

A270928


Number of ways to write n = x*(x1)/2 + y*(y1)/2 + z*(z1)/2, where 0 < x <= y <= z, and one of x, y, z is prime.


2



1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 3, 2, 1, 2, 1, 3, 3, 2, 2, 2, 2, 2, 2, 1, 3, 4, 2, 2, 2, 2, 1, 4, 2, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 3, 1, 2, 5, 1, 2, 3, 3, 4, 3, 3, 1, 5, 1, 3, 2, 3, 3, 5, 2, 2, 4
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OFFSET

1,3


COMMENTS

Conjecture: (i) a(n) > 0 for n > 0 with the only exception n = 15^2 = 225. Also, a(n) = 1 only for n = 1, 2, 4, 5, 6, 8, 11, 14, 15, 18, 20, 29, 36, 50, 53, 60, 62, 96, 117, 119, 218, 411, 540, 645, 1125, 1590, 2346, 4068.
(ii) Any positive integer can be written as p*(p1)/2 + x*(x1)/2 + P(y) with p prime and x and y integers, where the polynomial P(y) is either of the following ones: y*(y1), y*(3*y+1)/2, y*(5*y+j)/2 (j = 1,3), y*(3*y+j) (j = 1,2), y*(7*y+3)/2, y*(9*y+j)/2 (j = 1,5,7), y*(5*y+j) (j = 1,3), y*(11*y+9)/2, 2*y*(3*y+j) (j = 1,2), y*(7*y+3).
(iii) Any positive integer can be written as p*(p1)/2 + P(x,y) with p prime and x and y integers, where the polynomial P(x,y) is either of the following ones: a*x*(x1)/2+y*(3*y+1)/2 (a = 2,3,4), x*(x1)+y*(5*y+3)/2, b*x^2+y*(3*y+1)/2 (b = 1,2,3), x^2+y*(5*y+j)/2 (j = 1,3), x^2+y*(3*y+1), x^2+y*(7*y+j)/2 (j = 1,3,5), x^2+y*(4*y+1).
(iv) Every positive integer can be written as p*(p1)/2+x*(3*x+1)/2+y*(3*y+1)/2 with p prime, x an nonnegative integer and y an integer. Also, for each r = 1,3, any positive integer n can be written as p*(p1)/2+x*(3*x1)/2+y*(5*y+r)/2, where p is a prime, and x and y are integers with x nonnegative.
Note that Gauss proved a classical assertion of Fermat which states that any natural number is the sum of three triangular numbers.
See also A270966 for a similar conjecture involving (p1)^2 with p prime.
The conjecture that a(n) > 0 except for n = 225 appeared as Conjecture 1.2(i) of the author's JNT paper in the links.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
Z.W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103113.
Z.W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 13671396.
Z.W. 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), 275283.


EXAMPLE

a(1) = 1 since 1 = 1*(11)/2 + 1*(11)/2 + 2*(21)/2 with 2 prime.
a(4) = 1 since 4 = 1*(11)/2 + 2*(21)/2 + 3*(31)/2 with 2 and 3 prime.
a(29) = 1 since 29 = 1*(11)/2 + 2*(21)/2 + 8*(81)/2 with 2 prime.
a(50) = 1 since 50 = 2*(21)/2 + 7*(71)/2 + 8*(81)/2 with 2 and 7 prime.
a(119) = 1 since 119 = 8*(81)/2 + 9*(91)/2 + 11*(111)/2 with 11 prime.
a(411) = 1 since 411 = 16*(161)/2 + 16*(161)/2 + 19*(191)/2 with 19 prime.
a(1125) = 1 since 1125 = 3*(31)/2 + 34*(341)/2 + 34*(341)/2 with 3 prime.
a(1590) = 1 since 1590 = 7*(71)/2 + 37*(371)/2 + 43*(431)/2 with 7, 37 and 43 prime.
a(2346) = 1 since 2346 = 6*(61)/2 + 16*(161)/2 + 67*(671)/2 with 67 prime.
a(4068) = 1 since 4068 = 7*(71)/2 + 34*(341)/2 + 84*(841)/2 with 7 prime.


MATHEMATICA

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[nx(x1)/2y(y1)/2]&&(PrimeQ[x]PrimeQ[y]PrimeQ[(Sqrt[8(nx(x1)/2y(y1)/2)+1]+1)/2]), r=r+1], {x, 1, (Sqrt[8n/3+1]+1)/2}, {y, x, (Sqrt[8(nx(x1)/2)/2+1]+1)/2}]; Print[n, " ", r]; Continue, {n, 1, 70}]


CROSSREFS

Cf. A000040, A000217, A000290, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270966.
Sequence in context: A054868 A065081 A212185 * A290257 A196942 A184304
Adjacent sequences: A270925 A270926 A270927 * A270929 A270930 A270931


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 26 2016


STATUS

approved



