A241016 -id:A241016 - cp's OEIS frontend

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

1, 4, 15, 32, 65, 108, 175, 256, 369, 500, 671, 864, 1105, 1372, 1695, 2048, 2465, 2916, 3439, 4000, 4641, 5324, 6095, 6912, 7825, 8788, 9855, 10976, 12209, 13500, 14911, 16384, 17985, 19652, 21455, 23328, 25345, 27436, 29679, 32000, 34481, 37044, 39775, 42592

Offset: 1

Stefano Spezia, Aug 01 2018

Terms are obtained as partial sums in an algorithm for the generation of the sequence of the fourth powers (A000583). Starting with the sequence of the positive integers (A000027), it is necessary to delete every 4th term and to consider the partial sums of the obtained sequence, then to delete every 3rd term, and lastly to consider again the partial sums (see References).
a(n) is the trace of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern as shown in the examples below. Specifically, M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even, and it has det(M(n)) = 0 for n > 2 (proved).
From Saeed Barari, Oct 31 2021: (Start)
Also the sum of the entries in an n X n matrix whose elements start from 1 and increase as they approach the center. For instance, in case of n=5, the entries of the following matrix sum to 65:
  1 2 3 2 1
  2 3 4 3 2
  3 4 5 4 3
  2 3 4 3 2
  1 2 3 2 1. (End)
The n X n square matrix of the preceding comment is defined as: A[i,j,n] = n - abs((n + 1)/2 - j) - abs((n + 1)/2 - i). - Stefano Spezia, Nov 05 2021

			For n = 1 the matrix M(1) is
  1
with trace Tr(M(1)) = a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with Tr(M(2)) = a(2) = 4.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with Tr(M(3)) = a(3) = 15.

Edward A. Ashcroft, Anthony A. Faustini, Rangaswami Jagannathan, and William W. Wadge, Multidimensional Programming, Oxford University Press 1995, p. 12.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
G. Polya, Mathematics and Plausible Reasoning: Induction and analogy in mathematics, Princeton University Press 1990, p. 118.
Shailesh Shirali, A Primer on Number Sequences, Universities Press (India) 2004, p. 106.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.7.3 on pages 122-123.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, München 1952. Sitzungsberichte: 1951,3.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A000583, A000027, A186424 (first differences).
Cf. A000578, A000035, A193356, A210378.
Cf. A006003, A317297, A001489, A322844.
Cf. related to the M matrices: A074147 (antidiagonals), A130130 (rank), A241016 (row sums), A317617 (column sums), A322277 (permanent), A323723 (subdiagonal sums), A323724 (superdiagonal sums).

a_n:=List([1..nmax], n->(1/2)*(n^3 + n*RemInt(n, 2)));

List([1..50],n->(1/2)*(n^3+n*(n mod 2))); # Muniru A Asiru, Aug 24 2018

[IsEven(n) select n^3/2 else (n^3+n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 07 2018

a:=n->(1/2)*(n^3+n*modp(n,2)): seq(a(n),n=1..50); # Muniru A Asiru, Aug 24 2018

CoefficientList[Series[1/4 E^-x (1 + 3 E^(2 x) + 6 E^(2 x) x + 2 E^(2 x) x^2), {x, 0, 45}], x]*Table[(k + 1)!, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)*(1 + x)^3), {x, 0, 45}], x]*Table[(k + 1)*(-1)^k, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)^3*(1 + x)), {x, 0, 45}], x]*Table[(k + 1), {k, 0, 45}]
From Robert G. Wilson v, Aug 01 2018: (Start)
a[i_, j_, n_] := If[OddQ@ i, j + n (i - 1), n*i - j + 1]; f[n_] := Tr[Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 45]
CoefficientList[Series[(x^4 + 2x^3 + 6x^2 + 2x + 1)/((x - 1)^4 (x + 1)^2), {x, 0, 45}], x]
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 4, 15, 32, 65, 108}, 45]
(End)

a(n):=(1/2)*(n^3 + n*mod(n,2))$ makelist(a(n), n, 1, nmax);

Vec(x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2) + O(x^40)) \\ Colin Barker, Aug 02 2018

M(i, j, n) = if (i % 2, j + n*(i-1), n*i - j + 1);
a(n) = sum(k=1, n, M(k, k, n)); \\ Michel Marcus, Aug 07 2018

for (n in 1:nmax){
   a <- (n^3+n*n%%2)/2
   output <- c(n, a)
   cat(output, "\n")
}
(MATLAB and FreeMat)
for(n=1:nmax); a=(n^3+n*mod(n,2))/2; fprintf('%d\t%0.f\n',n,a); end

a(n) = (1/2)*(A000578(n) + n*A000035(n)).
a(n) = A006003(n) - (n/2)*(1 - (n mod 2)).
a(n) = Sum_{k=1..n} T(n,k), where T(n,k) = ((n + 1)*k - n)*(n mod 2) + ((n - 1)*k + 1)*(1 - (n mod 2)).
E.g.f.: E(x) = (1/4)*exp(-x)*x*(1 + 3*exp(2*x) + 6*exp(2*x)*x + 2*exp(2*x)*x^2).
L.g.f.: L(x) = -x*(1 + x^2)/((-1 + x)*(1 + x)^3).
H.l.g.f.: LH(x) = -x*(1 + x^2)/((-1 + x)^3*(1 + x)).
Dirichlet g.f.: (1/2)*(Zeta(-3 + s) + 2^(-s)*(-2 + 2^s)*Zeta(-1 + s)).
From Colin Barker, Aug 02 2018: (Start)
G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = n^3/2 for n even.
a(n) = (n^3+n)/2 for n odd. (End)
a(2*n) = A317297(n+1) + A001489(n). - Stefano Spezia, Dec 28 2018
Sum_{n>0} 1/a(n) = (1/2)*(-2*polygamma(0, 1/2) + polygamma(0, (1-i)/2)+ polygamma(0, (1+i)/2)) + zeta(3)/4 approximately equal to 1.3959168891658447368440622669882813003351669... - Stefano Spezia, Feb 11 2019
a(n) = (A000578(n) + A193356(n))/2. - Stefano Spezia, Jun 27 2022
a(n) = A210378(n-1)/n. - Stefano Spezia, Jul 15 2024

A322277 Permanent of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern.

Original entry on oeis.org

1, 11, 490, 60916, 15745548, 7477647372, 5799397213200, 6925325038489152, 11958227405868674880, 28853103567727115409600, 93561657023119005869616000, 398720531811315564754326938880, 2174628314166392755825875267321600, 14941853448103858870808931238617312000

Offset: 1

Stefano Spezia, Dec 01 2018

nonn

M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even.
det(M(1)) = 1, det(M(2)) = -5 and det(M(n)) = 0 for n > 2 (proved).
The trace of the matrix M(n) is A317614(n).

			For n = 1 the matrix M(1) is
  1
with permanent a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with permanent a(2) = 11.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with permanent a(3) = 490.

Vaclav Kotesovec, Table of n, a(n) for n = 1..35

Cf. A317614 (trace of matrix M(n)).

Cf. A241016 (row sums of M matrices), A317617 (column sums of M matrices), A074147 (antidiagonals of M matrices).

with(LinearAlgebra):
a := n -> Permanent(Matrix(n, (i, j) -> 1-j+i*n+(-1+2*j-n)*modp(i,2))):
seq(a(n), n = 1 .. 20);

M[i_, j_, n_] := 1 - j + i n + (-1 + 2 j - n) Mod[i, 2]; a[n_] := Permanent[Table[M[i, j, n], {i, n}, {j, n}]]; Array[a, 20]

a(n) = matpermanent(matrix(n, n, i, j, if (i % 2, j + n*(i-1), n*i - j + 1)));
vector(20, n, a(n))

A317617 Triangle T read by rows: T(n, k) = (n^3 + n)/2 + (k - (n + 1)/2)*(n mod 2).

Original entry on oeis.org

1, 5, 5, 14, 15, 16, 34, 34, 34, 34, 63, 64, 65, 66, 67, 111, 111, 111, 111, 111, 111, 172, 173, 174, 175, 176, 177, 178, 260, 260, 260, 260, 260, 260, 260, 260, 365, 366, 367, 368, 369, 370, 371, 372, 373, 505, 505, 505, 505, 505, 505, 505, 505, 505, 505, 666

Offset: 1

Stefano Spezia, Aug 01 2018

T(n, k) is the sum of the terms of the k-th column of an n X n square matrix M formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern (proved). The n X n square matrix M is defined as M[i, j, n] = j + n*(i - 1) if i is odd and M[i, j, n] = n*i - j + 1 if i is even (see the examples below).

The rows of even indices of the triangle T are made of all the same repeating number.

			n\k|   1   2   3   4   5   6
---+------------------------
1  |   1
2  |   5   5
3  |  14  15  16
4  |  34  34  34  34
5  |  63  64  65  66  67
6  | 111 111 111 111 111 111
...
For n = 1 the matrix M is
  1
with column sum 1.
For n = 2 the matrix M is
  1, 2
  4, 3
with column sums 5, 5.
For n = 3 the matrix M is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with column sums 14, 15, 16.

Stefano Spezia, First 150 rows of the triangle, flattened

Cf. A006003, A000027, A000035, A037270 (row sums).
A317614(n): the trace of the n X n square matrix M.
A074147(n): the elements of the antidiagonal of the n X n square matrix M.
A241016(n): the triangle of the row sums of the n X n square matrix M.
A246697(n): the right diagonal of the triangle T.

A317617 := function(n)
local i, j, t;
for i in [1 .. n] do
   for j in [1 .. i] do
      t := (i^3 + i)/2 + (j - (i + 1)/2)*(i mod 2);
      Print(t, "\t");
   od;
   Print("\n");
od;
end;
A317617(11); # yields sequence in triangular form

Flat(List([1..11],n->List([1..n],k->(n^3+n)/2+(k-(n+1)/2)*(n mod 2)))); # Muniru A Asiru, Aug 24 2018

[[(n^3 + n)/2 + (k - (n + 1)/2)*(n mod 2): k in [1..n]]: n in [1..11]];

a:=(n,k)->(n^3+n)/2+(k-(n+1)/2)*modp(n,2): seq(seq(a(n,k),k=1..n),n=1..11); # Muniru A Asiru, Aug 24 2018

f[n_] := Table[SeriesCoefficient[(x*(x*(5 - 7*y) + x^4*(1 - 2*y) - x^3*(-3 + y) - 3*x^2*(-1 + y) + y))/((-1 + x)^4*(1 + x)^2*(-1 + y)^2), {x, 0, i}, {y, 0, j}], {i, n, n}, {j, 1, n}]; Flatten[Array[f, 11]]
T[i_, j_, n_] := If[OddQ@ i, j + n*(i - 1), n*i - j + 1]; f[n_] := Plus @@@ Transpose[ Table[T[i, j, n], {i, n}, {j, n}]]; Array[f, 11] // Flatten  (* Robert G. Wilson v, Aug 01 2018 *)
f[n_] := Table[SeriesCoefficient[1/4 E^(-x + y) (1 - x - 2 y + E^(2 x) (-1 + 3 x + 6 x^2 + 2 x^3 + 2 y)), {x, 0, i}, {y, 0, j}]*i!*j!, {i, n, n}, {j, 1, n}]; Flatten[Array[f, 11]] (* Stefano Spezia, Jan 10 2019 *)

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 1 thru n do disp(sjoin(makelist((i^3+i)/2+(j-(i+1)/2)*mod(i, 2), j, 1, i), " ")); display_triangle(10);

M(i,j,n) = if (i % 2, j + n*(i-1), n*i - j + 1);
T(n, k) = sum(i=1, n, M(i,k,n));
tabl(nn) = for(n=1, nn, for(k=1, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Aug 09 2018

# by formula
for (n in 1:11){
   t <- c(n, "")
   for(j in 1:n){
      t <- c(t, (n^3+n)/2+(j-(n+1)/2)*(n%%2), "")
   }
   cat(t, "\n")
} # yields sequence in triangular form
(MATLAB and FreeMat)
for(i=1:11);
   for(j=1:i);
      t=(i^3 + i)/2 + (j - (i + 1)/2)*mod(i,2);
      fprintf('%0.f\t', t);
   end
   fprintf('\n');
end % yields sequence in triangular form

T(n, k) = A006003(n) + (k - (A000027(n) + 1)/2)*A000035(n).
G.f.: x*(x*(5 - 7*y) + x^4*(1 - 2*y) - x^3*(- 3 + y) - 3*x^2*(- 1 + y) + y)/((-1 + x)^4*(1 + x)^2*(-1 + y)^2).
E.g.f.: (1/4)*exp(-x + y)*(1 - x - 2*y + exp(2*x)*(-1 + 3*x + 6*x^2 + 2*x^3 + 2*y)). - Stefano Spezia, Jan 10 2019

A340804 Triangle read by rows: T(n, k) = 1 + k(n - 1) + (2k - n - 1)*(k mod 2) with 0 < k <= n.

Original entry on oeis.org

1, 1, 3, 1, 5, 9, 1, 7, 11, 13, 1, 9, 13, 17, 25, 1, 11, 15, 21, 29, 31, 1, 13, 17, 25, 33, 37, 49, 1, 15, 19, 29, 37, 43, 55, 57, 1, 17, 21, 33, 41, 49, 61, 65, 81, 1, 19, 23, 37, 45, 55, 67, 73, 89, 91, 1, 21, 25, 41, 49, 61, 73, 81, 97, 101, 121, 1, 23, 27, 45, 53, 67, 79, 89, 105, 111, 131, 133

Offset: 1

Stefano Spezia, Jan 22 2021

T(n, k) is the k-th diagonal element of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern.

It includes exclusively all the odd numbers (A005408). Except the term 1, all the other odd numbers appear a finite number of times.

			1
1,  3
1,  5,  9,
1,  7, 11, 13
1,  9, 13, 17, 25
1, 11, 15, 21, 29, 31
1, 13, 17, 25, 33, 37, 49
...

Stefano Spezia, Table of n, a(n) for n = 1..11325 (first 150 rows of the triangle, flattened).

Cf. A005408, A317614 (row sums).

Cf. A000012 (1st column), A006010 (sum of the first n rows), A060747 (2nd column), A074147 (antidiagonals of M matrices), A241016 (row sums of M matrices), A317617 (column sums of M matrices), A322277 (permanent of M matrices), A323723 (subdiagonal sum of M matrices), A323724 (superdiagonal sum of M matrices).

Table[1+k(n-1)+(2k-n-1)Mod[k,2],{n,12},{k,n}]//Flatten

T(n, k) = 1 + k*(n - 1) + (2*k - n - 1)*(k % 2); \\ Michel Marcus, Jan 25 2021

O.g.f.: (1 + y - 3*y^2 + y^3 + x*(-1 - y + 5*y^2 + y^3))/((-1 + x)^2*(-1 + y)^2*(1+y)^2).

E.g.f.: exp(x - y)*(1 + x + 2*y + exp(2*y)*(1 + x*(-1 + 2*y)))/2.

A322844 a(n) = (1/12)n^2(3(1 + n^2) - 2(2 + n^2)*(n mod 2)).

Original entry on oeis.org

0, 0, 5, 6, 68, 50, 333, 196, 1040, 540, 2525, 1210, 5220, 2366, 9653, 4200, 16448, 6936, 26325, 10830, 40100, 16170, 58685, 23276, 83088, 32500, 114413, 44226, 153860, 58870, 202725, 76880, 262400, 98736, 334373, 124950, 420228, 156066, 521645, 192660, 640400, 235340

Offset: 0

Stefano Spezia, Dec 28 2018

Conjectures: (Start)
For n > 1, a(n) is the absolute value of the trace of the 2nd exterior power of an n X n square matrix M(n) defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even (see A317614). Equivalently, a(n) is the absolute value of the coefficient of the term [x^(n-2)] in the characteristic polynomial of the matrix M(n), or the absolute value of the sum of all principal minors of M(n) of size 2.
For k > 2, the trace of the k-th exterior power of the matrix M(n) is equal to zero.
(End)

Stefano Spezia, Table of n, a(n) for n = 0..10000
Wikipedia, Characteristic polynomial
Wikipedia, Exterior algebra
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-10,0,10,0,-5,0,1).

Cf. A317614 (trace of matrix M(n)).

Cf. A002415, A037270, A074147 (antidiagonals of M matrices), A241016 (row sums of M matrices), A317617 (column sums of M matrices), A322277 (permanent of matrix M(n)), A323723 (subdiagonal sum of M matrices), A323724 (superdiagonal sum of M matrices), A325516 (k-superdiagonal sum of M matrices), A325655 (k-subdiagonal sum of M matrices).

Flat(List([0..50], n->(1/12)*n^2*(3*(1 + n^2) - 2*(2 + n^2)*(n mod 2))));

[IsEven(n) select (1/4)*n^2*(1 + n^2) else (1/12)*(- 1 + n)*n^2*(1 + n): n in [0..50]];

a:=n->(1/12)*n^2*(3*(1 + n^2) - 2*(2 + n^2)*modp(n,2)): seq(a(n), n=0..50);

a[n_]:=(1/12)*n^2*(3*(1 + n^2) - 2*(2 + n^2)*Mod[n,2]); Array[a,50,0]
LinearRecurrence[{0,5,0,-10,0,10,0,-5,0,1},{0,0,5,6,68,50,333,196,1040,540},50] (* Harvey P. Dale, Aug 23 2025 *)

a(n):=(1/12)*n^2*(3*(1 + n^2) - 2*(2 + n^2)*mod(n,2))$ makelist(a(n), n, 0, 50);

a(n) = (1/12)*n^2*(3*(1 + n^2) - 2*(2 + n^2)*(n % 2));

a(n) = abs(polcoeff(charpoly(matrix(n, n, i, j, if (i %2, j + n*(i-1), n*i - j + 1))), n-2)); \\ Michel Marcus, Feb 06 2019

[int(n**2*(3*(1 + n**2) - 2*(2 + n**2)*pow(n, 1, 2))/12) for n in range(0,50)]

O.g.f.: -x^2*(5 + 6*x + 43*x^2 + 20*x^3 + 43*x^4 + 6*x^5 + 5*x^6)/((-1 + x)^5*(1 + x)^5).
E.g.f.: (1/(12*x^2))*exp(-x)*(24 - 60*exp(x) + 21*x + 9*x^2 + 2*x^3 + x^4 + exp(2*x)*(36 - 33*x + 15*x^2 - 4*x^3 + 2*x^4)).
a(n) = (1/4)*n^2*(1 + n^2) for n even.
a(n) = (1/2)*A037270(n) for n even.
a(n) = (1/12)*(-1 + n)*n^2*(1 + n) for n odd.
a(n) = A002415(n) for n odd.
a(2*n+1) = 5*a(2*n-1) - 10*a(2*n-3) + 10*a(2*n-5) - 5*a(2*n-7) + a(2*n-9), for n > 4.
a(2*n) = 5*a(2*n-2) - 10*a(2*n-4) + 10*a(2*n-6) - 5*a(2*n-8) + a(2*n-10), for n > 4.
O.g.f. for a(2*n+1): -x*(2*(3 + 10*x + 3*x^2))/(-1 + x)^5.
O.g.f. for a(2*n): x*(-5 - 43*x - 43*x^2 - 5*x^3)/(-1 + x)^5.
E.g.f. for a(2*n+1): (1/12)*(6*x*cosh(sqrt(x)) + sqrt(x)*(6 + x)*sinh(sqrt(x))).
E.g.f. for a(2*n): (1/4)*(x*(8 + x)*cosh(sqrt(x)) + 2*sqrt(x)*(1 + 3*x)*sinh(sqrt(x))).
Sum_{k>=1} 1/a(2*k) = (1/6)*(12 + Pi^2 - 6*Pi*coth(Pi/2)) = 0.21955691692893092525407699347398665248691900...
Sum_{k>=1} 1/a(2*k+1) = 3*(5 - Pi^2/2) = 0.1955933983659620717482635001857732970...
Sum_{k>=2} 1/a(k) = 17 - (4*Pi^2)/3 - Pi*coth(Pi/2) = 0.415150315294892997002340493659759949516369894...

A247327 Triangle read by rows: T(n,k) = sum of k-th row of n X n square filled with odd numbers 1 through 2*n^2-1 reading across rows left-to-right.

Original entry on oeis.org

1, 4, 12, 9, 27, 45, 16, 48, 80, 112, 25, 75, 125, 175, 225, 36, 108, 180, 252, 324, 396, 49, 147, 245, 343, 441, 539, 637, 64, 192, 320, 448, 576, 704, 832, 960, 81, 243, 405, 567, 729, 891, 1053, 1215, 1377, 100, 300, 500, 700, 900, 1100, 1300, 1500, 1700, 1900, 121, 363

Offset: 1

Kival Ngaokrajang, Sep 13 2014

See illustration in links. Column c(k) = (2*k - 1)*n^2. Diagonal d(m) = (2*n - 2*m + 1)*n^2.

			Triangle begins:
  1
  4   12
  9   27  45
  16  48  80 112
  25  75 125 175 225
  36 108 180 252 324 396
  49 147 245 343 441 539 637

Kival Ngaokrajang, Illustration of initial terms

Column: c(1) = A000290, c(2) = A033428, c(3) = A033429.
Diagonal: d(1) = A015237, d(2) = A015238, d(3) = A015240.
Rows sum: A000538.
Cf. A241016.

trg(nn) = {for (n=1, nn, mm = matrix(n, n, i, j, (2*j-1) + (2*n)*(i-1)); for (i=1, n, print1(sum(j=1, n, mm[i, j]), ", ");); print(););} \\ Michel Marcus, Sep 15 2014

A251630 Column sums of the n X n square array filled with numbers from 1 to n^2, row by row, from left to right.

Original entry on oeis.org

1, 4, 6, 12, 15, 18, 28, 32, 36, 40, 55, 60, 65, 70, 75, 96, 102, 108, 114, 120, 126, 154, 161, 168, 175, 182, 189, 196, 232, 240, 248, 256, 264, 272, 280, 288, 333, 342, 351, 360, 369, 378, 387, 396, 405, 460, 470, 480, 490, 500, 510, 520, 530

Offset: 1

Wolfdieter Lang, Dec 09 2014

This triangle has been considered by Kival Ngaokrajang as a companion of A241016. See the link given there, the second triangle.

			The n=4 square array is:
1   2  3  4
5   6  7  8
9  10 11 12
13 14 15 16
and the column sums are 28 32 36 40, which appear
in row n=4 of the triangle T.
The triangle T(n,k) begins:
n\k   1   2   3   4   5   6   7   8   9  10 ...
1:    1
2:    4   6
3:   12  15  18
4:   28  32  36  40
5:   55  60  65  70  75
6:   96 102 108 114 120 126
7:  154 161 168 175 182 189 196
8:  232 240 248 256 264 272 280 288
9:  333 342 351 360 369 378 387 396 405
10: 460 470 480 490 500 510 520 530 540 550
...

Cf. A002411 (main diagonal), A006000 (column k=1), A241016.

T(n, k) = sum(n*(j-1)+ k, j=1..n), n >= k >= 1.

T(n, k) = n*(binomial(n+1, 2) + (k-n)).

cp's OEIS Frontend

A317614 a(n) = (1/2)*(n^3 + n*(n mod 2)).

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A322277 Permanent of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern.

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A317617 Triangle T read by rows: T(n, k) = (n^3 + n)/2 + (k - (n + 1)/2)*(n mod 2).

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A340804 Triangle read by rows: T(n, k) = 1 + k*(n - 1) + (2*k - n - 1)*(k mod 2) with 0 < k <= n.

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A322844 a(n) = (1/12)*n^2*(3*(1 + n^2) - 2*(2 + n^2)*(n mod 2)).

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A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

A340804 Triangle read by rows: T(n, k) = 1 + k(n - 1) + (2k - n - 1)*(k mod 2) with 0 < k <= n.

A322844 a(n) = (1/12)n^2(3(1 + n^2) - 2(2 + n^2)*(n mod 2)).