login
A266548
Least prime p such that n + p^3 = x^2 + y^3 for some positive integers x and y, or 0 if no such prime p exists.
4
71, 2, 2, 5, 2, 3767, 3, 7, 7, 2, 3, 23, 53, 13, 17, 13, 2, 3, 2, 7, 2, 23, 11, 2, 17, 2, 7, 5, 2, 2, 3, 19, 257, 8039, 13, 2, 2, 59, 3, 5, 17, 3, 2, 61, 2, 3, 3, 37, 313, 2, 631, 17, 5, 3, 17, 2, 17, 2, 7, 97, 2, 47, 3, 29, 2, 2, 31, 47, 2, 7, 19
OFFSET
0,1
COMMENTS
Conjecture: (i) Any integer can be written as x^2 + y^3 - p^3, where x and y are positive integers, and p is a prime.
(ii) Each integer can be written as x^2 - y^3 + p^3, where x and y are positive integers, and p is a prime.
See also A266230 and A266277 for related conjectures.
Is every prime in this sequence? - David A. Corneth, Dec 30 2017
LINKS
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
EXAMPLE
a(0) = 71 since 0 + 71^3 = 588^2 + 23^3 with 71 prime.
a(3) = 5 since 3 + 5^3 = 8^2 + 4^3 with 5 prime.
a(5) = 3767 since 5 + 3767^3 = 214886^2 + 1938^3 with 3767 prime.
a(2966) = 68371 since 2966 + 68371^3 = 17867983^2 + 6992^3 with 68371 prime.
a(7880) = 51137 since 7880 + 51137^3 = 10176509^2 + 31128^3 with 51137 prime.
MATHEMATICA
p[n_]:=p[n]=Prime[n]
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[x=1; Label[bb]; Do[If[SQ[n+p[x]^3-y^3], Print[n, " ", p[x]]; Goto[aa]], {y, 1, (n+p[x]^3-1)^(1/3)}]; x=x+1; Goto[bb]; Label[aa]; Continue, {n, 0, 70}]
PROG
(PARI) isokp(p, n) = {my(s = n+p^3); for (k=1, sqrtnint(s, 3), if ((q=s-k^3) && issquare(q), return (1)); ); }
a(n) = {p = 2; while(!isokp(p, n), p = nextprime(p+1)); p; } \\ Michel Marcus, Jan 04 2016
(PARI) a(n, {plim = 2}) = forprime(p = plim, oo, c = n + p^3; for(i = 1, sqrtnint(c, 3), if(issquare(c - i^3) && c - i^3 > 0, return(p))))
first(n, {plim = 100}) = {my(res = vector(n), l = List(), s, i, c);
for(u=1, sqrtint((n+plim^3)\1-1), for(v=1, sqrtnint((n+plim^3)\1-u^2, 3), listput(l, u^2+v^3))); l = Set(l); forprime(p = 2, plim, s = 1; while(l[s] < p^3 + 1, s++); for(i = s, #l, c = l[i] - p^3; if(c <= n, if(res[c] == 0, res[c] = p)
, next(2)))); for(i = 1, n, if(res[i] == 0, res[i] = a(i, plim + 1))); concat([71], res)} \\ David A. Corneth, Dec 30 2017
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 31 2015
STATUS
approved