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A230299
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Define a sequence b_s by b_s(1)=s, b_s(k+1)=b_s(k)+(sum of digits of b_s(k)); a(n) is the number of steps needed for b_n to reach a term in one of b_0, b_1, b_3 or b_9, or a(n) = -1 if b_n never joins one of these four sequences.
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2
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0, 0, 0, 0, 0, 52, 0, 11, 0, 0, 51, 50, 0, 49, 10, 0, 0, 48, 0, 9, 50, 0, 49, 0, 0, 47, 48, 0, 0, 8, 0, 49, 46, 0, 47, 48, 0, 45, 0, 0, 7, 46, 7, 47, 6, 0, 45, 44, 6, 0, 46, 0, 5, 5, 0, 45, 44, 0, 43, 4, 5, 4, 0, 0, 4, 44, 4, 43, 3, 0, 0, 42, 0, 3, 3, 4, 43, 0
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OFFSET
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0,6
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COMMENTS
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We conjecture that a(n) is never -1.
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LINKS
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MAPLE
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read transforms; # to get digsum
M:=2000;
# f(s) returns the sequence k->k+digsum(k) starting at s
f:=proc(s) global M; option remember; local n, k, s1;
s1:=[s]; k:=s;
for n from 1 to M do k:=k+digsum(k);
s1:=[op(s1), k]; od: end;
# g(s) returns (x, p), where x = first number in common between
# f(s) and one of f(1), f(3), f(9) and p is the position where it occurred.
# If f(s) and all of f(1), f(3), f(9) are disjoint for M terms, returns (-1, -1)
S1:=convert(f(1), set):
S3:=convert(f(3), set):
S9:=convert(f(9), set):
g:=proc(s) global f, S1, S3, S9; local t1, p, T0, T1, T2;
T0:=f(s):
T1:=convert(T0, set);
if ((s mod 9) = 3) or ((s mod 9) = 6) then T2:= T1 intersect S3; t1:=min(T2); if (t1 = infinity) then RETURN(-1, -1); else member(t1, T0, 'p'); RETURN(t1, p-1); fi;
elif ((s mod 9) = 0) then T2:= T1 intersect S9; t1:=min(T2); if (t1 = infinity) then RETURN(-1, -1); else member(t1, T0, 'p'); RETURN(t1, p-1); fi;
else T2:= T1 intersect S1; t1:=min(T2); if (t1 = infinity) then RETURN(-1, -1); else member(t1, T0, 'p'); RETURN(t1, p-1); fi;
fi;
end;
[seq(g(n)[2], n=1..45)];
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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