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%I #9 Aug 06 2017 22:31:33
%S 23,29,33,39,43,49,53,59,63,69,73,79,83,89,93,99,133,136,137,192,194,
%T 195,222,223,226,229,261,263,267,313,316,331,332,333,334,336,339,391,
%U 392,397,441,443,449,621,623,661,662,663,666,669,912,914,915,931,932
%N Numbers n such that n modulo (product of digits of n) = (sum of digits of n).
%C Do consecutive numbers like 136, 137 occur frequently? Do many primes appear in the sequence?
%C Numbers n such that the positive integer remainder of n/A007954(n) equals A007953(n). [_R. J. Mathar_, Jan 03 2009]
%H B. D. Swan, <a href="/A153631/b153631.txt">Table of n, a(n) for n=1,...,20000</a>
%e For n = 83, (product of digits of n) = 24, (sum of digits of n) = 11 and 83 = 3*24 + 11.
%e For n = 93, (product of digits of n) = 27, (sum of digits of n) = 12 and 93 = 3*27 + 12.
%p sd := proc (n) options operator, arrow: add(convert(n, base, 10)[j], j = 1 .. nops(convert(n, base, 10))) end proc: pd := proc (n) options operator, arrow: product(convert(n, base, 10)[j], j = 1 .. nops(convert(n, base, 10))) end proc: a := proc (n) if 0 < pd(n) and `mod`(n, pd(n)) = sd(n) then n else end if end proc: seq(a(n), n = 1 .. 1000); # _Emeric Deutsch_, Jan 02 2009
%p A007953 := proc(n) local i ; add(i,i=convert(n,base,10)) ; end: A007954 := proc(n) local i ; mul(i,i=convert(n,base,10)) ; end: for n from 1 to 1200 do if A007954(n) > 0 then if irem(n,A007954(n))= A007953(n) then printf("%d,",n) ; fi; fi; od: # _R. J. Mathar_, Jan 03 2009
%K base,easy,nonn
%O 1,1
%A _J. M. Bergot_, Dec 29 2008
%E Edited, corrected (93 inserted) and extended beyond a(21) by _Klaus Brockhaus_, Jan 06 2009
%E Corrected and extended by _Emeric Deutsch_ and _R. J. Mathar_, Jan 02 2009
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