OFFSET
0,1
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).
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.
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.
EXAMPLE
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,
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
PROG
(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]
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Tom Edgar, May 19 2016
STATUS
approved