A321013 a(n) = how many of {6,7,8} divide n.
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 1
Offset: 1
Examples
a(24) = 2 because 24 is divisible 6 and 8, i.e., by 2 of the numbers in {6, 7, 8}. _David A. Corneth_, Nov 05 2018
References
- Senechal, Marjorie. "Introduction to lattice geometry." In M. Waldschmidt et al., eds., From Number Theory to Physics, pp. 476-495. Springer, Berlin, Heidelberg, 1992. See Cor. 3.7.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
Programs
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Maple
d3:=proc(n) local c; c:=0; if (n mod 6)=0 then c:=c+1; fi; if (n mod 7)=0 then c:=c+1; fi; if (n mod 8)=0 then c:=c+1; fi; c; end; [seq(d3(n),n=1..120)];
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Mathematica
a[n_]:=Sum[If[Mod[n,i]==0, 1, 0], {i ,6, 8}]; Array[a, 120] (* Stefano Spezia, Nov 05 2018 *) Table[Total[Boole[Divisible[n,{6,7,8}]]],{n,120}] (* Harvey P. Dale, Nov 09 2022 *) -
PARI
a(n) = sum(i = 6, 8, !(n%i)) \\ David A. Corneth, Nov 05 2018
Formula
a(n + 168) = a(n). - David A. Corneth, Nov 05 2018
Conjectures from Colin Barker, Nov 05 2018: (Start)
G.f.: x^6*(1 + 2*x + 4*x^2 + 5*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 9*x^7 + 9*x^8 + 6*x^9 + 6*x^10 + 3*x^11 + 3*x^12) / ((1 - x)*(1 + x)*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).
a(n) = -a(n-1) - 2*a(n-2) - 2*a(n-3) - 3*a(n-4) - 3*a(n-5) - 3*a(n-6) - 2*a(n-7) - a(n-8) + a(n-10) + 2*a(n-11) + 3*a(n-12) + 3*a(n-13) + 3*a(n-14) + 2*a(n-15) + 2*a(n-16) + a(n-17) + a(n-18) for n>18.
(End)
From David A. Corneth, Nov 05 2018: (Start)
The above conjectures are true. The sequence is periodic with period 168. Let f(x) be the g.f. above. Then f(x + 168) = f(x).
The expression for a(n) holds for 19 <= n <= 1000, more than sufficient for a proof. (End)