[go: up one dir, main page]

login
A300656
Triangle read by rows: T(n,k) = 30*k^2*(n-k)^2 + 1; n >= 0, 0 <= k <= n.
10
1, 1, 1, 1, 31, 1, 1, 121, 121, 1, 1, 271, 481, 271, 1, 1, 481, 1081, 1081, 481, 1, 1, 751, 1921, 2431, 1921, 751, 1, 1, 1081, 3001, 4321, 4321, 3001, 1081, 1, 1, 1471, 4321, 6751, 7681, 6751, 4321, 1471, 1, 1, 1921, 5881, 9721, 12001, 12001, 9721, 5881, 1921, 1
OFFSET
0,5
COMMENTS
From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(2, n, k).
Fifth power can be expressed as row sum of triangle T(n, k).
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)
FORMULA
From Kolosov Petro, Apr 12 2020: (Start)
T(n, k) = 30 * k^2 * (n-k)^2 + 1.
T(n, k) = 30 * A094053(n,k)^2 + 1.
T(n, k) = A158558((n-k) * k).
T(n+2, k) = 3*T(n+1, k) - 3*T(n, k) + T(n-1, k), for n >= k.
Sum_{k=1..n} T(n, k) = A000584(n).
Sum_{k=0..n-1} T(n, k) = A000584(n).
Sum_{k=0..n} T(n, k) = A002561(n).
Sum_{k=1..n-1} T(n, k) = A258807(n).
Sum_{k=1..n-1} T(n, k) = -A024003(n), n > 1.
Sum_{k=1..r} T(n, k) = A316349(2,r,0)*n^0 - A316349(2,r,1)*n^1 + A316349(2,r,2)*n^2. (End)
G.f.: (1 + 26*y + 336*y^2 + 326*y^3 + 31*y^4 + x^2*(1 + 116*y + 486*y^2 + 116*y^3 + y^4) + x*(-2 - 82*y - 882*y^2 - 502*y^3 + 28*y^4))/((-1 + x)^3*(-1 + y)^5). - Stefano Spezia, Oct 30 2018
EXAMPLE
Triangle begins:
--------------------------------------------------------------------------
k= 0 1 2 3 4 5 6 7 8 9 10
--------------------------------------------------------------------------
n=0: 1;
n=1: 1, 1;
n=2: 1, 31, 1;
n=3: 1, 121, 121, 1;
n=4: 1, 271, 481, 271, 1;
n=5: 1, 481, 1081, 1081, 481, 1;
n=6: 1, 751, 1921, 2431, 1921, 751, 1;
n=7: 1, 1081, 3001, 4321, 4321, 3001, 1081, 1;
n=8: 1, 1471, 4321, 6751, 7681, 6751, 4321, 1471, 1;
n=9: 1, 1921, 5881, 9721, 12001, 12001, 9721, 5881, 1921, 1;
n=10: 1, 2431, 7681, 13231, 17281, 18751, 17281, 13231, 7681, 2431, 1;
MAPLE
a:=(n, k)->30*k^2*(n-k)^2+1: seq(seq(a(n, k), k=0..n), n=0..9); # Muniru A Asiru, Oct 24 2018
MATHEMATICA
T[n_, k_] := 30 k^2 (n - k)^2 + 1; Column[
Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Apr 12 2020 *)
f[n_]:=Table[SeriesCoefficient[(1 + 26 y + 336 y^2 + 326 y^3 + 31 y^4 + x^2 (1 + 116 y + 486 y^2 + 116 y^3 + y^4) + x (-2 - 82 y - 882 y^2 - 502 y^3 + 28 y^4))/((-1 + x)^3 (-1 + y)^5), {x, 0, i}, {y, 0, j}], {i, n, n}, {j, 0, n}]; Flatten[Array[f, 11, 0]] (* Stefano Spezia, Oct 30 2018 *)
PROG
(PARI) t(n, k) = 30*k^2*(n-k)^2+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)
(GAP) T:=Flat(List([0..9], n->List([0..n], k->30*k^2*(n-k)^2+1))); # Muniru A Asiru, Oct 24 2018
(Magma) [[30*k^2*(n-k)^2+1: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Dec 14 2018
(Sage) [[30*k^2*(n-k)^2+1 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 14 2018
CROSSREFS
Various cases of L(m, n, k): A287326(m=1), This sequence (m=2), A300785(m=3). See comments for L(m, n, k).
Row sums give the nonzero terms of A002561.
Sequence in context: A040963 A040962 A040961 * A239633 A174692 A172302
KEYWORD
nonn,tabl,easy
AUTHOR
Kolosov Petro, Mar 10 2018
STATUS
approved