cp's OEIS Frontend

A328911 Irregular triangle read by rows: T(n,k) = number of solutions to Erdös's Last Equation x_1*...*x_n = n*(x_1+...+x_n), 0 < x_1 <= ... <= x_n, having k+1 components x_i > 1, 1 <= k <= 2*log_2(n).

%I A328911 #32 Jun 05 2020 04:27:00
%S A328911 2,0,3,3,0,4,3,1,0,4,4,0,0,6,7,4,0,0,6,5,3,0,0,5,7,4,2,1,0,8,13,5,1,0,
%T A328911 0,9,12,3,1,0,0,6,6,3,0,0,0,8,13,9,3,0,0,0,8,7,1,0,0,0,0,6,15,6,2,1,0,
%U A328911 0,12,16,12,3,0,0,0,12,15,11,4,2,1,0,0,6,8,2,2,0,0,0,0
%N A328911 Irregular triangle read by rows: T(n,k) = number of solutions to Erdös's Last Equation x_1*...*x_n = n*(x_1+...+x_n), 0 < x_1 <= ... <= x_n, having k+1 components x_i > 1, 1 <= k <= 2*log_2(n).
%C A328911 For n = 1 the equation is trivially solved by any integer, therefore we only consider n >= 2.
%C A328911 If any x_k = 0, then all x_i must be zero, so (0, ..., 0) would be the only additional solution in nonnegative integers. This solution is not considered here.
%C A328911 A vector (1, ..., 1, x_n) can never be a solution for n > 1. The number of components different from 1 must be k+1 >= 2 <=> k >= 1.
%C A328911 It can be shown that no solution can have 2^k > n^2, cf. the Shiu paper. Therefore row lengths are floor(2 log_2(n)) = (2, 3, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, ...) = A329202(n), n >= 2.
%C A328911 Row sums yield the total number of nontrivial solutions A328910(n), see there for more information.
%C A328911 T(n,k) is equal to |C_k(n)| in the Shiu paper, but some values given in the table on top of p. 803 are erroneous (pers. comm. from the author).
%H A328911 David A. Corneth, <a href="/A328911/b328911.txt">Table of n, a(n) for n = 2..14645</a> (first 899 rows flattened)
%H A328911 Peter Shiu, <a href="https://doi.org/10.1080/00029890.2019.1639466">On Erdös's Last Equation</a>, Amer. Math. Monthly, 126 (2019), 802-808.
%e A328911 The table starts:
%e A328911    n : T(n,k), 1 <= k <= 2*log_2(n)
%e A328911    2 :   2   0
%e A328911    3 :   3   3   0
%e A328911    4 :   4   3   1   0
%e A328911    5 :   4   4   0   0
%e A328911    6 :   6   7   4   0   0
%e A328911    7 :   6   5   3   0   0
%e A328911    8 :   5   7   4   2   1   0
%e A328911    9 :   8  13   5   1   0   0
%e A328911   10 :   9  12   3   1   0   0
%e A328911   11 :   6   6   3   0   0   0
%e A328911   12 :   8  13   9   3   0   0   0
%e A328911   13 :   8   7   1   0   0   0   0
%e A328911   14 :   6  15   6   2   1   0   0
%e A328911   15 :  12  16  12   3   0   0   0
%e A328911 For n = 2 variables, we have the equation x1*x2 = 2*(x1 + x2) with positive integer solutions (3,6) and (4,4): Both have k+1 = 2 components > 1, i.e., k = 1.
%e A328911 For n = 3, we have T(3,1) = 3 solutions with k+1 = 2 components > 1, {(1, 4, 15), (1, 5, 9), (1, 6, 7)}, and T(3,2) = 3 with k+1 = 3 components > 1, {(2, 2, 12), (2, 3, 5), (3,3,3)}.
%e A328911 For n = 4 we have the 8 solutions (1, 1, 5, 28), (1, 1, 6, 16), (1, 1, 7, 12), (1, 1, 8, 10), (1, 2, 3, 12), (1, 2, 4, 7), (1, 3, 4, 4) and (2, 2, 2, 6). Four of them have k+1 = 2 components > 1, i.e., k = 1, whence T(4,1) = 4. Three have k+1 = 3 <=> k = 2, so T(4,2) = 3. One has k+1 = 4, so T(4,3) = 1.
%e A328911 For n = 5, the solutions are, omitting initial components x_i = 1: {(6, 45), (7, 25), (9, 15), (10, 13), (2, 3, 35), (2, 5, 9), (3, 3, 10), (3, 5, 5)}. Therefore T(5,1..4) = (4, 4, 0, 0).
%e A328911 For n = 6, the solutions are (omitting x_i = 1): {(7, 66), (8, 36), (9, 26), (10, 21), (11, 18), (12, 16), (2, 4, 27), (2, 5, 15), (2, 6, 11), (2, 7, 9), (3, 3, 18), (3, 4, 10), (3, 6, 6), (2, 2, 2, 24), (2, 2, 3, 9), (2, 2, 4, 6), (2, 3, 3, 5)}. Therefore T(6,1..5) = (6, 7, 4, 0, 0).
%e A328911 For n = 9, the 27 solutions are (omitting '1's): {(10, 153), (11, 81), (12, 57), (13, 45), (15, 33), (17, 27), (18, 25), (21, 21), (2, 5, 117), (2, 6, 42), (2, 7, 27), (2, 9, 17), (2, 12, 12), (3, 4, 39), (3, 5, 21), (3, 6, 15), (3, 7, 12), (3, 9, 9), (4, 4, 18), (5, 5, 9), (6, 6, 6), (2, 2, 3, 36), (2, 2, 6, 9), (2, 3, 3, 13), (3, 3, 3, 7), (3, 3, 4, 5), (2, 3, 3, 3, 3)}. Therefore T(9,1..6) = (8, 13, 5, 1, 0, 0).
%o A328911 (PARI) A328911(n,k,show=1)={if( k<min(exponent(n^2)+1, n), my(s=0,t=n*(n-k-1),d); forvec(x=vector(k,i,[2,n\(sqrt(2)-1)]), (d=vecprod(x)-n)>0 && (n*vecsum(x)+t)%d==0 && (n*vecsum(x)+t)\d >= x[k] && s++&& show&& printf("%d,",concat(x,(n*vecsum(x)+t)\d)),1);s)}
%Y A328911 Cf. A328910, A329202.
%K A328911 nonn,tabf
%O A328911 2,1
%A A328911 _M. F. Hasler_, Nov 07 2019