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%I #8 Mar 30 2012 18:38:57
%S 1,1,1,49,2401,113060689,260871824431729,9708455965188246321478801,
%T 361304320362377236050632364626862769,
%U 3511057522394397982450601057907077808699210592028881
%N Let M_n be the n X n matrix M_(i,j)=1/(2^i+2^j), then a(n) is the numerator of det(M_n).
%C a(n) seems always to be a square and 7 seems to follow a rule in a(n) factorization. Maximal k such that 7^k divides a(n) are 0, 0, 0, 2, 4, 6, 10, 14, 18, 24, 30, 36, 44, 52, 60, 70, 80, 90, 102, 114, 126, 142, 158, 174, 192... Hence if b(n)=maximum exponent of 7 in factorization of a(n), b(3n+1)=A049450(n); b(3n+2)=A049450(n)+2*n; b(3n+3)=A049450(n)+4n
%H Vincenzo Librandi, <a href="/A069741/b069741.txt">Table of n, a(n) for n = 1..25</a>
%o (PARI) for(n=1,70,print1(numerator(matdet(matrix(n,n,i,j,1/(2^i+2^j)))),","))
%Y Cf. A069743.
%K easy,nonn
%O 1,4
%A _Benoit Cloitre_, Apr 21 2002
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