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A308597 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c*5^d, where a,b,c,d are nonnegative integers with a <= b and d <= 1.

%I A308597 #21 Jun 10 2019 12:07:22
%S A308597 1,2,2,3,4,3,4,6,3,5,7,5,4,7,4,6,8,5,5,8,6,8,8,5,6,11,4,5,8,6,7,11,7,
%T A308597 5,8,8,6,10,7,8,11,6,7,11,5,9,13,7,5,11,7,9,10,6,5,12,7,8,10,7,10,10,
%U A308597 7,6,10,10,8,11,7,9,14,5,6,13,8,10,12,8,4,13,8,12,11,7,10,15,6,10,13,4,10,13,9,6,13,13,8,12,8,8
%N A308597 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c*5^d, where a,b,c,d are nonnegative integers with a <= b and d <= 1.
%C A308597 Recall an observation of Euler: {a*(a+1)/2 + b*(b+1)/2: a,b = 0,1,...} = {x^2 + y*(y+1): x,y = 0,1,...}.
%C A308597 Conjecture: a(n) > 0 for all n > 0. Moreover, any integer n > 1 can be written as a*(a+1)/2 + b*(b+1)/2 + 2^c*5^d, where a,b,c,d are nonnegative integers with c > 0 and d < 2.
%C A308597 We have verified this for n up to 5*10^8.
%C A308597 Since 2^(k+1) = 2^k + 2^k and 5 = 2^2 + 2^0. the above conjecture implies the conjecture in A303233.
%C A308597 On my request, _Giovanni Resta_ found that a(n) = 0 for n = 1217712376, 4371119377. Thus the conjecture fails. - _Zhi-Wei Sun_, Jun 10 2019
%H A308597 Zhi-Wei Sun, <a href="/A308597/b308597.txt">Table of n, a(n) for n = 1..10000</a>
%H A308597 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 A308597 a(1) = 1 with 1 = 0*1/2 + 0*1/2 + 2^0*5^0.
%e A308597 a(3) = 2 with 3 = 0*1/2 + 1*2/2 + 2^1*5^0 = 1*2/2 + 1*2/2 + 2^0*5^0.
%t A308597 TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
%t A308597 tab={};Do[r=0;Do[If[TQ[n-5^k*2^m-x(x+1)/2],r=r+1],{k,0,Min[1,Log[5,n]]},{m,0,Log[2,n/5^k]},{x,0,(Sqrt[4(n-5^k*2^m)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y A308597 Cf. A000079, A000217, A303233, A308566, A308584.
%K A308597 nonn
%O A308597 1,2
%A A308597 _Zhi-Wei Sun_, Jun 09 2019