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A330893
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Numbers whose set of divisors contains a Pythagorean quadruple.
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5
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42, 72, 84, 126, 144, 156, 168, 198, 210, 216, 252, 288, 294, 312, 330, 336, 342, 360, 378, 396, 420, 432, 462, 468, 504, 546, 570, 576, 588, 594, 624, 630, 648, 660, 672, 684, 714, 720, 756, 780, 792, 798, 840, 864, 882, 900, 924, 930, 936, 966, 990, 1008, 1026
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refs;
listen;
history;
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internal format)
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OFFSET
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1,1
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COMMENTS
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A Pythagorean quadruple (x, y, z, m) is a set of positive integers that satisfy x^2 + y^2 + z^2 = m^2.
The corresponding number of quadruples of the sequence is 1, 1, 2, 2, 2, 1, 3, 1, 3, 2, 4, 3, 2, 2, 1, 4, 1, 2, 3, 2, 7, 4, ... (see the sequence A330894).
It is interesting to note that each set of divisors of a(n) contains m primitive Pythagorean quadruples for some n, m = 1, 2,...
Examples:
- The set of divisors of a(1)= 42 contains only one primitive Pythagorean quadruple: (2, 3, 6, 7).
- The set of divisors of a(9) = 210 contains two primitive Pythagorean quadruples: (2, 3, 6, 7) and (2, 5, 14, 15).
- The set of divisors of a(21) = 420 contains three primitive Pythagorean quadruples: (2, 3, 6, 7), (2, 5, 14, 15) and (4, 5, 20, 21).
If k is in the sequence then so is m*k for m > 1.
Assumes the elements (x,y,z,m) in a quadruple are distinct divisors, as otherwise 6 would be in the sequence with 1^2+2^2+2^2=3^2. - Chai Wah Wu, Nov 16 2020
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LINKS
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FORMULA
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a(n) == 0 (mod 6).
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EXAMPLE
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168 is in the sequence because the set of divisors {1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168} contains the Pythagorean quadruples {2, 3, 6, 7}, {4, 6, 12, 14} and {8, 12, 24, 28}. The first quadruple is primitive.
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MAPLE
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with(numtheory):
for n from 3 to 1200 do :
d:=divisors(n):n0:=nops(d):it:=0:
for i from 1 to n0-3 do:
for j from i+1 to n0-2 do :
for k from j+1 to n0-1 do:
for m from k+1 to n0 do:
if d[i]^2 + d[j]^2 + d[k]^2 = d[m]^2
then
it:=it+1:
else
fi:
od:
od:
od:
od:
if it>0 then
printf(`%d, `, n):
else fi:
od:
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MATHEMATICA
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nq[n_] := If[ Mod[n, 6]>0, 0, Block[{t, u, v, c = 0, d = Divisors[n], m}, m = Length@ d; Do[ t = d[[i]]^2 + d[[j]]^2; Do[u = t + d[[h]]^2; If[u > n^2, Break[]]; If[ Mod[n^2, u] == 0 && IntegerQ[v = Sqrt@ u] && Mod[n, v] == 0, c++], {h, j+1, m - 1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[ Range@ 1026, nq[#] > 0 &] (* Giovanni Resta, May 04 2020 *)
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PROG
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(PARI) isok(n) = {my(d=divisors(n), x); for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (issquare(d[i]^2 + d[j]^2 + d[k]^2, &x) && !(n % x), return(1)); ); ); ); } \\ Michel Marcus, Nov 16 2020
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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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