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 A064716 Smallest member of three consecutive numbers each of which is the sum of two nonzero squares (not necessarily different). 5
 72, 232, 288, 520, 584, 800, 808, 1096, 1152, 1224, 1312, 1600, 1664, 1744, 1800, 1872, 1960, 2248, 2312, 2384, 2592, 2600, 2824, 3328, 3392, 3528, 3600, 4112, 4176, 4328, 4624, 5120, 5328, 5408, 5904, 6056, 6120, 6272, 6352, 6408, 6568, 6920, 8080 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS a(n) == 0 (modulo 4) since no integer == 3 (modulo 4) can be represented as the sum of two squares. 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. - Joshua Zucker, Nov 01 2002 From Altug Alkan, Apr 13 2016 If n is in this sequence, so is n*(n+2). Proof: 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. 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. 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. n*(n+2) + 2 = (n+1)^2 + 1, that is obviously the sum of two nonzero squares. 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. 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) For k > 0, 25*k^2*(10*k+2)^2 and 8*A001080(k)^2 are terms. - Jinyuan Wang, Feb 23 2019 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 EXAMPLE 72 = 6^2 + 6^2, 73 = 3^2 + 8^2, 74 = 5^2 + 7^2. MAPLE N:= 10000: # to get all terms <= N S:= {seq(seq(a^2+b^2, b=1..floor(sqrt(N+2-a^2))), a=1..floor(sqrt(N+2)))}: sort(convert(S intersect map(`-`, S, 1) intersect map(`-`, S, 2), list)); # Robert Israel, Apr 14 2016 MATHEMATICA 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 ]] & ]]] Select[Range@ 8080, AllTrue[# + {0, 1, 2}, Length[ PowersRepresentations[#, 2, 2] /. {0, _} -> Nothing] > 0 &] &] (* Michael De Vlieger, Apr 13 2016, Version 10 *) PROG (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)); 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 CROSSREFS Cf. A000404, A001080. Sequence in context: A204485 A050498 A077535 * A073412 A250749 A019507 Adjacent sequences:  A064713 A064714 A064715 * A064717 A064718 A064719 KEYWORD nonn AUTHOR Robert G. Wilson v, Oct 13 2001 STATUS approved

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Last modified August 14 08:20 EDT 2020. Contains 336480 sequences. (Running on oeis4.)