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A353651
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Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 4.
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5
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2, 3, 7, 8, 9, 10, 11, 15, 19, 23, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125
(list;
graph;
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listen;
history;
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internal format)
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OFFSET
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1,1
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COMMENTS
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Numbers m such that the base-4 representation of (3*m-1) starts with 11 or 12 or 13 or ends with 0.
First differences give a run of 4^i 1's followed by a run of 4^i 4's, for i = 0, 1, 2, ...
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LINKS
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FORMULA
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For n in the range (2*4^i + 1)/3 < n <= (5*4^i + 1)/3, for i >= 0:
a(n) = n + 4^i.
a(n) = 1 + a(n-1).
Otherwise, for n in the range (5*4^i + 1)/3 < n <= (8*4^i + 1)/3, for i >= 0:
a(n) = 4*(n - 4^i) - 1.
a(n) = 4 + a(n-1).
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EXAMPLE
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a(6) = 10 because (2*4^1 + 1)/3 < 6 <= (5*4^1 + 1)/3, hence a(6) = 6 + 4^1 = 10;
a(9) = 19 because (5*4^1 + 1)/3 < 9 <= (8*4^1 + 1)/3, hence a(9) = 4*(9 - 4^1) - 1 = 19.
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MAPLE
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isA353651 := proc(n)
if modp(n, 4) = 3 then
true;
else
b4 := convert(3*n-1, base, 4) ;
if op(-1, b4) = 1 and op(-2, b4) <> 0 then
true ;
else
false;
end if;
end if;
end proc:
for n from 2 to 122 do
if isA353651(n) then
printf("%d, ", n) ;
end if;
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PROG
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(PARI) a(n) = my(n3=3*n, s=logint(n3>>1, 4)<<1); if(n3>>s < 5, n + 1<<s, 4*(n - 1<<s) - 1); \\ Kevin Ryde, Apr 15 2022
(C++)
/* program used to generate the b-file */
#include<iostream>
using namespace std;
int main(){
int cnt1=1, flag=0, cnt2=1, a=2;
for(int n=1; n<=10000; n++) {
cout<<n<<" "<<a<<endl;
if(cnt2==cnt1) {
flag=1-flag, cnt1=1;
if(flag) a+=1;
else {
a+=4;
cnt2*=4;
}
}
else {
cnt1++;
a+=(flag?4:1);
}
}
return 0;
}
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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