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%I #12 Sep 07 2018 04:41:50
%S 1,1,1,1,1,2,1,1,1,3,1,2,1,4,2,1,1,2,1,3,3,6,1,2,1,7,1,4,1,3,1,1,4,9,
%T 2,2,1,10,5,3,1,4,1,6,2,12,1,2,1,3,6,7,1,2,3,4,7,15,1,3,1,16,3,1,3,6,
%U 1,9,8,4,1,2,1,19,2,10,2,7,1,3,1,21,1,4,4,22,10,6,1,3,2,12,11,24,4,2,1,4,4
%N a(n) = ceiling(GPF(n)/LPF(n)) where GPF is greatest prime factor, LPF is least prime factor.
%C Given GPF(n) and LPF(n), the sum is A074320, the difference is A046665 and the product is A066048. a(n) = 1 iff n is p^k iff n is in A000961.
%H Antti Karttunen, <a href="/A120454/b120454.txt">Table of n, a(n) for n = 1..65537</a>
%F a(n) = ceiling(A006530(n)/A020639(n)).
%e a(26) = ceiling(GPF(26)/LPF(26)) = ceiling(13/2) = 7.
%p A120454 := proc(n) local ifs ; if n = 1 then RETURN(1) ; else ifs := ifactors(n)[2] ; RETURN( ceil(op(1,op(-1,ifs))/op(1,op(1,ifs))) ) ; fi ; end ; for n from 1 to 100 do printf("%d, ",A120454(n)) ; od ; # _R. J. Mathar_, Dec 16 2006
%o (PARI) A120454(n) = if(1==n,1, my(f = factor(n), lpf = f[1, 1], gpf = f[#f~, 1]); ceil(gpf/lpf)); \\ _Antti Karttunen_, Sep 06 2018
%Y Cf. A000040, A006530, A020639, A074320, A046665, A066048.
%K easy,nonn
%O 1,6
%A _Jonathan Vos Post_, Aug 16 2006
%E Corrected and extended by _R. J. Mathar_, Dec 16 2006
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