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A066282
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Numbers k such that k = (product of nonzero digits of k) * (sum of digits of k).
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7
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OFFSET
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1,3
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COMMENTS
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Suppose a term k has d digits, then k > 10^(d-1), the product of nonzero digits <= 9^d, and the sum of digits <= 9*d. Since for d >= 85 we have 10^(d-1) > 9^d * (9*d), it follows that d <= 84. That is, the sequence is finite. I've further verified that there are no other terms, that is, the sequence is complete. - Max Alekseyev, Jul 29 2024
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LINKS
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EXAMPLE
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(1+0+8+8) * (1*8*8) = 17*64 = 1088, so 1088 belongs to the sequence.
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MATHEMATICA
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Do[ d = Sort[ IntegerDigits[n]]; While[ First[d] == 0, d = Drop[d, 1]]; If[n == Apply[ Plus, d] Apply[ Times, d], Print[n]], {n, 0, 5*10^7} ]
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PROG
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(ARIBAS) function a066282(a, b: integer); var n, k, j, p, d: integer; s: string; begin for n := a to b do s := itoa(n); k := 0; p := 1; for j := 0 to length(s) - 1 do d := atoi(s[j..j]); k := k + d; if d > 0 then p := p*d; end; end; if n = p*k then write(n, ", "); end; end; end; a066282(0, 25000).
(PARI) a066282(a, b) = local(n, k, q, p, d); for(n=a, b, k=0; p=1; q=n; while(q>0, d=divrem(q, 10); q=d[1]; k=k+d[2]; p=p*max(1, d[2])); if(n==p*k, print1(n, ", ")))
a066282(0, 25000)
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CROSSREFS
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KEYWORD
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base,fini,full,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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