A178574 a(n) = 2*n*(9*n-1).

16, 68, 156, 280, 440, 636, 868, 1136, 1440, 1780, 2156, 2568, 3016, 3500, 4020, 4576, 5168, 5796, 6460, 7160, 7896, 8668, 9476, 10320, 11200, 12116, 13068, 14056, 15080, 16140, 17236, 18368, 19536, 20740, 21980, 23256, 24568, 25916, 27300, 28720, 30176, 31668, 33196, 34760, 36360

Offset: 1

Author

Paul Weisenhorn, Dec 24 2010

Keywords

• nonn
• easy

Comments

Numbers with ordered partitions that have periods of length 6.

From each ordered partition of the numbers (15+j) with 0

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

Examples

			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.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A037306, A092964.

Programs

Magma

[2*n*(9*n-1): n in [1..50]]; // Vincenzo Librandi, Jul 10 2012

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 *)

PARI

a(n)=2*n*(9*n-1) \\ Charles R Greathouse IV, Jun 17 2017

Sage

[2*n*(9*n-1) for n in (1..50)] # G. C. Greubel, Jan 30 2019

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

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

For j=1 and j=(k-1) the numbers have 1 period.

For 1A092964(k-4,j-1) periods.

G.f. (binomial(k,2)*(1+x) + j + (k-j)*x)/(1-x)^3.

E.g.f.: 2*exp(x)*x*(8 + 9*x). - Elmo R. Oliveira, Jan 28 2025

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