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A232566
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Let x(0)x(1)x(2)... x(q) denote the decimal expansion of n and sopf(n) the sum of the distinct prime divisors of n. Sequence lists the numbers n such that x(0)/sopf(n) + x(1)/sopf(n) + ... + x(q)/sopf(n) + x(0)*x(1)*x(2)*...*x(q)/sopf(n) is an integer.
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0
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2, 3, 4, 5, 7, 8, 9, 12, 14, 19, 29, 34, 49, 59, 64, 66, 74, 79, 89, 94, 117, 135, 144, 147, 155, 160, 175, 189, 192, 243, 250, 319, 375, 391, 448, 486, 512, 545, 627, 648, 657, 729, 735, 756, 784, 792, 825, 864, 875, 936, 968, 975, 1144, 1232, 1239, 1344, 1408
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OFFSET
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1,1
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COMMENTS
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The corresponding integers are 2, 2, 4, 2, 2, 8, 6, 1, 1, 1, 1, 1, 7, 1, 17, 3, 1, 1, 1, 1, 1, 3, 5, 4, 1, 1, 4, 9, 6, 11,...
The primes of this sequence are 2, 3, 5, 7, 19, 29, 59, 79, 89. It seems that this subsequence is probably finite (no further terms less than 10^7).
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LINKS
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EXAMPLE
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657 is in the sequence because the prime divisors of 657 are 3 and 73 => sopf(657) = 3+73 = 76 and 6/76 + 5/76 + 7/76 + 6*5*7/76 = 3 is integer.
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MAPLE
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with(numtheory):for n from 2 to 1500 do:x:=convert(n, base, 10):n1:=nops(x):y:=factorset(n):n2:=nops(y):p:=1:s:=0:for i from 1 to n2 do:s:=s+y[i]:od:s1:=0:for j from 1 to n1 do:s1:=s1+x[j]/s:p:=p*x[j]:od:s1:=s1+p/s:if s1=floor(s1) then printf(`%d, `, n):else fi:od:
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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