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A117789
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Lucas numbers which are divisible by the sum of their digits.
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1
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1, 3, 4, 7, 18, 322, 5778, 505019158607, 84722519070079276, 1473646213395791149646646123, 105249261265075663875711417309855979021650214636
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OFFSET
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1,2
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COMMENTS
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a(12) has 83 digits and it is too large to include in the data section. - Amiram Eldar, Feb 08 2021
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LINKS
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FORMULA
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EXAMPLE
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322 is in the sequence because it is a Lucas number and it is divisible by the sum of its digits, 3+2+2 = 7.
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MATHEMATICA
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Select[LinearRecurrence[{1, 1}, {1, 3}, 230], Divisible[#, Plus @@ IntegerDigits[#]] &] (* Amiram Eldar, Feb 08 2021 *)
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PROG
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(PARI) {m=370; a=1; b=3; print1(a, ", ", b, ", "); for(n=3, m, c=b+a; a=b; b=c; s=0; k=b; while(k>0, d=divrem(k, 10); k=d[1]; s=s+d[2]); if(b%s==0, print1(b, ", ")))} \\ Klaus Brockhaus, Apr 17 2006
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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Luc Stevens (lms022(AT)yahoo.com), Apr 15 2006
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EXTENSIONS
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STATUS
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approved
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