

A267826


Numbers not of the form w^3 + 2*x^3 + 3*y^3 + 4*z^3, where w, x, y and z are nonnegative integers.


3



18, 22, 39, 60, 63, 74, 76, 77, 100, 103, 106, 107, 117, 126, 178, 180, 201, 215, 228, 230, 245, 271, 289, 291, 295, 315, 341, 356, 357, 393, 413, 419, 420, 480, 481, 523, 559, 606, 616, 671, 673, 705, 854, 855, 963, 980, 981, 998, 1103, 1121, 1130, 1298, 1484, 1510, 1643, 1729, 1849, 1916, 1934, 1946
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

Conjecture: The sequence has exactly 122 terms the last of which is a(122) = 41405.
We have verified that there are no terms between 41406 and 2*10^5.
The conjecture implies that {P(v)+w^3+2*x^3+3*y^3+4*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(v) is among the polynomials a*v^3 (a = 1,5,6,7,9,10,12,15,18), b*v^4 (b = 1,2,3,5,6,12,18), c*v^5 (c = 1,2,5,12) and d*v^k (d = 5,12; k = 6,7). Moreover, it also implies that {8*t+w^3+2*x^3+3*y^3+4*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}. If a,b,c,d and m are positive integers with {m*t+a*w^3+b*x^3+c*y^3+d*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}, then we must have m = 8 and {a,b,c,d} = {1,2,3,4}.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..122
ZhiWei Sun, Universal sums u^3+a*v^3+b*x^3+c*y^3+d*z^3 with u, v, x, y, z nonnegative integers, a message to Number Theory Mailing List, April 3, 2016.


EXAMPLE

a(1) = 18 since it is the first nonnegative integer not in the set {w^3 + 2*x^3 + 3*y^3 + 4*z^3: w,x,y,z = 0,1,2,...}.


MATHEMATICA

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
n=0; Do[Do[If[CQ[m4*z^33y^32x^3], Goto[aa]], {z, 0, (m/4)^(1/3)}, {y, 0, ((m4z^3)/3)^(1/3)}, {x, 0, ((m4z^33y^3)/2)^(1/3)}]; n=n+1; Print[n, " ", m]; Label[aa]; Continue, {m, 0, 1946}]


CROSSREFS

Cf. A000578, A002804, A022566, A267861, A271099, A271169, A271237.
Sequence in context: A250737 A290172 A031407 * A339473 A002505 A182438
Adjacent sequences: A267823 A267824 A267825 * A267827 A267828 A267829


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 07 2016


STATUS

approved



