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A153631
Numbers n such that n modulo (product of digits of n) = (sum of digits of n).
1
23, 29, 33, 39, 43, 49, 53, 59, 63, 69, 73, 79, 83, 89, 93, 99, 133, 136, 137, 192, 194, 195, 222, 223, 226, 229, 261, 263, 267, 313, 316, 331, 332, 333, 334, 336, 339, 391, 392, 397, 441, 443, 449, 621, 623, 661, 662, 663, 666, 669, 912, 914, 915, 931, 932
OFFSET
1,1
COMMENTS
Do consecutive numbers like 136, 137 occur frequently? Do many primes appear in the sequence?
Numbers n such that the positive integer remainder of n/A007954(n) equals A007953(n). [R. J. Mathar, Jan 03 2009]
EXAMPLE
For n = 83, (product of digits of n) = 24, (sum of digits of n) = 11 and 83 = 3*24 + 11.
For n = 93, (product of digits of n) = 27, (sum of digits of n) = 12 and 93 = 3*27 + 12.
MAPLE
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
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
CROSSREFS
Sequence in context: A277206 A214754 A344375 * A082943 A061754 A375595
KEYWORD
base,easy,nonn
AUTHOR
J. M. Bergot, Dec 29 2008
EXTENSIONS
Edited, corrected (93 inserted) and extended beyond a(21) by Klaus Brockhaus, Jan 06 2009
Corrected and extended by Emeric Deutsch and R. J. Mathar, Jan 02 2009
STATUS
approved