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A024362
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Number of primitive Pythagorean triangles with hypotenuse n.
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19
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0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0
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OFFSET
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1,65
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COMMENTS
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Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives number of times C takes value n.
If the formula given below is used one is sure to find all a(n) values for hypotenuses n <= N if the summation indices r and s are cut off at rmax(N) = floor((sqrt(N-4)+1)/2) and smax(N) = floor(sqrt(N-1)/2). a(n) is the number of primitive Pythagorean triples with hypotenuse n modulo catheti exchange. - Wolfdieter Lang, Jan 10 2016
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 116-117, 1966.
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LINKS
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FORMULA
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a(n) = [q^n] T(q), n >= 1, where T(q) = Sum_{r>=1,s>=1} rpr(2*r-1, 2*s)*q^c(r,s), with rpr(k,l) = 1 if gcd(k,l) = 1, otherwise 0, and c(r,s) = (2*r-1)^2 + (2s)^2. - Wolfdieter Lang, Jan 10 2016
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MAPLE
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f:= proc(n) local F;
F:= numtheory:-factorset(n);
if map(t -> t mod 4, F) <> {1} then return 0 fi;
2^(nops(F)-1)
end proc:
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MATHEMATICA
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Table[a0=IntegerExponent[n, 2]; If[n==1 || a0>0, cnt=0, m=n/2^a0; p=Transpose[FactorInteger[m]][[1]]; c=Count[p, _?(Mod[#, 4]==1 &)]; If[c==Length[p], cnt=2^(c-1), 0]]; cnt, {n, 100}]
a[n_] := If[n==1||EvenQ[n]||Length[Select[FactorInteger[n], Mod[#[[1]], 4]==3 &]] >0, 0, 2^(Length[FactorInteger[n]]-1)]; Array[a, 100] (* Frank M Jackson, Jan 28 2018 *)
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PROG
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(Haskell)
a024362 n = sum [a010052 y | x <- takeWhile (< nn) $ tail a000290_list,
let y = nn - x, y <= x, gcd x y == 1]
where nn = n ^ 2
(PARI) a(n)={my(m=0, k=n, n2=n*n, k2, l2);
while(1, k=k-1; k2=k*k; l2=n2-k2; if(l2>k2, break); if(issquare(l2), if(gcd(n, k)==1, m++))); return(m); } \\ Stanislav Sykora, Mar 23 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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