A212410 Let f(n) = n + floor(log(n)). Then a(n) is the smallest number of iterations of f on n such that the perfect square is obtained, or 0 if no such square exists.

1, 0, 1, 8, 7, 6, 5, 4, 11, 3, 10, 2, 9, 1, 8, 4, 7, 3, 6, 2, 5, 1, 13, 4, 8, 12, 3, 7, 11, 2, 6, 10, 1, 5, 9, 13, 4, 8, 12, 3, 7, 11, 2, 6, 10, 1, 5, 9, 35, 4, 8, 34, 3, 7, 33, 2, 6, 38, 32, 1, 5, 37, 31, 9, 4, 36, 30, 8, 3, 35, 29, 7, 2, 34, 28, 6, 1, 33, 27

Offset

1,4

Comments

a(2) = 0 because f(f(f(….f(2)))…) = 2.

a(n)=1 for n = {1, 3, 14, 22, 33, 46, 60, 77, 96, …}

References

T. Johnson and F. Jedrzejewski, Looking at Numbers, Springer, 2014

Links

Michel Lagneau, Table of n, a(n) for n = 1..10000

Example

a(5) = 7 because:
f(5)=6;
f(f(5))=7;
f(f(f(5)))=8;
f(f(f(f(5))))=10;
f(f(f(f(f(5)))))=12;
f(f(f(f(f(f(5))))))=14;
f(f(f(f(f(f(f(5)))))))=16.
The first square number in this sequence 6,7,8,10,12,14,16 is on the seventh place and therefore a(5)=7.

Maple

with(numtheory): for n from 1 to 100 do:n0:=n:i:=0:for k from 1 to 1000 while(i=0) do:n1:=n0+floor(log(n0)): n2:=sqrt(n1):if n2 = floor(n2) then printf(`%d, `, k):i:=1:else n0:=n1:fi:od:if i=0 then printf(`%d, `, 0):else fi:od:

Mathematica

f[n_] := Length@NestWhileList[#+Floor[Log[#]]&, n, !IntegerQ[Sqrt[#]] || #==n&] - 1; Table[f[n], {n, 3, 100}]

Crossrefs

Sequence in context: A090915 A194756 A132672 * A328759 A132037 A247095

Adjacent sequences: A212407 A212408 A212409 * A212411 A212412 A212413

Keyword

nonn,look

Author

Michel Lagneau, May 15 2012

Status

approved