A240588
a(1) = 1, a(2) = 2; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.
Original entry on oeis.org
1, 2, 3, 41, 7, 9, 137131, 61, 2023244487101, 13, 19, 11, 143, 142733, 21, 17, 193, 37, 3907, 1290366811360047359, 1805030483980039, 3803623, 123, 369, 27, 23, 58271, 47609, 523, 79, 307, 179, 73, 57, 18032419296851, 29, 31, 3281881401611107, 69, 171, 60244474373, 197, 97
Offset: 1
a(1)=1 and a(2)=2. a(1) U a(2) = 12 and its divisors are 1, 2, 3, 4, 6, 12. Therefore 3 is the least number not yet present in the sequence which divides 12. Again, a(1) U a(2) U a(3) = 123 and its divisors are 1, 3, 41, 123. Therefore a(4)=41. Etc.
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with(numtheory):
T:=proc(t) local x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:
P:=proc(q) local a,b,c,k,n; b:=12; print(1); print(2); c:=[1,2];
for n from 1 to q do a:=sort([op(divisors(b))]); for k from 2 to nops(a) do
if not member(a[k],c) then c:=[op(c),a[k]]; b:=a[k]+b*10^T(a[k]); print(a[k]); break;
fi; od; od; end: P(19);
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a = {1, 2}; While[Length[a] < 22,
n = ToExpression[StringJoin[ToString /@ a]];
AppendTo[a, SelectFirst[Sort[Divisors[n]], FreeQ[a, #] &]]
]; a
A096098
a(1) = 2, a(2) = 1; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.
Original entry on oeis.org
2, 1, 3, 71, 7, 21, 599, 173, 11, 23, 161, 49, 13, 9, 131, 19, 33, 17, 1489, 331, 3989, 69, 3097350956401900335673788279883089441874368101, 349387, 5651, 443, 29, 51, 479470832244949, 661, 1129, 1873, 181, 1544577973887516219070997863, 521
Offset: 1
a(6) = 21 as 213717 = 3*7*10177, and 3 = a(3) and 7 = a(4), hence 3*7 = 21 is the least number dividing 213717 not included earlier in the sequence.
A241811
a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.
Original entry on oeis.org
1, 0, 2, 3, 11, 71, 29, 9, 683, 67, 7, 743, 739, 1933, 23, 161, 21, 37, 19, 17, 119, 49, 332534262883, 13, 39, 13739483941387, 83, 111, 79853560395691, 5431567, 70610371, 69, 51, 4112497, 28384496881337963, 353, 77, 1531, 42787, 63, 27, 41, 709, 33, 81, 487, 139697
Offset: 1
a(1)=1 and a(2)=0. a(1) U a(2) = 10 and its divisors are 1, 2, 5, 10. Therefore 2 is the least number not yet present in the sequence which divides 10. Again, a(1) U a(2) U a(3) = 102 and its divisors are 1, 2, 3, 6, 17, 34, 51, 102. Therefore a(4)=3, etc.
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with(numtheory):
T:=proc(t) local x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:
P:=proc(q) local a,b,c,k,n; b:=10; print(1); print(0); c:=[0,1];
for n from 1 to q do a:=sort([op(divisors(b))]); for k from 2 to nops(a) do
if not member(a[k],c) then c:=[op(c),a[k]]; b:=a[k]+b*10^T(a[k]); print(a[k]); break;
fi; od; od; end: P(30);
A240229
a(n) is the shortest concatenation of the Fibonacci numbers F(1), F(2), ..., divisible by F(n) = A000045(n), n >= 1. a(n) = 0 if there is no such concatenation.
Original entry on oeis.org
1, 1, 112, 11235, 11235, 112, 1123581321345589144233, 11235, 11235813213455891442333776109871597258, 11235813213455891442333776109871597258441816765, 1123581321345589144233377610987159725844181676, 1123581321345589144233377610987159725844181676510946177112865746368
Offset: 1
a(3) = 112 because neither 1 nor 11 are divisible by F(3) = 2, but 112, the concatenation of F(1), F(2) and F(3) is.
Showing 1-4 of 4 results.
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