OFFSET
1,2
COMMENTS
Watson showed that a(n) <= 8 for all n.
It is conjectured that a(n) <= 5 for all n.
LINKS
Lars Blomberg, Table of n, a(n) for n = 1..10000
H. E. Salzer and N. Levine, Proof that every integer <= 452,479,659 is a sum of five numbers of the form Q_x = (x^3+5x)/6, x>= 0, Math. Comp., (1968), 191-192.
N. J. A. Sloane, Transforms
G. L. Watson, Sums of eight values of a cubic polynomial, J. London Math. Soc., 27 (1952), 217-224.
MAPLE
# LAGRANGE transform of a sequence {a(n)}
# Suggested by Lagrange's theorem that at most 4 squares are needed to sum to n.
# Returns b(n) = minimal number of terms of {a} needed to sum to n for 1 <= n <= M.
# C = maximal number of terms of {a} to try to build n
# M = upper limit on n
# Internally, the initial terms of both a and b are taken to be 0, but since this is a number-theoretic function, the output starts at n=1
LAGRANGE:=proc(a, C, M)
local t1, ip, i, j, a1, a2, b, c, N1, N2, Nc;
if whattype(a) <> list then RETURN([]); fi:
# sort a, remove duplicates, include 0
t1:=sort(a);
a1:=sort(convert(convert(a, set), list));
if not member(0, a1) then a1:=[0, op(a1)]; fi;
N1:=nops(a1);
b:=Array(1..M+1, -1);
for i from 1 to N1 while a1[i]<=M do b[a1[i]+1]:=1; od;
a2:=a1; N2:=N1;
for ip from 2 to C do
c:={}:
for i from 1 to N1 while a1[i] <= M do
for j from 1 to N2 while a1[i]+a2[j] <= M do
c:={op(c), a1[i]+a2[j]};
od;
od;
c:=sort(convert(c, list));
Nc:=nops(c);
for i from 1 to Nc do
if b[c[i]+1] = -1 then b[c[i]+1]:= ip; fi;
od;
a2:=c; N2:=Nc;
od;
[seq(b[i], i=2..M+1)];
end;
Q:=[seq((m^3+5*m)/6, m=0..20)];
LAGRANGE(Q, 8, 120);
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jul 15 2011
STATUS
approved