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%I #21 Sep 15 2022 02:17:29
%S 40,68,72,104,148,180,320,392,468,544,612,648,720,788,832,900,936,968,
%T 1040,1044,1156,1192,1256,1300,1332,1508,1732,1796,1800,1832,1872,
%U 1940,2056,2196,2308,2336,2372,2448,2664,2696,2740,2804,2848,2880,3060,3200,3280
%N Numbers k such that both k and k^2 + 2 can be written as the sum of two nonzero squares.
%C The terms in this sequence can be considered as a solution to the "near miss" problem which occurs frequently while solving Diophantine equations. It is known that if a number k can be written as the sum of two nonzero distinct squares then so can k^2 and k^2+1. Thus, finding numbers k such that k^2+2 satisfies the same property makes it quite interesting.
%e 40 is a term since 40 = 2^2 + 6^2 as well as 40^2 + 2 = 1602 = 9^2 + 39^2.
%e 320 is a term since 320 = 8^2 + 16^2 as well as 320^2 + 2 = 102402 = 201^2 + 249^2.
%o (PARI) is1(n)= for( i=1, #n=factor(n)~%4, n[1, i]==3 && n[2, i]%2 && return); n && ( vecmin(n[1, ])==1 || (n[1, 1]==2 && n[2, 1]%2)); \\ A000404
%o isok(k) = is1(k) && is1(k^2+2); \\ _Michel Marcus_, Jul 18 2022
%o (Python)
%o from itertools import count, islice
%o from sympy import factorint
%o def A355778_gen(startvalue=1): # generator of terms >= startvalue
%o for n in count(max(startvalue, 1)):
%o c = False
%o for p in (f:=factorint(n)):
%o if (q:= p & 3)==3 and f[p]&1:
%o break
%o elif q == 1:
%o c = True
%o else:
%o if c or f.get(2, 0)&1:
%o c = False
%o for p in (f:=factorint(n**2+2)):
%o if (q:= p & 3)==3 and f[p]&1:
%o break
%o elif q == 1:
%o c = True
%o else:
%o if c or f.get(2, 0)&1:
%o yield n
%o A355778_list = list(islice(A355778_gen(),30)) # _Chai Wah Wu_, Sep 14 2022
%Y Cf. A000290, A000404, A000415.
%K nonn
%O 1,1
%A _Angad Singh_, Jul 16 2022