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Numbers of the form ceiling(A179896(j)/A018252(j)) where A179896(j) mod A018252(j) <> 0.
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%I #16 Nov 29 2019 04:24:06

%S 5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59,62,65,68,71,

%T 74,77,80,83,86,89,92,95,98,101,104,107,110,113,116,119,122,125,128,

%U 131,134,137,140,143,146,149,152,155,158,161,164,167,170,173,176,179,182

%N Numbers of the form ceiling(A179896(j)/A018252(j)) where A179896(j) mod A018252(j) <> 0.

%C Previous name was "Ceiling(A179896 / n) for n > 0 and remainder <> 0".

%F Conjectures from _Colin Barker_, Nov 25 2019: (Start)

%F G.f.: x*(5 - 2*x) / (1 - x)^2.

%F a(n) = 2*a(n-1) - a(n-2) for n>2.

%F a(n) = 2 + 3*n.

%F (End)

%p From _R. J. Mathar_, Sep 19 2010: (Start)

%p A018252 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+1 do if not isprime(a) then return a; end if; end do; end if; end proc:

%p A045943 := proc(n) 3*n*(n+1)/2 ; end proc:

%p A179896 := proc(n) A045943(A018252(n)-1) ; end proc:

%p for n from 1 to 130 do a := A179896(n) ; c := A018252(n) ; if a mod c <> 0 then printf("%d,",ceil(a/c)) ; end if; end do: (End)

%Y Cf. A000040, A016789, A179545, A179896.

%K easy,nonn

%O 1,1

%A _Odimar Fabeny_, Sep 07 2010

%E Previous name replaced (with a Sep 19 2010 comments entry from _R. J. Mathar_) by _Jon E. Schoenfield_, Nov 28 2019