This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(1) = 0, since 4/1 = 4 cannot be expressed as the sum of three reciprocals. a(2) = 1 because 4/2 = 1/1 + 1/2 + 1/2, and there are no other solutions. a(3) = 3 since 4/3 = 1 + 1/4 + 1/12 = 1 + 1/6 + 1/6 = 1/2 + 1/2 + 1/3. a(4) = 3 = A002966(3).
A192787 := proc(n) local t,a,b,t1,count; t:= 4/n; count:= 0; for a from floor(1/t)+1 to floor(3/t) do t1:= t - 1/a; for b from max(a,floor(1/t1)+1) to floor(2/t1) do if type( 1/(t1 - 1/b),integer) then count:= count+1; end if end do end do; count; end proc; # Robert Israel, Feb 19 2013
f[n_] := Length@ Solve[ 4/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[f, 70] (* Allan C. Wechsler and Robert G. Wilson v, Jul 19 2013 *)
a(n, show=0)=my(t=4/n, t1, s, c); for(a=1\t+1, 3\t, t1=t-1/a; for(b=max(1\t1+1, a), 2\t1, c=1/(t1-1/b); if(denominator(c)==1&&c>=b, s++; show&&print("4/",n," = 1/",a," + 1/",b," + 1/",c)))); s \\ variant with print(...) added by Robert Munafo, Feb 19 2013, both combined through option "show" by M. F. Hasler, Jul 02 2022
a(3) = 12 because 4/(3rd prime) = 4/5 can be expressed in the following 12 ways: 4/5 = 1/2 + 1/4 + 1/20 4/5 = 1/2 + 1/5 + 1/10 4/5 = 1/2 + 1/10 + 1/5 4/5 = 1/2 + 1/20 + 1/4 4/5 = 1/4 + 1/2 + 1/20 4/5 = 1/4 + 1/20 + 1/2 4/5 = 1/5 + 1/2 + 1/10 4/5 = 1/5 + 1/10 + 1/2 4/5 = 1/10 + 1/2 + 1/5 4/5 = 1/10 + 1/5 + 1/2 4/5 = 1/20 + 1/2 + 1/4 4/5 = 1/20 + 1/4 + 1/2
checkmult[a_, b_, c_] := If[Denominator[c] == 1, If[a == b && a == c && b == c, Return[1], If[a != b && a != c && b != c, Return[6], Return[3]]], Return[0]];
a292581[n_] := Module[{t, t1, s, a, b, c, q = Quotient}, t = 4/n; s = 0; For[a = q[1, t] + 1, a <= q[3, t], a++, t1 = t - 1/a; For[b = Max[q[1, t1] + 1, a], b <= q[2, t1], b++, c = 1/(t1 - 1/b); s += checkmult[a, b, c]]]; Return[s]];
Reap[For[n = 1, n <= 54, n++, Print[n, " ", an = a292581[Prime[n]]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)
checkmult (a,b,c) =
{
if(denominator(c)==1,
if(a==b && a==c && b==c,
return(1),
if(a!=b && a!=c && b!=c,
return(6),
return(3)
)
),
return(0)
)
}
a292624(n) =
{
local(t, t1, s, a, b, c);
t = 4/prime(n);
s = 0;
for (a=1\t+1, 3\t,
t1=t-1/a;
for (b=max(1\t1+1,a), 2\t1,
c=1/(t1-1/b);
s+=checkmult(a,b,c);
)
);
return(s);
}
for (n=1,54,print1(a292624(n),", "))
a(1) = 1 because 4/prime(1) = 1/1 + 1/2 + 1/2.
a:= n-> A192787(ithprime(n)): seq(a(n), n=1..70);
a[n_] := a[n] = Module[{a, b, c, r}, r = Reduce[1 <= a <= b <= c && 4/Prime[n] == 1/a + 1/b + 1/c, {a, b, c}, Integers]; Which[Head[r] === Or, Length[r], Head[r] === And, 1, r === False, 0, True, Print["error: ", r]]];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 84}] (* Jean-François Alcover, Nov 22 2017 *)
a(n)=my(t=4/prime(n), t1, s, c); for(a=1\t+1, 3\t, t1=t-1/a; for(b=1\t1+1, 2\t1, c=1/(t1-1/b); if(denominator(c)==1&&c>=b, s++))); s
a(6)=6 because it is the least denominator z in the A192787(6)=8 solutions [x, y, z]: [2, 7, 42], [2, 8, 24], [2, 9, 18], [2, 10, 15], [2, 12, 12], [3, 4, 12], [3, 6, 6], [4, 4, 6]; a(13)=52 because the minimum of z in the A192787(13)=4 solutions is 52: [4, 18, 468], [4, 20, 130], [4, 26, 52], [5, 10, 130].
f:= proc(n) local z,x,y;
for z from floor(n/4)+1 do
for x from floor(n*z/(4*z-n))+1 to z do
y:= n*x*z/(4*x*z-n*x-n*z);
if y::posint and y >= x and y <= z then return z fi
od od
end proc:
map(f, [$2..100]); # Robert Israel, Oct 14 2020
a[n_] := For[z = Floor[n/4] + 1, True, z++, For[x = Floor[n(z/(4z - n))] + 1, x <= z, x++, y = n x z/(4 x z - n x - n z); If[IntegerQ[y] && x <= y <= z, Print[z]; Return [z]]]]; a /@ Range[2, 100] (* Jean-François Alcover, Oct 23 2020, after Robert Israel *)
a337432(n)={my(target=4/n,a,b,c,m=oo);for(a=1\target+1,3\target,my(t=target-1/a);for(b=max(1\t+1,a),2\t,c=1/(t-1/b);if(denominator(c)==1,m=min(m,max(a,max(b,c))))));m};
for(k=2,67,print1(a337432(k),", "))
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