A374490 Greatest common divisor of sums of n consecutive cubes.
1, 1, 9, 4, 5, 9, 7, 8, 27, 5, 11, 36, 13, 7, 45, 16, 17, 27, 19, 20, 63, 11, 23, 72, 25, 13, 81, 28, 29, 45, 31, 32, 99, 17, 35, 108, 37, 19, 117, 40, 41, 63, 43, 44, 135, 23, 47, 144, 49, 25, 153, 52, 53, 81, 55, 56, 171, 29, 59, 180, 61
Offset: 1
Examples
For n=3, the sum of 3 consecutive cubes is S(x) = x^3 + (x+1)^3 + (x+2)^3 which has S(0) = 9 and thereafter remains a multiple of 9 since S(x) - S(x-1) = 9*(x^2 + x + 1), so that the GCD of all S(x) is a(3) = 9.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
- Wikipedia, Integer-valued polynomial
- Kaleb Williams, Proof that 3*n^3 + 15*n is divisible 9 by induction
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,-1).
Programs
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PARI
f(n,x='x)=n*x^3 + (3/2*n^2 - 3/2*n)*x^2 + (n^3 - 3/2*n^2 + 1/2*n)*x + (1/4*n^4 - 1/2*n^3 + 1/4*n^2) Polya(P)=my(x=variable(P),D=poldegree(P),f=D!,t=0); forstep(d=D,0,-1, my(c=polcoef(P,d,x)*d!); P-=c*binomial(x,d); t=gcd(t,c); f/=max(d,1)); t a(n)=Polya(f(n)) \\ Charles R Greathouse IV, Jul 09 2024
Formula
From Stefano Spezia, Jul 10 2024: (Start)
G.f.: x*(1 + x + 9*x^2 + 4*x^3 + 5*x^4 + 9*x^5 + 7*x^6 + 8*x^7 + 27*x^8 + 5*x^9 + 11*x^10 + 36*x^11 + 11*x^12 + 5*x^13 + 27*x^14 + 8*x^15 + 7*x^16 + 9*x^17 + 5*x^18 + 4*x^19 + 9*x^20 + x^21 + x^22)/((1 - x)^2*(1 + x)^2*(1 + x^2)^2*(1 - x + x^2)^2*(1 + x + x^2)^2*(1 - x^2 + x^4)^2).
a(n) = n/2 if n is 2 or 10 mod 12; a(n) = 3n if n is 0, 3, or 9 mod 12; a(n) = 3n/2 if n = 6 mod 12; and a(n) = n otherwise (if n is 1, 4, 5, 7, 8, or 11 mod 12). In particular, n/2 <= a(n) <= 3n. - Charles R Greathouse IV, Jul 11 2024
Extensions
a(41)-a(61) from Charles R Greathouse IV, Jul 09 2024
Comments