A116731 -id:A116731 - cp's OEIS frontend

A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531

Offset: 0

Klaus Brockhaus, Nov 14 2009

a(n) = ((n*(n+1)*(n+2))+(n+(n+1)+(n+2)))/3, n >= 0.
Equals A006527 without initial term 0: a(n) = A006527(n+1).
Binomial transform of A167876.
Inverse binomial transform of A080930.
a(n) = A007290(n+2)+n+1.
a(n) = A014820(n)/(n+1) for n > 0.
a(n) = A116731(n+2)-1.
a(n) = A033547(n+1)-n.
a(n) = A054602(n)/3.
a(n) = A086514(n+3)-2.
a(n) = A002061(n+1)+a(n-1) for n > 0.
a(n) = A005894(n)-a(n-1) for n > 0.
First bisection is A057813.
Second differences are in A004277.
a(n) = A177342(n)*(-1)+a(n-1)*5 with n>0. For n=8, a(8)=-A177342(8)+a(7)*5=-631+176*5=249. - Bruno Berselli, May 18 2010

			a(0) = (0*1*2+0+1+2)/3 = (0+3)/3 = 1.
a(1) = (1*2*3+1+2+3)/3 = (6+6)/3 = 4.
a(6)-4*a(5)+6*a(4)-4*a(3)+a(2) = 119-4*76+6*45-4*24+11 = 0. - _Bruno Berselli_, May 26 2010

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. A001477 (nonnegative integers),
A006527 ((n^3+2*n)/3),
A167876 (1, 3, 4, 2, 0, 0, 0, 0, ...),
A080930,
A007290 (2*C(n, 3)),
A014820 ((1/3)*(n^2+2*n+3)*(n+1)^2),
A116731,
A033547 (n*(n^2+5)/3),
A054602 (Sum_{d|3} phi(d)*n^(3/d)),
A086514 ((n^3-6*n^2+14*n-6)/3),
A002061 (n^2-n+1),
A005894 (centered tetrahedral numbers),
A057813 ((2*n+1)*(4*n^2+4*n+3)/3),
A004277 (1 and the positive even numbers),
A028387 (n+(n+1)^2),
A166941, A166942, A166943.

[ (&*s + &+s)/3 where s is [n..n+2]: n in [0..42] ];

Select[Table[(n*(n+1)*(n+2)+n+(n+1)+(n+2))/3,{n,0,5!}],IntegerQ[#]&] (* Vladimir Joseph Stephan Orlovsky, Dec 04 2010 *)
(Times@@#+Total[#])/3&/@Partition[Range[0,65],3,1]  (* Harvey P. Dale, Mar 14 2011 *)

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

a(n) = (n^3+3*n^2+5*n+3)/3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+2 for n > 3; a(0)=1, a(1)=4, a(2)=11, a(3)=24.
G.f.: (1+x^2)/(1-x)^4.
a(n) = SUM(A109613(k)*A005408(n-k): 0<=k<=n). - Reinhard Zumkeller, Dec 05 2009
a(n)-4*a(n-1)+6*a(n-2)-4*a(n-3)+a(n-4)=0 for n>3. - Bruno Berselli, May 26 2010

A130154 Triangle read by rows: T(n, k) = 1 + 2(n-k)(k-1) (1 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 9, 7, 1, 1, 9, 13, 13, 9, 1, 1, 11, 17, 19, 17, 11, 1, 1, 13, 21, 25, 25, 21, 13, 1, 1, 15, 25, 31, 33, 31, 25, 15, 1, 1, 17, 29, 37, 41, 41, 37, 29, 17, 1, 1, 19, 33, 43, 49, 51, 49, 43, 33, 19, 1, 1, 21, 37, 49, 57, 61, 61, 57, 49, 37, 21, 1

Offset: 1

Emeric Deutsch, May 22 2007

Column k, except for the initial k-1 0's, is an arithmetic progression with first term 1 and common difference 2(k-1). Row sums yield A116731. First column of the inverse matrix is A129779.
Studied by Paul Curtz circa 1993.
From Rogério Serôdio, Dec 19 2017: (Start)
T(n, k) gives the number of distinct sums of 2(k-1) elements in {1,1,2,2,...,n-1,n-1}. For example, T(6, 2) = the number of distinct sums of 2 elements in {1,1,2,2,3,3,4,4,5,5}, and because each sum from the smallest 1 + 1 = 2 to the largest 5 + 5 = 10 appears, T(6, 2) = 10 - 1 = 9. [In general: 2*(Sum_{j=1..(k-1)} n-j) - (2*(Sum_{j=1..k-1} j) - 1) = 2*n*(k-1) - 4*(k-1)*k/2 + 1 = 2*(k-1)*(n-k) + 1 = T(n, k). - Wolfdieter Lang, Dec 20 2017]
T(n, k) is the number of lattice points with abscissa x = 2*(k-1) and even ordinate in the closed region bounded by the parabola y = x*(2*(n-1) - x) and the x axis. [That is, (1/2)*y(2*(k-1)) + 1 = T(n, k). - Wolfdieter Lang, Dec 20 2017]
Pascal's triangle (A007318, but with apex in the middle) is formed using the rule South = West + East; the rascal triangle A077028 uses the rule South = (West*East + 1)/North; the present triangle uses a similar rule: South = (West*East + 2)/North. See the formula section for this recurrence. (End)

			The triangle T(n, k) starts:
  n\k  1  2  3  4  5  6  7  8  9 10 ...
  1:   1
  2:   1  1
  3:   1  3  1
  4:   1  5  5  1
  5:   1  7  9  7  1
  6:   1  9 13 13  9  1
  7:   1 11 17 19 17 11  1
  8:   1 13 21 25 25 21 13  1
  9:   1 15 25 31 33 31 25 15  1
 10:   1 17 29 37 41 41 37 29 17  1
 ... reformatted. - _Wolfdieter Lang_, Dec 19 2017

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A116731, A129779. A007318, A077028.

Column sequences (no leading zeros): A000012, A016813, A016921, A017077, A017281, A017533, A131877, A158057, A161705, A215145.

Flat(List([1..12], n-> List([1..n], k-> 1 + 2*(n-k)*(k-1) ))); # G. C. Greubel, Nov 25 2019

[1 + 2*(n-k)*(k-1): k in [1..n], n in [1..12]]; // G. C. Greubel, Nov 25 2019

T:=proc(n,k) if k<=n then 2*(n-k)*(k-1)+1 else 0 fi end: for n from 1 to 14 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form

Flatten[Table[1+2(n-k)(k-1),{n,0,20},{k,n}]] (* Harvey P. Dale, Jul 13 2013 *)

T(n, k) = 1 + 2*(n-k)*(k-1) \\ Iain Fox, Dec 19 2017

first(n) = my(res = vector(binomial(n+1,2)), i = 1); for(r=1, n, for(k=1, r, res[i] = 1 + 2*(r-k)*(k-1); i++)); res \\ Iain Fox, Dec 19 2017

[[1 + 2*(n-k)*(k-1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Nov 25 2019

T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).
G.f.: G(t,z) = t*z*(3*t*z^2 - z - t*z + 1)/((1-t*z)*(1-z))^2.
Equals = 2 * A077028 - A000012 as infinite lower triangular matrices. - Gary W. Adamson, Oct 23 2007
T(n, 1) = 1 and  T(n, n) = 1 for n >= 1; T(n, k) = (T(n-1, k-1)*T(n-1, k) + 2)/T(n-2, k-1), for n > 2 and 1 < k < n. See a comment above. - Rogério Serôdio, Dec 19 2017
G.f. column k (with leading zeros): (x^k/(1-x)^2)*(1 + (2*k-3)*x), k >= 1. See the g.f. of the triangle G(t,z) above: (d/dt)^k G(t,x)/k!|{t=0}. - _Wolfdieter Lang, Dec 20 2017

Edited by Wolfdieter Lang, Dec 19 2017

A129779 a(1) = 1, a(2) = -1, a(3) = 2; for n > 3, a(n) = -(2n-5)a(n-1).

Original entry on oeis.org

1, -1, 2, -6, 30, -210, 1890, -20790, 270270, -4054050, 68918850, -1309458150, 27498621150, -632468286450, 15811707161250, -426916093353750, 12380566707258750, -383797567925021250, 12665319741525701250

Offset: 1

Paul Curtz, May 17 2007

sign

Sequence is also the first column of the inverse of the infinite lower triangular matrix M, where M(j,k) = 1+2*(k-1)*(j-k) for k < j, M(j,k) = 1 for k = j, M(j,k) = 0 for k > j.
Upper left 6 X 6 submatrix of M is
  [ 1  0  0  0  0  0 ]
  [ 1  1  0  0  0  0 ]
  [ 1  3  1  0  0  0 ]
  [ 1  5  5  1  0  0 ]
  [ 1  7  9  7  1  0 ]
  [ 1  9 13 13  9  1 ],
and upper left 6 X 6 submatrix of M^-1 is
  [    1    0    0    0    0    0 ]
  [   -1    1    0    0    0    0 ]
  [    2   -3    1    0    0    0 ]
  [   -6   10   -5    1    0    0 ]
  [   30  -50   26   -7    1    0 ]
  [ -210  350 -182   50   -9    1 ].
Row sums of M are 1, 2, 5, 12, 25, 46, ... (see A116731); diagonal sums of M are 1, 1, 2, 4, 7, 13, 20, 32, 45, 65, 86, 116, 147, 189, ... with first differences 0, 1, 2, 3, 6, 7, 12, 13, 20, 21, 30, 31, 42, ... and second differences 1, 1, 1, 3, 1, 5, 1, 7, 1, 9, 1, 11, ... (see A093178).

G. C. Greubel, Table of n, a(n) for n = 1..400

Cf. A093178, A097801, A116731.

F:=Factorial;; Concatenation([1,-1], List([3..25], n-> (-1)^(n+1)*F(2*n-5)/(2^(n-4)*F(n-3)) )); # G. C. Greubel, Nov 25 2019

m:=19; M:=Matrix(IntegerRing(), m, m, [< j, k, Maximum(0, 1+2*(k-1)*(j-k)) > : j, k in [1..m] ] ); Transpose(ColumnSubmatrix(M^-1, 1, 1)); // Klaus Brockhaus, May 21 2007

F:=Factorial; [1,-1] cat [(-1)^(n+1)*F(2*n-5)/(2^(n-4)*F(n-3)): n in [3..25]]; // G. C. Greubel, Nov 25 2019

seq(`if`(n<3, (-1)^(n-1), (-1)^(n-1)*(2*n-5)!/(2^(n-4)*(n-3)!)), n=1..25); # G. C. Greubel, Nov 25 2019

a[n_]:= -(2*n-5)*a[n-1]; a[1]=1; a[2]=-1; a[3]=2; Array[a, 20] (* Robert G. Wilson v *)
Table[If[n<3, (-1)^(n-1), (-1)^(n+1)*(2*n-5)!/(2^(n-4)*(n-3)!)], {n,25}] (* G. C. Greubel, Nov 25 2019 *)

{m=19; print1(1, ",", -1, ","); print1(a=2, ","); for(n=4, m, k=-(2*n-5)*a; print1(k, ","); a=k)} \\ Klaus Brockhaus, May 21 2007

{print1(1, ",", -1, ","); for(n=3, 19, print1((-1)^(n-1)*(2*(n-2))!/((n-2)!*2^(n-3)), ","))} \\ Klaus Brockhaus, May 21 2007

{m=19; M=matrix(m, m, j, k, if(k>j, 0, if(k==j, 1, 1+2*(k-1)*(j-k)))); print((M^-1)[, 1]~)} \\ Klaus Brockhaus, May 21 2007

f=factorial; [1,-1]+[(-1)^(n+1)*f(2*n-5)/(2^(n-4)*f(n-3)) for n in (3..25)] # G. C. Greubel, Nov 25 2019

a(n) = (-1)^(n-1)*A097801(n-2) = (-1)^(n-1)*(2*(n-2))!/((n-2)!*2^(n-3)) for n > 2, with a(1)=1, a(2)=-1.

G.f.: 1 + x - x*W(0) , where W(k) = 1 + 1/( 1 - x*(2*k+1)/( x*(2*k+1) - 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 22 2013

Edited and extended by Klaus Brockhaus and Robert G. Wilson v, May 21 2007

A227182 Simple self-inverse permutation of natural numbers: List each block of n^2 - n + 1 numbers from ((n-1)^3 + 2(n-1))/3 + 1 to (n^3 + 2n)/3 in reverse order.

Original entry on oeis.org

0, 1, 4, 3, 2, 11, 10, 9, 8, 7, 6, 5, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46

Offset: 0

Antti Karttunen, Jul 04 2013

nonn

In other words, after zero, list each block of A002061(n) numbers from A116731(n) to A006527(n) in reverse order.

Antti Karttunen, Table of n, a(n) for n = 0..9951
Index entries for sequences that are permutations of the natural numbers

Cf. A227177, A227179, A227147.

Flatten[(Reverse/@( (-1+#) (9-5 #+#^2)/3+Range[(#-2) (#-1)+1]-1)&)[Range[7]]] (* Peter J. C. Moses, Jul 11 2013 *)

(define (A227182 n) (- (A006527 (A227177 n)) (A227179 n)))

a(n) = A006527(A227177(n)) - A227179(n).

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.

A173809 a(2n+1) = 1+A131941(2n+1). a(2n) = A131941(2n).

Original entry on oeis.org

2, 3, 9, 16, 30, 47, 73, 104, 146, 195, 257, 328, 414, 511, 625, 752, 898, 1059, 1241, 1440, 1662, 1903, 2169, 2456, 2770, 3107, 3473, 3864, 4286, 4735, 5217, 5728, 6274, 6851, 7465, 8112, 8798, 9519, 10281, 11080

Offset: 1

Paul Curtz, Feb 25 2010

An array T(n,k) of a(n) and its successive differences is T(1,k) = a(k), T(n,k) = T(n-1,k+1)-T(n-1,k) and starts:
2, 3, 9, 16, 30, 47, 73, 104, 146, 195, 257, 328,...
1, 6, 7, 14, 17, 26, 31, 42, 49, 62, 71, 86, 97, 114,... interleaved A056220 and A051890
5, 1, 7, 3, 9, 5, 11, 7, 13, 9, 15, 11, 17, 13, 19, 15.... A158552
-4, 6, -4, 6, -4, 6, -4, 6, -4, 6, -4, 6, -4, 6, -4,... A010711
10, -10, 10, -10, 10, -10, 10, -10, 10, -10, 10,..

Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

LinearRecurrence[{3,-2,-2,3,-1},{2,3,9,16,30},60] (* Harvey P. Dale, Aug 10 2021 *)

G.f.: -x*(-2+3*x+x^3-4*x^2) / ( (1+x)*(x-1)^4 ). - R. J. Mathar, Jan 13 2011
a(2n) = A172482(n-1).
a(n)+a(n+1) = A116731(n+2). - R. J. Mathar, Jan 13 2011

A329854 Triangle read by rows: T(n,k) = ((n - k)*(n + k - 1) + 2)/2, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 4, 3, 1, 7, 7, 6, 4, 1, 11, 11, 10, 8, 5, 1, 16, 16, 15, 13, 10, 6, 1, 22, 22, 21, 19, 16, 12, 7, 1, 29, 29, 28, 26, 23, 19, 14, 8, 1, 37, 37, 36, 34, 31, 27, 22, 16, 9, 1, 46, 46, 45, 43, 40, 36, 31, 25, 18, 10, 1, 56, 56, 55, 53, 50, 46, 41, 35, 28, 20, 11, 1

Offset: 0

Werner Schulte, Nov 22 2019

This triangle equals A309559 with reversed rows and supplemented main diagonal (all terms are 1).
There are two lower triangular matrices M and N so that the matrix product M * N equals T (seen as a matrix).
           / 1             \                 / 1             \
           | 0 1           |                 | 1 1           |
           | 0 1 1         |                 | 1 1 1         |
  M(n,k) = | 0 1 2 1       |        N(n,k) = | 1 1 1 1       |
           | 0 1 2 3 1     |                 | 1 1 1 1 1     |
           | 0 1 2 3 4 1   |                 | 1 1 1 1 1 1   |
           \ . . . . . . . /                 \ . . . . . . . /
  The matrix product N * M equals the rascal triangle A077028 (seen as a matrix).

			The triangle T(n,k) starts:
n \ k :   0    1    2    3    4    5    6    7    8    9   10   11
==================================================================
   0  :   1
   1  :   1    1
   2  :   2    2    1
   3  :   4    4    3    1
   4  :   7    7    6    4    1
   5  :  11   11   10    8    5    1
   6  :  16   16   15   13   10    6    1
   7  :  22   22   21   19   16   12    7    1
   8  :  29   29   28   26   23   19   14    8    1
   9  :  37   37   36   34   31   27   22   16    9    1
  10  :  46   46   45   43   40   36   31   25   18   10    1
  11  :  56   56   55   53   50   46   41   35   28   20   11    1
etc.

Row sums equal A116731(n+1).
Row sums apart from column 0 equal A081489.
Cf. A077028, A309559.

O.g.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = ((t^2+(1-t)^2) * (1-x*t) + x * t^2 * (1-t)) / ((1-t)^3 * (1-x*t)^2).
G.f. of column k: Sum_{n>=k} T(n,k) * t^n = t^k * (t^2/(1-t)^3 + 1/(1-t) + k*t/(1-t)^2) for k >= 0.
T(n,k) = 1 + T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) for 0 < k < n with initial values T(n,0) = (n*(n-1)+2)/2 and T(n,n) = 1 for n >= 0.
T(n,k) = (2 + T(n-1,k-1) * T(n-1,k+1)) / T(n-2,k) for 0 < k < n-1 with initial values given above and T(n,n-1) = n for n > 0.
Referring to the triangle M(n,k) (see comments), we get:
  (1) Sum_{k=0..n} (k+1) * M(n,k) = A116731(n+1) for n >= 0;
  (2) Sum_{k=1..n} k * M(n,k) = A081489(n) for n >= 1.
T(n,k) = T(n-1,k-1) + n-k for 0 < k <= n with initial values T(n,0) = (n*(n-1)+2)/2 for n >= 0.
T(n,k) = 2 * T(n-1,k-1) - T(n-2,k-2) for 1 < k <= n with initial values T(0,0) = 1 and T(n,0) = T(n,1) = (n*(n-1)+2)/2 for n > 0.

cp's OEIS Frontend

A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

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A130154 Triangle read by rows: T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).

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A129779 a(1) = 1, a(2) = -1, a(3) = 2; for n > 3, a(n) = -(2*n-5)*a(n-1).

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A227182 Simple self-inverse permutation of natural numbers: List each block of n^2 - n + 1 numbers from ((n-1)^3 + 2*(n-1))/3 + 1 to (n^3 + 2*n)/3 in reverse order.

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

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A173809 a(2n+1) = 1+A131941(2n+1). a(2n) = A131941(2n).

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A130154 Triangle read by rows: T(n, k) = 1 + 2(n-k)(k-1) (1 <= k <= n).

A129779 a(1) = 1, a(2) = -1, a(3) = 2; for n > 3, a(n) = -(2n-5)a(n-1).

A227182 Simple self-inverse permutation of natural numbers: List each block of n^2 - n + 1 numbers from ((n-1)^3 + 2(n-1))/3 + 1 to (n^3 + 2n)/3 in reverse order.