1,1

Intersection of A004214 with products of pairs of terms of A000408.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

a(2) = 87 = 3 * 29 = (1^2+1^2+1^2) * (4^2+3^2+2^2)

87 does not have a partition as a sum x^2+y^2+z^2 with x,y,z>0

63=3*21; 87=3*29; 135=3*45; 156=6*26; 572=22*26;

isA000408 := proc(n) local a, b, c2 ; a:=1; while a^2<n do b:=1 ; while b<=a and a^2+b^2<n do c2 := n-a^2-b^2 ; if issqr(c2) then RETURN(true) ; fi ; b := b+1 ; od ; a := a+1 ; od ; RETURN(false) ; end: isA125112 := proc(n) local d, i; if isA000408(n) then RETURN(false) ; else d := numtheory[divisors](n) ; for i from 1 to nops(d) do if isA000408(op(i, d)) and isA000408(n/op(i, d)) then RETURN(true) ; fi ; od ; RETURN(false) ; fi ; end: for an from 1 to 1600 do if isA125112(an) then printf("%d, ", an) ; fi ; od ; # R. J. Mathar, Nov 23 2006

isA000408[n_] := Module[{a, b, c2}, a = 1; While[a^2 < n, b = 1; While[b <= a && a^2 + b^2 < n, c2 = n - a^2 - b^2; If[IntegerQ@Sqrt@c2, Return[True]]; b++]; a++]; Return[False]];

isA125112[n_] := Module[{d, i}, If[isA000408[n], Return[False], d = Divisors[n]; For[i = 1, i <= Length[d], i++, If[isA000408[d[[i]]] && isA000408[n/d[[i]]], Return[True]]]; Return[False]]];

Select[Range[1600], isA125112] (* Jean-François Alcover, Jul 22 2024, after R. J. Mathar *)

Cf. A000408 (sums of 3 nonzero squares), A004214 (not sums of 3 nonzero squares).

Sequence in context: A320066 A372557 A372558 * A043186 A039363 A043966

Adjacent sequences: A125109 A125110 A125111 * A125113 A125114 A125115

Artur Jasinski, Nov 21 2006

Edited and extended by R. J. Mathar and Ray Chandler, Nov 23 2006

approved