1 code reversible cube 960 examples.
Craig Knecht, magic space.
Craig Knecht, Most-perfect space.
Walter Trump, Most-Perfect magic cube.
Walter Trump, 6 unique neighbors for the most-perfect magic cube.
Wikipedia, Most-perfect magic square translated to a cube via the Hilbert space filling curve.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002411, A059986, A118659.

Cf. A011379, A181617, A270205 (sigma irregularities of maximal k-degenerate graphs).

[3*n^3 - 6*n^2 + 3*n: n in [0..50]]; // Wesley Ivan Hurt, Mar 13 2016

A270205:=n->3*n^3-6*n^2+3*n: seq(A270205(n), n=0..50); # Wesley Ivan Hurt, Mar 13 2016

Table[3*n^3 - 6*n^2 + 3*n, {n, 0, 50}] (* Wesley Ivan Hurt, Mar 13 2016 *)
CoefficientList[Series[(6 (x^2 + 2 x^3))/(-1 + x)^4, {x, 0, 32}], x] (* Michael De Vlieger, Mar 15 2016 *)

concat([0, 0], Vec(6*x^2*(1+2*x)/(x-1)^4 + O(x^100))) \\ Altug Alkan, Mar 14 2016

a(n) = 3*n^3 - 6*n^2 + 3*n \\ Charles R Greathouse IV, Mar 15 2016

a(n) = 3*n^3 - 6*n^2 + 3*n.
From Wesley Ivan Hurt, Mar 13 2016: (Start)
G.f.: 6*x^2*(1+2*x)/(x-1)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
E.g.f.: 3*x^2*(1+x)*exp(x). - G. C. Greubel, May 10 2016
a(n) = 6 * A002411(n-1) for n>=1. - Joerg Arndt, May 11 2016
a(n) = A118659((n-1)^3), n>1. - Mohammed Yaseen, Aug 22 2021
From Amiram Eldar, Jul 02 2023: (Start)
Sum_{n>=2} 1/a(n) = Pi^2/18 - 1/3.
Sum_{n>=2} (-1)^n/a(n) = Pi^2/36 - 2*log(2)/3 + 1/3. (End)

A279221 Expansion of Product_{k>=1} 1/(1 - x^(k^2*(k+1)/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 12, 12, 12, 12, 13, 13, 16, 16, 16, 16, 17, 17, 20, 20, 20, 20, 21, 21, 25, 25, 25, 25, 27, 27, 31, 31, 31, 31, 33, 33, 37, 37, 37, 37, 39, 39, 44, 44, 44, 45, 48, 48, 53, 53, 54, 55, 58, 58, 63, 63, 64, 65, 68, 68, 74

Offset: 0

Ilya Gutkovskiy, Dec 08 2016

nonn

Number of partitions of n into nonzero pentagonal pyramidal numbers (A002411).

			a(7) = 2 because we have [6, 1] and [1, 1, 1, 1, 1, 1, 1].

M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
Eric Weisstein's World of Mathematics, Pentagonal Pyramidal Number
Index to sequences related to pyramidal numbers
Index entries for related partition-counting sequences

Cf. A002411, A068980, A279220, A279222, A279223, A279224.

nmax=90; CoefficientList[Series[Product[1/(1 - x^(k^2 (k + 1)/2)), {k, 1, nmax}], {x, 0, nmax}], x]

G.f.: Product_{k>=1} 1/(1 - x^(k^2*(k+1)/2)).

A003777 a(n) = n^3 + n^2 - 1.

Original entry on oeis.org

1, 11, 35, 79, 149, 251, 391, 575, 809, 1099, 1451, 1871, 2365, 2939, 3599, 4351, 5201, 6155, 7219, 8399, 9701, 11131, 12695, 14399, 16249, 18251, 20411, 22735, 25229, 27899, 30751, 33791, 37025, 40459, 44099, 47951, 52021, 56315, 60839, 65599, 70601, 75851

Offset: 1

N. J. A. Sloane, Jun 14 1998

This sequence in related to A095794 by a(n) = n*A095794(n) - Sum_{i=1..n-1} A095794(i), for n > 1. - Bruno Berselli, Dec 28 2010
a(n) is the area of a triangle with vertices at points (n-1,(n-1)^2), (n,n^2), and ((n+1)^2,n+1). - J. M. Bergot, Jun 03 2014
Old name was: "Number of stacks of n pikelets, distance 3 flips from a well-ordered stack".

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A001107, A002411, A028294, A028295, A095794.

[(n^3+n^2-1): n in [1..50]]; // Vincenzo Librandi, Apr 06 2011

A003777:=n->n^3+n^2-1; seq(A003777(n), n=1..50); # Wesley Ivan Hurt, Jun 04 2014

Table[n^3+n^2-1,{n,100}] (* Vladimir Joseph Stephan Orlovsky, Mar 09 2011 *)
CoefficientList[Series[(1 + 7 x - 3 x^2 + x^3)/(1-x)^4, {x,0,50}], x] (* Vincenzo Librandi, Jun 20 2013 *)
LinearRecurrence[{4,-6,4,-1},{1,11,35,79},50] (* Harvey P. Dale, Jul 20 2024 *)

a(n)=n^3+n^2-1 \\ Charles R Greathouse IV, Oct 07 2015

[n^3+n^2-1 for n in range(1,51)] # G. C. Greubel, Jan 03 2024

G.f.: x*(1+7*x-3*x^2+x^3)/(1-x)^4. Also, a(n) = 2*A002411(n) - 1 = A000578(n-1) + A001107(n). - Bruno Berselli, Dec 28 2010
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4. - Wesley Ivan Hurt, Oct 08 2017
E.g.f.: 1 + (-1 + 2*x + 4*x^2 + x^3)*exp(x). - G. C. Greubel, Jan 03 2024

More terms from Wesley Ivan Hurt, Jun 04 2014

Entry revised by N. J. A. Sloane, Jun 15 2014

A100119 a(n) = n-th centered n-gonal number.

Original entry on oeis.org

1, 2, 7, 19, 41, 76, 127, 197, 289, 406, 551, 727, 937, 1184, 1471, 1801, 2177, 2602, 3079, 3611, 4201, 4852, 5567, 6349, 7201, 8126, 9127, 10207, 11369, 12616, 13951, 15377, 16897, 18514, 20231, 22051, 23977, 26012, 28159, 30421, 32801, 35302

Offset: 0

Jonathan Vos Post, Dec 26 2004

a(n) is n times the n-th triangular number plus 1. - Thomas M. Green, Nov 16 2009
From Gary W. Adamson, Jul 31 2010: (Start)
Equals (1, 2, 3, 4, ...) convolved with (1, 0, 4, 7, 10, 13, ...).
Example: a(5) = 76 = (6, 5, 4, 3, 2, 1) dot (1, 0, 4, 7, 10, 13) = (6 + 0 + 16 + 21 + 20 + 13). (End)

			a(2) = 2*3 + 1 = 7, a(3) = 3*6 + 1 = 19, a(4) = 4*10 + 1 = 41. - _Thomas M. Green_, Nov 16 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

See also A101357 (Cumulative sums of the n-th n-gonal numbers).
A diagonal of A101321.
Cf. A060354, A098547, A101357.
Cf. A006003, A005920.

I:=[1, 2, 7, 19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 25 2012

Table[(n^3+n^2)/2+1,{n,0,6!}] (* Vladimir Joseph Stephan Orlovsky, Mar 06 2010 *)
LinearRecurrence[{4,-6,4,-1},{1,2,7,19},40] (* Vincenzo Librandi, Jun 25 2012 *)

a(n) = n^2*(n+1)/2+1; \\ Altug Alkan, Sep 21 2018

a(n) = 1 + n*(n + n^2)/2 = 1 + (1/2)*n^2 + (1/2) * n^3 = 1 + mean(n^2, n^3). - Joshua Zucker, May 03 2006
Equals A002411(n) + 1. - Olivier Gérard, Jun 20 2007
G.f.: (1 - 2*x + 5*x^2 - x^3) / (x-1)^4. - R. J. Mathar, Apr 04 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 25 2012
a(n) = (A098547(n)+1)/2. - Richard Turk, Jul 18 2017
a(n) = A060354(n+2) - A000290(n+1) = A006003(n+1) - A005563(n) and for n>0 A005920(n) - A068601(n+1). - Bruce J. Nicholson, Jun 23 2018

Corrected and extended by Joshua Zucker, May 03 2006

A110427 The r-th term of the n-th row of the following array contains the sum of r successively decreasing integers beginning from n. 0 < r <= n. Sequence contains the leading diagonal.

Original entry on oeis.org

1, 1, -3, -14, -35, -69, -119, -188, -279, -395, -539, -714, -923, -1169, -1455, -1784, -2159, -2583, -3059, -3590, -4179, -4829, -5543, -6324, -7175, -8099, -9099, -10178, -11339, -12585, -13919, -15344, -16863, -18479, -20195, -22014, -23939, -25973, -28119, -30380, -32759, -35259, -37883

Offset: 1

Amarnath Murthy, Aug 01 2005

			E.g., the row corresponding to 4 contains 4, (3+2),{(1) +(0)+(-1)}, {(-2)+(-3)+(-4)+(-5)} ----> 4,5,0,-14
  1
  2  1
  3  3 -3
  4  5  0 -14
  5  7  3 -10 -35
  6  9  6  -6 -30 -69
  ...
Sequence contains the diagonal.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A110425, A110426.

A110427[n_] := n*(1 - (n - 2)*n)/2; Array[A110427, 50] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {1, 1, -3, -14}, 50] (* Paolo Xausa, Aug 25 2025 *)

From R. J. Mathar, Jul 10 2009: (Start)
a(n) = n*(1 + 2*n - n^2)/2 = n - A002411(n-1).
G.f.: x*(1 - 3*x - x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
E.g.f.: -x*(-1 + x)*(2 + x)*exp(x)/2. - Elmo R. Oliveira, Aug 24 2025

More terms from Joshua Zucker, May 10 2006

A213772 Principal diagonal of the convolution array A213771.

Original entry on oeis.org

1, 11, 42, 106, 215, 381, 616, 932, 1341, 1855, 2486, 3246, 4147, 5201, 6420, 7816, 9401, 11187, 13186, 15410, 17871, 20581, 23552, 26796, 30325, 34151, 38286, 42742, 47531, 52665, 58156, 64016, 70257, 76891, 83930, 91386, 99271, 107597, 116376, 125620, 135341

Offset: 1

Clark Kimberling, Jul 04 2012

Zhu Shijie gives in his Magnus Opus "Jade Mirror of the Four Unknowns" the problem: "The total number of apples in a pile in the form of a cone is 932, and the number of layers is an odd number." Zhu Shijie assumed the rational sequence s(k) = (k*(k+1)*(2*k+1)+k+1)/8 for the total number of apples in k layers, with n = (k+1)/2 is the solution 932 = a((15+1)/2) with k = 15. Zhu Shijie gave the solution polynomial: "Let the element tian be the number of layers. From the statement we have 7455 for the negative shi, 2 for the positive fang, 3 for the positive first lian, and 2 for the positive yu." This translates into the polynomial equation: 2*x^3 + 3*x^2 + 2*x - 7455 = 0. - Thomas Scheuerle, Feb 10 2025

Zhu Shijie, Jade Mirror of the Four Unknowns (Siyuan yujian), Book III Guo Duo Die Gang (Piles of Fruit), Problem number 7, (1303).

Clark Kimberling, Table of n, a(n) for n = 1..1000
Zhu Shijie, Jade Mirror of the Four Unknowns 2, Translation by Library of Chinese classics, original from 1303.
Wikipedia, Jade Mirror of the Four Unknowns.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000326, A002411, A085473, A213771, A220084 (for a list of numbers of the form n*P(k,n) - (n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number).

Cf. A260260 (comment). [Bruno Berselli, Jul 22 2015]

(See A213771.)
LinearRecurrence[{4,-6,4,-1},{1,11,42,106},70] (* Harvey P. Dale, Mar 29 2025 *)

a(n) = (4*n^3-3*n^2+n)/2; \\ Altug Alkan, Dec 16 2017

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: x*(1 + 7*x + 4*x^2)/(1 - x)^4.
a(n) = (4*n^2 - 3*n + 1)*n/2 = n*A002411(n) - (n-1)*A002411(n-1). - Bruno Berselli, Dec 11 2012
a(n) = n*A000326(n) + Sum_{i=0..n-1} A000326(i). - Bruno Berselli, Dec 18 2013
a(n) - a(n-1) = A085473(n-1). - R. J. Mathar, Mar 02 2025
E.g.f.: exp(x)*x*(1 + 4*x)*(2 + x)/2. - Elmo R. Oliveira, Aug 08 2025

cp's OEIS Frontend

A093375 Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.

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A125232 Triangle T(n,k) read by rows: the (n-k)-th term of the k-fold iterated partial sum of the pentagonal numbers.

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A181617 Molecular topological indices of the complete graph K_n.

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A220084 a(n) = (n + 1)*(20*n^2 + 19*n + 6)/6.

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A270205 Number of 2 X 2 planar subsets in an n X n X n cube.

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A279221 Expansion of Product_{k>=1} 1/(1 - x^(k^2*(k+1)/2)).

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A220084 a(n) = (n + 1)(20n^2 + 19*n + 6)/6.