OFFSET
1,3
COMMENTS
Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: If f(x) is one of the polynomials x^2, x*(5x+1), x*(5x+1)/2, x*(7x+1)/2, x*(7x+5)/2, then any positive integers n can be written as (2^a*5^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 3: Any positive integers n can be written as (2^a*7^b)^2 + c*(7c+5)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
See also A308640 for similar conjectures.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), No. 7, 1367-1396.
EXAMPLE
a(12) = 1 with 12 = (2^1*5^0)^2 + (-1)*(3*(-1)+1)/2 + 2*(3*2+1)/2.
a(22) = 1 with 22 = (2^2*5^0)^2 + (-1)*(3*(-1)+1)/2 + (-2)*(3*(-2)+1)/2.
a(50) = 1 with 50 = (2^2*5^0)^2 + (-3)*(3*(-3)+1)/2 + (-4)*(3*(-4)+1)/2.
a(330) = 1 with 330 = (2^2*5^0)^2 + (-8)*(3*(-8)+1)/2 + 12*(3*12+1)/2.
a(8650) = 1 with 8650 = (2^5*5^0)^2 + 8*(3*8+1)/2 + (-71)*(3*(-71)+1)/2.
a(29440) = 1 with 29440 = (2*5)^2 + (-80)*(3*(-80)+1)/2 + (-115)*(3*(-115)+1)/2.
a(48459) = 1 with 48459 = (2^7*5^0)^2 + 20*(3*20+1)/2 + (-145)*(3*(-145)+1)/2.
a(153035) = 1 with 153035 = (2*5^2)^2 + 35*(3*35+1)/2 + (-315)*(3*(-315)+1)/2.
a(164043) = 1 with 164043 = (2^2*5^2)^2 + (-46)*(3*(-46)+1)/2 + 317*(3*317+1)/2.
MATHEMATICA
PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={}; Do[r=0; Do[If[PenQ[n-4^a*25^b-c(3c+1)/2], r=r+1], {a, 0, Log[4, n]}, {b, 0, Log[25, n/4^a]}, {c, -Floor[(Sqrt[12(n-4^a*25^b)+1]+1)/6], (Sqrt[12(n-4^a*25^b)+1]-1)/6}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jun 13 2019
STATUS
approved