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A316156
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The lexicographically earliest increasing sequence such that a(n) divides the sum of the first a(n)+1 terms.
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4
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1, 2, 3, 6, 7, 8, 9, 13, 15, 17, 18, 19, 20, 31, 32, 39, 40, 43, 55, 59, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 86, 105, 106, 107, 108, 109, 110, 111, 118, 135, 136, 137, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 184, 185, 186, 187, 195
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OFFSET
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1,2
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COMMENTS
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Sequence b(n) of the sums of the first a(n)+1 terms of a(n) = Sum_{k=1..a(n)+1} a(k): 3, 6, 12, 36, 49, 64, 81, 169, 240, 323, 378, 437, 500, 1271, 1376, 2145, 2280, 2709, 4675, 5428, ... = A318872(1+a(n)).
Sequence c(n) of quotients when a(n) is calculated = (Sum_{k=1..a(n)+1} a(k) ) / a(n): 3, 3, 4, 6, 7, 8, 9, 13, 16, 19, 21, 23, 25, 41, 43, 55, 57, 63, 85, 92, ...
Is there a lexicographically earliest bijective sequence such that a(n) divides the sum of the first a(n)+1 terms?
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LINKS
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FORMULA
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a(1) = 1; for n > 1, if n-1 is not in the sequence, a(n) = a(n-1)+1, otherwise, a(n) is the least k > a(n-1) such that A318872(n-1)+k is a multiple of n-1. - Antti Karttunen, Sep 16 2018
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EXAMPLE
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a(1) = 1 because 1 divides the sum of the first 2 (i.e., a(1) + 1) terms (a(1) + a(2)) for whatever term a(2) > a(1).
a(2) = 2 because 2 is the smallest number > a(1) and 2 divides the sum of the first 3 (i.e., a(2) + 1) terms (a(1) + a(2) + a(3)) for whatever term a(3) > a(2) such that 2 divides the sum a(1) + a(2) + a(3); the smallest number > a(2) with this property for a(3) is 3.
a(3) = 3.
a(4) = 6 because 6 is the smallest number > a(3) such that term a(3) = 3 divides the sum of the first 4 (i.e., a(3) + 1) terms.
a(5) = 7 and a(6) = 8 because a(4) < a(5) < a(6) and 6 divides sum of the first 7 (i.e., a(4) + 1) terms (a(1) + a(2) + ... + a(7)) for whatever term a(7) > a(6) such that 6 divides the sum a(1) + a(2) + ... + a(7); the smallest number with this property for a(7) is 9.
a(7) = 9.
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PROG
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(PARI)
povisin(v, n) = { forstep(j=n, 1, -1, if(v[j] == n, return(j))); (0); }; \\ Here: povisin = position_of_n_in_strictly_increasing_v
A316156list(up_to) = { my(v316156 = vector(up_to), v318872 = vector(up_to), k, s); v316156[1] = v318872[1] = 1; for(n=2, up_to, k = 1+v316156[n-1]; if(povisin(v316156, n-1), s = v318872[n-1]; while((s+k)%(n-1), k++)); v316156[n] = k; v318872[n] = v318872[n-1] + v316156[n]); (v316156); }; \\ Antti Karttunen, Sep 16 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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