A308834 - OEIS

%I #24 Feb 04 2024 18:32:38

%S 14,49,94,113,46,-191,399,64,-657,545,-466,-721,-145,1328,270,-2751,

%T 719,-751,1118,-1376,-1041,1839,1310,1663,815,5184,-306,9104,863,1455,

%U 4320,7024,-5105,4289,11504,64,-12016,2816,10799,-11200,6094,-2671,-226,20753

%N a(n) is the smallest error in trying to solve n^4 = x^4 + y^4. That is, for each n from 2 on, find positive integers x and y, x <= y < n such that |n^4 - x^4 - y^4| is minimal and let a(n) = n^4 - x^4 - y^4.

%C This sequence was suggested to me by _Moshe Shmuel Newman_.

%C This is A135998 with the exponent 4 replacing 3.

%C From _M. F. Hasler_, Feb 03 2024: (Start)

%C Can it happen that (x,y) and (x',y') yield the same minimal absolute difference, but with opposite signs? If so, how is a(n) defined in this case?

%C Without the condition y < n, the trivial "solution" (x, y) = (1, n) would always yield a(n) = -1. With the condition, there is no admissible pair (x,y) for n = 1, whence a(1) is undefined. (End)

%e Here are the calculations for the first few values.

%e For 2, the only possible values for x and y are 1,1, so we have

%e a(2) = 2^4 - 1^4 - 1^4 = 16 - 2 = 14.

%e For 3, y can be 1 or 2. if y is 1, x is 1 as well, and if y=2, then x can be 1 or 2.

%e 3^4 - 1^4 - 1^4 = 79

%e 3^4 - 1^4 - 2^4 = 64

%e 3^4 - 2^4 - 2^4 = 49.

%e The smallest absolute value is in the last case, so a(3) = 49.

%t nend = 100;For[n = 2, n <= nend, n++, a[n] = 0];For[n = 2, n <= nend, n++, min = n^4; For[y = 1, y <= n - 1, y++,  For [x = y, x <= n - 1, x++,   changed = False;   sol = n^4 - x^4 - y^4;   If[(sol < min) && (sol > 0), min = sol; changed = True];   If[(Abs[sol] < min) && (sol < 0), min = -sol; changed = True];   If[changed, a[n] = sol]]]]; Print[t = Table[a[i], {i, 2, nend}]] (* or *)

%t a[n_] := SortBy[n^4 - Flatten[Table[x^4 + y^4, {x, n-1}, {y, x}]], Abs][[1]]; Array[a, 99, 2] (* _Giovanni Resta_, Jul 05 2019 *)

%o (PARI) A308834(n, p=4) = { my(np=n^p, m=np); for(y=max(sqrtnint(np\2, p), 1), n-1, my(x = sqrtnint(np - y^p, p), dy = np-y^p, d = if(dy-x^p > (x+1)^p-dy && x < n-1, dy-(x+1)^p, dy-x^p)); abs(d) < abs(m) && abs(m=d) < 2 && break); m} \\ _M. F. Hasler_, Feb 03 2024

%Y Cf. A135998 (equivalent for 3rd powers).

%K sign

%O 2,1

%A _David S. Newman_, Jun 27 2019