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A343897 Number of ways to write n as 2*x + y + z with x,y,z positive integers such that 16*x^2*y^2 + 19*y^2*z^2 + 29*z^2*x^2 is a square. 3
0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 2, 4, 2, 1, 3, 1, 3, 1, 5, 2, 2, 6, 1, 1, 5, 5, 6, 2, 2, 4, 2, 4, 5, 6, 3, 2, 3, 2, 5, 2, 7, 10, 4, 1, 3, 3, 10, 9, 2, 5, 5, 10, 6, 7, 6, 7, 8, 8, 4, 7, 4, 5, 8, 2, 4, 4, 13, 9, 5, 6, 10, 11, 6, 11, 6, 6, 5, 4, 4, 10, 9, 8, 8, 8, 8, 9, 16, 5, 5, 6, 4, 7, 3, 12, 7, 11, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,10
COMMENTS
Conjecture: a(n) > 0 for all n > 3.
We have verified a(n) > 0 for all n = 4..10000. The conjecture holds if a(p) > 0 for each odd prime p.
It seems that a(n) = 1 only for n = 4..9, 13, 17, 19, 21, 26, 27, 47.
LINKS
EXAMPLE
a(4) = 1 with 4 = 2*1 + 1 + 1 and 16*1^2*1^2 + 19*1^2*1^2 + 29*1^2*1^2 = 8^2.
a(6) = 1 with 6 = 2*1 + 2 + 2 and 16*1^2*2^2 + 19*2^2*2^2 + 29*2^2*1^2 = 22^2.
a(9) = 1 with 9 = 2*2 + 4 + 1 and 16*2^2*4^2 + 19*4^2*1^2 + 29*1^2*2^2 = 38^2.
a(13) = 1 with 13 = 2*5 + 2 + 1 and 16*5^2*2^2 + 19*2^2*1^2 + 29*1^2*5^2 = 49^2.
a(19) = 1 with 19 = 2*2 + 14 + 1 and 16*2^2*14^2 + 19*14^2*1^2 + 29*1^2*2^2 = 128^2.
a(47) = 1 with 47 = 2*13 + 13 + 8 and 16*13^2*13^2 + 19*13^2*8^2 + 29*8^2*13^2 = 988^2.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[16(x*y)^2+(n-2x-y)^2*(19*y^2+29x^2)], r=r+1], {x, 1, (n-2)/2}, {y, 1, n-1-2x}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
CROSSREFS
Sequence in context: A163530 A114409 A340680 * A193585 A245436 A285581
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 03 2021
STATUS
approved

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Last modified March 29 03:51 EDT 2024. Contains 371264 sequences. (Running on oeis4.)