A064716 Smallest member of three consecutive numbers each of which is the sum of two nonzero squares (not necessarily different).

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

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. See A254371 - Joshua Zucker, Nov 01 2002

From Altug Alkan, Apr 13 2016: (Start)

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.

Cf. A254371 \ {0, 8} (a subsequence).

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