A178572 Numbers with ordered partitions that have periods of length 5.

11, 47, 108, 194, 305, 441, 602, 788, 999, 1235, 1496, 1782, 2093, 2429, 2790, 3176, 3587, 4023, 4484, 4970, 5481, 6017, 6578, 7164, 7775, 8411, 9072, 9758, 10469, 11205, 11966, 12752, 13563, 14399, 15260, 16146, 17057, 17993, 18954, 19940, 20951

Offset

1,1

Comments

From each ordered partition of the numbers (10+j) with 0<j<5 one removes the first part z(1) and adds 1 to the next z(1) parts to get a new partition until a period is reached.

The a(n) sequence begins with 11 and each member has 1 period; the b(n) = A022282(n) sequence begins with 12 and each member has 2 periods; the c(n) = A022283(n) sequence begins with 13 and each member has 2 periods; the d(n) = n*(25*n + 3)/2 sequence begins with 14 and each member has 1 period of length 5.

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

G.f. for a(n): (11 + 14*x)/(1-x)^3.

for b(n): (12 + 13*x)/(1-x)^3.

for c(n): (13 + 12*x)/(1-x)^3.

for d(n): (14 + 11*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) = 10 + j, s(2) = 45 + 2*j and 0<j<5.

s(n) = n*(25*n - 5 + 2*j)/2 and 0<j<5.

The general formula for numbers with periods of length k: a(k,j,n) = n*(k^2*n - k + 2*j)/2 with 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

For n=11 the period is [(4,3,2,1,1), (4,3,2,2), (4,3,3,1), (4,4,2,1), (5,3,2,1)].
For n=47 the period is [(9,8,7,6,6,4,3,2,1,1), (9,8,7,7,5,4,3,2,2), (9,8,8,6,5,4,3,3,1), (9,9,7,6,5,4,4,2,1), (10,8,7,6,5,5,3,2,1)].
For n=12 the 2 periods are [(4,3,2,2,1), (4,3,3,2), (4,4,3,1), (5,4,2,1), (5,3,2,1,1)] and [(4,3,3,1,1), (4,4,2,2), (5,3,3,1), (4,4,2,1,1), (5,3,2,2)].
For n=49 the 2 periods are [(9,8,7,7,6,4,3,2,2,1), (9,8,8,7,5,4,3,3,2), (9,9,8,6,5,4,4,3,1), (10,9,7,6,5,5,4,2,1), (10,8,7,6,6,5,3,2,1,1)] and [(9,8,8,6,6,4,3,3,1,1), (9,9,7,7,5,4,4,2,2),(10,8,8,6,5,5,3,3,1), (9,9,7,6,6,4,4,2,1,1), (10,8,7,7,5,5,3,2,2)].

Mathematica

LinearRecurrence[{3, -3, 1}, {11, 47, 108}, 50] (* Harvey P. Dale, Jan 14 2019 *)
Table[n*(25*n-3)/2, {n, 1, 50}] (* G. C. Greubel, Jan 30 2019 *)

Prog

(PARI) a(n)=n*(25*n-3)/2 \\ Charles R Greathouse IV, Jun 18 2017
(Magma) [n*(25*n-3)/2: n in [1..50]]; // G. C. Greubel, Jan 30 2019
(Sage) [n*(25*n-3)/2 for n in (1..50)] # G. C. Greubel, Jan 30 2019
(GAP) List([1..50], n -> n*(25*n-3)/2); # G. C. Greubel, Jan 30 2019

Crossrefs

Cf. A092964, A037306.

Sequence in context: A158463 A143830 A339504 * A036489 A076306 A219079

Adjacent sequences: A178569 A178570 A178571 * A178573 A178574 A178575

Keyword

nonn,easy

Author

Paul Weisenhorn, Dec 24 2010

Status

approved