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%I #32 Oct 26 2021 16:54:50
%S 72,232,288,520,584,800,808,1096,1152,1224,1312,1600,1664,1744,1800,
%T 1872,1960,2248,2312,2384,2592,2600,2824,3328,3392,3528,3600,4112,
%U 4176,4328,4624,5120,5328,5408,5904,6056,6120,6272,6352,6408,6568,6920,8080
%N Smallest member of three consecutive numbers each of which is the sum of two nonzero squares (not necessarily different).
%C a(n) == 0 (modulo 4) since no integer == 3 (modulo 4) can be represented as the sum of two squares.
%C This sequence has as a subsequence 72, 288, 800, 1800, ... which is 8 * (triangular numbers)^2. Proof: If x = 8*(n(n+1)/2)^2 then x = (n(n+1))^2 + (n(n+1))^2, x+1 = ((n-1)(n+1))^2 + (n(n+2))^2 and x+2 = (n^2+n-1)^2 + (n^2+n+1)^2. See A254371 - _Joshua Zucker_, Nov 01 2002
%C From _Altug Alkan_, Apr 13 2016: (Start)
%C If n is in this sequence, so is n*(n+2). Proof:
%C If n is in this sequence, then n = a^2 + b^2, n+1 = c^2 + d^2, n+2 = e^2 + f^2 for a, b, c, d, e, f being nonzero integers.
%C So, n*(n+2) = (a^2 + b^2)*(e^2 + f^2) = (a*e + b*f)^2 + (a*f - b*e)^2. Note that a*f cannot be equal to b*e because of their definitions.
%C n*(n+2) + 1 = n^2 + 2*n + 1 = (n+1)^2. Since we know that n mod 4 = 0, then n+1 cannot be of the form 2*k^2, that is, c and d must be different. So (n+1)^2 is the sum of two nonzero squares because n+1 = c^2 + d^2.
%C n*(n+2) + 2 = (n+1)^2 + 1, that is obviously the sum of two nonzero squares.
%C So if n is in this sequence, then n*(n+2), n*(n+2) + 1 and n*(n+2) + 2 are the sums of two nonzero squares, that is n*(n+2) must also be member of this sequence.
%C Note that it can be produced by repeating of this result and n*(n+2)*(n*(n+2)+2)*(n*(n+2)*(n*(n+2)+2)+2)... is always a member, if n is a member. (End)
%C For k > 0, 25*k^2*(10*k+2)^2 and 8*A001080(k)^2 are terms. - _Jinyuan Wang_, Feb 23 2019
%H Robert Israel, <a href="/A064716/b064716.txt">Table of n, a(n) for n = 1..10000</a>
%e 72 = 6^2 + 6^2, 73 = 3^2 + 8^2, 74 = 5^2 + 7^2.
%p N:= 10000: # to get all terms <= N
%p S:= {seq(seq(a^2+b^2, b=1..floor(sqrt(N+2-a^2))),a=1..floor(sqrt(N+2)))}:
%p sort(convert(S intersect map(`-`,S,1) intersect map(`-`,S,2),list)); # _Robert Israel_, Apr 14 2016
%t a = Table[n^2, {n, 1, 100}]; c = {}; Do[ c = Append[c, a[[i]] + a[[j]]], {i, 1, 100}, {j, 1, i} ]; c = Union[c]; c[[ Select[ Range[ Length[c] - 2], c[[ # ]] + 2 == c[[ # + 2 ]] & ]]]
%t Select[Range@ 8080, AllTrue[# + {0, 1, 2}, Length[ PowersRepresentations[#, 2, 2] /. {0, _} -> Nothing] > 0 &] &] (* _Michael De Vlieger_, Apr 13 2016, Version 10 *)
%o (PARI) is(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));
%o lista(nn) = {for(n=1,nn,if(is(n)==1&&is(n+1)==1&&is(n+2)==1,print1(n,", ")))}; \\ _Jinyuan Wang_, Feb 23 2019
%Y Cf. A000404, A001080.
%Y Cf. A254371 \ {0, 8} (a subsequence).
%K nonn
%O 1,1
%A _Robert G. Wilson v_, Oct 13 2001
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