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A375249
Integers which cannot be partitioned into the sum of a hexagonal number plus a pentagonal number, nor a hexagonal number plus a square, nor a pentagonal number plus a square.
1
3, 8, 34, 43, 56, 59, 62, 68, 72, 73, 83, 89, 90, 97, 104, 110, 111, 114, 131, 138, 139, 148, 152, 163, 164, 167, 168, 186, 193, 200, 203, 205, 207, 222, 227, 228, 229, 233, 244, 249, 250, 252, 258, 269, 273, 279, 299, 300, 305, 306, 308, 309, 318, 319, 321, 333, 343, 344, 356, 364, 365
OFFSET
1,1
COMMENTS
Inspired by A160324 and analogous to A020757.
C. F. Gauss proved that all positive integers can be expressed as the sum of three triangular numbers. However, Zhi-Wei Sun (2009) has shown that there are 95 candidates for universal triples. This sequence looks at the {p4, p5, p6} triples and ask which integers require all three members to satisfy the sum.
Obviously, a(n) cannot be either a square, a pentagonal number, nor a hexagonal number.
There are more terms less than some integer in A020757 than in this sequence. In a sense, a square plus a pentagonal plus a hexagonal number is more efficient than the sum of three triangular numbers.
a(n) =~ 5.14 n^(.97).
LINKS
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.
EXAMPLE
7 is not in the sequence since the third hexagonal number 6 plus the second square or pentagonal number sum to 7;
8 is in the sequence because s = {0, 1, 4}, p = {0, 1, 5}, and h = {0, 1, 6} with no two sets having members which sum to 8.
MATHEMATICA
planeFiguratePi[r_, n_] := Floor[((r -4) +Sqrt[(r -4)^2 + 8n (r -2)])/(2 (r - 2))];
h = Table[PolygonalNumber[6, n], {n, 0, planeFiguratePi[6, 500]}];
p = Table[PolygonalNumber[5, n], {n, 0, planeFiguratePi[5, 500]}];
s = Table[PolygonalNumber[4, n], {n, 0, planeFiguratePi[4, 500]}];
Complement[ Range@ 500, Flatten[{Outer[Plus, h, p], Outer[Plus, h, s], Outer[Plus, p, s]} ]]
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Robert G. Wilson v, Aug 07 2024
STATUS
approved