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A193138
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Number of square satins of order n.
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6
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0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0
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OFFSET
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3,63
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COMMENTS
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This sequence is conjectured to coincide with the multiplicities of the representation of n >= 3 as primitive sums of two squares. Neither the order of the squares nor the signs of the numbers to be squared are taken into account. a(n) = 0 if no such representation exists. Checked for n = 3,4, ..., 1000 (using the program below). The two squares are in each case nonzero and distinct. If one includes also 0 as a square in the primitive sum of two squares one could take a(0) = 0, a(1) = 1, a(2) = 1. If only nonzero squares are considered, then one could take a(0) = 0, a(1) = 0, a(2) = 1.
For the numbers n with a(n) > 0 (in this conjectured interpretation of a(n)) see A008784. - Wolfdieter Lang, Apr 17 2013
The stated conjecture is true because it follows immediately from Theorem 3.22, p. 165, of the Niven-Zuckerman-Montgomery reference. There r(n) gives the number of primitive solutions of n = x^2 + y^2 with ordered and signed pairs of integers x,y. Because x and y are distinct if n >= 3 one needs here a(n) = r(n)/2^3. This then coincides with the formula for u(n) given in the Grünbaum-Shephard Theorem 5. - Wolfdieter Lang, Apr 18 2013
The equality noted by R. J. Mathar above indeed holds for all n > 2. Regarding n = 2 case: if we consider periodic twills as satins (which seems more consistent), we'll get a(2) = 1 from the plain weave; otherwise (following Grünbaum and Shephard), a(1) = a(2) = 0 (so we get A157228). In the former case, the all-black pattern can formally be counted as a(1) = 1, but physically it is dubious (this pattern corresponds to unweaved warp and weft). - Andrey Zabolotskiy, May 09 2018
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REFERENCES
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Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
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LINKS
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FORMULA
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Take the prime number factorization (symbolically) as n = 2^a*product(p^b)*product(q^c) with primes p == 1(mod 4) and primes q == 3(mod 4) and n>=3. If a = 0 or 1 and all c's vanish then a(n) = 2^(t-1) with t the number of distinct primes congruent 1(mod 4). Otherwise a(n) = 0. (See the Niven-Zuckerman-Montgomery reference, Theorem 3.22, p. 165, and the Grünbaum-Shephard Theorem 5 formula for u(n)). - Wolfdieter Lang, Apr 18 2013
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EXAMPLE
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Primitive sums of two squares stated as a comment above: a(3) = 0 because 3 is not a sum of two squares. a(5) = 1 because 5 = 1^2 + 2^2, denoted by the unique (primitive) doublet [1, 2]. a(65) = 2 from the two (primitive) doublets [1, 8] and [4, 7]. a(85) = 2 with the (primitive) doublets [2, 9] and [6, 7]. a(8) = 0 because the doublet [2, 2] is imprimitive. - Wolfdieter Lang, Apr 18 2013
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MAPLE
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U:=proc(n) local nop, p3, i, t1, t2, al, even;
t1:=ifactors(n)[2];
t2:=nops(t1);
if (n mod 2) = 0 then even:=1; al:=t1[1][2]; else even:=0; al:=0; fi;
nop:=t2-even;
p3:=0;
for i from 1 to t2 do if t1[i][1] mod 4 = 3 then p3:=1; fi; od:
if (al >= 2) or (p3=1) then RETURN(0) else RETURN(2^(nop-1)); fi;
end;
[seq(U(n), n=3..120)];
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MATHEMATICA
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a[n_] := Select[ PowersRepresentations[n, 2, 2], GCD @@ # == 1 &] // Length; a[2] = 0; Table[a[n], {n, 3, 120}] (* Jean-François Alcover, Apr 18 2013 *)
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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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