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From Robert Israel, Aug 11 2019: (Start)
a(n) = 1 + n/2 if n is even, since 0 < 1+n/2 < A001844(n+1) and (1+n/2)*A001844(n)-1 = (n/2)*A001844(n+1).
a(n) = n^2 + 7/2*(n+1) if n is odd, since 0 < n^2+7/2*(n+1) < A001844(n+1) and (n^2+7/2*(n+1))*A001844(n)-1 = (n^2+3*k/2+1/2)*A001844(n+1).
Colin Barker's conjectures easily follow. (End)
Robert Israel, <a href="/A308217/b308217.txt">Table of n, a(n) for n = 0..10000</a>
A001844:= n -> 2*n*(n+1)+1:
seq(1/A001844(n) mod A001844(n+1), n=0..100); # Robert Israel, Aug 11 2019
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editing
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approved
editing
proposed
The sequence explores the relationship between the terms of A001844, the sums of consecutive squares. The sequence is an interleaving of A033951 (a number spiral arm) and the natural numbers. The gap between the lower values of A308215 and the upper values of A308217 increase at by 3n; each successive gap increasing by 6.
proposed
editing
editing
proposed
The sequence explores the relationship between the terms of A001844, the sums of consecutive squares. The sequence is an interleaving of A033951 (a number spiral arm) and the natural numbers. The gap between the lower values of A308215 and the upper values of A308217 at odd n increases increase at 6n3n; each successive gap increasing by 6.
proposed
editing