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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). 6

%I

%S 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 18,4,1,1,159330691,297976,3644,168,18,4,1,1

%N 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 Conjecture: T(2n + m, n + m) = T(2n, n) ( = A156870(n) ) if and only if m >= 0.

%C 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 The next term will be a(37) = A002966(9). - _M. F. Hasler_, Feb 20 2009

%e Triangle begins:

%e n=1: 1

%e n=2: 1, 1

%e n=3: 3, 1, 1

%e n=4: 14, 4, 1, 1

%e n=5: 147, 17, 4, 1, 1

%e n=6: 3462, 164, 18, 4, 1, 1

%e n=7: 294314, 3627, 167, 18, 4, 1, 1

%e 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 (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 Cf. A002966 (column k=1), A156871 (row sums), A280519, A280520.

%Y T(2n, n) = A156870(n).

%K more,nonn,tabl

%O 1,4

%A _Jens Voß_, Feb 17 2009

%E a(21)-a(36) from _M. F. Hasler_, Feb 20 2009

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Last modified April 18 16:38 EDT 2019. Contains 322209 sequences. (Running on oeis4.)