A391077 a(n) is the conjectured largest number such that both a(n) and a(n) - n are 5-smooth numbers, or 0 if no such number exists.

81, 162, 243, 324, 405, 486, 250, 648, 729, 810, 75, 972, 256, 500, 1215, 1296, 125, 1458, 144, 1620, 750, 150, 2048, 1944, 2025, 512, 2187, 1000, 125, 2430, 256, 2592, 225, 250, 1250, 2916, 32805, 288, 768, 3240, 81, 1500, 243, 300, 3645, 4096, 128, 3888, 625

Offset

1,1

Comments

a(1) is just A002072(3) + 1, which is proven by Størmer's theorem. This provides a(n) is not zero where n is 5-smooth. All other terms are conjectured, but the abc-conjecture can help.

a(n) can be less than n and this first occurs at a(281) = 256 because there is no answer to >281. 256 = 2^8 and 256-281 = -25 = -(5^2). 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..49.

Formula

If a(n) or a(p*n) exists then a(p*n) >= p * a(n) for p in {2, 3, 5}. - David A. Corneth, Jan 13 2026

Example

a(31) = 256 since both 256 and 256 - 31 = 225 are 5-smooth numbers, and it is conjectured that no number k larger than 256 such that both k and k - 31 are 5-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, 5)); 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))<=5&return(t))
A391077(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. A051037, A219794, A392256.

Sequence in context: A206065 A139705 A226643 * A353456 A044251 A043454

Adjacent sequences: A391074 A391075 A391076 * A391078 A391079 A391080

Keyword

nonn

Author

Zhicheng Wei, Jan 04 2026

Status

approved