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

Author

Paul Weisenhorn, Dec 24 2010

Keywords

• nonn
• easy

Comments

From each ordered partition of the numbers (10+j) with 0

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.

Examples

			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)].
		

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).

Crossrefs

Cf. A092964, A037306.

Programs

GAP

List([1..50], n -> n*(25*n-3)/2); # G. C. Greubel, Jan 30 2019

Magma

[n*(25*n-3)/2: n in [1..50]]; // G. C. Greubel, Jan 30 2019

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 *)

PARI

a(n)=n*(25*n-3)/2 \\ Charles R Greathouse IV, Jun 18 2017

Sage

[n*(25*n-3)/2 for n in (1..50)] # G. C. Greubel, Jan 30 2019

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

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

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

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