A264100 Sum of the lengths of the arithmetic progressions in {1,2,3,...,n}, including trivial arithmetic progressions of lengths 1 and 2.

0, 1, 4, 12, 26, 50, 82, 130, 188, 265, 358, 473, 602, 765, 944, 1151, 1386, 1657, 1948, 2284, 2642, 3048, 3490, 3972, 4480, 5056, 5666, 6322, 7022, 7784, 8578, 9452, 10360, 11337, 12366, 13453, 14592, 15831, 17110, 18453, 19856, 21357, 22902, 24551

Offset: 0

Author

Gionata Neri, Nov 03 2015

Keywords

• nonn

Comments

Conjecture: the second differences give A007503(n+1), the sum of the divisors (A000203) plus the number of divisors (A000005) of n+1.

The first differences trivially are the total length of such sequences that end in n+1. Mapping each sequence to a different sequence by adding 1 to each term, we see that the second differences are the number of sequences up to n+2 that include both 1 and n+2. For each divisor d of n+1, there is a single such sequence of length d+1 (with increment (n+1)/d). The second difference is then sum_{d|n+1} d+1, which is sigma(n+1) + tau(n+1), as claimed. - Franklin T. Adams-Watters, Nov 05 2015

Examples

			For n = 3 the arithmetic progressions are (1), (2), (3), (1, 2), (1, 3), (2, 3), (1, 2, 3) and the respective lengths are (1), (1), (1), (2), (2), (2), (3), so a(3) = 1 + 1 + 1 + 2 + 2 + 2 + 3 = 12.
The first difference at 2, sequences ending with 3, are (3), (1, 3), (2, 3), and (1, 2, 3), total length 8 = 12-4. The second difference at 2, sequences starting with 1 and ending with 4 are (1, 4) and (1, 2, 3, 4), total length 6 = 26 - 2*12 +4.
		

Crossrefs

Cf. A000005, A000203, A007503, A257644.

Programs

PARI

vector(50, n, n--; n + sum(k=2, n, k*floor((n-1)/(k-1))*(2*n-(k-1)*floor((n+k-2)/(k-1)))/2)) \\ Altug Alkan, Nov 04 2015

Formula

a(n) = n + Sum_{k=2..n} k*floor((n-1)/(k-1))*(2*n-(k-1)*floor((n+k-2)/(k-1)))/2.