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A192717
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Positive integers of the form (p^e)(k^2) for p prime congruent to 3 (mod 8), e congruent to 1 (mod 4), and k an odd integer coprime to p.
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2
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3, 11, 19, 43, 59, 67, 75, 83, 99, 107, 131, 139, 147, 163, 171, 179, 211, 227, 243, 251, 275, 283, 307, 331, 347, 363, 379, 387, 419, 443, 467, 475, 491, 499, 507, 523, 531, 539, 547, 563, 571, 587, 603, 619, 643, 659, 683, 691, 739, 747, 787, 811, 827
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OFFSET
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1,1
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COMMENTS
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This sequence is equivalent to all of the following sets (written in increasing order):
- all integers the form (p^e)(k^2) for p prime congruent to 3 (mod 8), e congruent to 1 (mod 4), and k an odd number coprime to p;
- all integers with an odd number of representations as x^2 + 2y^2 for odd x and y; and
- elements of A192628 which are congruent to 3 (mod 8).
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LINKS
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EXAMPLE
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3 is in the sequence since 3 = (3^1)(1^2); 3 is prime and congruent to 3 (mod 8), 1 is congruent to 1 (mod 4), and 1 is an odd integer coprime to 3.
6 is not in the sequence: since it is squarefree, k must be 1, but 6 cannot be written as p^e.
27 is not in the sequence: the only possible values for k are 1 and 3. In the k=1 case, 27 = (3^3)(1^2) does not work since e = 3 is not congruent to 1 (mod 4), and in the k=3 case, 27 = (3^1)(3^2), k=3 and p=3 are not coprime.
243 is in the sequence since 243 = (3^5)(1^2); 3 is prime and congruent to 3 (mod 8), 5 is congruent to 1 (mod 4), and 1 is an odd integer coprime to 3.
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MATHEMATICA
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ofTheFormQ[n_] := If[Length[fin = FactorInteger[n]] == 1 && Mod[fin[[1, 1]], 8] == 3 && Mod[fin[[1, 2]], 4] == 1, True, pe = Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ fin); k = Sqrt[n/pe]; fip = FactorInteger[pe]; Length[fip] == 1 && Mod[p = fip[[1, 1]], 8] == 3 && Mod[e = fip[[1, 2]], 4] == 1 && OddQ[k] && CoprimeQ[k, p]]; Select[Range[1, 999, 2], ofTheFormQ] (* Jean-François Alcover, Jan 22 2013 *)
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PROG
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(Sage)
prec = 2^10
L = []
for n in range(1, prec, 2):
n = Integer(n)
sfp = n.squarefree_part()
if mod(sfp, 8) == 3 and sfp.is_prime() and mod(n.ord(sfp), 4) == 1:
L.append(n)
print(L)
(Sage)
def BPS(n): #binary power series
return sum([q^s for s in n])
prec = 2^14
R = PowerSeriesRing(GF(2), 'q', default_prec = prec)
q = R.gen()
dList = [(2*n+1)^2 for n in range(0, (sqrt(prec)-1)/2)]
dSeries = BPS(dList)
print((dSeries^3).exponents()[:128])
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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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