

A115878


a(n) is the number of positive solutions of the Diophantine equation x^2 = y(y+n).


8



0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 4, 2, 1, 2, 1, 1, 4, 1, 1, 4, 2, 1, 3, 1, 1, 4, 1, 3, 4, 1, 4, 2, 1, 1, 4, 4, 1, 4, 1, 1, 7, 1, 1, 7, 2, 2, 4, 1, 1, 3, 4, 4, 4, 1, 1, 4, 1, 1, 7, 4, 4, 4, 1, 1, 4, 4, 1, 7, 1, 1, 7, 1, 4, 4, 1, 7, 4, 1, 1, 4, 4, 1, 4, 4, 1, 7, 4, 1, 4, 1, 4, 10, 1, 2, 7, 2, 1, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,9


COMMENTS

Number of divisors d of n^2 such that d^2 < n^2 and n^2/d == d (mod 4).  Antti Karttunen, Oct 06 2018, based on Robert Israel's Jun 27 2014 comment in A115880.
For odd n, a(n) can be computed from the prime signature.  David A. Corneth, Oct 07 2018


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..18480
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000


FORMULA

From David A. Corneth, Oct 07 2018: (Start)
a((2k+1) * 2^m) = floor(tau((2k + 1) ^ 2) / 2) for m <= 2.
a((2k+1) * 2^m) = (2m  3) * a(2k+1) + (m2) for m > 2. (End)


EXAMPLE

a(15) = 4 since there are 4 solutions (x,y) to x^2 = y(y+15), namely (4,1), (10,5), (18, 12) and (56, 49).
Note how each x is obtained from each such divisor pair n2/d and d of n2 as (n2/d  d)/4, when their difference is a positive multiple of four, thus in case of n2 = 15^2 = 225 we get (225/1  1)/4 = 56, (225/3  3)/4 = 18, (225/5  5) = 10 and (225/9  9)/4 = 4.  Antti Karttunen, Oct 06 2018
a(96) = 10. We compute P, the largest power of 2 dividing n = 96. Then compute min(P, 4) and divide n by it. This gives 96/4 = 24. Then find the number of divisors of 24^2, which is 21. Dividing by 2 rounding down to the nearest integer gives 10, the value of a(96).  David A. Corneth, Oct 06 2018


MATHEMATICA

a[n_] := Sum[Boole[d^2 < n^2 && Mod[n^2/dd, 4] == 0], {d, Divisors[n^2]}];
Array[a, 102] (* JeanFrançois Alcover, Feb 27 2019, from PARI *)


PROG

(PARI) A115878(n) = { my(n2 = n^2); sumdiv(n2, d, ((d*d)<n2)&&(0==(((n2/d)d)%4))); }; \\ Antti Karttunen, Oct 06 2018
(PARI) a(n) = my(v=min(2, valuation(n, 2))); numdiv((n>>v)^2)>>1 \\ David A. Corneth, Oct 06 2018


CROSSREFS

Cf. A067721, A115879, A115880, A115881.
Sequence in context: A129192 A062540 A173636 * A127125 A256671 A114171
Adjacent sequences: A115875 A115876 A115877 * A115879 A115880 A115881


KEYWORD

nonn


AUTHOR

Giovanni Resta, Feb 02 2006


STATUS

approved



