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Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a square minus 1.
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%I #19 Mar 27 2016 07:57:06

%S 1,2,2,3,3,3,4,2,2,5,5,3,2,3,4,4,4,5,7,5,3,6,5,3,7,8,5,4,5,7,8,6,2,4,

%T 5,5,10,7,5,7,6,4,3,5,8,10,6,2,3,5,6,10,9,5,7,6,4,4,5,6,8,5,3,8,7,5,7,

%U 5,6,11,9,5,3,5,5,4,4,3,8,9,7,10,7,5,11,10,8,5,1

%N Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a square minus 1.

%C Conjecture: a(n) > 0 for all n > 0. In other words, for any positive integer n, either n or n + 1 can be written as the sum of a fourth power, a square and a positive triangular number.

%C We also guess that a(n) = 1 only for n = 1, 89, 244, 464, 5243, 14343.

%H Zhi-Wei Sun, <a href="/A262956/b262956.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127(2007), 103-113.

%e a(1) = 1 since 1 = 0^4 + 1*2/2 + 0^2.

%e a(89) = 1 since 89 = 2^4 + 4*5/2 + 8^2 - 1.

%e a(244) = 1 since 244 = 2^4 + 2*3/2 + 15^2.

%e a(464) = 1 since 464 = 2^4 + 22*23/2 + 14^2 - 1.

%e a(5243) = 1 since 5243 = 0^4 + 50*51/2 + 63^2 - 1.

%e a(14343) = 1 since 14343 = 2^4 + 163*164/2 + 31^2.

%t SQ[n_]:=IntegerQ[Sqrt[n]]||IntegerQ[Sqrt[n+1]]

%t Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000217, A000290, A000583, A262941, A262944, A262945, A262954, A262955, A262959.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Oct 05 2015