A181475 a(n) = 3*n^4 + 6*n^3 - 3*n + 1.

1, 7, 91, 397, 1141, 2611, 5167, 9241, 15337, 24031, 35971, 51877, 72541, 98827, 131671, 172081, 221137, 279991, 349867, 432061, 527941, 638947, 766591, 912457, 1078201, 1265551, 1476307, 1712341, 1975597, 2268091, 2591911, 2949217, 3342241, 3773287, 4244731

Offset: 0

Author

Bruno Berselli, Oct 25 2010 - Oct 29 2010

Keywords

• nonn
• easy

Comments

If gcd(n,7) = gcd(n+1,7) = gcd(2*n+1,7) = 1 then a(n) == 0 (mod 7) (E. Picutti, see References).

References

• Ettore Picutti, Sul numero e la sua storia, Feltrinelli Economica, 1977, p. 208.

Links

• Bruno Berselli, Table of n, a(n) for n = 0..10000.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Subsequence of A003215.

Programs

Magma

[3*n^4+6*n^3-3*n+1: n in [0..31]];

Mathematica

Table[3 n^4 + 6 n^3 - 3 n + 1, {n, 0, 40}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,7,91,397,1141},40] (* Harvey P. Dale, Jul 12 2022 *)

Formula

G.f.: (1 + 2*x + 66*x^2 + 2*x^3 + x^4)/(1-x)^5.

a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 6*12.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6*A008594(n-1).

a(n) = 2*a(n-1) - a(n-2) + 6*A003154(n).

a(n) = a(n-1) + 6*A007588(n).

a(n) = 1 + 6*A062392(n).

a(n) = 7*A000540(n)/A000330(n) = A154105(A000096(n-1)) for n > 0.

Sum_{i=0..n} a(i) = (3*n^5 + 15*n^4 + 20*n^3 - 3*n + 5)/5.

a(n) = 7*(3*n^2 + 3*n - 1)*(Sum_{k=1..n} k^6)/(5*Sum_{k=1..n} k^4), n > 0. - Gary Detlefs, Oct 18 2011

Extensions

Formula, program and crossref added by Bruno Berselli, Aug 22 2011