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A306382
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Positive integers not representable as Pen(x) + Pen(y) + 2*Pen(z), where x, y, z are nonnegative integers, and Pen(k) denotes the pentagonal number k*(3k-1)/2.
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3
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9, 18, 21, 28, 31, 39, 43, 55, 69, 74, 89, 90, 98, 109, 111, 113, 134, 135, 144, 193, 202, 214, 230, 243, 260, 265, 273, 275, 310, 510, 553, 698, 699, 749, 773, 780, 865, 878, 945, 965, 1219, 1398, 1413, 2153, 2168, 2335, 2828, 3178, 3793
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OFFSET
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1,1
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COMMENTS
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Conjecture 1: This sequence only has 49 terms as listed.
Conjecture 2: Any integer n > 6471 can be written as Pen(x) + Pen(y) + 3*Pen(z) with x,y,z nonnegative integers. Any integer n > 7727 can be written as Pen(x) + 2*Pen(y) + 2*Pen(z) with x,y,z nonnegative integers. Also, any integer n > 4451 can be written as Pen(x) + 2*Pen(y) + 4*Pen(z) with x,y,z nonnegative integers.
Conjecture 3: Let N(1) = 5928, N(4) = 2761, N(8) = 8868 and N(11) = 9929. For each c among 1, 4, 8, 11, any integer n > N(c) can be written as x*(3x+1)/2 + y*(3y+1) + c*z*(3z+1)/2 with x,y,z nonnegative integers.
Conjecture 4: Any integer n > 5544 can be written as x*(3x+1)/2 + y*(3y+1)/2 + 3z*(3z+1)/2 with x,y,z nonnegative integers. Any integer n > 7093 can be written as x*(3x+1)/2 + 3*y*(3y+1)/2 + 2z*(3z+1) with x,y,z nonnegative integers. Also, any integer n > 8181 can be written as x*(3x+1)/2 + 2y*(3y+1) + 3z*(3z+1) with x,y,z nonnegative integers.
Conjecture 5: For each positive integer m, there are only finitely many positive integers not representable as x*(x+1)/2 + y*(y+1)/2 + z*(z+1)/2 with x,y,z in the set {m, m+1, ...}.
See also A306383 for similar conjectures.
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LINKS
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EXAMPLE
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a(1) = 9 since the set {Pen(x) + Pen(y) + 2*Pen(z): x,y,z = 0,1,2,...} contains 1..8 but it does not contain 9.
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MATHEMATICA
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PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1, 6]==0);
tab={}; Do[Do[If[PenQ[m-x(3x-1)-y(3y-1)/2], Goto[aa]], {x, 0, (Sqrt[12m+1]+1)/6}, {y, 0, (Sqrt[12(m-x(3x-1))+1]+1)/6}]; tab=Append[tab, m]; Label[aa], {m, 1, 5000}]; Print[tab]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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