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A159639 Last divisors at which integral quotients occur consecutively.

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%I A159639 #10 Jul 22 2025 06:23:50
%S A159639 154,2183,4002,8635,19203,93017,96298,122414,166762,182090,201354,
%T A159639 241237,337645,346495,406813,456729,574678,668811,781635,799006,
%U A159639 929442,952150,1014194,1379625,1455259,1513549,1558110,1573089,1938354,2028842
%N A159639 Last divisors at which integral quotients occur consecutively.
%C A159639 Considering dividend/divisor=quotient, in a(1) of A116536, A159578, and A159579, 3=30/10; sometimes integral quotients appear n times in succession -- see A159580 where a(3)indicates that 5 integral quotients appear one after another. A159581 and A159639 give the first and last values of the divisors producing these integral quotients.
%C A159639 Many of the associated sequences submitted by this author were wrong. Should be recomputed. The UBASIC program should be regarded with suspicion. - _N. J. A. Sloane_, Oct 02 2011.
%e A159639 a(1)=154 because it is the last or second of two divisors (125 being the first) where integral quotients are produced in succession (one after the other): 5577321750/125=44618574, integral; and 161742330750/154=1050274875, integral. See a(4) in A116536, A159578, and A159579.
%o A159639 (UBASIC) 10 'product of cons primes divided by sum cons primes 20 N=3:Q=2*N:R=R+N:R=R+2 30 A=3:S=sqrt(N) 40 B=N/A 50 if int(B)*A=N then 110 60 A=A+2:if A<S then 40 70 Q=Q*N:R=R+N 90 if Q/R=int(Q/R) then print R;T;:C=C+1:print C:stop:else C=0 100 T=R 110 N=N+2:goto 30
%Y A159639 Cf. A116536, A159578 A159579 A159580 A159581
%K A159639 easy,nonn
%O A159639 1,1
%A A159639 _Enoch Haga_, Apr 17 2009, Apr 21 2009