OFFSET
0,2
COMMENTS
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 by 3n; each successive gap increasing by 6.
LINKS
Robert Israel, Table of n, a(n) for n = 0..10000
Daniel Hoyt, Graph of A308215 and A308217 in relation to A001844
FORMULA
a(n) satisfies a(n)*(2*n*(n-1)+1) == 1 (mod 2*n*(n+1)+1).
Conjectures from Colin Barker, May 16 2019: (Start)
G.f.: (1 + 8*x - x^2 - x^3 + x^5) / ((1 - x)^3*(1 + x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>5.
a(n) = (9 - 5*(-1)^n + (8-6*(-1)^n)*n - 2*(-1+(-1)^n)*n^2) / 4.
(End)
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)
MAPLE
A001844:= n -> 2*n*(n+1)+1:
PROG
(PARI) f(n) = 2*n*(n+1)+1; \\ A001844
a(n) = lift(1/Mod(f(n), f(n+1))); \\ Michel Marcus, May 16 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Daniel Hoyt, May 15 2019
STATUS
approved