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%I #36 Nov 04 2023 13:27:45
%S 71,347,1193,2139,2709,17823,18337,26057,32847,34037,65793,87519,
%T 159541,245573,383037,421957,489731,520547,574841,800589,1291333,
%U 2010341,2113003,2990187,4528667,7430553,8284063,8402417,8520567,9220519,9865989,10621507,11961043,12335203,16405581,17648561,22224647,22918853,24171273
%N Positive integers m such that 2*m^2 - 1 = x^4 + y^4 for some nonnegative integers x and y with |x - y| > 1.
%C Conjecture: The sequence has infinitely many terms.
%C Clearly all the terms must be odd and not divisible by 5. Note also that 2*(n^2+n+1)^2 - 1 = n^4 + (n+1)^4.
%C See also A343917 for a similar conjecture.
%H Robert Israel, <a href="/A343913/b343913.txt">Table of n, a(n) for n = 1..112</a> (all terms < 10^10; first 53 terms from Zhi-Wei Sun)
%e a(1) = 71, and 2*71^2 - 1 = 10^4 + 3^4 with |10 - 3| > 1.
%e a(53) = 99532937, and 2*99532937^2 - 1 = 19813611095691937 = 11337^4 + 7576^4 with |11337 - 7576| > 1.
%p N:= 10^18: # for all terms <= sqrt(N)
%p R:= {}: count:= 0:
%p for x from 1 while 2*x^4 < 2*N-1 do
%p for y from x+3 by 2 do
%p v:= (x^4 + y^4 + 1)/2;
%p if v > N then break fi;
%p if issqr(v) then
%p m:= sqrt(v);
%p if not member(m,R) then
%p count:= count+1; R:= R union {m};
%p fi fi
%p od od:
%p sort(convert(R,list)); # _Robert Israel_, May 04 2021
%t QQ[n_]:=IntegerQ[n^(1/4)];
%t n=0;Do[Do[If[QQ[2*m^2-1-(2x)^4]&&Abs[2x-(2*m^2-1-(2x)^4)^(1/4)]>1,n=n+1;Print[n," ",m];Goto[aa]],{x,0,((2m^2-1)^(1/4))/2}];Label[aa],{m,1,25000000}]
%Y Cf. A000290, A000583, A003336, A056220, A343917.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, May 03 2021
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