A389684 Number of representations of n as binomial(i,4) + binomial(j,3) + binomial(k,2) + binomial(l,1) with i,j,k,l>=0.

24, 50, 59, 72, 87, 100, 118, 130, 135, 138, 163, 180, 184, 188, 193, 218, 236, 239, 243, 251, 268, 292, 302, 307, 309, 326, 338, 340, 354, 362, 372, 380, 385, 389, 391, 413, 442, 453, 466, 471, 481, 498, 503, 508, 509, 533, 550, 553, 558, 564, 579, 588, 589, 592

Offset

0,1

Comments

This is the combinatorial number system of degree t = 4, but without the standard ordering requirement i>j>k>l>=0. Instead, we only require i,j,k,l>=0.

Exactly one of the a(n) representations of n always satisfies i>j>k>l>=0. For example, of the 1062 representations of 100, only {i,j,k,l}={8,6,5,0} is strictly decreasing.

References

D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.3, Eq. (20), p. 360.

Links

David Dewan, Table of n, a(n) for n = 0..1000

David Dewan, Plot of A389683 and A389684.

Wikipedia, Combinatorial number system.

Formula

G.f.: Sum_{i>=0} z^binomial(i,4) * Sum_{j>=0} z^binomial(j,3) * Sum_{k>=0} z^binomial(k,2) / (1-z). - Robert Israel, Oct 13 2025

Example

a(10)=163 because the 163 tuples {i,j,k,l} = {0,0,0,10}, {0,0,1,10}, {0,0,2,9},...{5,4,0,1}, {5,4,1,1}, {5,4,2,0} each represent 10.

Maple

N:= 100: # for a(0) .. a(N)
A:= Array(0..N):
for i from 0 do
  s1:= binomial(i, 4);
  if s1 > N then break fi;
  for j from 0 do
    s2:= binomial(j, 3) + s1;
    if s2 > N then break fi;
    for k from 0 do
      s3:= binomial(k, 2) + s2;
      if s3 > N then break fi;
      for l from 0 do
        s4:= l + s3;
        if s4 > N then break fi;
        A[s4]:= A[s4]+1
od od od od:
convert(A, list); # Robert Israel, Oct 13 2025

Mathematica

maxN=100; rangeI=Range[0, Ceiling[(24*maxN)^(1/4)]+1];
rangeJ=Range[0, Ceiling[CubeRoot[6*maxN]+1]];
rangeK=Range[0, Ceiling[Sqrt[2*maxN]+1]];
rangeL=Range[0, maxN]; allQuads=Tuples[{rangeI, rangeJ, rangeK, rangeL}];
conv[x_List]:=Total[MapThread[Binomial, {x, {4, 3, 2, 1}}]]
assoc=GroupBy[allQuads, conv[#]&]; assoc=KeySort[KeySelect[assoc, #<=maxN&]];
valueLengths=Length/@Values[assoc]

Crossrefs

Cf. A389683, A194882, A194883, A194884.

Sequence in context: A042150 A087355 A376422 * A132765 A274898 A025137

Adjacent sequences: A389681 A389682 A389683 * A389685 A389686 A389687

Keyword

nonn

Author

David Dewan, Oct 10 2025

Status

approved