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A246042
Numbers n such that there is both a square and a triangular number strictly between n^3 and n^3+n.
1
15774, 67880, 101340, 482434, 91710898, 268436480, 877362668, 4832460192
OFFSET
1,1
COMMENTS
Is the sequence infinite?
a(9) > 10^10. - Hiroaki Yamanouchi, Sep 03 2014
a(9) > 10^12. - Jon E. Schoenfield, Dec 23 2014
Under certain assumptions, a heuristic argument can be made that the sequence is infinite. Given a value of n, let I_n be the interval consisting of the n-1 integers lying strictly between n^3 and n^3+n. The density of squares in the vicinity of a large number C is 1/(2*sqrt(C)), so if we assume that we may treat the event that I_n includes a square as a random event, its probability of occurrence should be about P_S = (n-1)/(2*sqrt(n^3)), which approaches 1/(2*sqrt(n)) for large n. Similarly, the density of triangular numbers in the vicinity of a large number C is 1/sqrt(2*C), so if we assume that we may treat the event that I_n includes a triangular number as a random event, its probability of occurrence should be about P_T = (n-1)/sqrt(2*n^3), which approaches 1/sqrt(2*n) for large n. If we further assume that these two events can be modeled as occurring independently of each other, then the probability that I_n will include both a square and a triangular number approaches P = P_S * P_T = sqrt(1/8)/n. Since sum_{n=1..inf} sqrt(1/8)/n diverges, the sequence should have an infinite number of terms. (The expected number of D-digit terms would approach log(10)/sqrt(8) = 0.814... as D increases; as it turns out, the average number of D-digit terms for D in the interval [1,10] is 8/10 = 0.8.) - Jon E. Schoenfield, Oct 09 2014
For a simple algorithm that uses only integer arithmetic, see the Magma program link. - Jon E. Schoenfield, Oct 09 2014
LINKS
PROG
(C)
#include <stdio.h>
#include <stdlib.h>
typedef unsigned long long U64;
U64 isqrt(U64 a) {
U64 sr = 1ULL<<31;
while (a<sr*sr) sr>>=1;
U64 b = sr>>1;
while (b) {
U64 s = sr + b;
if (a>=s*s) sr = s;
b>>=1;
}
return sr;
}
int main()
{
U64 tn, t, n, cu;
for (t = n = 0; n < 2642246; ++n) {
for (cu = n*n*n; tn <= cu; ++t)
tn += t;
if (tn < cu+n && isqrt(cu) != isqrt(cu+n-1))
printf("%llu, ", n);
}
}
(PARI) isok(n) = {st = 0; ss = 0; for (i=n^3+1, n^3+n-1, if (issquare(i), ss++); if (ispolygonal(i, 3), st++)); (st == 1) && (ss == 1); } \\ Michel Marcus, Aug 24 2014
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Alex Ratushnyak, Aug 23 2014
EXTENSIONS
a(5)-a(8) from Hiroaki Yamanouchi, Sep 03 2014
STATUS
approved