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A254668 Number of ways to write n as the sum of a square, a second pentagonal number, and a hexagonal number. 2
1, 2, 2, 2, 2, 1, 2, 3, 3, 3, 2, 2, 3, 1, 1, 3, 5, 6, 2, 3, 1, 2, 4, 2, 4, 3, 4, 3, 3, 2, 4, 7, 4, 4, 2, 2, 4, 3, 3, 4, 3, 5, 5, 3, 6, 3, 5, 4, 2, 4, 4, 6, 5, 3, 2, 6, 5, 7, 4, 3, 2, 4, 4, 4, 7, 3, 8, 4, 5, 3, 5, 6, 8, 3, 2, 3, 4, 9, 2, 8, 3, 7, 7, 4, 5, 5, 4, 4, 4, 6, 5, 4, 6, 7, 9, 2, 8, 4, 3, 4, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 5, 13, 14, 20, 112, 125.

Compare this conjecture with the conjecture in A160324.

The conjecture that a(n) > 0 for all n = 0,1,2,... appeared in Conjecture 1.2(ii) of the author's JNT paper in the links. - Zhi-Wei Sun, Oct 03 2016

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

EXAMPLE

a(20) = 1 since 20 = 2^2 + 3*(3*3+1)/2 + 1*(2*1-1).

a(112) = 1 since 112 = 7^2 + 6*(3*6+1)/2 + 2*(2*2-1).

a(125) = 1 since 125 = 5^2 + 8*(3*8+1)/2 + 0*(2*0-1).

MATHEMATICA

SQ[n_]:=IntegerQ[Sqrt[n]]

Do[r=0; Do[If[SQ[n-y(3y+1)/2-z(2z-1)], r=r+1], {y, 0, (Sqrt[24n+1]-1)/6}, {z, 0, (Sqrt[8(n-y(3y+1)/2)+1]+1)/4}];

Print[n, " ", r]; Continue, {n, 0, 100}]

CROSSREFS

Cf. A000290, A000384, A005449, A160324, A160325, A254661.

Sequence in context: A289641 A209312 A054715 * A306250 A145443 A192056

Adjacent sequences:  A254665 A254666 A254667 * A254669 A254670 A254671

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Feb 04 2015

STATUS

approved

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Last modified May 15 07:17 EDT 2021. Contains 343909 sequences. (Running on oeis4.)