A392256 a(n) is the conjectured largest number such that both a(n) and a(n) - n are 11-smooth numbers, or 0 if no such number exists. a(n) can be less than n.

9801, 19602, 29403, 39204, 49005, 58806, 68607, 78408, 88209, 98010, 107811, 117612, 151263, 137214, 147015, 156816, 5120, 176418, 43923, 196020, 205821, 215622, 3773, 235224, 245025, 302526, 264627, 274428, 19712, 294030, 41503, 313632, 323433, 10240, 343035, 352836, 117649

Offset

1,1

Comments

a(1) is just A002072(5) + 1, which is proven by Størmer's theorem. All other terms are conjectured, but the abc-conjecture can help.

For "a(n) can be less than n", the first example will be a(9007) = 9000 because there is no answer to >9007. 9000 = 2^3*3^2*5^3 and 9000-9007 = -7 = -7. Here we also consider negative numbers as smooth numbers.

There are no larger possible a(n) under 10^30 for every n.

Links

Table of n, a(n) for n=1..37.

Example

a(19) = 43923 since both 43923 and 43923 - 19 = 43904 are 11-smooth numbers, and it is conjectured that no number k larger than 43923 such that both k and k - 19 are 11-smooth numbers.

Prog

(PARI)
M(v, u, lim)={vecsort(concat(vector(#v, i, my(m=lim\v[i]); v[i]*select(t->t<=m, u))))}
Gen(lim, k)={my(v=[1]); forprime(p=2, k, v=M(v, vector(logint(lim, p)+1, e, p^(e-1)), lim)); v}
lista(n, lim=10^30)={my(v=vector(n), G=Gen(lim, 11)); for(i=1, #G, my(t=G[i], k=i+1); while(k<=#G && G[k]-t<=n, v[G[k]-t]=G[k]; k++)); v}
A006530(n)=if(n>1, vecmax(factor(n)[, 1]), 1)
b(n)=for(t=1, n, A006530(t*(n-t))<=11&return(t))
A392256(n)=if(lista(n)[n]!=0, return(lista(n)[n]); lista(n)[n]==0&b(n)<n, return(n-b(n)); lista(n)[n]==0&b(n)==n, return(0))

Crossrefs

Cf. A051038.

Sequence in context: A145209 A035911 A069333 * A222814 A156735 A227489

Adjacent sequences: A392253 A392254 A392255 * A392257 A392258 A392259

Keyword

nonn

Author

Zhicheng Wei, Jan 04 2026

Status

approved