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A051336
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Number of arithmetic progressions in {1,2,3,...,n}, including trivial arithmetic progressions of lengths 1 and 2.
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4
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1, 3, 7, 13, 22, 33, 48, 65, 86, 110, 138, 168, 204, 242, 284, 330, 381, 434, 493, 554, 621, 692, 767, 844, 929, 1017, 1109, 1205, 1307, 1411, 1523, 1637, 1757, 1881, 2009, 2141, 2282, 2425, 2572, 2723, 2882, 3043, 3212, 3383, 3560, 3743, 3930, 4119
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OFFSET
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1,2
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COMMENTS
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The number of arithmetic subsequences of [1, ..., n] with successive-term increment i and length k is (n-i*(k-1))(i > 0, k > 0, n > i*(k-1)). - Robert E. Sawyer (rs.1(AT)mindspring.com)
The best known algorithm to generate a(n) from scratch is O(sqrt(n)) (see below). If a(n-1) is known, it reduces to O(n^(1/3)). - Daniel Hoying, May 20 2020
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LINKS
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FORMULA
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Theorem: the second differences give tau(n+1), the number of divisors of n+1 (A000005).
a(n) = n + Sum_{ i=1..n-1, j=1..floor(n/i) } (n - i*j). - Robert E. Sawyer (rs.1(AT)mindspring.com)
a(n+1) = a(n) + 1 + Sum_{i=1..n} tau(i).
a(n+1) = (n + 1)*(1 + Sum_{i=1..n} floor(n/i)) - Sum_{i=1..n} i*tau(i).
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EXAMPLE
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a(1): [1];
a(2): [1],[2],[1,2];
a(3): [1],[2],[3],[1,2],[1,3],[2,3],[1,2,3].
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MATHEMATICA
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nmax = 48; t = Table[ DivisorSigma[0, n], {n, 1, nmax}]; Accumulate[ Accumulate[t]+1] - Accumulate[t] (* Jean-François Alcover, Nov 08 2011 *)
With[{c=Accumulate[DivisorSigma[0, Range[50]]]}, Accumulate[c+1]-c] (* Harvey P. Dale, Dec 23 2015 *)
nmax = 50; RecurrenceTable[{a[n] == a[n-1]+1+p[n], p[n] == p[n-1]+DivisorSigma[0, n-1], a[1] == 1, p[1] == 0}, {a, p}, {n, 1, nmax}][[All, 1]] (* Daniel Hoying, May 16 2020 *)
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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