A143689 -id:A143689 - cp's OEIS frontend

A104249 a(n) = (3*n^2 + n + 2)/2.

1, 3, 8, 16, 27, 41, 58, 78, 101, 127, 156, 188, 223, 261, 302, 346, 393, 443, 496, 552, 611, 673, 738, 806, 877, 951, 1028, 1108, 1191, 1277, 1366, 1458, 1553, 1651, 1752, 1856, 1963, 2073, 2186, 2302, 2421, 2543, 2668, 2796, 2927, 3061, 3198, 3338, 3481

Offset: 0

Thomas Wieder, Feb 26 2005

Second differences are all 3.
Related to the sequence of odd numbers A005408 since for these numbers the first differences are all 2.
Column 2 of A114202. - Paul Barry, Nov 17 2005
Equals third row of A167560 divided by 2. - Johannes W. Meijer, Nov 12 2009
A242357(a(n)) = n + 1. - Reinhard Zumkeller, May 11 2014
Also, this sequence is related to A011379, for n>0, by A011379(n) = n*a(n) - Sum_{i=0..n-1} a(i). - Bruno Berselli, Jul 08 2015
The number of Hamiltonian nonisomorphic unfoldings in an n-gonal Archimedean antiprism. See sequence A284647. - Rick Mabry, Apr 10 2021

			The sequence of first differences delta_a(n) = a(n+1) - a(n) is 2, 5, 8, 11, 14, 17, 20, 23, 26, ...
The sequence of second differences delta_delta_a(n) = a(n+2) - 2*a(n+1) + a(n) is: 3, 3, 3, 3, 3, 3, 3, ... E.g., 78 - 2*58 + 41 = 3.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Rick Mabry, Fibonacci Numbers, Integer Compositions, and Nets of Antiprisms, The American Mathematical Monthly, Vol. 126 (2019), no. 9, pp. 786-801.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001399, A002597, A005408, A011379, A016777, A143689.

Counts special cases of A284647.

a104249 n = n*(3*n+1) `div` 2 + 1 -- Reinhard Zumkeller, May 11 2014

[(3*n^2+n+2)/2: n in [0..50]]; // Vincenzo Librandi, May 09 2011

a := proc (n) local i, u; option remember; u[0] := 1; u[1] := 3; u[2] := 8; for i from 3 to n do u[i] := -(4*u[i-3]-8*u[i-2]-2*u[i-1]+(-2*u[i-3]+2*u[i-2]-u[i-1])*i)/i end do; [seq(u[i],i = 0 .. n)] end proc;

A104249[n_] := (3*n^2 + n + 2)/2; Table[A104249[n], {n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Jul 22 2011 *)
LinearRecurrence[{3,-3,1},{1,3,8},70] (* Harvey P. Dale, Jul 21 2023 *)

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

G.f.: (1 + 2*x^2)/(1 - x)^3.
Recurrence: (n+3)*u(n+3) + (-5-n)*u(n+2)*(-2+2*n)*u(n+1) + (-2-2*n)*u(n) = 0 for n >= 0 with u(0) = 1, u(1) = 3, and u(2) = 8.
From Paul Barry, Nov 17 2005: (Start)
a(0) = 1, a(n) = a(n-1) + 3*n - 1 for n > 0;
a(n) = Sum_{k=0..n} C(n, k)*C(2, k)*J(k+1), where J(n) = A001045(n). (End)
Binomial transform of [1, 2, 3, 0, 0, 0, ...]. - Gary W. Adamson, Apr 23 2008
E.g.f.: exp(x)*(2 + 4*x + 3*x^2)/2. - Stefano Spezia, Apr 10 2021

A033292 A Connell-like sequence: take 1 number = 1 (mod Q), 2 numbers = 2 (mod Q), 3 numbers = 3 (mod Q), etc., where Q = 3.

Original entry on oeis.org

1, 2, 5, 6, 9, 12, 13, 16, 19, 22, 23, 26, 29, 32, 35, 36, 39, 42, 45, 48, 51, 52, 55, 58, 61, 64, 67, 70, 71, 74, 77, 80, 83, 86, 89, 92, 93, 96, 99, 102, 105, 108, 111, 114, 117, 118, 121, 124, 127, 130, 133, 136, 139, 142, 145, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 177, 180

Offset: 1

Gary E. Stevens

Left border of the triangle (1, 2, 6, 13, 23, 36, ...) = A143689 = A000326(n) - 3n, where A000326 = the pentagonal numbers, right border. - Gary W. Adamson, Aug 29 2008
Row sums = A143690: (1, 7, 27, 70, 145, 261, 427, 652, ...). - Gary W. Adamson, Aug 29 2008
Central terms = A136392. - Reinhard Zumkeller, Jan 18 2012

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
Douglas E. Iannucci and Donna Mills-Taylor, On Generalizing the Connell Sequence, J. Integer Sequences, Vol. 2, 1999, #99.1.7.
Gary E. Stevens, A Connell-Like Sequence, J. Integer Sequences, 1 (1998), #98.1.4.

a033292 n k = a033292_tabl !! (n-1) !! (k-1)
a033292_row n = a033292_tabl !! (n-1)
a033292_tabl = f 1 [1..] where
   f k xs = ys : f (k+1) (dropWhile (<= last ys) xs) where
     ys  = take k $ filter ((== 0) . (`mod` 3) . (subtract k)) xs
-- Reinhard Zumkeller, Jan 18 2012 2011

row[n_] := n*(3*n-1)/2 + Range[1, 3*n+1, 3]; Flatten[ Table[ row[n], {n, 0, 11}]] (* Jean-François Alcover, Aug 03 2012 *)

More terms from jeroen.lahousse(AT)icl.com

Offset changed by Reinhard Zumkeller, Jan 18 2012

A242357 Crescendo trapezoidal.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 2, 3, 3, 3, 2, 1, 1, 2, 3, 4, 4, 4, 4, 3, 2, 1, 1, 2, 3, 4, 5, 5, 5, 5, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 6, 6, 6, 6, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 7, 7, 7, 7, 7, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 7

Offset: 1

Reinhard Zumkeller, May 11 2014

nonn

a(A000326(n)) = a(A143689(n-1)) = 1;

for all n: a(k) = n, A104249(n-1) <= k <= A005448(n).

			. Initial values
. seen trapezoidal:                                  666666
.                                    55555          5......5
.                       4444        4.....4        4........4
.             333      3....3      3.......3      3..........3
.      22    2...2    2......2    2.........2    2............2
.  1  1..1  1.....1  1........1  1...........1  1..............1 .

Reinhard Zumkeller, Table of n, a(n) for n = 1..1001

Cf. A004737.

import Data.List (inits)
a242357 n = a242357_list !! (n-1)
a242357_list = concatMap f $ tail $ inits [1..] where
   f us = (init us) ++ (take v [v, v ..]) ++ vs
          where (v:vs) = reverse us

Table[Join[Range[n-1],PadRight[{},n,n],Range[n-1,1,-1]],{n,9}]//Flatten (* Harvey P. Dale, Oct 23 2020 *)

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 1, 0, 2, 6, 6, 2, 0, 3, 13, 22, 18, 7, 1, 0, 4, 23, 56, 75, 60, 29, 8, 1, 0, 5, 36, 115, 215, 261, 215, 121, 45, 10, 1, 0, 6, 52, 206, 495, 806, 938, 798, 496, 220, 66, 12, 1, 0, 7, 71, 336, 987, 2016, 3031, 3452, 3010, 2003, 1001, 364, 91

Offset: 0

Franck Maminirina Ramaharo, Feb 28 2018

			The triangle T(n, k) begins:
n\k  0  1   2    3    4     5     6     7     8     9    10   11  12  13 14
0:   0  0   1
1:   0  1   2    1
2:   0  2   6    6    2
3:   0  3  13   22   18     7     1
4:   0  4  23   56   75    60    29     8     1
5:   0  5  36  115  215   261   215   121    45    10     1
6:   0  6  52  206  495   806   938   798   496   220    66   12   1
7:   0  7  71  336  987  2016  3031  3452  3010  2003  1001  364  91  14  1

M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996.

Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
A. M. Mathai and P. N. Rathie, Enumeration of almost cubic maps, Journal of Combinatorial Theory, Series B, Vol 13 (1972), 83-90.
Franck Ramaharo, A one-variable bracket polynomial for some Turk's head knots, arXiv:1807.05256 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.

Row sums: A000302 (powers of 4).

Cf. A034870, A034871, A032443.

T := (n, k) -> binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k);
for n from 0 to 10 do seq(T(n, k), k = 0 .. max(2*n, n + 2)) od;

T(n, k) := binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k)$
a : []$
for n:0 thru 10 do
  a : append(a, makelist(T(n, k), k, 0, max(2*n, n + 2)))$
a;

row(n) = Vecrev((x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n); \\ Michel Marcus, Nov 12 2022

T(n,k) = binomial(2*n,k) + binomial(n,k-2) - binomial(n,k).
T(n,k) = T(n-1,k-1)+ T(n-1,k) + A034871(n-1,k-1), with T(n,0) = T(0,1) = 0 and T(0,2) = 1
T(n,1) = A001477(n).
T(n,2) = A143689(n).
T(n,3) = n + A002492(n-1) - A000292(n-2).
T(n,n) = A247493(n+1,n).
T(n,n+1) = n + A001791(n).
T(n,n+2) = 1 + A002694(n), n >= 2.
T(n,n+k) = binomial(2*n, n-k) = A094527(n,k), for k >= 3 and n>=k.
G.f.: 1/(1 - y*(x^2 + 2*x + 1)) + (x^2 - 1)/(1 - y*(x + 1)).

A192136 a(n) = (5n^2 - 3n + 2)/2.

Original entry on oeis.org

1, 2, 8, 19, 35, 56, 82, 113, 149, 190, 236, 287, 343, 404, 470, 541, 617, 698, 784, 875, 971, 1072, 1178, 1289, 1405, 1526, 1652, 1783, 1919, 2060, 2206, 2357, 2513, 2674, 2840, 3011, 3187, 3368, 3554, 3745, 3941, 4142, 4348, 4559, 4775, 4996, 5222, 5453, 5689

Offset: 0

Eric Werley, Jun 24 2011

Binomial transform of [1, 1, 5, 0, 0, 0, 0, 0, ...]. - Johannes W. Meijer, Jul 07 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000124, A000217, A002522, A006137, A130883, A140063, A140064, A143689.

A192136:=func< n | (5*n^2-3*n+2)/2 >; [ A192136(n): n in [0..50] ]; // Klaus Brockhaus, Jun 27 2011

Table[(5n^2-3n+2)/2,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,2,8},50] (* Harvey P. Dale, Aug 08 2016 *)

a(n)=n*(5*n-3)/2+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = (5*n^2 - 3*n + 2)/2.
a(n) = 2*a(n-1) - a(n-2) + 5.
a(n) = a(n-1) + 5*n - 4.
a(n) = 5*binomial(n+2,2) - 9*n - 4.
a(n) = A000217(n+1) - A000217(n) + 5*A000217(n-1); triangular numbers. - Johannes W. Meijer, Jul 07 2011
O.g.f.: (1-x+5*x^2)/(1-x)^3.
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: exp(x)*(2 + 2*x + 5*x^2)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A238410 a(n) = floor((3(n-1)^2 + 1)/2).

Original entry on oeis.org

0, 2, 6, 14, 24, 38, 54, 74, 96, 122, 150, 182, 216, 254, 294, 338, 384, 434, 486, 542, 600, 662, 726, 794, 864, 938, 1014, 1094, 1176, 1262, 1350, 1442, 1536, 1634, 1734, 1838, 1944, 2054, 2166, 2282, 2400, 2522, 2646, 2774, 2904, 3038, 3174, 3314, 3456, 3602, 3750, 3902, 4056, 4214, 4374, 4538, 4704

Offset: 1

Emeric Deutsch, Feb 27 2014

a(n) = the eccentric connectivity index of the path P[n] on n vertices. The eccentric connectivity index of a simple connected graph G is defined to be the sum over all vertices i of G of the product E(i)D(i), where E(i) is the eccentricity and D(i) is the degree of vertex i. For example, a(4)=14 because the vertices of P[4] have degrees 1,2,2,1 and eccentricities 3,2,2,3; we have 1*3 + 2*2 + 2*2 + 1*3 = 14.
From Paul Curtz, Feb 23 2023: (Start)
East spoke of the hexagonal spiral using A004526 with a single 0:
.
            43  42  42  41  41  40
          43  28  28  27  27  26  40
        44  29  17  16  16  15  26  39
      44  29  17   8   8   7  15  25  39
    45  30  18   9   3   2   7  14  25  38
  45  30  18   9   3   0---2---6--14--24--38-->
    31  19  10   4   1   1   6  13  24  37
      31  19  10   4   5   5  13  23  37
        32  20  11  11  12  12  23  36
          32  20  21  21  22  22  36
            33  33  34  34  35  35
.
Other spokes are A032528, A005449, A227017, A045943, A005448, A212959, A000326, A282513, A143689, and A104249. (End)

Matthew House, Table of n, a(n) for n = 1..10000
M. J. Morgan, S. Mukwembi and H. C. Swart, On the eccentric connectivity index of a graph, Discrete Math., 311, 2011, 1229-1234.
B. Zhou and Zh. Du, On eccentric connectivity index, Comm. Math. Comp. Chem. (MATCH), 63, 2010, 181-198.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A004526, A077043.

(Cf. A056105, A056107.)

a := proc (n) options operator, arrow: floor((3/2)*(n-1)^2+1/2) end proc: seq(a(n), n = 1 .. 70);

Table[Floor[(3(n-1)^2+1)/2],{n,80}]  (* or *) LinearRecurrence[{2,0,-2,1},{0,2,6,14},80] (* Harvey P. Dale, Apr 30 2022 *)

a(n)=(3*(n-1)^2 + 1)\2 \\ Charles R Greathouse IV, Feb 15 2017

a(n) = (3*n)^2/6 for n even and a(n) = ((3*n)^2 + 3)/6 for n odd. - Miquel Cerda, Jun 17 2016
From Ilya Gutkovskiy, Jun 17 2016: (Start)
G.f.: 2*x^2*(1 + x + x^2)/((1 - x)^3*(1 + x)).
a(n) = (6*n^2 - 12*n + 7 + (-1)^n)/4.
a(n) = 2* A077043(n-1). (End)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Matthew House, Feb 15 2017
Sum_{n>=2} 1/a(n) = Pi^2/36 + tanh(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)). - Amiram Eldar, Mar 12 2023

A273465 Numbers generated by starting at 1 and adding twice and subtracting once following the sequence of positive integers.

Original entry on oeis.org

1, 3, 6, 2, 7, 13, 6, 14, 23, 13, 24, 36, 23, 37, 52, 36, 53, 71, 52, 72, 93, 71, 94, 118, 93, 119, 146, 118, 147, 177, 146, 178, 211, 177, 212, 248, 211, 249, 288, 248, 289, 331, 288, 332, 377, 331, 378, 426, 377, 427

Offset: 1

Mattias Dooreman, May 23 2016

nonn

This sequence is generated by listing the successive results of 1 + 2 + 3 - 4 + 5 + 6 - 7 + 8 + 9 - 10 + ... etc.

By adding twice and subtracting once, the results at positions i, i+2 and i+4 are linked by the relationship: n, n+1, n.

			n1 = 0 + 1 = 1
n2 = 1 + 2 = 3
n3 = 1 + 2 + 3 = 6
n4 = 1 + 2 + 3 - 4 = 2
n5 = 1 + 2 + 3 - 4 + 5 = 7
n6 = 1 + 2 + 3 - 4 + 5 + 6 = 13
n7 = 1 + 2 + 3 - 4 + 5 + 6 - 7 = 6
...
->
(n3, n5, n7) = (6, 7, 6)
(n6, n8, n10) = (13, 14, 13)
...
->
n9 = 2.n6 - n7 + 3 = 23

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

Cf. A143689.

Accumulate@ Table[If[Mod[n, 3] == 1 && n > 1, -n, n], {n, 50}] (* Michael De Vlieger, May 23 2016 *)
LinearRecurrence[{1,0,2,-2,0,-1,1},{1,3,6,2,7,13,6},80] (* Harvey P. Dale, Oct 13 2024 *)

Let the sequence be written 1, 3, a, 2, a+1, b, a, b+1, c, b, c+1, d, c, ... The (a, a+1, a) triple at positions i, i+2, i+4 can be determined from the previous two sequences: (c = 2*b - a + 3). This new sequence starts at position i+3.
G.f.: (-2*z^6 - z^4 + 6*z^3 - 3*z^2 - 2*z - 1)/((z - 1)^3*(z^2 + z + 1)^2).
a(3*n-2) = A143689(n-1).
From Franck Maminirina Ramaharo, Dec 27 2018: (Start)
a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7), n >= 8.
E.g.f.: (1/18)*(exp(3*x/2)*(32 + 6*x + 3*x^2) + (4 - 12*x)*cos(sqrt(3)*x/2) - 4*sqrt(3)*(1 + x)*sin(sqrt(3)*x/2))/exp(x/2). (End)

a(25) corrected by Seiichi Manyama, May 26 2016

A345462 Triangle T(n,k) (n >= 1, 0 <= k <= n-1) read by rows: number of distinct permutations after k steps of the "first transposition" algorithm.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 24, 13, 4, 1, 120, 67, 23, 5, 1, 720, 411, 146, 36, 6, 1, 5040, 2921, 1067, 272, 52, 7, 1, 40320, 23633, 8800, 2311, 456, 71, 8, 1, 362880, 214551, 81055, 21723, 4419, 709, 93, 9, 1, 3628800, 2160343, 825382, 224650, 46654, 7720, 1042, 118, 10, 1

Offset: 1

Olivier Gérard, Jun 20 2021

The first transposition algorithm is: if the permutation is sorted, then exit; otherwise, exchange the first unsorted letter with the letter currently at its index. Repeat.
At each step at least 1 letter (possibly 2) is sorted.
If one counts the steps necessary to reach the identity, this gives the Stirling numbers of the first kind (reversed).

			Triangle begins:
      1;
      2,     1;
      6,     3,    1;
     24,    13,    4,    1;
    120,    67,   23,    5,   1;
    720,   411,  146,   36,   6,  1;
   5040,  2921, 1067,  272,  52,  7, 1;
  40320, 23633, 8800, 2311, 456, 71, 8, 1;
  ...

D. E. Knuth, The Art of Computer Programming, Vol. 3 / Sorting and Searching, Addison-Wesley, 1973.

Alois P. Heinz, Rows n = 1..150, flattened

Cf. A321352, A345461 (same idea for other sorting algorithms).
Cf. A180191 (second column, k=1).
Cf. A107111 a triangle with some common parts.
Cf. A143689 (diagonal T(n,n-3)).
Cf. A000142, A000166, A000255, A000274, A000313, A010027, A123513.

b:= proc(n, k) option remember; (k+1)!*
      binomial(n, k)*add((-1)^i/i!, i=0..k+1)/n
    end:
T:= proc(n, k) option remember;
     `if`(k=0, n!, T(n, k-1)-b(n, n-k+1))
    end:
seq(seq(T(n, k), k=0..n-1), n=1..10);  # Alois P. Heinz, Aug 11 2021

b[n_, k_] := b[n, k] = (k+1)!*Binomial[n, k]*Sum[(-1)^i/i!, {i, 0, k+1}]/n;
T[n_, k_] := T[n, k] = If[k == 0, n!, T[n, k-1] - b[n, n-k+1]];
Table[Table[T[n, k], {k, 0, n - 1}], {n, 1, 10}] // Flatten (* Jean-François Alcover, Mar 06 2022, after Alois P. Heinz *)

T(n,0) = n!; T(n,n-3) = (3*(n-1)^2 - n + 3)/2.
From Alois P. Heinz, Aug 11 2021: (Start)
T(n,k) = T(n,k-1) - A010027(n,n-k) for k >= 1.
T(n,k) - T(n,k+1) = A123513(n,k).
T(n,0) - T(n,1) = A000255(n-1) for n >= 2.
T(n,1) - T(n,2) = A000166(n) for n >= 3.
T(n,2) - T(n,3) = A000274(n) for n >= 4.
T(n,3) - T(n,4) = A000313(n) for n >= 5. (End)

A173247 a(0) = -1 and a(n) = (-1)^n(n - 4 - 3n^2)/2 for n >= 1.

Original entry on oeis.org

-1, 3, -7, 14, -24, 37, -53, 72, -94, 119, -147, 178, -212, 249, -289, 332, -378, 427, -479, 534, -592, 653, -717, 784, -854, 927, -1003, 1082, -1164, 1249, -1337, 1428, -1522, 1619, -1719, 1822, -1928, 2037, -2149, 2264, -2382, 2503, -2627, 2754

Offset: 0

Roger L. Bagula, Feb 13 2010

Index entries for linear recurrences with constant coefficients, signature (-3,-3,-1).

Cf. A089071, A105811, A143689.

[-1] cat [(-1)^n*(n-4-3*n^2)/2: n in [1..50]]; // Vincenzo Librandi, Apr 20 2015

p[x_] = (x^3 - x^2 - 1)/(x + 1)^3;
a = Table[SeriesCoefficient[ Series[p[x], {x, 0, 50}], n], {n, 0, 50}]

Vec((x^3 - x^2 - 1)/(x + 1)^3 + O(x^50)) \\ Michel Marcus, Apr 20 2015

G.f.: (x^3 - x^2 - 1)/(x + 1)^3.
a(n) = -3*a(n-1) -3*a(n-2) -a(n-3).
From Franck Maminirina Ramaharo, Dec 27 2018: (Start)
a(n) = (A143689(n) + 1)*(-1)^(n + 1), n >= 1.
E.g.f.: 1 - (1/2)*(4 - 2*x + 3*x^2)*exp(-x). (End)

Definition simplified by the Assoc. Editors of the OEIS, Feb 21 2010

Incorrect comment removed by Joerg Arndt, Dec 27 2018

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

A104249 a(n) = (3*n^2 + n + 2)/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Formula

A033292 A Connell-like sequence: take 1 number = 1 (mod Q), 2 numbers = 2 (mod Q), 3 numbers = 3 (mod Q), etc., where Q = 3.

Views

Author

Keywords

Comments

Links

Programs

Haskell

Mathematica

Extensions

A242357 Crescendo trapezoidal.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Maple

Maxima

PARI

Formula

A192136 a(n) = (5*n^2 - 3*n + 2)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A238410 a(n) = floor((3(n-1)^2 + 1)/2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A273465 Numbers generated by starting at 1 and adding twice and subtracting once following the sequence of positive integers.

Views

Author

Keywords

Comments

Examples

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

A192136 a(n) = (5n^2 - 3n + 2)/2.

A173247 a(0) = -1 and a(n) = (-1)^n(n - 4 - 3n^2)/2 for n >= 1.