A194883 Write n = C(i,4)+C(j,3)+C(k,2)+C(l,1) with i>j>k>l>=0; sequence gives j values.

2, 2, 3, 3, 3, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 0

Author

N. J. A. Sloane, Sep 04 2011

Keywords

• nonn

Comments

Each n >= 0 has a unique representation as n = C(i,4)+C(j,3)+C(k,2)+C(l.1) with i>j>k>l>=0. This is the combinatorial number system of degree t = 4, where we get [A194882, A194883, A194884, A127324]. For degree t = 3 see A194847.

References

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

Crossrefs

Equals A127322 + 2.

Cf. A194882-A194884, A127324, A194885; A194847, A194848, A056558, A194849.