A376007 For integers n>=4, greatest integer that can satisfy sqrt((n^2-c)*b^2 + c*(b+1)^2) where b and c are positive integers and c < n^2.

7, 8, 27, 19, 61, 42, 125, 83, 211, 137, 343, 204, 505, 299, 729, 428, 991, 578, 1331, 749, 1717, 964, 2197, 1229, 2731, 1523, 3375, 1846, 4081, 2229, 4913, 2678, 5815, 3164, 6859, 3687, 7981, 4286, 9261, 4967, 10627, 5693, 12167, 6464, 13801, 7327, 15625

Offset: 4

Author

Charles L. Hohn, Sep 05 2024

Keywords

• nonn

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).

Crossrefs

Cf. A375336, A376005.

Programs

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

Formula

a(n) = A375336(n, A376005(n)).

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)