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 A054552 (a number spiral arm) and (A001844-n). The gap between the lower values of A308215 and the upper values of A308217 increase by 3n; each successive gap increasing by 6.
LINKS
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.: x*(2 + 12*x + 5*x^2 + 3*x^3 + x^4 + x^5) / ((1 - x)^3*(1 + x)^3).
a(n) = (3 + (-1)^n + 2*(2+(-1)^n)*n + 2*(3+(-1)^n)*n^2) / 4 for n>0.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>6.
(End)
PROG
(Python)
import gmpy2
sos = [] # sum of squares
a=0
b=1
for i in range(50):
c = a**2 + b**2
sos.append(c)
a +=1
b +=1
ls = []
for i in range(len(sos)-1):
c = gmpy2.invert(sos[i+1], sos[i])
ls.append(int(c))
print(ls)
(PARI) f(n) = 2*n*(n+1)+1; \\ A001844
a(n) = lift(1/Mod(f(n+1), f(n))); \\ Michel Marcus, May 16 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Daniel Hoyt, May 15 2019
STATUS
approved