

A002966


Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n where 0 < x_1 <= ... <= x_n.
(Formerly M2981)


54




OFFSET

1,3


COMMENTS

All denominators in the expansion 1 = 1/x_1 + ... + 1/x_n are bounded by A000058(n1), i.e., 0 < x_1 <= ... <= x_n < A000058(n1). Furthermore, for a fixed n, x_i <= (n+1i)*(A000058(i1)1).  Max Alekseyev, Oct 11 2012
This is the leading edge of the triangle A156869. This is also the row n=1 of an array T(n,m) which gives the number of ways to write 1/n as a sum over m (not necessarily distinct) unit fractions:
1, 1, 3, 14, 147, 3462, 294314, ...
1, 2, 10, 108, 2892, 270332, ...
1, 2, 21, 339, 17253, ...
1, 3, 28, 694, 51323, ...
...


REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, D11.
D. Singmaster, The number of representations of one as a sum of unit fractions, unpublished manuscript, 1972.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA



EXAMPLE

For n=3 the 3 solutions are {2,3,6}, {2,4,4}, {3,3,3}.
For n=4 the solutions are: {2,3,7,42}, {2,3,8,24}, {2,3,9,18}, {2,3,10,15}, {2,3,12,12}, {2,4,5,20}, {2,4,6,12}, {2,4,8,8}, {2,5,5,10}, {2,6,6,6}, {3,3,4,12}, {3,3,6,6}, {3,4,4,6}, {4,4,4,4}. [Neven Juric, May 14 2008]


PROG

(PARI) a(n, rem=1, mn=1)=if(n==1, return(numerator(rem)==1)); sum(k=max(1\rem+1, mn), n\rem, a(n1, rem1/k, k)) \\ Charles R Greathouse IV, Jan 04 2015


CROSSREFS



KEYWORD

nonn,nice,hard,more


AUTHOR



EXTENSIONS

a(8) from John Dethridge (jcd(AT)ms.unimelb.edu.au) and Jacques Le Normand (jacqueslen(AT)sympatico.ca), Jan 06 2004


STATUS

approved



