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%I #16 Mar 11 2016 06:25:51
%S 2,5,6,11,12,13,14,25,26,37,38,61,62,73,74,85,86,97,98,121,122,133,
%T 134,145,146,157,158,181,182,221,222,253,254,325,326,337,338,365,366,
%U 397,398,445,446,613,614,625,626,697,698,721,722,793,794,865,866
%N Integers n such that the n-th golden rectangle number is the sum of 2 nonzero squares.
%C Corresponding golden rectangle numbers are 2, 40, 104, 12816, 33552, 87841, 229970, 9107509825, 23843770274, 944284833567073, 2472169789339634, ...
%C Initial terms of first differences are 3, 1, 5, 1, 1, 1, 11, 1, 11, 1, 23, 1, 11, 1, 11, 1, 11, 1, 23, ...
%e 5 is a term because 1^2 + 1^2 + 2^2 + 3^2 + 5^2 = 5*8 = 40 = 2^2 + 6^2.
%e 6 is a term because 1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 8^2 = 8*13 = 104 = 2^2 + 10^2.
%t Rest@ Select[Range@ 200, SquaresR[2, #] > 0 &[Fibonacci[#] Fibonacci[# + 1]] &] (* _Michael De Vlieger_, Mar 09 2016 *)
%o (PARI) isA000404(n)= for( i=1, #n=factor(n)~%4, n[1, i]==3 && n[2, i]%2 && return); n && ( vecmin(n[1, ])==1 || (n[1, 1]==2 && n[2, 1]%2))
%o a001654(n) = fibonacci(n)*fibonacci(n+1);
%o for(n=1, 1e2, if(isA000404(a001654(n)), print1(n, ", ")));
%o (PARI) has(f)=for(i=1,#f~, if(f[i,1]%4==3 && f[i,2]%2, return(0))); 1
%o isA009003(f)=for(i=1,#f~,if(f[i,1]%4==1, return(1))); 0
%o is(n)=my(f,g); has(f=factor(fibonacci(n))) && has(g=factor(fibonacci(n+1))) && (n%3!=1 || isA009003(f) || isA009003(g)) \\ _Charles R Greathouse IV_, Mar 08 2016
%Y Cf. A000404, A001654, A009003.
%K nonn
%O 1,1
%A _Altug Alkan_, Mar 08 2016
%E a(36)-a(55) from _Charles R Greathouse IV_, Mar 08 2016
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