cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3).

Original entry on oeis.org

0, 1, 3, 3, 1, 0, 7, 14, 9, 2, 0, 13, 37, 43, 26, 8, 1, 0, 19, 72, 129, 141, 98, 42, 10, 1, 0, 25, 119, 291, 463, 504, 378, 192, 63, 12, 1, 0, 31, 178, 553, 1156, 1716, 1848, 1452, 825, 330, 88, 14, 1, 0, 37, 249, 939, 2432, 4576, 6435, 6864, 5577, 3432, 1573
Offset: 0

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Author

Keywords

Comments

The triangle is related to the Kauffman bracket polynomial evaluated at the shadow diagram of the two-bridge knot with Conway's notation C(2n,3).

Examples

			The triangle T(n,k) begins:
n\k| 0   1    2    3     4     5     9     7     8     9    10   11   12  13 14
-------------------------------------------------------------------------------
0  | 0   1    3    3     1
1  | 0   7   14    9     2
2  | 0  13   37   43    26     8     1
3  | 0  19   72  129   141    98    42    10     1
4  | 0  25  119  291   463   504   378   192    63    12     1
5  | 0  31  178  553  1156  1716  1848  1452   825   330    88   14    1
6  | 0  37  249  939  2432  4576  6435  6864  5577  3432  1573  520  117  16  1
...
		

Crossrefs

Programs

  • Maple
    T := proc (n, k) if k = 1 then 6*n + 1 else binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) end if end proc:
    for n from 0 to 12 do seq(T(n, k), k = 0 .. max(4, 2*(n + 1))) od;
  • Mathematica
    row[n_] := CoefficientList[(x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3), x];
    Array[row, 12, 0] // Flatten
  • Maxima
    T(n, k) := binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) - kron_delta(1, k)$
    for n:0 thru 12 do print(makelist(T(n, k), k, 0, max(4, 2*(n + 1))));

Formula

T(n,1) = A016921(n) and T(n,k) = C(2*n+3,k+1) + (C(2*n+1,k)*(2*k - 2*n) + C(4,k)*(2*k - 3))/(k + 1) for k > 1.
T(n,2) = A173247(2*n+1) = A300401(2*n,3).
T(n,3) = 2*A099721(n) + 3.
T(n,4) = A244730(n) - A002412(n) + 1.
T(n,k) = A093560(2*n,k) for n > 2 and k > 4.
G.f.: (x^2 + 4*x + 3)/(1 - y*(x + 1)^2) + (x^4 + 3*x^3 + 2*x^2 - 3*x - 3)/(1 - y).

A269792 a(n) = 5*n^4.

Original entry on oeis.org

0, 5, 80, 405, 1280, 3125, 6480, 12005, 20480, 32805, 50000, 73205, 103680, 142805, 192080, 253125, 327680, 417605, 524880, 651605, 800000, 972405, 1171280, 1399205, 1658880, 1953125, 2284880, 2657205, 3073280, 3536405, 4050000, 4617605, 5242880, 5929605
Offset: 0

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Author

Ilya Gutkovskiy, Mar 31 2016

Keywords

Comments

More generally, the ordinary generating function for the sequences of the form k*n^m, is k*Sum_{j>=1}x^j*j^m (when abs(x)<1).
More generally, the ordinary generating function for the values of quartic polynomial p*n^4 + q*n^3 + k*n^2 + m*n + r, is (r + (p + q + k + m - 4*r)*x + (11*p + 3*q - k - 3*m + 6*r)*x^2 + (11*p - 3*q - k + 3*m - 4*r)*x^3 + (p - q + k - m + r)*x^4)/(1 - x)^5.

Crossrefs

Cf. similar sequences of the form k*n^m, for k = 1...5, m = 1...10: A001477(k = 1, m = 1), A005843 (k = 2, m = 1), A008585 (k = 3, m = 1), A008586 (k = 4, m = 1), A008587 (k = 5, m = 1), A000290 (k = 1, m = 2), A001105 (k = 2, m = 2), A033428 (k = 3, m = 2), A016742 (k = 4, m = 2), A033429 (k = 5, m = 2), A000578 (k = 1, m = 3), A033431 (k = 2, m = 3), A117642 (k = 3, m = 3), A033430 (k = 4, m = 3), A244725 (k = 5, m = 3), A000583 (k = 1, m = 4), A244730 (k = 2, m = 4), A219056 (k = 3, m = 4), A141046 (k = 4, m = 4), this sequence(k = 5, m = 4), A000584 (k = 1, m = 5), A001014 (k = 1, m = 6), A106318 (k = 2, m = 6), A001015 (k = 1, m = 7), A001016 (k = 1, m = 8), A001017 (k = 1, m = 9), A008454 (k = 1, m = 10).

Programs

  • Maple
    A269792:=n->5*n^4: seq(A269792(n), n=0..50); # Wesley Ivan Hurt, Apr 28 2017
  • Mathematica
    Table[5 n^4, {n, 0, 33}]
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 5, 80, 405, 1280}, 34]
  • PARI
    x='x+O('x^99); concat(0, Vec(5*x*(1+11*x+11*x^2+x^3)/(1-x)^5)) \\ Altug Alkan, Mar 31 2016

Formula

G.f.: 5*x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5.
E.g.f.: 5*exp(x)^x*x*(1 + 7*x + 6*x^2 + x^3).
a(n) = 5*a(n-1) - 10*(9n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 5*A000583(n) = A008587(n)*A000578(n).
Sum_{n>=1} 1/a(n) = Pi^4/450 = (1/450)*A092425 = 0.216464646742...
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