%I #68 Aug 12 2019 02:24:50

%S 1,8,2,23,3,46,4,77,5,116,6,163,7,218,8,281,9,352,10,431,11,518,12,

%T 613,13,716,14,827,15,946,16,1073,17,1208,18,1351,19,1502,20,1661,21,

%U 1828,22,2003,23,2186,24,2377,25,2576,26,2783,27,2998,28,3221,29,3452

%N a(n) is the multiplicative inverse of A001844(n) modulo A001844(n+1); where A001844 is the sequence of centered square numbers.

%C 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.

%H Robert Israel, <a href="/A308217/b308217.txt">Table of n, a(n) for n = 0..10000</a>

%H Daniel Hoyt, <a href="/A308217/a308217_3.png">Graph of A308215 and A308217 in relation to A001844</a>

%F a(n) satisfies a(n)*(2*n*(n-1)+1) == 1 (mod 2*n*(n+1)+1).

%F Conjectures from _Colin Barker_, May 16 2019: (Start)

%F G.f.: (1 + 8*x - x^2 - x^3 + x^5) / ((1 - x)^3*(1 + x)^3).

%F a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>5.

%F a(n) = (9 - 5*(-1)^n + (8-6*(-1)^n)*n - 2*(-1+(-1)^n)*n^2) / 4.

%F (End)

%F From _Robert Israel_, Aug 11 2019: (Start)

%F 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).

%F 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).

%F Colin Barker's conjectures easily follow. (End)

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

%p seq(1/A001844(n) mod A001844(n+1),n=0..100); # _Robert Israel_, Aug 11 2019

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

%o a(n) = lift(1/Mod(f(n), f(n+1))); \\ _Michel Marcus_, May 16 2019

%Y Cf. A001844, A054552, A033951, A308215.

%K nonn

%O 0,2

%A _Daniel Hoyt_, May 15 2019