A053717 a(n) = n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n + 1.
1, 8, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 11111111, 21435888, 39089245, 67977560, 113522235, 183063616, 286331153, 435984840, 648232975, 943531280, 1347368421, 1891142968, 2613136835, 3559590240, 4785883225, 6357828776, 8353082583
Offset: 0
Examples
a(3) = 3280 because 11111111 base 3 = 2187+729+243+81+27+9+3+1 = 3280.
Links
- Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
- Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).
Programs
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Mathematica
Table[FromDigits["11111111",n],{n,1,30}] (* or *) Table[n^7+n^6+n^5+n^4+n^3+n^2+n+1,{n,1,60}] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)
Formula
a(n) = (n^8-1)/(n-1) for n != 1.
G.f.: 1 -x*(x^7-8*x^6-57*x^5-1016*x^4-2297*x^3-1464*x^2-191*x-8)/(x-1)^8. - Colin Barker, Oct 29 2012
E.g.f.: exp(x)*(1 + 7*x + 120*x^2 + 423*x^3 + 426*x^4 + 156*x^5 + 22*x^6 + x^7). - Stefano Spezia, Oct 03 2024
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8). - Wesley Ivan Hurt, Jun 19 2025
Extensions
a(0)=1 prepended by Alois P. Heinz, May 04 2021
Comments