0,2

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.

Robert Israel, Table of n, a(n) for n = 0..10000

Daniel Hoyt, Graph of A308215 and A308217 in relation to A001844

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

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)

E.g.f.: ((2 + 9*x)*cosh(x) + (7 + x + 2*x^2)*sinh(x))/2. - Stefano Spezia, Dec 06 2024

A001844:= n -> 2*n*(n+1)+1:

seq(1/A001844(n) mod A001844(n+1), n=0..100); # Robert Israel, Aug 11 2019

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 8, 2, 23, 3, 46}, 30] (* Georg Fischer, Dec 06 2024 *)

(PARI) f(n) = 2*n*(n+1)+1; \\ A001844

a(n) = lift(1/Mod(f(n), f(n+1))); \\ Michel Marcus, May 16 2019

Cf. A001844, A054552, A033951, A308215.

Sequence in context: A161593 A008866 A180728 * A248297 A340008 A006708

Adjacent sequences: A308214 A308215 A308216 * A308218 A308219 A308220

nonn,easy

Daniel Hoyt, May 15 2019

approved