cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A224815 Number of subsets of {1,2,...,n-8} without differences equal to 4 or 8.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 24, 36, 54, 81, 108, 144, 192, 256, 384, 576, 864, 1296, 1944, 2916, 4374, 6561, 9477, 13689, 19773, 28561, 41743, 61009, 89167, 130321, 192052, 283024, 417088, 614656, 900032, 1317904, 1929788, 2825761
Offset: 0

Views

Author

Vladimir Baltic, May 18 2013

Keywords

Comments

a(n) is the number of permutations (p(1), p(2), ..., p(n)) satisfying -k <= p(i)-i <= r and p(i)-i in the set I, i=1..n, with k=4, r=8, I={-4,0,8}.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^3 + x^4 - x^5 - x^6 - 3*x^7 + 3*x^8 - 2*x^9 - x^10 - 5*x^11 - 3*x^12 - 2*x^13 + 3*x^15 - 3*x^16 - 3*x^18 + 3*x^19 - 3*x^20 + 3*x^21 + 3*x^23 + 6*x^24 - 3*x^25 - 2*x^26 - 4*x^27 - x^29 - x^30 - 2*x^31 - x^32 + x^33 + x^35 - x^36 + x^37 + x^39)/((1 - x - x^3)*(1 + x^4 + x^6)*(1 + x^4 - x^6)*(1 - x^4 - x^12)*(1 + x^4 + 6*x^8 - 3*x^12 + 2*x^20 + x^24)), {x, 0, 50}], x] (* G. C. Greubel, Apr 28 2017 *)

Formula

a(n) = a(n-1)+a(n-3)-2*a(n-4)+2*a(n-5)+2*a(n-7)-6*a(n-8)+6*a(n-9)+6*a(n-11) +a(n-12)-a(n-13)-a(n-15)+13*a(n-16)-13*a(n-17)-13*a(n-19)+15*a(n-20)-15*a(n-21)-15*a(n-23)-6*a(n-24)+6*a(n-25)+6*a(n-27)+3*a(n-28)-3*a(n-29)-3*a(n-31)-2*a(n-32)+2*a(n-33)+2*a(n-35)+8*a(n-36)-8*a(n-37)-8*a(n-39)+3*a(n-40)-3*a(n-41)-3*a(n-43)-a(n-44)+a(n-45)+a(n-47)-a(n-48)+a(n-49)+a(n-51).
G.f.: ( 1-x^3+x^4-x^5-x^6-3*x^7+3*x^8-2*x^9-x^10-5*x^11-3*x^12-2*x^13 +3*x^15-3*x^16-3*x^18+3*x^19-3*x^20+3*x^21+3*x^23+6*x^24-3*x^25-2*x^26-4*x^27-x^29-x^30-2*x^31-x^32+x^33+x^35-x^36+x^37+x^39 ) / ((1-x-x^3)*(1+x^4+x^6)*(1+x^4-x^6)*(1-x^4-x^12)*(1+x^4+6*x^8-3*x^12+2*x^20+x^24)).
a(4*k) = (A000930(k))^4,
a(4*k+1) = (A000930(k))^3 * A000930(k+1),
a(4*k+2) = (A000930(k))^2 * (A000930(k+1))^2,
a(4*k+3) = A000930(k) * (A000930(k+1))^3.