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A100073
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Number of representations of n as the difference of two positive squares.
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9
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0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 2, 0, 1, 2, 1, 0, 2, 1, 1, 0, 1, 2, 2, 0, 2, 1, 1, 0, 2, 2, 1, 0, 1, 1, 3, 0, 1, 3, 1, 0, 2, 1, 1, 0, 2, 2, 2, 0, 1, 2, 1, 0, 3, 2, 2, 0, 1, 1, 2, 0, 1, 3, 1, 0, 3, 1, 2, 0, 1, 3, 2, 0, 1, 2, 2, 0, 2, 2, 1, 0, 2, 1, 2, 0, 2, 4, 1, 0, 3, 1, 1, 0, 1, 2, 4
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OFFSET
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1,15
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COMMENTS
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Note that for odd n, a(n) = 1 iff n is a prime, or a prime squared.
A decomposition n = a^2 - b^2 = (a-b)(a+b) = d*(n/d) is given for each divisor d less than (as to exclude b = 0) but having the same parity as n/d. For even n this implies that d and n/d must be even, i.e., 4 | n. This leads to the given formula, a(n) = floor(numdiv(n)/2) for odd n, floor(numdiv(n/4)/2) for n = 4k, 0 else. - M. F. Hasler, Jul 10 2018
a(n) is the number of self-conjugate partitions of n into parts of 2 different sizes, i.e., the order of the set of partitions obtained by the intersection of the partitions in A000700 and A002133. See A270060. - R. J. Mathar, Jun 15 2022
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LINKS
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FORMULA
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a(n) = A056924(n) for odd n, a(n) = A056924(n/4) if 4|n, otherwise a(n) = 0.
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EXAMPLE
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a(15) = 2 because 15 = 16 - 1 = 64 - 49.
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MAPLE
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if n::odd then floor(numtheory:-tau(n)/2)
elif (n/2)::odd then 0
else floor(numtheory:-tau(n/4)/2)
fi
end proc:
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MATHEMATICA
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nn=150; a=Table[0, {nn}]; Do[y=x-1; While[d=x^2-y^2; d<=nn&&y>0, a[[d]]++; y-- ], {x, 1+nn/2}]; a
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PROG
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(PARI) a(n) = if (n % 2, ceil((numdiv(n)-1)/2), if (!(n%4), ceil((numdiv(n/4)-1)/2), 0)); \\ Michel Marcus, Mar 07 2016
(PARI) A100073(n)=if(bittest(n, 0), numdiv(n)\2, !bittest(n, 1), numdiv(n\4)\2) \\ or shorter: a(n)=if(n%4!=2, numdiv(n\4^!(n%2))\2) \\ - M. F. Hasler, Jul 10 2018
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CROSSREFS
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Cf. A056924 (number of divisors of n that are less than sqrt(n)), A016825 (numbers not the difference of two squares), A034178 (number of representations of n as the difference of two squares).
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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