cp's OEIS Frontend

A156869 Triangle read by rows: T(n,k) = number of nondecreasing sequences of n positive integers with reciprocals adding up to k (1 <= k <= n).

%I A156869 #17 Mar 13 2017 04:22:46
%S A156869 1,1,1,3,1,1,14,4,1,1,147,17,4,1,1,3462,164,18,4,1,1,294314,3627,167,
%T A156869 18,4,1,1,159330691,297976,3644,168,18,4,1,1
%N A156869 Triangle read by rows: T(n,k) = number of nondecreasing sequences of n positive integers with reciprocals adding up to k (1 <= k <= n).
%C A156869 Conjecture: T(2n + m, n + m) = T(2n, n) ( = A156870(n) ) if and only if m >= 0.
%C A156869 Yes, the diagonals are constant for n <= 2k. Any such sequence must have at least one 1; remove that 1, and you get a sequence for n-1,k-1. - _Franklin T. Adams-Watters_, Feb 20 2009
%C A156869 The next term will be a(37) = A002966(9). - _M. F. Hasler_, Feb 20 2009
%e A156869 Triangle begins:
%e A156869 n=1:      1
%e A156869 n=2:      1,    1
%e A156869 n=3:      3,    1,   1
%e A156869 n=4:     14,    4,   1,  1
%e A156869 n=5:    147,   17,   4,  1, 1
%e A156869 n=6:   3462,  164,  18,  4, 1, 1
%e A156869 n=7: 294314, 3627, 167, 18, 4, 1, 1
%e A156869 For n = 4 and k = 2, the T(4, 2) = 4 sequences are (1, 2, 3, 6), (1, 2, 4, 4), (1, 3, 3, 3) and (2, 2, 2, 2) because 1/1 + 1/2 + 1/3 + 1/6 = 1/1 + 1/2 + 1/4 + 1/4 = 1/1 + 1/3 + 1/3 + 1/3 = 1/2 + 1/2 + 1/2 + 1/2 = 2.
%o A156869 (PARI) { A156869(n,k,m=1) = n==1 & return(numerator(k)==1 & denominator(k)>=m); sum( i=max(m,1\k+1),n\k, A156869(n-1, k-1/i, i)); } \\ _M. F. Hasler_, Feb 20 2009
%Y A156869 Cf. A002966 (column k=1), A156871 (row sums), A280519, A280520.
%Y A156869 T(2n, n) = A156870(n).
%K A156869 more,nonn,tabl
%O A156869 1,4
%A A156869 _Jens Voß_, Feb 17 2009
%E A156869 a(21)-a(36) from _M. F. Hasler_, Feb 20 2009