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%I #14 Dec 13 2014 00:48:50
%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,16,18,0,0,32,100,0,0,424,510,0,0,
%T 2792,5988,0,0,29058,45106,0,0,276828,473854,0,0,2455340,4777436,0,0,
%U 27466324,46429640,0,0,280395282,526489336,0,0,3193589950,5661226928,0,0
%N Number of solutions to +-1+-2^4+-3^4+-4^4...+-n^4=0.
%C Constant term in the expansion of (x + 1/x)(x^16 + 1/x^16)..(x^n^4 + 1/x^n^4).
%C a(n)=0 for any n=1 (mod 4) or n=2 (mod 4).
%C Andrica & Tomescu give a more general integral formula than the one below. The asymptotic formula below is a conjecture by Andrica & Ionascu; it remains unproven. - _Jonathan Sondow_, Nov 11 2013
%H D. Andrica and E. J. Ionascu, <a href="http://www.dorinandrica.ro/files/presentation-INTEGERS-2013.pdf">Variations on a result of Erdős and Surányi</a>INTEGERS 2013 slides.
%H Dorin Andrica and Ioan Tomescu, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL5/Tomescu/tomescu4.html">On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance</a>, J. Integer Sequences, 5 (2002), Article 02.2.4.
%F Integral representation: a(n) = ((2^n)/Pi)*int_0^pi prod_{k=1}^n cos(x*k^4) dx.
%F Asymptotic formula: a(n) = (2^n)*sqrt(18/(Pi*n^9))*(1+o(1)) as n->infinity; n=-1 or 0 (mod 4).
%e For n=16 the a(16) = 2 solutions are +1 +16 +81 +256 -625 -1296 -2401 +4096 +6561 +10000 +14641 +20736 -28561 -38416 -50625 +65536 = 0 and the opposite.
%p N:=32: p:=1 a:=[]: for n from 32 to N do p:=expand
%p (p*(x^(n^4)+x^(-n^4))): a:=[op(a), coeff(p,x,0)]: od:a;
%Y Cf. A063865, A158092, A158118, A158380, A019568.
%Y A111253(n) = a(n)/2. - _Alois P. Heinz_, Oct 31 2011
%K nonn
%O 1,16
%A _Pietro Majer_, Mar 19 2009
%E a(35)-a(58) from _Alois P. Heinz_, Oct 31 2011
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