OFFSET
1,1
COMMENTS
Numbers with ordered partitions that have periods of length 6.
From each ordered partition of the numbers (15+j) with 0<j<6 one remove the first part z(1) and add 1 to the next z(1) parts to get a new partition until a period is reached.
The a(n) sequence begins with 16 and each member has 1 period; the b(n) sequence begins with 17 and each member has 2 periods; the c(n) sequence begins with 18 and each member has 3 periods; the d(n) sequence begins with 19 and each member has 2 periods; the e(n) sequence begins with 20 and each member has 1 period of length 6
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f. for a(n): (16 + 20*x)/(1-x)^3.
for b(n): (17 + 19*x)/(1-x)^3.
for c(n): (18 + 18*x)/(1-x)^3.
for d(n): (19 + 17*x)/(1-x)^3.
for e(n): (20 + 16*x)/(1-x)^3.
all sequences have the same recurrence:
s(n+3) = 3*s(n+2) - 3*s(n+1) + s(n);
with s(0) = 0, s(1) = 15 + j, s(2) = 66 + 2*j and 0<j<6.
a(n) = n*(18*n-2) = 4*A022266(n).
b(n) = n*(18*n-1) = a(n) + n.
c(n) = 18*n^2 = a(n) + 2*n.
d(n) = n*(18*n+1) = a(n) + 3*n.
e(n) = n*(18*n+2) = a(n) + 4*n.
The general formula for numbers with periods of length k: a(k,j,n) = n*(k^2*n - k + 2*j)/2 and 0<j<k.
For j=1 and j=(k-1) the numbers have 1 period.
For 1<j<(k-1) the numbers have A092964(k-4,j-1) periods.
G.f. (binomial(k,2)*(1+x) + j + (k-j)*x)/(1-x)^3.
EXAMPLE
a(5) = 5*(18*5-2) = 440; b(5) = a(5) + 5 = 445; c(5) = a(5)*2*5 = 450; d(5) = a(5) + 3*5 = 455; e(5) = a(5) + 4*5 = 460.
MATHEMATICA
Table[2n(9n-1), {n, 1, 50}] (* Vincenzo Librandi, Jul 10 2012 *)
LinearRecurrence[{3, -3, 1}, {16, 68, 156}, 50] (* Harvey P. Dale, Feb 15 2016 *)
PROG
(Magma) [2*n*(9*n-1): n in [1..50]]; // Vincenzo Librandi, Jul 10 2012
(PARI) a(n)=2*n*(9*n-1) \\ Charles R Greathouse IV, Jun 17 2017
(Sage) [2*n*(9*n-1) for (1..50)] # G. C. Greubel, Jan 30 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Weisenhorn, Dec 24 2010
STATUS
approved