A273317 Irregular table read by rows: T(0,0) = 2 and T(n,2k) = T(n-1,k)+1, T(n,2k+1) = T(n-1,k)*(T(n-1,k)+1) for 0 <= k < 2^(n-1).

2, 3, 6, 4, 12, 7, 42, 5, 20, 13, 156, 8, 56, 43, 1806, 6, 30, 21, 420, 14, 182, 157, 24492, 9, 72, 57, 3192, 44, 1892, 1807, 3263442, 7, 42, 31, 930, 22, 462, 421, 176820, 15, 210, 183, 33306, 158, 24806, 24493, 599882556, 10, 90, 73, 5256, 58, 3306, 3193, 10192056

Offset: 0

Author

Tom Edgar, May 19 2016

Keywords

• nonn
• tabf

Comments

The first entry in row n is n+2.

The second entry in row n (n>0) is the A002378(n+2).

No number appears twice in the same row, so row n has 2^n distinct terms.

Row n and row n+1 have no elements in common.

There are infinitely many mutually disjoint rows; this fact can be used to show that the harmonic series diverges since the sum of reciprocals of entries in every row equals 1/2. This also allows a proof that every positive rational number is the sum of a finite number of distinct Egyptian fractions.

Let S(0) = {2} and for n>=1 define S(n) = {a | a = c+1 or a = c*(c+1) for some c in S(n-1)}; then row n contains the elements of S(n).

Examples

			The table begins:
2,
3, 6,
4, 12, 7, 42,
5, 20, 13, 156, 8, 56, 43, 1806,
6, 30, 21, 420, 14, 182, 157, 24492, 9, 72, 57, 3192, 44, 1892, 1807, 3263442,
		

Links

• Steven J. Kifowit, More Proofs of Divergence of the Harmonic Series.
• J. C. Owings, Jr., Another Proof of the Egyptian Fraction Theorem, Amer. Math. Monthly, 75(7) (1968), 777-778.

Crossrefs

Cf. A002378, A002061.

Programs

Maple

A273317 := proc(n,j)
    if n = 0 then
        2 ;
    elif type(j,'even') then
        1+procname(n-1,j/2) ;
    else
        procname(n-1,floor(j/2)) ;
        %*(%+1) ;
    end if;
end proc: # R. J. Mathar, May 20 2016

Sage

def T(n,j):
    if n==0:
        return 2
    if j%2==0:
        return T(n-1,floor(j/2))+1
    else:
        t=T(n-1,floor(j/2))
        return t*(t+1)
S=[[T(n,k) for k in [0..2^n-1]] for n in [0..10]]
[x for sublist in S for x in sublist]

Formula

T(0,0) = 2, and T(n,2k) = T(n-1,k)+1, T(n,2k+1) = T(n-1,k)*(T(n-1,k)+1) for 0 <= k < 2^(n-1).

Sum_{a in row(n)} 1/a = 1/2 for all n.