A051936 -id:A051936 - cp's OEIS frontend

A055998 a(n) = n*(n+5)/2.

0, 3, 7, 12, 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, 150, 168, 187, 207, 228, 250, 273, 297, 322, 348, 375, 403, 432, 462, 493, 525, 558, 592, 627, 663, 700, 738, 777, 817, 858, 900, 943, 987, 1032, 1078, 1125, 1173, 1222, 1272

Offset: 0

Barry E. Williams, Jun 14 2000

If X is an n-set and Y a fixed (n-3)-subset of X then a(n-3) is equal to the number of 2-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Bisection of A165157. - Jaroslav Krizek, Sep 05 2009
a(n) is the number of (w,x,y) having all terms in {0,...,n} and w=x+y-1. - Clark Kimberling, Jun 02 2012
Numbers m >= 0 such that 8m+25 is a square. - Bruce J. Nicholson, Jul 26 2017
a(n-1) = 3*(n-1) + (n-1)*(n-2)/2 is the number of connected, loopless, non-oriented, multi-edge vertex-labeled graphs with n edges and 3 vertices. Labeled multigraph analog of A253186. There are 3*(n-1) graphs with the 3 vertices on a chain (3 ways to label the middle graph, n-1 ways to pack edges on one of connections) and binomial(n-1,2) triangular graphs (one way to label the graphs, pack 1 or 2 or ...n-2 on the 1-2 edge, ...). - R. J. Mathar, Aug 10 2017
a(n) is also the number of vertices of the quiver for PGL_{n+1} (see Shen). - Stefano Spezia, Mar 24 2020
Starting from a(2) = 7, this is the 4th column of the array: natural numbers written by antidiagonals downwards. See the illustration by Kival Ngaokrajang and the cross-references. - Andrey Zabolotskiy, Dec 21 2021

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Karl Dilcher and Larry Ericksen, Polynomials and algebraic curves related to certain binary and b-ary overpartitions, arXiv:2405.12024 [math.CO], 2024. See p. 10.
Milan Janjic, Two Enumerative Functions.
Kival Ngaokrajang, Illustration from A000027 (contains errors).
Linhui Shen, Duals of semisimple Poisson-Lie groups and cluster theory of moduli spaces of G-local systems, arXiv:2003.07901 [math.RT], 2020. See p. 8.
Leo Tavares, Illustration: Truncated Point Triangles.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

a(n) = A095660(n+1, 2): third column of (1, 3)-Pascal triangle.
Cf. A000096, A000217, A001477, A002522.
Row n=2 of A255961.
Cf. other rows, columns and diagonals of A000027 written as a table: A034856, A046691, A052905, A055999, A155212, A051936, A056000, A183897, A056115, A051938; A000124, A022856, A152950, A145018, A077169, A166136, A167487, A173036; A059993, A090288, A054000, A142463, A056220, A001105, A001844, A058331, A051890, A097080, A093328, A137882.

[n*(n+5)/2: n in [0..50]]; // G. C. Greubel, Apr 05 2019

f[n_]:=n*(n+5)/2; f[Range[0,50]] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2011 *)

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

[n*(n+5)/2 for n in (0..50)] # G. C. Greubel, Apr 05 2019

G.f.: x*(3-2*x)/(1-x)^3.
a(n) = A027379(n), n > 0.
a(n) = A126890(n,2) for n > 1. - Reinhard Zumkeller, Dec 30 2006
a(n) = A000217(n) + A005843(n). - Reinhard Zumkeller, Sep 24 2008
If we define f(n,i,m) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-m-j), then a(n) = -f(n,n-1,3), for n >= 1. - Milan Janjic, Dec 20 2008
a(n) = A167544(n+8). - Philippe Deléham, Nov 25 2009
a(n) = a(n-1) + n + 2 with a(0)=0. - Vincenzo Librandi, Aug 07 2010
a(n) = Sum_{k=1..n} (k+2). - Gary Detlefs, Aug 10 2010
a(n) = A034856(n+1) - 1 = A000217(n+2) - 3. - Jaroslav Krizek, Sep 05 2009
Sum_{n>=1} 1/a(n) = 137/150. - R. J. Mathar, Jul 14 2012
a(n) = 3*n + A000217(n-1) = 3*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = Sum_{i=3..n+2} i. - Wesley Ivan Hurt, Jun 28 2013
a(n) = 3*A000217(n) - 2*A000217(n-1). - Bruno Berselli, Dec 17 2014
a(n) = A046691(n) + 1. Also, a(n) = A052905(n-1) + 2 = A055999(n-1) + 3 for n>0. - Andrey Zabolotskiy, May 18 2016
E.g.f.: x*(6+x)*exp(x)/2. - G. C. Greubel, Apr 05 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/5 - 47/150. - Amiram Eldar, Jan 10 2021
From Amiram Eldar, Feb 12 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = -5*cos(sqrt(33)*Pi/2)/(4*Pi).
Product_{n>=1} (1 + 1/a(n)) = 15*cos(sqrt(17)*Pi/2)/(2*Pi). (End)

A028313 Elements in the 5-Pascal triangle (by row).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 6, 6, 1, 1, 7, 12, 7, 1, 1, 8, 19, 19, 8, 1, 1, 9, 27, 38, 27, 9, 1, 1, 10, 36, 65, 65, 36, 10, 1, 1, 11, 46, 101, 130, 101, 46, 11, 1, 1, 12, 57, 147, 231, 231, 147, 57, 12, 1, 1, 13, 69, 204, 378, 462, 378, 204, 69, 13, 1, 1, 14, 82, 273, 582, 840, 840, 582, 273, 82, 14, 1

Offset: 0

Mohammad K. Azarian

			Triangle begins as:
  1;
  1,  1;
  1,  5,  1;
  1,  6,  6,   1;
  1,  7, 12,   7,   1;
  1,  8, 19,  19,   8,   1;
  1,  9, 27,  38,  27,   9,   1;
  1, 10, 36,  65,  65,  36,  10,  1;
  1, 11, 46, 101, 130, 101,  46, 11,  1;
  1, 12, 57, 147, 231, 231, 147, 57, 12,  1;

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

Cf. A000007, A005009, A010892, A022112, A028275, A028314, A028315.
Cf. A028316, A028317, A028318, A028319, A028320, A028321, A028322.
Cf. A028323, A028324, A028325, A029653, A051472, A051936, A051937.
Cf. A178915.

[n le 1 select 1 else Binomial(n,k) +3*Binomial(n-2,k-1): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 05 2024

Table[If[n<2, 1, Binomial[n,k] +3*Binomial[n-2,k-1]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 05 2024 *)

def A028313(n,k): return 1 if n<2 else binomial(n,k) + 3*binomial(n-2,k-1)
flatten([[A028313(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 05 2024

From Ralf Stephan, Jan 31 2005: (Start)
T(n, k) = C(n, k) + 3*C(n-2, k-1), with T(0, k) = T(1, k) = 1.
G.f.: (1 + 3*x^2*y)/(1 - x*(1+y)). (End)
From G. C. Greubel, Jan 05 2024: (Start)
T(n, n-k) = T(n, k).
T(n, n-1) = n + 3*(1 - [n=1]) = A178915(n+3), n >= 1.
T(n, n-2) = A051936(n+2), n >= 2.
T(n, n-3) = A051937(n+1), n >= 3.
T(2*n, n) = A028322(n).
Sum_{k=0..n} T(n, k) = A005009(n-2) - (3/4)*[n=0] - (3/2)*[n=1].
Sum_{k=0..n} (-1)^k * T(n, k) = A000007(n) - 3*[n=2].
Sum_{k=0..floor(n/2)} T(n-k, k) = A022112(n-2) + 3*([n=0] - [n=1]).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = 4*A010892(n) - 3*([n=0] + [n=1]). (End)

More terms from Sam Alexander (pink2001x(AT)hotmail.com)

A181510 Number of permutations of the multiset {1,1,2,2,3,3,...,n+1,n+1} avoiding the permutation patterns {132, 231, 2134}.

Original entry on oeis.org

6, 18, 34, 54, 78, 106, 138, 174, 214, 258, 306, 358, 414, 474, 538, 606, 678, 754, 834, 918, 1006, 1098, 1194, 1294, 1398, 1506, 1618, 1734, 1854, 1978, 2106, 2238, 2374, 2514, 2658, 2806, 2958, 3114, 3274, 3438, 3606, 3778, 3954, 4134, 4318, 4506, 4698, 4894

Offset: 1

Lara Pudwell, Oct 25 2010

a(n) is also the surface ares of the n-th solid in the following recursive construction:
The first solid is a unit cube (hence a(1)=6).
To form the n-th solid from the (n-1)st solid, construct a row of 2n-1 cubes, then center the (n-1)st solid on top of this row. (For example, the second solid is a row of 3 unit cubes, with a single unit cube centered on top of the middle cube. This construction has surface area a(2)=18.)
The sequence provides all nonnegative integers m such that 2*m + 13 is a square. - Bruno Berselli, Mar 01 2013

			For n=1, the permutations of {1,1,2,2} avoiding the patterns {132, 231, 2134} are {1122, 1212, 1221, 2112, 2121, 2211}.
For n=2, the permutations of {1,1,2,2,3,3} avoiding the patterns {132, 231, 2134} are {112233, 121233, 122133, 211233, 212133, 221133, 311223, 312123, 312213, 321123, 321213, 322113, 331122, 331212, 331221, 332112, 332121, 332211}.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Lara K. Pudwell, Stacking Blocks and Counting Permutations, Mathematics Magazine, Vol. 83 (2010), pp. 297-302.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014209, A051936, A082111.

A181510:= func< n | ((2*n+3)^2 -13)/2 >;
[A181510(n): n in [1..60]]; // G. C. Greubel, Jan 21 2025

a[n_] := 2*n^2 + 6*n - 2; Array[a, 60] (* Amiram Eldar, Dec 23 2022 *)
LinearRecurrence[{3,-3,1},{6,18,34},50] (* Harvey P. Dale, May 10 2023 *)
((2*Range[1,60]+3)^2 -13)/2 (* G. C. Greubel, Jan 21 2025 *)

a(n)=2*n^2+6*n-2 \\ Charles R Greathouse IV, Jun 17 2017

def A181510(n): return (pow(2*n+3,2) -13)//2
print([A181510(n) for n in range(1,61)]) # G. C. Greubel, Jan 21 2025

a(n) = 2*n^2 + 6*n - 2.
From Bruno Berselli, Oct 29 2010: (Start)
G.f.: 2*x*(3-x^2)/(1-x)^3.
a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 for n > 3.
a(n) = 2*A014209(n) = 2*A082111(n-1) + 4 = A051936(2*n+2) + n + 4. (End)
Sum_{n>=1} 1/a(n) = 2/3 + Pi*tan(sqrt(13)*Pi/2)/(2*sqrt(13)). - Amiram Eldar, Dec 23 2022
E.g.f.: 2*(exp(x)*(x^2 + 4*x - 1) + 1). - Elmo R. Oliveira, Nov 17 2024

A028314 Elements in the 5-Pascal triangle A028313 that are not 1.

Original entry on oeis.org

5, 6, 6, 7, 12, 7, 8, 19, 19, 8, 9, 27, 38, 27, 9, 10, 36, 65, 65, 36, 10, 11, 46, 101, 130, 101, 46, 11, 12, 57, 147, 231, 231, 147, 57, 12, 13, 69, 204, 378, 462, 378, 204, 69, 13, 14, 82, 273, 582, 840, 840, 582, 273, 82, 14, 15, 96, 355, 855, 1422, 1680, 1422, 855, 355, 96, 15

Offset: 0

Mohammad K. Azarian

			Triangle begins as:
   5;
   6,  6;
   7, 12,   7;
   8, 19,  19,   8;
   9, 27,  38,  27,   9;
  10, 36,  65,  65,  36,  10;
  11, 46, 101, 130, 101,  46,  11;
  12, 57, 147, 231, 231, 147,  57,  12;
  13, 69, 204, 378, 462, 378, 204,  69,  13;

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

Cf. A000027, A010892, A022112, A028313, A028322,

Cf. A051936, A051937, A121262, A176448.

A028314:= func< n,k | Binomial(n+2,k+1) + 3*Binomial(n,k) >;
[A028314(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 06 2024

A028314[n_, k_]:= Binomial[n+2,k+1] + 3*Binomial[n,k];
Table[A028314[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 06 2024 *)

def A028314(n,k): return binomial(n+2,k+1) + 3*binomial(n,k)
flatten([[A028314(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 06 2024

From G. C. Greubel, Jan 06 2024: (Start)
T(n, k) = binomial(n+2, k+1) + 3*binomial(n, k).
T(n, n-k) = T(n, k).
T(n, 0) = T(n, n) = A000027(n+5).
T(n, 1) = T(n, n-1) = A051936(n+4).
T(n, 2) = T(n, n-2) = A051937(n+3).T(2*n, n) = A028322(n+1).
Sum_{k=0..n} T(n, k) = A176448(n).
Sum_{k=0..n} (-1)^k * T(n, k) = 1 + (-1)^n + 3*[n=0].
Sum_{k=0..n} T(n-k, k) = A022112(n+1) - (3-(-1)^n)/2.
Sum_{k=0..n} (-1)^k * T(n-k, k) = 4*A010892(n) - 2*A121262(n+1) - (3 - (-1)^n)/2. (End)
G.f.: (5 - 4*x - 4*x*y + 3*x^2*y)/((1 - x)*(1 - x*y)*(1 - x - x*y)). - Stefano Spezia, Dec 06 2024

More terms from James Sellers

A101853 a(n) = n(20 + 15n + n^2)/6.

Original entry on oeis.org

6, 18, 37, 64, 100, 146, 203, 272, 354, 450, 561, 688, 832, 994, 1175, 1376, 1598, 1842, 2109, 2400, 2716, 3058, 3427, 3824, 4250, 4706, 5193, 5712, 6264, 6850, 7471, 8128, 8822, 9554, 10325, 11136, 11988, 12882, 13819, 14800

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004

4th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 3: 1,4,1. The 1,4,1 is the left column, A101101 the second column, A008458 the third, A003215 the fourth column etc of the array in the example. a(n) is the 4th row.

			Left column the third row of A008292, and subsequent columns defined as partial sums along their preceding neighbors:
1 1  1   1   1    1    1     1     1     1     1
4 5  6   7   8    9   10    11    12    13    14
1 6 12  19  27   36   46    57    69    82    96  A051936
0 6 18  37  64  100  146   203   272   354   450  A101853
0 6 24  61 125  225  371   574   846  1200  1650  A101854
0 6 30  91 216  441  812  1386  2232  3432  5082  A101855
0 6 36 127 343  784 1596  2982  5214  8646 13728
0 6 42 169 512 1296 2892  5874 11088 19734 33462
0 6 48 217 729 2025 4917 10791 21879 41613 75075
...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Row n=3 of A255961.

Cf. A003215, A008292, A008458, A101101.

I:=[6, 18, 37, 64]; [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..40]]; // Vincenzo Librandi, Jun 26 2012

LinearRecurrence[{4,-6,4,-1},{6,18,37,64},40] (* or *) CoefficientList[Series[(6-6*x+x^2)/(x-1)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 26 2012 *)

a(n)=n*(20+15*n+n^2)/6 \\ Charles R Greathouse IV, Oct 16 2015

G.f.: x*(6 - 6*x + x^2)/(x - 1)^4. - R. J. Mathar, Dec 06 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 26 2012
E.g.f.: exp(x)*x*(36 + 18*x + x^2)/6. - Stefano Spezia, Oct 14 2022

A101883 Write four numbers, skip one, write four, skip two, write four, skip three... and so on.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 15, 19, 20, 21, 22, 27, 28, 29, 30, 36, 37, 38, 39, 46, 47, 48, 49, 57, 58, 59, 60, 69, 70, 71, 72, 82, 83, 84, 85, 96, 97, 98, 99, 111, 112, 113, 114, 127, 128, 129, 130, 144, 145, 146, 147, 162, 163, 164, 165, 181, 182, 183, 184, 201

Offset: 1

Candace Mills (scorpiocand(AT)yahoo.com), Dec 19 2004

Union of A051936, A051936+1, A051936+2, and A051936+3. - Ivan Neretin, Aug 03 2016

First terms of the four repeated terms belong to A051936. - Michael De Vlieger, Aug 03 2016

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

Cf. A000217, A051936, A101881, A101882.

LinearRecurrence[{1,0,0,2,-2,0,0,-1,1},{1,2,3,4,6,7,8,9,12},80] (* or *) Rest[CoefficientList[Series[x*(1+x+x^2+x^3-x^5-x^6-x^7)/((1+x)^2(x^2+1)^2(1-x)^3),{x,0,80}],x]](* Harvey P. Dale, Jan 19 2016 *)
Table[Range[#, # + 3] &[n (n + 1)/2 - 9], {n, 4, 20}] // Flatten (* Michael De Vlieger, Aug 03 2016 *)

G.f.: x*(1+x+x^2+x^3-x^5-x^6-x^7)/ ((1+x)^2 * (x^2+1)^2 * (1-x)^3). [Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009; checked and corrected by R. J. Mathar, Sep 16 2009]

A302537 a(n) = (n^2 + 13*n + 2)/2.

Original entry on oeis.org

1, 8, 16, 25, 35, 46, 58, 71, 85, 100, 116, 133, 151, 170, 190, 211, 233, 256, 280, 305, 331, 358, 386, 415, 445, 476, 508, 541, 575, 610, 646, 683, 721, 760, 800, 841, 883, 926, 970, 1015, 1061, 1108, 1156, 1205, 1255, 1306, 1358, 1411, 1465, 1520, 1576

Offset: 0

Franck Maminirina Ramaharo, Apr 09 2018

Binomial transform of [1, 7, 1, 0, 0, 0, ...].

Numbers m > 0 such that 8*m + 161 is a square.

			Illustration of initial terms (by the formula a(n) = A052905(n) + 3*n):
.                                                                    o
.                                                                  o o
.                                                    o           o o o
.                                                  o o         o o o o
.                                      o         o o o       o o o o o
.                                    o o       o o o o     o o o o o o
.                          o       o o o     o o o o o   o . . . . . o
.                        o o     o o o o   o . . . . o   o . . . . . o
.                o     o o o   o . . . o   o . . . . o   o . . . . . o
.              o o   o . . o   o . . . o   o . . . . o   o . . . . . o
.        o   o . o   o . . o   o . . . o   o . . . . o   o . . . . . o
.      o o   o . o   o . . o   o . . . o   o . . . . o   o . . . . . o
.  o   o o   o o o   o o o o   o o o o o   o o o o o o   o o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
----------------------------------------------------------------------
.  1     8      16        25          35            46              58

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

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

Sequences whose n-th terms are of the form binomial(n, 2) + n*k + 1:
A152947 (k = 0); A000124 (k = 1); A000217 (k = 2); A034856 (k = 3);
A052905 (k = 4); A051936 (k = 5); A246172 (k = 6).
Cf. A000578, A002939, A004120, A008589, A056119, A060544, A162261.

A302537:= func< n | ((n+1)^2 +12*n +1)/2 >;
[A302537(n): n in [0..50]]; // G. C. Greubel, Jan 21 2025

a := n -> (n^2 + 13*n + 2)/2;
seq(a(n), n = 0 .. 100);

Table[(n^2 + 13 n + 2)/2, {n, 0, 100}]
CoefficientList[ Series[(5x^2 - 5x - 1)/(x - 1)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 8, 16}, 51] (* Robert G. Wilson v, May 19 2018 *)

makelist((n^2 + 13*n + 2)/2, n, 0, 100);

a(n) = (n^2 + 13*n + 2)/2; \\ Altug Alkan, Apr 12 2018

def A302537(n): return (n**2 + 13*n + 2)//2
print([A302537(n) for n in range(51)]) # G. C. Greubel, Jan 21 2025

a(n) = binomial(n + 1, 2) + 6*n + 1 = binomial(n, 2) + 7*n + 1.
a(n) = a(n-1) + n + 6.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, where a(0) = 1, a(1) = 8 and a(2) = 16.
a(n) = 2*a(n-1) - a(n-2) + 1.
a(n) = A004120(n+1) for n > 1.
a(n) = A056119(n) + 1.
a(n) = A152947(n+1) + A008589(n).
a(n) = A060544(n+1) - A002939(n).
a(n) = A000578(n+1) - A162261(n) for n > 0.
G.f.: (1 + 5*x - 5*x^2)/(1 - x)^3.
E.g.f.: (1/2)*(2 + 14*x + x^2)*exp(x).
Sum_{n>=0} 1/a(n) = 24097/45220 + 2*Pi*tan(sqrt(161)*Pi/2) / sqrt(161) = 1.4630922534498496... - Vaclav Kotesovec, Apr 11 2018

A103145 a(n) = (1/6)(n^3 + 21n^2 + 74*n + 18).

Original entry on oeis.org

3, 19, 43, 76, 119, 173, 239, 318, 411, 519, 643, 784, 943, 1121, 1319, 1538, 1779, 2043, 2331, 2644, 2983, 3349, 3743, 4166, 4619, 5103, 5619, 6168, 6751, 7369, 8023, 8714, 9443, 10211, 11019, 11868, 12759, 13693, 14671, 15694, 16763

Offset: 0

Creighton Dement, Mar 17 2005

A floretion-generated sequence relating truncated triangle and pyramidal numbers. The following reasoning suggests that (a(n)) may not be the result of some "arbitrary" addition of these sequences--it may possess some geometric meaning of its own: The FAMP identity: "jesrightfor + jesleftfor = jesfor" holds and was used to find the relation a(n) = 2*A051936(n+4)_4 + A051937(n+4)_4 . In the above case, "jesfor" returns the truncated triangular numbers (times -1); "jesrightfor" returns the truncated pyramidal numbers; and (a(n)) is given by "jesleftfor" (times -1). All sequences result from a Force transform of the sequence c(n) = n + 5 (c was not chosen arbitrarily, for details see program code). Specifically, the sequence (a(n)) is the (ForType 1A) jesleftfor-transform of the sequence c(n) = n + 5 with respect to the floretion given in the program code.

Floretion Algebra Multiplication Program, FAMP Code: 1jesleftfor[A*B] with A = .25'i - .25i' - .25'ii' + .25'jj' + .25'kk' + .25'jk' + .25'kj' - .25e and B = + 'i + .5j' + .5k' + .5'ij' + .5'ik'; 1vesfor[A*B](n) = n + 5. ForType: 1A Alternative description: 1jesleftfor[A*B], ForType: 1A, LoopType: tes (first iteration after transforming the zero-sequence A000004).

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

Cf. A051936, A051937.

I:=[3, 19, 43, 76]; [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..45]]; // Vincenzo Librandi, Jun 26 2012

CoefficientList[Series[(3-2*x)*(1+3*x-3*x^2)/(1-x)^4,{x,0,40}],x] (* or *) LinearRecurrence[ {4,-6,4,-1},{3,19,43,76},50] (* Vincenzo Librandi, Jun 26 2012 *)
Table[(n^3+21n^2+74n+18)/6,{n,0,50}] (* Harvey P. Dale, Jun 18 2024 *)

a(n) = (n^3+21*n^2+74*n+18)/6; \\ Altug Alkan, Sep 23 2018

def A103145(n): return (n*(n*(n+21)+74)+18)//6 # Chai Wah Wu, Mar 07 2024

a(n) = 2*A051936(n+4)_4 + A051937(n+4)_4 (for n = 0, 1, 2, 3) or a(m) = (1/6)*(m^3 + 9m^2 - 46m - 6) = 2*A051936(m) + A051937(m) (for m = 4, 5, 6).
G.f.: (3-2*x)*(1 + 3*x - 3*x^2)/(1-x)^4. - Colin Barker, Apr 30 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 26 2012

A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 6, 7, 7, 1, 5, 8, 10, 11, 11, 1, 6, 10, 13, 15, 16, 16, 1, 7, 12, 16, 19, 21, 22, 22, 1, 8, 14, 19, 23, 26, 28, 29, 29, 1, 9, 16, 22, 27, 31, 34, 36, 37, 37, 1, 10, 18, 25, 31, 36, 40, 43, 45, 46, 46, 1, 11, 20, 28, 35, 41

Offset: 0

Franck Maminirina Ramaharo, Apr 20 2018

Columns are linear recurrence sequences with signature (3,-3,1).
8*T(n,k) + A166147(k-1) are squares.
Columns k are binomial transforms of [1, k, 1, 0, 0, 0, ...].
Antidiagonals sums yield A116731.

			The array T(n,k) begins
1    1    1    1    1    1    1    1    1    1    1    1    1  ...  A000012
1    2    3    4    5    6    7    8    9   10   11   12   13  ...  A000027
2    4    6    8   10   12   14   16   18   20   22   24   26  ...  A005843
4    7   10   13   16   19   22   25   28   31   34   37   40  ...  A016777
7   11   15   19   23   27   31   35   39   43   47   51   55  ...  A004767
11  16   21   26   31   36   41   46   51   56   61   66   71  ...  A016861
16  22   28   34   40   46   52   58   64   70   76   82   88  ...  A016957
22  29   36   43   50   57   64   71   78   85   92   99  106  ...  A016993
29  37   45   53   61   69   77   85   93  101  109  117  125  ...  A004770
37  46   55   64   73   82   91  100  109  118  127  136  145  ...  A017173
46  56   66   76   86   96  106  116  126  136  146  156  166  ...  A017341
56  67   78   89  100  111  122  133  144  155  166  177  188  ...  A017401
67  79   91  103  115  127  139  151  163  175  187  199  211  ...  A017605
79  92  105  118  131  144  157  170  183  196  209  222  235  ...  A190991
...
The inverse binomial transforms of the columns are
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    1    2    3    4    5    6    7    8    9   10   11   12  ...
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
...
T(k,n-k) = A087401(n,k) + 1 as triangle
1
1   1
1   2   2
1   3   4   4
1   4   6   7   7
1   5   8  10  11  11
1   6  10  13  15  16  16
1   7  12  16  19  21  22  22
1   8  14  19  23  26  28  29  29
1   9  16  22  27  31  34  36  37  37
1  10  18  25  31  36  40  43  45  46  46
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

Cf. A085475, A086270, A086271, A086272, A086273, A130154, A159798, A162609, A162610, A300401.

T := (n, k) -> binomial(n, 2) + k*n + 1;
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

Table[With[{n = m - k}, Binomial[n, 2] + k n + 1], {m, 0, 11}, {k, m, 0, -1}] // Flatten (* Michael De Vlieger, Apr 21 2018 *)

T(n, k) := binomial(n, 2)+ k*n + 1$
for n:0 thru 20 do
    print(makelist(T(n, k), k, 0, 20));

T(n,k) = binomial(n, 2) + k*n + 1;
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 17 2018

G.f.: (3*x^2*y - 3*x*y + y - 2*x^2 + 2*x - 1)/((x - 1)^3*(y - 1)^2).
E.g.f.: (1/2)*(2*x*y + x^2 + 2)*exp(y + x).
T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k), with T(0,k) = 1, T(1,k) = k + 1 and T(2,k) = 2*k + 2.
T(n,k) = T(n-1,k) + n + k - 1.
T(n,k) = T(n,k-1) + n, with T(n,0) = 1.
T(n,0) = A152947(n+1).
T(n,1) = A000124(n).
T(n,2) = A000217(n).
T(n,3) = A034856(n+1).
T(n,4) = A052905(n).
T(n,5) = A051936(n+4).
T(n,6) = A246172(n+1).
T(n,7) = A302537(n).
T(n,8) = A056121(n+1) + 1.
T(n,9) = A056126(n+1) + 1.
T(n,10) = A051942(n+10) + 1, n > 0.
T(n,11) = A101859(n) + 1.
T(n,12) = A132754(n+1) + 1.
T(n,13) = A132755(n+1) + 1.
T(n,14) = A132756(n+1) + 1.
T(n,15) = A132757(n+1) + 1.
T(n,16) = A132758(n+1) + 1.
T(n,17) = A212427(n+1) + 1.
T(n,18) = A212428(n+1) + 1.
T(n,n) = A143689(n) = A300192(n,2).
T(n,n+1) = A104249(n).
T(n,n+2) = T(n+1,n) = A005448(n+1).
T(n,n+3) = A000326(n+1).
T(n,n+4) = A095794(n+1).
T(n,n+5) = A133694(n+1).
T(n+2,n) = A005449(n+1).
T(n+3,n) = A115067(n+2).
T(n+4,n) = A133694(n+2).
T(2*n,n) = A054556(n+1).
T(2*n,n+1) = A054567(n+1).
T(2*n,n+2) = A033951(n).
T(2*n,n+3) = A001107(n+1).
T(2*n,n+4) = A186353(4*n+1) (conjectured).
T(2*n,n+5) = A184103(8*n+1) (conjectured).
T(2*n,n+6) = A250657(n-1) = A250656(3,n-1), n > 1.
T(n,2*n) = A140066(n+1).
T(n+1,2*n) = A005891(n).
T(n+2,2*n) = A249013(5*n+4) (conjectured).
T(n+3,2*n) = A186384(5*n+3) = A186386(5*n+3) (conjectured).
T(2*n,2*n) = A143689(2*n).
T(2*n+1,2*n+1) = A143689(2*n+1) (= A030503(3*n+3) (conjectured)).
T(2*n,2*n+1) = A104249(2*n) = A093918(2*n+2) = A131355(4*n+1) (= A030503(3*n+5) (conjectured)).
T(2*n+1,2*n) = A085473(n).
a(n+1,5*n+1)=A051865(n+1) + 1.
a(n,2*n+1) = A116668(n).
a(2*n+1,n) = A054569(n+1).
T(3*n,n) = A025742(3*n-1), n > 1 (conjectured).
T(n,3*n) = A140063(n+1).
T(n+1,3*n) = A069099(n+1).
T(n,4*n) = A276819(n).
T(4*n,n) = A154106(n-1), n > 0.
T(2^n,2) = A028401(n+2).
T(1,n)*T(n,1) = A006000(n).
T(n*(n+1),n) = A211905(n+1), n > 0 (conjectured).
T(n*(n+1)+1,n) = A294259(n+1).
T(n,n^2+1) = A081423(n).
T(n,A000217(n)) = A158842(n), n > 0.
T(n,A152947(n+1)) = A060354(n+1).
floor(T(n,n/2)) = A267682(n) (conjectured).
floor(T(n,n/3)) = A025742(n-1), n > 0 (conjectured).
floor(T(n,n/4)) = A263807(n-1), n > 0 (conjectured).
ceiling(T(n,2^n)/n) = A134522(n), n > 0 (conjectured).
ceiling(T(n,n/2+n)/n) = A051755(n+1) (conjectured).
floor(T(n,n)/n) = A133223(n), n > 0 (conjectured).
ceiling(T(n,n)/n) = A007494(n), n > 0.
ceiling(T(n,n^2)/n) = A171769(n), n > 0.
ceiling(T(2*n,n^2)/n) = A046092(n), n > 0.
ceiling(T(2*n,2^n)/n) = A131520(n+2), n > 0.

cp's OEIS Frontend

A055998 a(n) = n*(n+5)/2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A028313 Elements in the 5-Pascal triangle (by row).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A181510 Number of permutations of the multiset {1,1,2,2,3,3,...,n+1,n+1} avoiding the permutation patterns {132, 231, 2134}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Formula

A028314 Elements in the 5-Pascal triangle A028313 that are not 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A101853 a(n) = n*(20 + 15*n + n^2)/6.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A101883 Write four numbers, skip one, write four, skip two, write four, skip three... and so on.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A302537 a(n) = (n^2 + 13*n + 2)/2.

Views

A101853 a(n) = n(20 + 15n + n^2)/6.

A103145 a(n) = (1/6)(n^3 + 21n^2 + 74*n + 18).