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A086767 Last coefficient of the last term in the numerator of the simplified expansion of the solutions of FLT for n=2 for FLT n=1,2,3,..

%I A086767 #19 Feb 20 2025 06:42:16
%S A086767 1,1,1,1,1,3,1,1,1,5,1,3,1,7,1,1,1,9,1,5,1,11,1,3,1,13,1,7,1,15,1,1,1,
%T A086767 17,1,9,1,19,1,5,1,21,1,11,1,23,1,3,1,25,1,13,1,27,1,7,1,29,1,15,1,31,
%U A086767 1,1,1,33,1,17,1,35,1,9,1,37,1,19,1,39,1,5,1,41,1,21,1,43,1,11,1,45,1,23
%N A086767 Last coefficient of the last term in the numerator of the simplified expansion of the solutions of FLT for n=2 for FLT n=1,2,3,..
%C A086767 Integers a > b form the solution to FLT n = 2 as follows. (2ab)^2 = (a^2-b^2)^2 - (a^2+b^2)^2. The sequence is the coefficient c of the last b term in the numerator for the simplified expansion of the solution for n=2 as verification of FLT for n=1, 2, ...
%H A086767 Anonymous, <a href="https://web.archive.org/web/20050220082825/http://mathforum.org/discuss/sci.math/t/525751">Fermat's Theorem for Pythagorean Triples</a>.
%F A086767 a(n) = A004731(n+1)/A004730(n). - _Flávio V. Fernandes_, Feb 13 2025
%e A086767 b/a
%e A086767 1
%e A086767 (3*a^4 + b^4)/(4*b*a^3)
%e A086767 (a^4 + b^4)/(2*b^2*a^2)
%e A086767 (5*a^8 + 10*b^4*a^4 + b^8)/(16*b^3*a^5)
%e A086767 (3*a^8 + 10*b^4*a^4 + 3*b^8)/(16*b^4*a^4)
%e A086767 (7*a^12 + 35*b^4*a^8 + 21*b^8*a^4 + b^12)/(64*b^5*a^7)
%e A086767 (a^12 + 7*b^4*a^8 + 7*b^8*a^4 + b^12)/(16*b^6*a^6)
%e A086767 (9*a^16 + 84*b^4*a^12 + 126*b^8*a^8 + 36*b^12*a^4 + b^16)/(256*b^7*a^9)
%e A086767 (5*a^16 + 60*b^4*a^12 + 126*b^8*a^8 + 60*b^12*a^4 + 5*b^16)/(256*b^8*a^8)
%e A086767 ........
%e A086767 (K + cb^m)/2^m1b^m2c^m3
%e A086767 Seq = c for integers K,b,m1,m2,m3,n = 1,2,3...
%p A086767 sigma := proc(n) local i; add(i,i=convert(n,base,2)) end:
%p A086767 a := proc(n) if n=0 or type(n,odd) then 1 else if type(iquo(n,2),odd) then n/2 else n/2^(1-sigma(n)+sigma(n-1)) fi fi end: # _Peter Luschny_, Aug 03 2009
%o A086767 (PARI) \ verification of general solution in integers \ a>b,x = 2ab,y=a^2-b^2,z=a^2+b^2 \ or FLT n=2 x^n+y^n <> z^n = (2ab)^n + (a^2-b^2)^n <> \(a^2+b^2)^n for n > 2 flt(n,a1,b1) = for(x=0,n,print(f(x,a1,b1))) f(n,a,b) = simplify(((a^2+b^2)^n - (a^2-b^2)^n)/(2*a*b)^n) coeffb(m) = { for(y=1,m, n=y; if(n%2,x=1, while(n%2==0,n=n/2); x=n; ); print1(x",") ) }
%Y A086767 Cf. A004731, A004730.
%K A086767 easy,nonn,uned
%O A086767 0,6
%A A086767 _Cino Hilliard_, Aug 02 2003