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.

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, 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, 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

Offset

1,2

Comments

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,...}.

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.

We have verified this for n up to 5*10^8.

Since 2^(k+1) = 2^k + 2^k and 5 = 2^2 + 2^0. the above conjecture implies the conjecture in A303233.

On my request, Giovanni Resta found that a(n) = 0 for n = 1217712376, 4371119377. Thus the conjecture fails. - Zhi-Wei Sun, Jun 10 2019

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Example

a(1) = 1 with 1 = 0*1/2 + 0*1/2 + 2^0*5^0.
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.

Mathematica

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
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]

Crossrefs

Cf. A000079, A000217, A303233, A308566, A308584.

Sequence in context: A357589 A336200 A032355 * A205153 A300302 A091257

Adjacent sequences: A308594 A308595 A308596 * A308598 A308599 A308600

Keyword

nonn

Author

Zhi-Wei Sun, Jun 09 2019

Status

approved