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.

Previous Showing 11-14 of 14 results.

A268347 Number of partitions of (4, n) into a sum of distinct pairs.

Original entry on oeis.org

2, 7, 14, 27, 46, 74, 116, 174, 254, 363, 510, 703, 957, 1285, 1706, 2244, 2924, 3777, 4844, 6168, 7802, 9813, 12272, 15267, 18902, 23295, 28584, 34935, 42532, 51592, 62369, 75150, 90265, 108102, 129094, 153743, 182627, 216395, 255792, 301672, 354994, 416851
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 02 2016

Keywords

Crossrefs

Column 4 of A054242.

Programs

  • Mathematica
    max=50; col=4; s1=Series[Product[(1+x^(n-k)*y^k), {n, 1, max+2}, {k, 0, n}], {y, 0, col}]//Normal; s2=Series[s1, {x, 0, max+1}]; a[n_]:=SeriesCoefficient[s2, {x, 0, n}, {y, 0, col}]; Table[a[n], {n, 0, max}] (* after Jean-François Alcover *)
    nmax = 50; CoefficientList[Series[((2 + 3*x - x^3 - 4*x^4 - 2*x^5 + x^6 + x^7 + 2*x^8 - x^9) / ((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)))*Product[1 + x^k, {k, 1, nmax}], {x, 0, nmax}], x]

Formula

a(n) ~ 3^(3/4) * n^(5/4) * exp(Pi*sqrt(n/3)) / (2*Pi^4).

A268348 Number of partitions of (5, n) into a sum of distinct pairs.

Original entry on oeis.org

3, 10, 21, 42, 74, 123, 197, 303, 452, 659, 943, 1323, 1830, 2496, 3363, 4485, 5922, 7748, 10058, 12958, 16578, 21077, 26637, 33476, 41855, 52077, 64496, 79536, 97683, 119505, 145671, 176948, 214225, 258542, 311085, 373227, 446553, 532873, 634265, 753118
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 02 2016

Keywords

Crossrefs

Column 5 of A054242.

Programs

  • Mathematica
    max=50; col=5; s1=Series[Product[(1+x^(n-k)*y^k), {n, 1, max+2}, {k, 0, n}], {y, 0, col}]//Normal; s2=Series[s1, {x, 0, max+1}]; a[n_]:=SeriesCoefficient[s2, {x, 0, n}, {y, 0, col}]; Table[a[n], {n, 0, max}] (* after Jean-François Alcover *)
    nmax = 50; CoefficientList[Series[((3 + 4*x + x^2 - 4*x^4 - 5*x^5 - 4*x^6 + 2*x^8 + 3*x^9 + 3*x^10 - x^12 - 2*x^13 + x^14) / ((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)*(1 - x^5)))*Product[1 + x^k, {k, 1, nmax}], {x, 0, nmax}], x]

Formula

a(n) ~ 3^(5/4) * n^(7/4) * exp(Pi*sqrt(n/3)) / (5*Pi^5).

A378126 Array read by antidiagonals: T(n, m) is the maximal size of partitions of (n, m) into sums of distinct pairs of nonnegative integers, excluding (0, 0).

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 4, 4, 4, 3, 3, 3, 4, 4, 4, 4, 4, 4, 3, 3, 4, 4, 4, 5, 4, 4, 4, 3, 3, 4, 5, 5, 5, 5, 5, 5, 4, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 5, 5, 5, 4, 4, 5, 5, 6, 6, 6, 6
Offset: 0

Views

Author

Jimin Park, Nov 17 2024

Keywords

Comments

A378379(n) is the least number x such that T(x, x) >= n, and as the growth rate of A378379(n) is Theta(n^(3/2)), the growth rate of T(n, m) is O((n+m)^(2/3)).

Examples

			Table begins:
  0 1 1 2 2 2 3 3 3 3 ...
  1 2 2 3 3 3 4 4 4 4 ...
  1 2 3 3 4 4 4 5 5 5 ...
  2 3 3 4 4 4 5 5 5 6 ...
  2 3 4 4 5 5 5 6 6 6 ...
  2 3 4 4 5 5 6 6 6 7 ...
  3 4 4 5 5 6 6 6 7 7 ...
  3 4 5 5 6 6 6 7 7 7 ...
  3 4 5 5 6 6 7 7 7 8 ...
  3 4 5 6 6 7 7 7 8 8 ...
  ...
T(9, 5) = 7, as (9, 5) can be expressed as the sum (0, 1) + (0, 2) + (1, 0) + (1, 1) + (2, 0) + (2, 1) + (3, 0), which is the longest for (9, 5).
		

Crossrefs

Maximal size among partitions considered by A054242 and A201377.
The first row is T(n, 0) = A003056(n).

Programs

  • Python
    import functools
    @functools.cache
    def A378126(n: int, m: int, t: tuple[int, int] = (0, 0)) -> int:
      if (n, m) <= t: return 0
      v = 1
      for nn in range(t[0], n//2+1):
        for nm in range(m+1):
          if (nn, nm) <= t: continue
          rn, rm = n-nn, m-nm
          if (rn, rm) <= (nn, nm): continue
          nv = 1 + A378126(rn, rm, (nn, nm))
          if nv > v: v = nv
      return v

Formula

T(n, m) = T(m, n).
T(n, m) >= T(n, 0) + T(0, m).
T(n, m) = A086435(p^n * q^m) for any distinct primes p and q.
T(n, 0) = A003056(n).
T(n, 1) = T(n, 0) + 1.
T(n, 2) = T(n-1, 0) + 2, for n >= 1.
T(n, m) = O((n+m)^(2/3)).

A378379 Minimal x such that there is a partition of (x, x) into sums of distinct pairs of nonnegative integers with size at least n, excluding (0, 0).

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 7, 9, 10, 12, 14, 16, 18, 20, 23, 25, 28, 30, 33, 35, 38, 41, 44, 47, 50, 53, 56, 60, 63, 67, 70, 74, 77, 81, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 125, 129, 134, 138, 143, 147, 152, 156, 161, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220
Offset: 1

Views

Author

Jimin Park, Nov 24 2024

Keywords

Comments

For (n, n), there is at least one maximal partition P that's symmetric: (x, y) in P <=> (y, x) in P. This can be proven by manipulating integer sequences c(i) (i >= 1) such that 0 <= c(i) <= i+1 for all i and Sum_{i > 0} i*c(i) = 2n, which correspond to partitions P of (n, n) with size |P| = Sum_{i > 0} c(i), where c(i) is equal to number of (x, y) in P such that x+y = i.

Examples

			For n = 8, a(n) = 9, as (9, 9) can be expressed as the sum (0, 1) + (0, 2) + (0, 3) + (1, 0) + (2, 0) + (3, 0) + (1, 2) + (2, 1), but the longest sum for (8, 8) has 7 pairs.
		

Crossrefs

Maximal size among partitions considered by A054242 and A201377.
Minimal x such that A378126(x, x) >= n.
Cf. A086435.

Programs

  • Python
    import math
    def A378379(n: int) -> int:
      l = (math.isqrt(1+8*n)-1)//2 # l = A003056(n), min. possible largest pair norm
      r = n - (l-1)*(l+2)//2 # r = n - A000096(l-1), number of pairs with norm l
      return ((l-1)*l*(l+1)//3 + l*r + 1)//2 # ceil((A007290(l+1) + l*r) / 2)

Formula

a(n*(n+3)/2) = n*(n+1)*(n+2)/6.
Previous Showing 11-14 of 14 results.