cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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%I A264100 #22 Nov 10 2015 14:18:12
%S A264100 0,1,4,12,26,50,82,130,188,265,358,473,602,765,944,1151,1386,1657,
%T A264100 1948,2284,2642,3048,3490,3972,4480,5056,5666,6322,7022,7784,8578,
%U A264100 9452,10360,11337,12366,13453,14592,15831,17110,18453,19856,21357,22902,24551
%N A264100 Sum of the lengths of the arithmetic progressions in {1,2,3,...,n}, including trivial arithmetic progressions of lengths 1 and 2.
%C A264100 Conjecture: the second differences give A007503(n+1), the sum of the divisors (A000203) plus the number of divisors (A000005) of n+1.
%C A264100 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
%F A264100 a(n) = n + Sum_{k=2..n} k*floor((n-1)/(k-1))*(2*n-(k-1)*floor((n+k-2)/(k-1)))/2.
%e A264100 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.
%e A264100 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.
%o A264100 (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
%Y A264100 Cf. A000005, A000203, A007503, A257644.
%K A264100 nonn
%O A264100 0,3
%A A264100 _Gionata Neri_, Nov 03 2015