0,1

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).

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.

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.

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,

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

(SageMath)

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]

Cf. A002378, A002061.

Sequence in context: A372127 A372031 A225642 * A328443 A122866 A097275

Adjacent sequences: A273314 A273315 A273316 * A273318 A273319 A273320

nonn,tabf

Tom Edgar, May 19 2016

approved