A079772 Let C(n) be the n-th composite number; then a(n) is the smallest number > C(n) and not coprime to C(n).

6, 8, 10, 12, 12, 14, 16, 18, 18, 20, 22, 24, 24, 26, 30, 28, 30, 30, 32, 34, 36, 36, 40, 38, 40, 42, 42, 44, 46, 48, 48, 50, 56, 52, 54, 54, 56, 60, 58, 60, 60, 62, 64, 66, 66, 70, 68, 70, 72, 72, 74, 76, 78, 78, 84, 80, 82, 84, 84, 86, 90, 88, 90, 90, 92, 98, 94, 96, 96, 100

Offset

1,1

Comments

For n >= 1, if p = A020639(A002808(n)) = A056608(n), then a(n) = A002808(n) + p satisfies a(n) - p = A002808(n) and p divides a(n). In contrast with A085271, the first term a(1) = 6 is already composite. It corresponds to A002808(1) = 4 and satisfies 6 - 2 = 4. - Charles Kusniec, Jul 03 2026

Links

Table of n, a(n) for n=1..70.

Formula

a(n) = C(n) + smallest prime divisor of C(n).

a(n) = A002808(n) + A056608(n). - Vladeta Jovovic, Jan 31 2003

n + n/log n + n/log^2 n < a(n) < n + n/log n + 4n/log^2 n for n >= 4. - Charles R Greathouse IV, Jul 05 2026

Example

C(4) = 9 and C(5) = 10 hence a(4) = a(5) = 12.
For 91=7*13 we have 91+7 = 98; for 92=2*2*23 we have 92+2 = 94.
Since A002808(15) = 25 and A020639(25) = 5, we have a(15) = 25 + 5 = 30. Thus, a(15) - 5 = 25, and 5 divides both 25 and a(15). Here, 30 = 5*(5 + 1) is the upper oblong adjacent to 25 = 5^2.

Prog

(PARI) cminusp31(n) = { for(x=2, n, forprime(p=2, floor(sqrt(x)), if(x%p==0 & isprime(x)==0, print1(x+p, ", "); break); ) ) }

Crossrefs

Cf. A002808, A056608.

Cf. A020639, A036690, A085271 (difference between the n-th composite number and its smallest prime divisor).

Sequence in context: A171562 A018221 A071278 * A080731 A080257 A331201

Adjacent sequences: A079769 A079770 A079771 * A079773 A079774 A079775

Keyword

nonn

Author

Amarnath Murthy, Jan 31 2003

Extensions

More terms from Cino Hilliard, Aug 12 2003

Status

approved