A230864 -id:A230864 - cp's OEIS frontend

A001069 Log2*(n) (version 2): take log_2 of n this many times to get a number < 2.

0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 1

N. J. A. Sloane and J. H. Conway

From Hieronymus Fischer, Apr 08 2012: (Start)
In terms of A010096 the definition could read: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 2".
With the only difference in the termination criterion, the definition is essentially the same as A010096. If we change the definition to " ...number < 1" we get A010096. Therefore we get A010096 when adding 1 to each term. (End)

			a(n)=1, 2, 3, 4, 5, ... for n=2, 2^2, 2^2^2, 2^2^2^2, 2^2^2^2^2, ... =2, 4, 16, 65536, 2^65536, ...

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A010096 (version 1), A230864 (version 3).

f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 2 &] - 1; Array[f, 105] (* Robert G. Wilson v, Apr 19 2012 *)

From Hieronymus Fischer, Apr 08 2012: (Start)
a(n) = A010096(n)-1.
With the exponentiation definition E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n-1)^(c(n)))...))); E_{i=1..0} := 1; example: E_{i=1..4} 2 = 2^(2^(2^2)) = 2^16, we get:
a(E_{i=1..n} 2) = a(E_{i=1..n-1} 2) +1, for n>=1.
G.f.: g(x) = 1/(1-x)*Sum_{k >= 1} x^(E_{i=1..k} 2).
The explicit first terms of this g.f. are
g(x) = (x^2+x^4+x^16+x^65536+...)/(1-x).  (End)

A010096 log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Leonid Broukhis

From Hieronymus Fischer, Apr 08 2012: (Start)
A possibly simpler definition could be: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 1".
Changing "< 1" to "<= 1" produces version 3, A230864.
With the only difference in the termination criterion, the definition is essentially the same as version 2, A001069. If we change the definition to "floor(log_2(... = 1" we get A001069. Therefore we get A001069 when subtracting 1 from each term. (End)

			Becomes 5 at 65536, 6 at 2^65536, etc.

Cf. A063510, A000523, A001069 (version 2), A230864 (version 3).

a010096 = length . takeWhile (/= 0) . iterate a000523
-- Reinhard Zumkeller, Mar 16 2012

f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 1 &] - 1; Array[f, 105] (* Robert G. Wilson v, Apr 19 2012 *)

a(n)=if(n<1,0,1+a(log(n)\log(2))) \\ Charles R Greathouse IV, Apr 17 2012

a(n)=if(n<1,0,1+a(logint(n,2))) \\ Charles R Greathouse IV, Oct 23 2015

From Hieronymus Fischer, Apr 08 2012: (Start)
a(n) = A001069(n) + 1.
With the exponentiation definition E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n-1)^(c(n)))...))); E_{i=1..0} := 1; example: E_{i=1..4} 2 = 2^(2^(2^2)) = 2^16, we get:
a(E_{i=1..n} 2) = a(E_{i=1..n-1} 2) +1, for n >= 1.
G.f.: g(x) = 1/(1-x)*Sum_{k>=0} x^(E_{i=1..k} 2).
The explicit first terms of this g.f. are
g(x) = (x + x^2 + x^4 + x^16 + x^65536 + ...)/(1-x). (End)

Edited by Hieronymus Fischer, Apr 08 2012

Edited by N. J. A. Sloane, Nov 03 2013

A230303 Let M(1)=0 and for n >= 2, let B(n)=M(ceiling(n/2))+M(floor(n/2))+2, M(n)=2^B(n)+M(floor(n/2))+1; sequence gives M(n).

Original entry on oeis.org

0, 5, 129, 4102, 87112285931760246646623899502532662132742, 1852673427797059126777135760139006525652319754650249024631321344126610074239106

Offset: 1

N. J. A. Sloane, Oct 24 2013; Mar 26 2014

nonn

M(n) is the smallest value of k such that A228085(k) = n. For example, 129 is the first time a 3 appears in A228085 (and is therefore the first term in A230092). M(4) = 4102 is the first time a 4 appears in A228085 (and is therefore the first term in A227915).

			The terms are a(1) = 0, a(2) = 2^2+0+1, a(3) = 2^7+0+1, a(4) = 2^12+5+1, a(5) = 2^136+5+1, a(6) = 2^160+129+1, a(7) = 2^4233+129+1, a(8) = 2^8206+4102+1, a(9) = 2^k+4102+1 with k=2^136+4110, ... .
The length (in bits) of the n-th term is A230302(n)+1.

Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory 12:3 (2022), 115-155.
Index entries for Colombian or self numbers and related sequences

Cf. A228085, A230092, A227915, A230093, A230302 (for B(n)), A230864.

Smallest number m such that u + (sum of base-b digits of u) = m has exactly n solutions, for bases 2 through 10: A230303, A230640, A230638, A230867, A238840, A238841, A238842, A238843, A006064.

f:=proc(n) option remember; local B, M;
if n<=1 then RETURN([0,0]);
else
if (n mod 2) = 0 then B:=2*f(n/2)[2]+2;
else B:=f((n+1)/2)[2]+f((n-1)/2)[2]+2; fi;
M:=2^B+f(floor(n/2))[2]+1; RETURN([B,M]); fi;
end proc;
[seq(f(n)[2],n=1..6)];

Define i by 2^(i-1) < n <= 2^i. Then it appears that
a(n) = 2^2^2^...^2^x
a tower of height i+3, containing i+2 2's, where x is in the range 0 < x <= 1.
For example, if n=7, i=3, and
a(7) = 2^4233+130 = 2^2^2^2^2^.88303276...
Note also that i+2 = A230864(a(n)).

a(1)-a(8) were found by Donovan Johnson, Oct 22 2013.

A293668 First differences of A292046.

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Andrey Zabolotskiy, Oct 14 2017

a(n) is also the length of n-th run of consecutive integers in the complement of A292046, starting from the 1st run "4, 5".

This sequence is invariant under the following transform: subtract 1 from every term, eliminate zeros. Other sequences with this property include A001511 and other generalized ruler functions, A002260, A272729.

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A001511, A002260, A230864, A255308, A272729, A292046.

A293668(n) = { my(k=1); while(n && !bitand(n,n-1),n = valuation(n,2); k++); (k); }; \\ Antti Karttunen, Sep 30 2018

a(0) = 1, a(n) = A292046(n+1)-A292046(n) for n>0.
If n = 2^k, a(n) = a(k)+1; otherwise a(n) = 1.
a(n) = A255308(n) + 1.
a(n) = O(log*(n)), where log* is the iterated logarithm. More precisely, a(n) <= A230864(n+1)+1.

A230863 a(1)=0; thereafter a(n) = 2^(a(ceiling(n/2)) + a(floor(n/2))).

Original entry on oeis.org

0, 1, 2, 4, 8, 16, 64, 256, 4096, 65536, 16777216, 4294967296, 1208925819614629174706176, 340282366920938463463374607431768211456, 2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576

Offset: 1

N. J. A. Sloane, Nov 02 2013; revised Mar 26 2014

nonn

a(16) = 2^512
  = 134078079299425970995740249982058461274793658205923933777235\
    614437217640300735469768018742981669034276900318581864860508537538828119465\
    69946433649006084096.

Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory 12:3 (2022), 115-155.

Cf. A230864, A230874, A230875.

f:=proc(n) option remember;
if n=1 then 0 else 2^(f(ceil(n/2))+f(floor(n/2))); fi; end;
[seq(f(n),n=1..16)];

In general, for n >= 11, define i by 9*2^(i-1) < n <= 9*2^i. Then it appears that a(n) = 2^2^2^...^2^x, a tower of height i+5, containing i+4 2's, where x is in the range 0 < x <= 1.
For example, if n=18, i=1, and a(18) = 2^8192 = 2^2^2^2^2^0.91662699..., of height 6.
Note also that i+5 = A230864(a(n)).

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A001069 Log2*(n) (version 2): take log_2 of n this many times to get a number < 2.

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A010096 log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.

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A230303 Let M(1)=0 and for n >= 2, let B(n)=M(ceiling(n/2))+M(floor(n/2))+2, M(n)=2^B(n)+M(floor(n/2))+1; sequence gives M(n).

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A293668 First differences of A292046.

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A230863 a(1)=0; thereafter a(n) = 2^(a(ceiling(n/2)) + a(floor(n/2))).

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