%I #16 May 07 2019 07:04:25
%S 3655,3740,4510,4760,5244,5434,5915,7230,7574,8415,11055,11900,12524,
%T 14905,17484,18879,19005,19855,20449,20510,21790,22806,23681,25580,
%U 25585,27230,27420,28985,31395,34224,37114,39606,41685,42419,44919,45435,45955,48026,48139,48225,49015,53941,57164,62006
%N Numbers that can be written as the sum of four or more consecutive squares in more than one way.
%C Numbers that are in A174071 in two or more ways.
%C The first number with more than two representations as a sum of four or more consecutive positive squares is 147441 = 18^2 + ... + 76^2 = 29^2 + ... + 77^2 = 85^2 + ... + 101^2.
%C If x = 2*A049629(n) and y = A007805(n) for n >= 1 (satisfying the Pell equation x^2 - 5*y^2 = -1), then the sequence contains 5*x^2+10 = Sum_{(5*y-3)/2 <= i <= (5*y+3)/2} i^2 = Sum_{x-2 <= i <= x+2} i^2 = 25*y^2 + 5.
%H Robert Israel, <a href="/A307937/b307937.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 3655 is in the sequence because 3655 = 8^2 + ... + 22^2 = 25^2 + ... + 29^2.
%p N:= 10^5: # to get all terms <= N
%p R:= 'R':
%p dups:= NULL:
%p for m from 4 while m*(m+1)*(2*m+1)/6 <= N do
%p for k from 1 do
%p v:= m*(6*k^2 + 6*k*m + 2*m^2 - 6*k - 3*m + 1)/6;
%p if v > N then break fi;
%p if assigned(R[v]) then
%p dups:= dups, v;
%p else
%p R[v]:= [k, k+m-1];
%p fi;
%p od od:
%p sort(convert({dups},list));
%t M = 10^5;
%t dups = {}; Clear[rQ]; rQ[_] = False;
%t For[m = 4, m(m+1)(2m+1)/6 <= M, m++, For[k = 1, True, k++, v = m(6k^2 + 6k m + 2m^2 - 6k - 3m + 1)/6; If[v > M, Break[]]; If[rQ[v], AppendTo[dups, v], rQ[v] = True]]];
%t dups // Sort (* _Jean-François Alcover_, May 07 2019, after _Robert Israel_ *)
%Y Cf. A007805, A049629, A174071.
%K nonn
%O 1,1
%A _Robert Israel_, May 06 2019