A308217 a(n) is the multiplicative inverse of A001844(n) modulo A001844(n+1); where A001844 is the sequence of centered square numbers.

1, 8, 2, 23, 3, 46, 4, 77, 5, 116, 6, 163, 7, 218, 8, 281, 9, 352, 10, 431, 11, 518, 12, 613, 13, 716, 14, 827, 15, 946, 16, 1073, 17, 1208, 18, 1351, 19, 1502, 20, 1661, 21, 1828, 22, 2003, 23, 2186, 24, 2377, 25, 2576, 26, 2783, 27, 2998, 28, 3221, 29, 3452

Offset: 0

Author

Daniel Hoyt, May 15 2019

Keywords

• nonn
• easy

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

Crossrefs

Cf. A001844, A054552, A033951, A308215.

Programs

Maple

A001844:= n -> 2*n*(n+1)+1:
seq(1/A001844(n) mod A001844(n+1),n=0..100); # Robert Israel, Aug 11 2019

Mathematica

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

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)

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