

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
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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 = ((n1)(n+1))^2 + (n(n+2))^2 and x+2 = (n^2+n1)^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+2a^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: A050498 A077535 A339994 * A073412 A250749 A019507
Adjacent sequences: A064713 A064714 A064715 * A064717 A064718 A064719


KEYWORD

nonn


AUTHOR

Robert G. Wilson v, Oct 13 2001


STATUS

approved



