cp's OEIS Frontend

A383654 a(n) is the number k such that A383653(n)^4 is the sum of squares of k consecutive integers.

%I A383654 #33 May 14 2025 15:44:57
%S A383654 2,2,169,242,177,352,1536,2401,40898,163607,230121,60625,218089,
%T A383654 185761,19512097,47761921,1170329056,1224370081,7957888849,
%U A383654 10842382346,11474926944,208152552417,12230369281,190412616875,497818686976,72899460001,1384334025217,313455536641
%N A383654 a(n) is the number k such that A383653(n)^4 is the sum of squares of k consecutive integers.
%H A383654 Zhining Yang, <a href="https://bbs.emath.ac.cn/forum.php?mod=viewthread&amp;tid=39808">Can be expressed as the fourth power of the sum of squares of consecutive positive integers</a>, Chinese BBS.
%e A383654 Case a(1)=2: 13^4 = 119^2 + 120^2, 1^4 = 0^2 + 1^2.
%e A383654 Case a(3)=169: 26^4 = (-67+1)^2 + (-67+2)^2 + ... + (-67+168)^2 + (-67+169)^2.
%e A383654 Case a(5)=177: 295^4 = (6452+1)^2 + (6452+2)^2 + ... + (6452+176)^2 + (6452+177)^2.
%e A383654 ...
%e A383654 Case a(10)=163607: 5546^4 = (-22206+1)^2 + (-22206+2)^2 + ... + (-22206+163606)^2 + (-22206+163607)^2.
%t A383654 lst={};Monitor[Do[mm=6 m^4;div=TakeWhile[Divisors[mm][[2;;-2]],2mm/#+1>#^2&];
%t A383654 ans=Select[div,IntegerQ[Sqrt[(2mm/#+1-#^2)/3]]&&Mod[#-Sqrt[(2mm/#+1-#^2)/3],2]==1&];
%t A383654 If[Length[ans]>0,tmp={m,{#,q=Sqrt[(2mm/#+1-#^2)/3],p=(q+1-#)/2}&/@ans};Print[tmp];
%t A383654 AppendTo[lst,tmp]],{m,1,10^4}],m];lst
%Y A383654 Cf. A000330, A097812, A189173, A383359, A383367, A383653.
%K A383654 nonn,more
%O A383654 1,1
%A A383654 _Xianwen Wang_, May 04 2025