OFFSET
4,1
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1,0,0,0,2,0,-4,0,2,0,0,0,-1,0,2,0,-1).
FORMULA
a(n) = (n^3-4*n+8)/8 for n%4==0.
a(n) = (n^3-9*n+24)/16 for n%8==1 or n%8==7.
a(n) = (n^3)/8 for n%4==2.
a(n) = (n^3-n+8)/16 for n%8==3 or n%8==5.
From Chai Wah Wu, Sep 27 2024: (Start)
a(n) = 2*a(n-2) - a(n-4) + 2*a(n-8) - 4*a(n-10) + 2*a(n-12) - a(n-16) + 2*a(n-18) - a(n-20) for n > 23.
G.f.: x^4*(-2*x^19 - x^18 + 3*x^17 + x^16 - 2*x^15 + 2*x^14 + 4*x^13 + 2*x^12 + 7*x^11 + 20*x^10 - 3*x^9 + 8*x^8 + 18*x^7 + 30*x^6 + 12*x^5 + 14*x^4 + 3*x^3 + 13*x^2 + 8*x + 7)/((x - 1)^4*(x + 1)^4*(x^2 + 1)^2*(x^4 + 1)^2). (End)
PROG
(PARI) a(n)=my(d=n^2, t=ceil(n^3/8)); while(t>=n, my(b=floor(sqrt(t^2/d)), r=t^2-d*b^2); if (r && r%(b*2+1)==0, return(t)); t--)
for(n=4, 100, print(n, " ", a(n)))
(PARI) a(n)=if(n%4==2, 2*n^3, n%4==0, 2*n^3-8*n+16, abs(n%8-4)==1, n^3-n+8, n^3-9*n+24)/16
for(n=4, 100, print(n, " ", a(n)))
(Python)
def A376007(n):
if (m:=n&7)==0 or m==4:
return n*(n**2-4)+8>>3
elif m==1 or m==7:
return n*(n**2-9)+24>>4
elif m==2 or m==6:
return n**3>>3
else:
return n*(n**2-1)+8>>4 # Chai Wah Wu, Sep 27 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Charles L. Hohn, Sep 05 2024
STATUS
approved