|
%I #16 Feb 15 2024 08:19:42
%S 1,0,0,1,1,-1,1,1,2,-1,1,-2,2,1,3,-1,1,2,3,-2,1,-3,2,-3,3,1,4,-1,1,3,
%T 4,-3,1,-4,3,2,3,-4,5,-2,4,1,5,-1,5,2,7,-2,4,3,7,-3,1,5,6,1,6,-5,7,-1,
%U 3,4,7,-4,1,-6,5,-6
%N Irregular triangle read by rows: row n gives the pairs of proper solutions (X, Y), with gcd(X, Y) = 1 and X >= 0, of the Diophantine equation 2*X^2 + 2*X*Y + 3*Y^2 = A344232(n), for n >= 1.
%C The length of row n is r(n) = 2*A343240(b(n)), if A344232(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, ...).
%C See A344231 for references and links on parallel forms and half-reduced right neighbor forms (R-transformations), and also for the remark on the equivalent reduced form [2, -2, 3].
%C The number of proper solutions (X, Y), with X > 0, is 1 for n = 1 and 4. X = 0 only for n = 2, but the solution (0, -1) = (-0, -1) is not listed here.
%C For other n each distinct odd prime from {1, 3, 7, 9} (mod 20), i.e., from A139513, that divides A344232(n) contributes a factor of 2 to the listed number of solutions. See A343238 and A343240 for the multiplicities.
%C Only solutions with nonnegative X are listed. There is also the corresponding solution (-X, -Y). Hence the total number of signed solution is twice the number considered here.
%F T(n, m) gives for m = 2^j - 1, the nonnegative X(n, j) solution, and for m = 2*j the Y(n, j) solution of 2*T(n, 2*j-1)^2 + 2*T(n, 2*j-1)*T(n, 2*j) + 3*T(n, 2*j)^2 = A344232(n), for j = 1, 2, ..., r(n), for n >= 1. For n = 2 the solution (0, -1) is not listed here.
%e The irregular triangle T(n, m) begins (A(n) = A344232(n)):
%e n A(n) \ m 1 2 3 4 5 6 7 8 ...
%e 1, 2: 1 0
%e 2, 3: 0 1 1 -1
%e 3, 7: 1 1 2 -1
%e 4, 10: 1 -2
%e 5, 15: 2 1 3 -1
%e 6, 18: 1 2 3 -2
%e 7, 23: 1 -3 2 -3
%e 8, 27: 3 1 4 -1
%e 9, 35: 1 3 4 -3
%e 10, 42: 1 -4 3 2 3 -4 5 -2
%e 11, 43: 4 1 5 -1
%e 12, 47: 2 3 5 -3
%e 13, 58: 1 4 5 -4
%e 14, 63: 2 -5 3 -5 5 1 6 -1
%e 15, 67: 1 -5 4 -5
%e 16, 82: 5 2 7 -2
%e 17, 83: 4 3 7 -3
%e 18, 87: 1 5 6 1 6 -5 7 -1
%e 19, 90: 3 4 7 -4
%e 20, 98: 1 -6 5 -6
%e ...
%e n = 2: The prime 3 is a member of A139513, hence 2^1 = 2 solutions are listed. There are also the corresponding (-X, -Y) solutions.
%e n = 4: 10 = A344232(4) = A343238(8) = 2*5, A343240(8) = 1, hence there is 1 pair of proper solution with X >= 0. This is because neither 2 nor 5 are primes from A139513. There is also the solution (-1, 2).
%e n = 6: Prime 3 is a member of A139513, not prime 2. This there are 2 solutions listed. The solution (3, 0) does not appear; it is not proper.
%e n = 10: 42 = A344232(10) = A343238(19) = 2*3*7, A343240(19) = 2^2 = 4, hence there are 4 pairs of proper solution with X >= 0. 3 and 7 are primes from A139513.
%Y Cf. A343238, A343240, A344232.
%K sign,tabf,easy
%O 1,9
%A _Wolfdieter Lang_, May 17 2021
|
