OFFSET
0,2
COMMENTS
Relationship with tetrahedral numbers: a(4) = (first term + last term)*(6*Tetra_n + n^3) = (82+255)*(6*10+27) = (337)*(87) = 29319.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8, -28, 56, -70, 56, -28, 8, -1).
FORMULA
a(n) = 4*n^7 + 14*n^6 + 28*n^5 + 34*n^4 + 26*n^3 + 11*n^2 + 2*n.
G.f.: (119*x +2152*x^2 +7819*x^3 +7800*x^4 +2141*x^5 +128*x^6 +x^7)/(x-1)^8. - Harvey P. Dale, Apr 23 2011
EXAMPLE
Between 3^4 and 4^4 we have 82+83+...+254+255 which is 29319 or a(4).
MAPLE
A048397:=n->4*n^7 + 14*n^6 + 28*n^5 + 34*n^4 + 26*n^3 + 11*n^2 + 2*n; seq(A048397(n), n=0..40); # Wesley Ivan Hurt, Feb 10 2014
MATHEMATICA
Table[Total[Range[n^4+1, (n+1)^4-1]], {n, 0, 40}] (* or *) Table[4n^7+ 14n^6+28n^5+34n^4+26n^3+11n^2+2n, {n, 0, 40}] (* Harvey P. Dale, Apr 23 2011 *)
LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 119, 3104, 29319, 162104, 643535, 2040744, 5502959}, 40] (* Harvey P. Dale, Jul 31 2021 *)
PROG
(PARI) a(n)=n*(2*n^2 + 3*n + 2)*(2*n^4 + 4*n^3 + 6*n^2 + 4*n + 1) \\ Charles R Greathouse IV, Jan 24 2022
(Python)
def A048397(n): return n*(n*((n<<1)+3)+2)*(n*(n*(n*((n+2)<<1)+6)+4)+1) # Chai Wah Wu, Oct 19 2024
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
Patrick De Geest, Mar 15 1999
STATUS
approved