

A159639


Last divisors at which integral quotients occur consecutively


2



154, 2183, 4002, 8635, 19203, 93017, 96298, 122414, 166762, 182090, 201354, 241237, 337645, 346495, 406813, 456729, 574678, 668811, 781635, 799006, 929442, 952150, 1014194, 1379625, 1455259, 1513549, 1558110, 1573089, 1938354, 2028842
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

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.
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.


LINKS

Table of n, a(n) for n=1..30.


EXAMPLE

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.


PROG

(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


CROSSREFS

Cf. A116536, A159578 A159579 A159580 A159581
Sequence in context: A230804 A200552 A160854 * A160863 A160840 A160841
Adjacent sequences: A159636 A159637 A159638 * A159640 A159641 A159642


KEYWORD

easy,nonn


AUTHOR

Enoch Haga, Apr 17 2009, Apr 21 2009


STATUS

approved



