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A302717
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Start with a(0) = 0, then append the terms in [x, 2*x+1, x*(x+1)] which do not occur earlier, for x = 1, 2, ...
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1
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0, 1, 3, 2, 5, 6, 7, 12, 4, 9, 20, 11, 30, 13, 42, 15, 56, 8, 17, 72, 19, 90, 10, 21, 110, 23, 132, 25, 156, 27, 182, 14, 29, 210, 31, 240, 16, 33, 272, 35, 306, 18, 37, 342, 39, 380, 41, 420, 43, 462, 22, 45, 506, 47, 552, 24, 49, 600, 51, 650, 26, 53, 702, 55, 756, 28, 57, 812, 59, 870, 61, 930, 63, 992, 32, 65, 1056, 67, 1122, 34, 69, 1190
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OFFSET
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0,3
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COMMENTS
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A permutation of the nonnegative integers.
If a(n) is in A024701 (i.e., of the form (prime^2-1)/4), then a(n-1) is prime. Indeed, A024701(m) = k*(k+1) with k = (prime(m+1)-1)/2, and any term k*(k+1) > 0 is preceded by 2*k+1 = prime(m+1). [Edited and proof added by M. F. Hasler, Apr 13 2018]
The term x*(x+1) will always be appended since it is larger than all preceding terms (except for x = 1), and also 2*x+1 cannot occur earlier because it is odd while x*(x+1) is always even. So only the term x will be inserted (or not) in a somewhat irregular pattern, namely whenever x is an even but not oblong number (A002378). We see that this is the case for x = 4, 8, 10, 14, 16, 18, 22, ...; recognizable by the fact that a(n) = (a(n+1)-1)/2 and equivalently, there are two and not only one smaller number between two larger "records" x*(x+1).
If we count the terms added from each 4-tuple during each iteration we find that either two or three terms are added: 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, ... where the set of three twos (2, 2, 2) appears with decreasing frequency.
A302906 is the sequence of starting indices of these sets.
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LINKS
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EXAMPLE
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Repeatedly take consecutive numbers a and b and append to the sequence any of {a, a+b, a*b, b} not already in the sequence. Beginning with a=0 and b=1:
(0,1) -> {0, 0+1, 0*1, 1} -> [0,1]
(1,2) -> {1, 1+2, 1*2, 2} -> [0,1,3,2]
(2,3) -> {2, 2+3, 2*3, 3} -> [0,1,3,2,5,6]
(3,4) -> {3, 3+4, 3*4, 4} -> [0,1,3,2,5,6,7,12,4]
etc.
In the above construction, we always have b = a+1. Thus [a, a+b, a*b, b] = [a, 2*a+1, a*(a+1), a+1], and a simpler description is to consider only { a, 2*a+1, a*(a+1) }, the 4th term being equal to the 1st term of the next 4-tuple. To ensure we have a permutation of the integers >= 0 starting at index 0 and not a list stating at index 1, we can fix a(0) = 0 explicitly and then go on with a = x = 1, 2, 3, ... to get the same sequence.
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PROG
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(PARI) u=[]; (do(x)=setsearch(u, x)||print1(x", ")||u=setunion(u, [x])); for(a=0, 199, do(a); do(2*a+1); do(a^2+a)) \\ M. F. Hasler, Apr 12 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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