A197424 Number of subsets of {1, 2, ..., 4*n + 2} which do not contain two numbers whose difference is 4.
4, 36, 225, 1600, 10816, 74529, 509796, 3496900, 23961025, 164249856, 1125736704, 7716041281, 52886200900, 362488284900, 2484529385121, 17029223715904, 116720020119616, 800010960336225, 5483356589096100, 37583485459535236, 257601040852192129
Offset: 0
References
- F. Alayont and E. Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs; submitted.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Mathramz Problem Solving Group, Solution for Problem 1854 in Mathematics Magazine Vol. 83, No.4, October 2010, February 28, 2011
- Marian Tetiva, Problem 1854, Mathematics Magazine, 84 (2011) 300.
- Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).
Programs
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Mathematica
Table[(1/25) (LucasL[2 (2 n + 5)] - 2 (-1)^n LucasL[2 n + 5] - 1), {n, 0, 20}] (* Michael De Vlieger, Mar 27 2016 *)
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PARI
Vec((4+16*x-15*x^2-5*x^3+x^4) / ((1-x)*(1-7*x+x^2)*(1+3*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 26 2016
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PARI
a(n) = (fibonacci(n+2)*fibonacci(n+3))^2; \\ Altug Alkan, Mar 26 2016
Formula
G.f.: ( -4-16*x+15*x^2+5*x^3-x^4 ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Oct 15 2011
Empirical: a(n) = A189145(2n+3). - R. J. Mathar, Oct 15 2011
For L=Lucas, a(n) = (1/25)*(L(2*(2*n+5)) - 2*(-1)^n*L(2*n+5) - 1), an instance of (F(n+p)*F(n+q))^2 = (1/25)*(L(2*(2*n+p+q)) - 2*(-1)^(n+q)*L(p-q)*L(2*n+p+q) + L(2*(p-q)) + 4*(-1)^(p-q)) which follows from squaring a specialization of identity 17b in the Vajda reference at A000045, F(n+p)*F(n+q) = (1/5)*(L(2*n+p+q) - (-1)*(n+q)*L(p-q)), then applying Vajda 17c, L(n)^2 = L(2*n) + 2*(-1)^n, to the expansion. - Ehren Metcalfe, Mar 26 2016
a(n) = A060635(n+2)/2. - Alois P. Heinz, Jul 03 2025
Comments