A193138 Number of square satins of order n.

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

Offset: 3

Author

N. J. A. Sloane, Jul 16 2011

Keywords

• nonn

Comments

a(n) = A157228(n) for all entries known. - R. J. Mathar, Aug 10 2011

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

Examples

			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
		

References

• 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.

Links

• B. Grünbaum and G. C. Shephard, Satins and twills: an introduction to the geometry of fabrics, Math. Mag., 53 (1980), 139-161. See Theorem 5, p. 152.

Crossrefs

Cf. A193139, A193140, A157228.

Programs

Maple

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)];

Mathematica

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 *)

Formula

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