%I #43 Nov 16 2021 12:00:22

%S 5,15,20,34,39,41,45,60,65,80,85,111,125,135,136,145,150,156,164,175,

%T 180,194,219,240,245,255,260,265,306,313,320,325,340,351,353,369,371,

%U 375,405,410,444,445,455,500,505,514,540,544,580,585,600,605,609,624,629,656,671,674,689

%N Integers whose square is of the form x^2 + y^4, with x,y > 0 (see A111925).

%C If z is in this sequence, z*k^2 is, for k > 0. Also note that since there are no fourth powers in A111925, there are no squares in this sequence. - _Altug Alkan_, Apr 10 2016

%C From _Karl-Heinz Hofmann_, Oct 22 2021: (Start)

%C Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).

%C If the terms additionally have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too. (End)

%C From _Jon E. Schoenfield_, Nov 15 2021: (Start)

%C Apparently, the vast majority of the terms of this sequence can be expressed as x^2 + y^4 with x,y > 0 in only one way (A345645 lists those terms), but some can be so expressed in exactly two, three, four, five, six, or more ways. Among the first 3976926961 terms of this sequence,

%C 3948648229 are in A345645 (exactly 1 way),

%C 25415062 are in A345700 (exactly 2 ways),

%C 2697713 are in A345968 (exactly 3 ways),

%C 161543 are in A346110 (exactly 4 ways),

%C 3989 are in A348655 (exactly 5 ways),

%C 424 are in A349324 (exactly 6 ways),

%C and just 1 -- a(3976926961) = 2474052064291275 = A346115(7) -- is a number whose square can be written as x^2 + y^4 with x,y > 0 in exactly 7 ways. The ratios of successive counts above, i.e., 3948648229/25415062, 25415062/2697713, 2697713/161543, 161543/3989, 3989/424, and 424/1, are approximately 155.4, 9.4, 16.7, 40.5, 9.4, and 424.0. What is it that (over the interval [1, 2474052064291275], at least) makes numbers whose squares can be written as x^2 + y^4 with x,y > 0 in more than 6 ways so much rarer than those that can be written thus in exactly 6 ways? (End)

%H Chai Wah Wu, <a href="/A271576/b271576.txt">Table of n, a(n) for n = 1..10000</a>

%e 5^2 = 25 = 9 + 16 = 3^2 + 2^4, so 5 is a term.

%t Select[Range@ 200, Resolve[Exists[{x, y}, Reduce[#^2 == x^2 + y^4, {x, y}, Integers], And[x > 0, y > 0]]] &] (* _Michael De Vlieger_, Apr 10 2016 *)

%o (PARI) isok(n) = n = n^2; for (k=1, n-1, if (issquare(k) && (p=ispower(n-k, 4)), return (1)))

%o (PARI) is(n)=my(n2=n^2); for(b=sqrtnint(2*n-2,4)+1, sqrtint(n-1), if(issquare(n2-b^4), return(1))); 0 \\ _Charles R Greathouse IV_, Nov 16 2021

%Y Cf. A111925, A271577, A345645 (in exactly 1 way).

%Y Cf. A345700 (in exactly 2 ways), A345968 (in exactly 3 ways).

%Y Cf. A346110 (in exactly 4 ways), A349324 (in exactly 6 ways), A346115 (the least solutions).

%Y Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

%K nonn

%O 1,1

%A _Michel Marcus_, Apr 10 2016