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A086767
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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,..
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0
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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, 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, 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
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OFFSET
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0,6
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LINKS
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FORMULA
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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, ...
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EXAMPLE
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b/a
1
(3*a^4 + b^4)/(4*b*a^3)
(a^4 + b^4)/(2*b^2*a^2)
(5*a^8 + 10*b^4*a^4 + b^8)/(16*b^3*a^5)
(3*a^8 + 10*b^4*a^4 + 3*b^8)/(16*b^4*a^4)
(7*a^12 + 35*b^4*a^8 + 21*b^8*a^4 + b^12)/(64*b^5*a^7)
(a^12 + 7*b^4*a^8 + 7*b^8*a^4 + b^12)/(16*b^6*a^6)
(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)
(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)
........
(K + cb^m)/2^m1b^m2c^m3
Seq = c for integers K,b,m1,m2,m3,n = 1,2,3...
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MAPLE
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sigma := proc(n) local i; add(i, i=convert(n, base, 2)) end:
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
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PROG
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(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", ") ) }
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CROSSREFS
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KEYWORD
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easy,nonn,uned
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AUTHOR
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STATUS
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approved
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