A114377
Perfect powers equal to the sum of three factorial numbers.
Original entry on oeis.org
4, 8, 9, 27, 32, 36, 49, 128, 841, 5184
Offset: 1
a(1) = 4 because 4 = 1! + 1! + 2!;
a(2) = 8 because 8 = 1! + 1! + 3!;
a(3) = 9 because 9 = 1! + 2! + 3!.
A082920
Squares that are the sum of four factorials.
Original entry on oeis.org
4, 9, 16, 169, 361, 729, 961, 1444, 10201, 403225, 725904
Offset: 1
These appear to be the only solutions to a! + b! + c! + d! = n^2:
a b c d n
0 0 0 0 4
0 0 0 1 4
0 0 0 3 9
0 0 1 1 4
0 0 1 3 9
0 1 1 1 4
0 1 1 3 9
0 2 3 6 729
0 4 4 5 169
0 4 8 9 403225
0 5 5 5 361
0 5 5 6 961
0 5 7 7 10201
1 1 1 1 4
1 1 1 3 9
1 2 3 6 729
1 4 4 5 169
1 4 8 9 403225
1 5 5 5 361
1 5 5 6 961
1 5 7 7 10201
2 2 3 3 16
2 2 6 6 1444
4 5 9 9 725904
1!+2!+3!+6! = 729 = 27^2. This shows that 4 factorials can add to a cube.
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S:= {}:
N:= 100: # for terms < (N+1)!
F:= [seq(i!,i=1..N)]:
for a from 1 to N do
for b from a to N do
for c from b to N do
for d from c to N do
if issqr(F[a]+F[b]+F[c]+F[d]) then
S:= S union {F[a]+F[b]+F[c]+F[d]}
fi od od od od:
sort(convert(S,list)); # Robert Israel, Dec 23 2024
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e = 75; a = Union[ Flatten[ Table[a! + b! + c! + d!, {a, 1, e}, {b, a, e}, {c, b, e}, {d, c, e}]]]; l = Length[a]; Do[ If[ IntegerQ[ Sqrt[ a[[i]] ]], Print[ a[[i]] ]], {i, 1, l}]
Select[Union[Total/@Tuples[Range[10]!,4]],IntegerQ[Sqrt[#]]&] (* Harvey P. Dale, Aug 23 2014 *)
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sum4factsq(n) = { for(a1=0,n, for(a2=a1,n, for(a3=a2,n, for(a4=a3,n, z = a1!+a2!+a3!+a4!; if(issquare(z),print(a1" "a2" "a3" "a4" "z)) ) ) ) ) }
A162681
Numbers k such that k^2 is a sum of three factorials.
Original entry on oeis.org
2, 3, 6, 7, 29, 72
Offset: 1
2^2 = 1! + 1! + 2!;
3^2 = 1! + 2! + 3!;
6^2 = 3! + 3! + 4!;
7^2 = 1! + 4! + 4!;
29^2 = 1! + 5! + 6!;
72^2 = 4! + 5! + 7!.
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s := 10^40 ; sqr := s^2 : for a from 1 do if a! > sqr then break; fi; for b from a do if a!+b! > sqr then break; fi; for c from b do if a!+b!+c! > sqr then break; fi; if issqr(a!+b!+c!) then print( sqrt(a!+b!+c!)); fi; od: od: od: # R. J. Mathar, Jul 16 2009
w := 7: f := proc (x, y, z) options operator, arrow: sqrt(factorial(x)+factorial(y)+factorial(z)) end proc: A := {}: for x to w do for y to w do for z to w do if type(f(x, y, z), integer) = true then A := `union`(A, {f(x, y, z)}) else end if end do end do end do: A; # Emeric Deutsch, Aug 03 2009
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$MaxExtraPrecision=Infinity; lst={};Do[Do[Do[x=(a!+b!+c!)^(1/2);If[x==IntegerPart[x], AppendTo[lst,x]],{c,b,2*4!}],{b,a,2*4!}],{a,2*4!}];Union[lst]
Showing 1-3 of 3 results.
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