A079696 Numbers one more than composite numbers.

5, 7, 9, 10, 11, 13, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29, 31, 33, 34, 35, 36, 37, 39, 40, 41, 43, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 61, 63, 64, 65, 66, 67, 69, 70, 71, 73, 75, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96

Offset: 1

Author

Vladeta Jovovic, Jan 31 2003

Keywords

• nonn
• easy

Comments

From Hieronymus Fischer, Mar 27 2014: (Start)

Numbers m such that m == 1 mod j and m > j^2 for any j > 1.

Example: m == 6 mod 10 is a term for m > 6, since m = 6 + 10k = 1 + (2k+1)*5, and m > (2k+1)^2 (for k := 1, m = 16), and m > 5^2 (for k > 1, m > 16).

A187813 and this sequence have no terms in common; this means that for each term a(n) there exists a base b such that the base-b digit sum is b.

Example: m = 1 + 3k, k > 3, is a term, since m > 3(1+3) > 3^2, thus the base-b-digit sum of (m) is = b for any b > 1 (here the base b is k+1 since 1+3k = 2(k+1) + k-1).

In general: Given a term a(n) there are p and q with p >= q > 1 such that a(n) = 1 + p*q. With b := p + 1 we get a(n) = (q-1)*b + b - (q-1), where 1 <= q-1 < b, which implies that the base-b digital sum of a(n) is = q-1 + b - (q-1) = b.

This sequence is the complement of the disjunction of A187813 with A239708. This means that a number m is a term if and only if there is a base b > 2 such that the base-b digit sum of m is b.

(End)

Links

• Hieronymus Fischer, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A072668.

Cf. A007953, A187813, A239703, A239708.

Programs

Python

from sympy import composite
def A079696(n): return composite(n)+1 # Chai Wah Wu, Mar 19 2025

Formula

a(n) = A002808(n) + 1.

A239703(a(n)) > 0. - Hieronymus Fischer, Apr 10 2014

Extensions

Edited by Charles R Greathouse IV, Mar 19 2010