login
A276106
Number of compositions of n into parts 1, 7, and 8.
3
1, 1, 1, 1, 1, 1, 1, 2, 4, 6, 8, 10, 12, 14, 17, 23, 33, 47, 65, 87, 113, 144, 184, 240, 320, 432, 584, 784, 1041, 1369, 1793, 2353, 3105, 4121, 5489, 7314, 9724, 12886, 17032, 22490, 29716, 39326, 52129, 69167, 91777, 121695, 161217, 213423, 282465, 373920, 495216, 656160, 869632, 1152544, 1527184
OFFSET
0,8
COMMENTS
Suppose A is a subset of {1..n} having the following property: if A includes an integer k, then A includes none of the integers k+2, k+3, k+4, k+5 or k+6. The number of subsets having this property is a(n+6).
The terms of this sequence also give us the answer to the following coloring problem: suppose that, given an n-section board, if we paint the k-th section, we can't paint the (k+2)-th, (k+3)-th, (k+4)-th, (k+5)-th or (k+6)-th section. In how many different ways can we paint this n-section board (where painting none of the sections is considered one of the ways)? Similarly the answer is a(n+6).
FORMULA
a(n) = a(n-1) + a(n-7) + a(n-8) for n > 7.
G.f.: 1/(1-x-x^7-x^8).
EXAMPLE
G.f. = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + 2*x^7 + 4*x^8 + 6*x^9 + 8*x^10 + ...
For n=3 so {1,2,3}, the answer is a(3+6) = a(9) = 6. Here are the subsets: {},{1},{2},{3},{1,2},{2,3}.
For n=4, the number of ways of painting a 4-section board is a(4+6)=a(10)=8; here are the 8 situations: {},{1},{2},{3},{4},{1,2},{2,3},{3,4}.
situation 1: none
situation 2: painted only 1st section
situation 3: painted only 2nd section
situation 4: painted only 3rd section
situation 5: painted only 4th section
situation 6: painted 1st and 2nd sections
situation 7: painted 2nd and 3rd sections
situation 8: painted 3rd and 4th sections
MATHEMATICA
CoefficientList[Series[1/(1 - x - x^7 - x^8), {x, 0, 54}], x] (* Michael De Vlieger, Aug 28 2016 *)
PROG
(PARI) Vec(1/(1-x-x^7-x^8) + O(x^50)) \\ Felix Fröhlich, Aug 27 2016
CROSSREFS
Sequence in context: A062417 A240114 A076828 * A321501 A238264 A098807
KEYWORD
nonn
AUTHOR
STATUS
approved