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A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n. 17
1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

The triangle is symmetric: T(n, k) = T(n-k, k).

From Kolosov Petro, Oct 07 2018: (Start)

T(n,k) is the case m = 1 in L(m,n,k) := Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (n-k)^j. For the cases m = 2 and m = 3 see A300656 and A300785, respectively.

For l = n, S1(m,l,n) = Sum_{k=1..l} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (n-k)^j. The result is: S(m,n,n) = n^(2m+1) for every positive integer m and n.

Also, for l = n, S0(m,l-1,n) = Sum_{k=0..l-1} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (n-k)^j. The result is: S(m,n,n) = n^(2m+1) for every positive integer m and n. (End)

LINKS

Kolosov Petro, Rows n = 0..30, flattened

Kolosov Petro, Another Power Identity involving Binomial Theorem and Faulhaber's formula, arXiv:1603.02468 [math.NT], 2016-2018.

FORMULA

T(n, k) = 6*k*(n-k) + 1.

Sum_{k=1..n} T(n,k) = A000578(n).

G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.

From Kolosov Petro, Jan 09 2018: (Start)

Sum_{k=0..n-1} T(A000124(k), 1) = A000578(n).

2*T(n, k) = T(A294317(n,k), k) + T(A294317(n,k), 0).

T(n, k) = 6*A077028(n,k) - 5.

T(n, 2) - 1 = A008458(n).

2*T(n, k) = T(n+1, k) + T(n-1, k), for n >= k. (End)

G.f.: (-1 + 8*y + 5*y^2 + x*(1 - 14*y + y^2))/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 09 2018

From Kolosov Petro, Oct 30 2018: (Start)

Sum_{k=0..n-1} T(n,k) = A000578(n).

Sum_{k=1..n-1} T(n,k) = A068601(n).

Sum_{k=0..n} T(n,k) = A001093(n). (End)

Sum_{k=1..n} T(k,1) = A000567(n). - Kolosov Petro, Nov 15 2018

EXAMPLE

Triangle begins:

----------------------------------------

k=    0   1   2   3   4   5   6   7   8

----------------------------------------

n=0:  1;

n=1:  1,  1;

n=2:  1,  7,  1;

n=3:  1, 13, 13,  1;

n=4:  1, 19, 25, 19,  1;

n=5:  1, 25, 37, 37, 25,  1;

n=6:  1, 31, 49, 55, 49, 31,  1;

n=7:  1, 37, 61, 73, 73, 61, 37,  1;

n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;

MAPLE

T := (n, k) -> 6*k*(n-k) + 1:

seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018

PROG

(Python)

def f(x):

    a=[]

    for k in range(x):

        for m in range(k+1):

            a.append(6*m*(k-m)+1)

    return a

print f(10)

(PARI) t(n, k) = 6*k*(n-k)+1

trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))

/* Print initial 9 rows of triangle as follows */

trianglerows(9) \\ Felix Fröhlich, Jan 09 2018

(GAP) Flat(List([0..11], n->List([0..n], k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018

(MAGMA) /* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018

CROSSREFS

Sequences A300656, A300785 represent the cases for m = 2, 3 in L(m,n,k), see comments, line 2.

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.

Sum of the n-th row terms of the triangle A287326 gives A001093.

T(2n,n) gives A227776.

T(A000124(n), 1) gives A003215.

Cf. A000578, A038593, A007318, A055012, A077028, A302971, A304042, A068601, A001093, A000567.

Sequence in context: A183352 A217510 A273506 * A131065 A081580 A082110

Adjacent sequences:  A287323 A287324 A287325 * A287327 A287328 A287329

KEYWORD

nonn,tabl,easy

AUTHOR

Kolosov Petro, Aug 31 2017

STATUS

approved

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Last modified February 20 20:55 EST 2019. Contains 320345 sequences. (Running on oeis4.)