A086484 Let k be the largest number such that n is a k-th power; then a(n) is the least positive number m such that m + n is a (k+1)st power.

1, 2, 1, 4, 4, 3, 2, 8, 18, 6, 5, 4, 3, 2, 1, 16, 8, 7, 6, 5, 4, 3, 2, 1, 2, 10, 54, 8, 7, 6, 5, 32, 3, 2, 1, 28, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 64, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 162, 18, 17, 16, 15, 14, 13, 12, 11, 10

Offset

1,2

Links

David A. Corneth, Table of n, a(n) for n = 1..10001

Example

From David A. Corneth, Sep 24 2023: (Start)
For n = 3, k = 1 as 3 = 3^1 is a 1st power (the gcd of the exponents is 1). Hence we look for m such that m + 3 is a (1+1)th power, i.e., a 2nd power. The next 2nd power after 3 is 4 and so m + 3 = 4 so a(n) = m = 4-3 = 1.
For n = 36, k = 2 as 36 = 2^2 * 3^2 is a 2nd power (the gcd of the exponents is 2). Hence we look for m such that m + 36 is a (2+1)th power, i.e., a 3rd power. The next 3rd power after 36 is 64 and so m + 36 = 64 so a(n) = m = 64 - 36 = 28. (End)

Prog

(PARI) print1(1", "); for (n = 2, 100, f = factor(n); g = f[1, 2]; for (i = 2, matsize(f)[1], g = gcd(g, f[i, 2])); x = sqrtn(n, g+1); print1(round(ceil(x))^(g + 1) - n, ", ")); \\ David Wasserman, Mar 07 2005
(PARI) a(n) = {my(f = factor(n), k = gcd(f[, 2])); (sqrtnint(n, k+1) + 1)^(k + 1) - n} \\ David A. Corneth, Sep 24 2023

Crossrefs

Cf. A052409.

Sequence in context: A095830 A193915 A101621 * A349572 A349571 A091335

Adjacent sequences: A086481 A086482 A086483 * A086485 A086486 A086487

Keyword

nonn

Author

Amarnath Murthy, Jul 26 2003

Extensions

More terms from David Wasserman, Mar 07 2005

Offset corrected by David A. Corneth, Sep 24 2023

Status

approved