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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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A321201 Irregular triangle T with the nontrivial solutions of 2*e2 + 3*e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.

Original entry on oeis.org

1, 0, 0, 1, 2, 0, 1, 1, 0, 2, 3, 0, 2, 1, 1, 2, 4, 0, 0, 3, 3, 1, 2, 2, 5, 0, 1, 3, 4, 1, 0, 4, 3, 2, 6, 0, 2, 3, 5, 1, 1, 4, 4, 2, 7, 0, 0, 5, 3, 3, 6, 1, 2, 4, 5, 2, 8, 0, 1, 5, 4, 3, 7, 1, 0, 6, 3, 4, 6, 2, 9, 0, 2, 5, 5, 3, 8, 1, 1, 6, 4, 4, 7, 2, 10, 0
Offset: 2

Views

Author

Wolfdieter Lang, Nov 05 2018

Keywords

Comments

The length of row n is 2*A(n), with A(n) = A008615(n+2) for n >= 2: 2*[1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, ...].
The trivial solution for n = 0 is [0, 0]. There is no solution for n = 1.
The row sums are given in A321202.
If a partition of n with parts 2 or 3 (with inclusive or) is written as 2^{e2} 3^{e3}, where e2 and e3 are nonnegative numbers, then in row n, all pairs [e2, e3] are given, for n >= 2, ordered with increasing values of e2.
The corresponding irregular triangle with the multinomial numbers n!/((n - (e2 + e3))!*e2!*e3!) is given in A321203. It gives the coefficients of x^n = x^{2*{e2} + 3*{e3}} of (1 + x^2 + x^3)^n, for n >= 2.

Examples

			The triangle T(n, k) begins (pairs are separated by commas):
  n\k  0  1   2  3   4  5   6  7 ...
  2:   1  0
  3:   0  1
  4:   2  0
  5:   1  1
  6:   0  2,  3  0
  7:   2  1
  8:   1  2,  4  0
  9:   0  3,  3  1
  10:  2  2,  5  0
  11:  1  3,  4  1
  12:  0  4,  3  2,  6  0
  13:  2  3,  5  1,
  14:  1  4,  4  2,  7  0
  15:  0  5,  3  3,  6  1
  16:  2  4,  5  2,  8  0
  17:  1  5,  4  3,  7  1
  18:  0  6,  3  4,  6  2,  9  0
  19:  2  5,  5  3,  8  1
  20:  1  6,  4  4,  7  2, 10  0
  ...
n=8: the two solutions of 2*e2 + 3*e3 = 8 are [e2, e3] = [1, 2] and = [4, 0], and 1 < 4, therefore row 8 is 1  2  4  0, with a comma after the first pair.

Crossrefs

Cf. A008615, A321202, A321203.

Programs

  • Mathematica
    row[n_] := Reap[Do[If[2 e2 + 3 e3 == n, Sow[{e2, e3}]], {e2, 0, n/2}, {e3, 0, n/3}]][[2, 1]];
Table[row[n], {n, 2, 20}] // Flatten (* Jean-François Alcover, Nov 23 2018 *)

Formula

T(n, k) gives all pairs [e2, e3] solving 2*e2 + 3*e3 = n, ordered with increasing value of e2, for n >= 2. The trivial solution [0, 0] for n = 0 is not recorded. There is no solution for n = 1.

Extensions

Missing row 2 inserted by Jean-François Alcover, Nov 23 2018

A307018 Total number of parts of size 3 in the partitions of n into parts of size 2 and 3.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 1, 2, 4, 2, 4, 6, 4, 6, 9, 6, 9, 12, 9, 12, 16, 12, 16, 20, 16, 20, 25, 20, 25, 30, 25, 30, 36, 30, 36, 42, 36, 42, 49, 42, 49, 56, 49, 56, 64, 56, 64, 72, 64, 72, 81, 72, 81, 90, 81, 90, 100, 90, 100, 110, 100, 110, 121, 110, 121, 132
Offset: 0

Views

Author

Andrew Ivashenko, Mar 19 2019

Keywords

Links

Crossrefs

Cf. A000041, A002620, A103221, A114209, A321202.
Cf. A008133, A069905, A008611.

Programs

  • GAP
    a:=[0,0,0,1,0,1,2,1];; for n in [9..80] do a[n]:=a[n-2]+2*a[n-3] -2*a[n-5]-a[n-6]+a[n-8]; od; a; # G. C. Greubel, Apr 03 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 80); [0,0,0] cat Coefficients(R!( x^3/((1-x^2)*(1-x^3)^2) )); // G. C. Greubel, Apr 03 2019
  • Mathematica
    LinearRecurrence[{0,1,2,0,-2,-1,0,1}, {0,0,0,1,0,1,2,1}, 80] (* G. C. Greubel, Apr 03 2019 *)
Table[(6n(2+n)-5-27(-1)^n+8(4+3n)Cos[2n Pi/3]-8Sqrt[3]n Sin[2n Pi/3])/216,{n,0,66}] (* Stefano Spezia, Apr 21 2022 *)
  • PARI
    my(x='x+O('x^80)); concat([0,0,0], Vec(x^3/((1-x^2)*(1-x^3)^2))) \\ G. C. Greubel, Apr 03 2019
  • Sage
    (x^3/((1-x^2)*(1-x^3)^2)).series(x, 80).coefficients(x, sparse=False) # G. C. Greubel, Apr 03 2019

Formula

a(n+2) = A321202(n) - A114209(n+1).
a(3n+1) = A002620(n+2).
a(3n+2) = A002620(n+1).
a(3n+3) = A002620(n+2).
G.f.: x^3/((1+x)*(1+x+x^2)^2*(1-x)^3). - Alois P. Heinz, Mar 19 2019
a(n) = a(n-2) + 2*a(n-3) - 2*a(n-5) - a(n-6) + a(n-8). - G. C. Greubel, Apr 03 2019
a(n) = (6*n*(2 + n) + 8*(4 + 3*n)*cos(2*n*Pi/3) - 8*sqrt(3)*n*sin(2*n*Pi/3) - 5 - 27*(-1)^n)/216. - Stefano Spezia, Apr 21 2022
From Ridouane Oudra, Nov 24 2024: (Start)
a(n) = (7*n/2 - 7*n^2/2 - 9*floor(n/2) + (6*n+4)*floor(2*n/3) + 4*floor(n/3))/18.
a(n) = A008133(n) - A069905(n-1).
a(n) = A002620(A008611(n)). (End)

Extensions

More terms from Alois P. Heinz, Mar 19 2019
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