A101925 -id:A101925 - cp's OEIS frontend

A101926 a(n) = 2^A101925(n).

2, 4, 16, 32, 256, 512, 2048, 4096, 65536, 131072, 524288, 1048576, 8388608, 16777216, 67108864, 134217728, 4294967296, 8589934592, 34359738368, 68719476736, 549755813888, 1099511627776, 4398046511104, 8796093022208

Offset: 0

Ralf Stephan, Dec 28 2004

nonn

a(n) is the numerator of 2^(2*n+1)*(n!)^2/(2*n+1)/(2*n)!. The corresponding denominator is A001803. - Daniel Suteu, Feb 03 2017
a(n) is the numerator of Integral_{x=-oo..oo} sech(x)^(2*n+2) dx. The corresponding denominator is A001803(n). - Mohammed Yaseen, Jul 25 2023
a(n) is the denominator of (1/Pi) * Integral_{x=0..Pi/2} sin(x)^(2*n) dx. The corresponding numerator is A001790(n). - Mohammed Yaseen, Sep 19 2023
a(n) = numerator(Pi*binomial(n, -1/2)). - Peter Luschny, Dec 05 2024

Index to divisibility sequences

Bisection of A036069 and of A086117.

Cf. A001803, A001790, A101925.

denom((binomial(2n,n)*4^-n)/2); # Stephen Crowley, Mar 05 2007

Table[Numerator[Beta[1, n + 1, 1/2]], {n, 0, 22}] (* Gerry Martens, Nov 13 2016 *)

More terms from Joshua Zucker, May 15 2006

A048881 a(n) = A000120(n+1) - 1 = wt(n+1) - 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3

Offset: 0

Wolfdieter Lang

Highest power of 2 dividing n-th Catalan number (A000108).
a(n) = 0 iff n = 2^k - 1, k=0,1,...
Appears to be number of binary left-rotations (iterations of A006257) to reach fixed point of form 2^k-1. Right-rotation analog is A063250. This would imply that for n >= 0, a(n)=f(n), recursively defined to be 0 for n=0, otherwise as f( ( (1-n)(1-p)(1-s) - (1-n-p-s) ) / 2) + p (s+1) / 2, where p = n mod 2 and s = - signum(n) (f(n<0) is A000120(-n)). - Marc LeBrun, Jul 11 2001
Let f(0) = 01, f(1) = 12, f(2) = 23, f(3) = 34, f(4) = 45, etc. Sequence gives concatenation of 0, f(0), f(f(0)), f(f(f(0))), ... Also f(f(...f(0)...)) converges to A000120. - Philippe Deléham, Aug 14 2003
C(n, k) is the number of occurrence of k in the n-th group of terms in this sequence read by rows: {0}; {0, 1}; {0, 1, 1, 2}; {0, 1, 1, 2, 1, 2, 2, 3}; {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 }; ... - Philippe Deléham, Jan 01 2004
Highest power of 2 dividing binomial(n,floor(n/2)). - Benoit Cloitre, Oct 20 2003
2^a(n) are numerators in the Maclaurin series for (sin x)^2. - Jacob A. Siehler, Nov 11 2009
Conjecture: a(n) is the sum of digits of the n-th word in A076478, for n >= 1; has been confirmed for n up to 20000. - Clark Kimberling, Jul 14 2021

			From _Omar E. Pol_, Mar 08 2011: (Start)
Sequence can be written in the following form (irregular triangle):
  0,
  0,1,
  0,1,1,2,
  0,1,1,2,1,2,2,3,
  0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,
  0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  ...
Row sums are A001787.
(End)

R. Alter and K. K. Kubota, Prime and Prime Power Divisibility of Catalan Numbers, J. Comb. Th. A 15 (1973) pp. 243-256.
E. Deutsch and B. E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, arXiv:math/0407326 [math.CO], 2004; J. Num. Theory 117 (2006), 191-215.

Cf. A000120, A006257, A007318, A063250.

First differences of A078903.

a048881 n = a048881_list !! n
a048881_list = c [0] where c (x:xs) = x : c (xs ++ [x,x+1])
-- Reinhard Zumkeller, Mar 07 2011
(Python 3.10+)
def A048881(n): return (n+1).bit_count()-1 # Chai Wah Wu, Nov 15 2022

A048881 := proc(n)
    A000120(n+1)-1 ;
end proc:
seq(A048881(n),n=0..200) ; # R. J. Mathar, Mar 12 2018

a[n_] := IntegerExponent[ CatalanNumber[n], 2]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Jun 21 2013 *)

{ a(n) = if( n<0, 0, n++; n /= 2^valuation(n,2); subst( Pol( binary( n ) ), x, 1) - 1 ) } /* Michael Somos, Aug 23 2007 */

{a(n) = if( n<0, 0, valuation( (2*n)! / n! / (n+1)!, 2 ) ) } /* Michael Somos, Aug 23 2007 */

a(n) = hammingweight(n+1) - 1; \\ Michel Marcus, Nov 15 2022

Writing n as 2^m+k with -1 <= k < 2^m-1, then a(n) = A000120(k+1). - Henry Bottomley, Mar 28 2000
a(n) = k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n).
a(2*n) = a(n-1)+1, a(2*n+1) = a(n). - Vladeta Jovovic, Oct 10 2002
G.f.: (1/(x-x^2)) * (x^2/(1-x) - Sum_{k>=1} x^(2^k)/(1-x^(2^k))). - Ralf Stephan, Apr 13 2002
a(n) = A000120(A129760(n+1)). - Reinhard Zumkeller, Jun 30 2010
a(n+k) = A240857(n,k), 0 <= k <= n; in particular: a(n) = A240857(n,0). - Reinhard Zumkeller, Apr 14 2014
a(n) = (n+1)*2 - A101925(n+1). - Gleb Ivanov, Jan 12 2022

Entry revised by N. J. A. Sloane, Jun 07 2009

A046699 a(1) = a(2) = 1, a(n) = a(n - a(n-1)) + a(n-1 - a(n-2)) if n > 2.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10, 11, 12, 12, 12, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 18, 18, 19, 20, 20, 20, 21, 22, 22, 23, 24, 24, 24, 24, 25, 26, 26, 27, 28, 28, 28, 29, 30, 30, 31, 32, 32, 32, 32, 32, 32, 33, 34, 34, 35, 36, 36, 36, 37

Offset: 1

R. K. Guy

Ignoring first term, this is the meta-Fibonacci sequence for s=0. - Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca)

Except for the first term, n occurs A001511(n) times. - Franklin T. Adams-Watters, Oct 22 2006

Sequence was proposed by Reg Allenby.
B. W. Conolly, "Meta-Fibonacci sequences," in S. Vajda, editor, Fibonacci and Lucas Numbers and the Golden Section. Halstead Press, NY, 1989, pp. 127-138. See Eq. (2).
Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 5, 1990, pp. 212-213, 1993.
S. Vajda, Fibonacci and Lucas Numbers and the Golden Section, Wiley, 1989, see p. 129.
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 129.

N. J. A. Sloane, Table of n, a(n) for n = 1..20000
Altug Alkan, On a Generalization of Hofstadter's Q-Sequence: A Family of Chaotic Generational Structures, Complexity (2018) Article ID 8517125.
Joerg Arndt, Matters Computational (The Fxtbook)
Stephen M. Buckley, Wiring switches to more light bulbs, arXiv:2410.02460 [math.CO], 2024. See pp. 3, 12, 22.
Joseph Callaghan, John J. Chew III, and Stephen M. Tanny, On the behavior of a family of meta-Fibonacci sequences, SIAM Journal on Discrete Mathematics 18.4 (2005): 794-824. See p. 794. - _N. J. A. Sloane_, Apr 16 2014
M. Celaya and F. Ruskey, Morphic Words and Nested Recurrence Relations, arXiv preprint arXiv:1307.0153 [math.CO], 2013.
C. Deugau and F. Ruskey, Complete k-ary Trees and Generalized Meta-Fibonacci Sequences, J. Integer Seq., Vol. 12. [This is a later version than that in the GenMetaFib.html link]
C. Deugau and F. Ruskey, Complete k-ary Trees and Generalized Meta-Fibonacci Sequences
A. Erickson, A. Isgur, B. W. Jackson, F. Ruskey and S. M. Tanny, Nested recurrence relations with Conolly-like solutions, See Table 2.
Nathan Fox, A Slow Relative of Hofstadter's Q-Sequence, arXiv:1611.08244 [math.NT], 2016.
Nathan Fox, Trees, Fibonacci Numbers, and Nested Recurrences, Rutgers University Experimental Math Seminar, Mar 07, 2019.
Nathan Fox, Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences, arXiv:2203.09340 [math.CO], 2022.
The IMO Compendium, Problem 5, 22nd Canadian Mathematical Olympiad 1990.
Abraham Isgur, Mustazee Rahman, and Stephen Tanny, Solving non-homogeneous nested recursions using trees, Annals of Combinatorics 17.4 (2013): 695-710. See p. 695. - _N. J. A. Sloane_, Apr 16 2014
A. Isgur, R. Lech, S. Moore, S. Tanny, Y. Verberne, and Y. Zhang, Constructing New Families of Nested Recursions with Slow Solutions, SIAM J. Discrete Math., 30(2), 2016, 1128-1147. (20 pages); DOI:10.1137/15M1040505.
B. Jackson and F. Ruskey, Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes
B. Jackson and F. Ruskey, Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes, Electronic Journal of Combinatorics, 13 (2006), #R26, 13 pages.
Oliver Kullmann and Xishun Zhao, Parameters for minimal unsatisfiability: Smarandache primitive numbers and full clauses, arXiv preprint, arXiv:1505.02318 [cs.DM], 2015.
Thomas M. Lewis and Fabian Salinas, Optimal pebbling of complete binary trees and a meta-Fibonacci sequence, arXiv:2109.07328 [math.CO], 2021.
Ramin Naimi and Eric Sundberg, A Combinatorial Problem Solved by a Meta-Fibonacci Recurrence Relation, arXiv:1902.02929 [math.CO], 2019.
Index entries for Hofstadter-type sequences.
Index to sequences related to Olympiads.

Cf. A001511, A005185, A005187, A007843, A055938, A079559, A080578, A101925, A182105, A213714, A226222, A234016, A275363, A324473, A324475, A324477.

Callaghan et al. (2005)'s sequences T_{0,k}(n) for k=1 through 7 are A000012, A046699, A046702, A240835, A241154, A241155, A240830.

a046699 n = a046699_list !! (n-1)
a046699_list = 1 : 1 : zipWith (+) zs (tail zs) where
   zs = map a046699 $ zipWith (-) [2..] a046699_list
-- Reinhard Zumkeller, Jan 02 2012

[ n le 2 select 1 else Self(n - Self(n-1)) + Self(n-1 -Self(n-2)):n in [1..80]]; // Marius A. Burtea, Oct 17 2019

a := proc(n) option remember; if n <= 1 then return 1 end if; if n <= 2 then return 2 end if; return add(a(n - i + 1 - a(n - i)), i = 1 .. 2) end proc # Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca)
a := proc(n) option remember; if n <= 2 then 1 else a(n - a(n-1)) + a(n-1 - a(n-2)); fi; end; # N. J. A. Sloane, Apr 16 2014

a[n_] := (k = 1; While[ !Divisible[(2*++k)!, 2^(n-1)]]; k); a[1] = a[2] = 1; Table[a[n], {n, 1, 72}] (* Jean-François Alcover, Oct 06 2011, after Benoit Cloitre *)
CoefficientList[ Series[1 + x/(1 - x)*Product[1 + x^(2^n - 1), {n, 6}], {x, 0, 80}], x] (* or *)
a[1] = a[2] = 1; a[n_] := a[n] = a[n - a[n - 1]] + a[n - 1 - a[n - 2]]; Array[a, 80] (* Robert G. Wilson v, Sep 08 2014 *)

a[1]:1$
a[2]:1$
a[n]:=a[n-a[n-1]]+a[n-1-a[n-2]]$
makelist(a[n],n,2,60); /* Martin Ettl, Oct 29 2012 */

a(n)=if(n<0,1,s=1;while((2*s)!%2^(n-1)>0,s++);s) \\ Benoit Cloitre, Jan 19 2007

from sympy import factorial
def a(n):
    if n<3: return 1
    s=1
    while factorial(2*s)%(2**(n - 1))>0: s+=1
    return s
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 11 2017, after Benoit Cloitre

First differences seem to be A079559. - Vladeta Jovovic, Nov 30 2003. This is correct and not too hard to prove, giving the generating function x + x^2(1+x)(1+x^3)(1+x^7)(1+x^15).../(1-x). - Paul Boddington, Jul 30 2004
G.f.: x + x^2/(1-x) * Product_{n=1}^{infinity} (1 + x^(2^n-1)). - Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca)
For n>=1, a(n)=w(n-1) where w(n) is the least k such that 2^n divides (2k)!. - Benoit Cloitre, Jan 19 2007
Conjecture: a(n+1) = a(n) + A215530(a(n) + n) for all n > 0. - Velin Yanev, Oct 17 2019
From Bernard Schott, Dec 03 2021: (Start)
a(n) <= a(n+1) <=  a(n) +1.
For n > 1, if a(n) is odd, then a(n+1) = a(n) + 1.
a(2^n+1) = 2^(n-1) + 1 for n > 0.
Results coming from the 5th problem proposed during the 22nd Canadian Mathematical Olympiad in 1990 (link IMO Compendium and Doob reference). (End)

A082850 Let S(0) = {}, S(n) = {S(n-1), S(n-1), n}; sequence gives S(infinity).

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 4, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 4, 5, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 4, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 4, 5, 6, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 4, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 4, 5, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1

Offset: 1

Benoit Cloitre, Apr 14 2003

Sequence counts up to successive values of A001511; i.e., apply the morphism k -> 1,2,...,k to A001511. If all 1's are removed from the sequence, the resulting sequence b has b(n) = a(n)+1. A101925 lists the positions of 1's in this sequence.

The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 30 2025

			S(1) = {1}, S(2) = {1,1,2}, S(3) = {1,1,2,1,1,2,3}, etc.

Alois P. Heinz, Table of n, a(n) for n = 1..65535

Cf. A082851 (partial sums).
Cf. A001511, A101925, A112302.
Cf. A215020.

Fold[Flatten[{#1, #1, #2}] &, {}, Range[5]] (* Birkas Gyorgy, Apr 13 2011 *)
Flatten[Table[Length@Last@Split@IntegerDigits[2 n, 2], {n, 20}] /. {n_ ->Range[n]}] (* Birkas Gyorgy, Apr 13 2011 *)

S = []; [S.extend(S + [n]) for n in range(1, 8)]
print(S) # Michael S. Branicky, Jul 02 2022

from itertools import count, islice
def A082850_gen(): # generator of terms
    S = []
    for n in count(1):
        yield from (m:=S+[n])
        S += m #
A082850_list = list(islice(A082850_gen(),20)) # Chai Wah Wu, Mar 06 2023

a(2^m - 1) = m.
If n = 2^m - 1 + k with 0 < k < 2^m, then a(n) = a(k). - Franklin T. Adams-Watters, Aug 16 2006
a(n) = log_2(A182105(n)) + 1. - Laurent Orseau, Jun 18 2019
a(n) = 1 + A215020(n). - Joerg Arndt, Mar 04 2025

A364675 Number of integer partitions of n whose nonzero first differences are a submultiset of the parts.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 7, 7, 10, 12, 15, 15, 26, 25, 35, 45, 55, 60, 86, 94, 126, 150, 186, 216, 288, 328, 407, 493, 610, 699, 896, 1030, 1269, 1500, 1816, 2130, 2620, 3029, 3654, 4300, 5165, 5984, 7222, 8368, 9976, 11637, 13771, 15960, 18978, 21896, 25815, 29915

Offset: 0

Gus Wiseman, Aug 04 2023

nonn

Conjecture: For subsets of {1..n} instead of partitions of n we have A101925.

Conjecture: The strict version is A154402.

			The partition y = (3,2,1,1) has first differences (1,1,0), and (1,1) is a submultiset of y, so y is counted under a(7).
The a(1) = 1 through a(8) = 10 partitions:
  (1)  (2)   (3)    (4)     (5)      (6)       (7)        (8)
       (11)  (21)   (22)    (221)    (33)      (421)      (44)
             (111)  (211)   (2111)   (42)      (2221)     (422)
                    (1111)  (11111)  (222)     (3211)     (2222)
                                     (2211)    (22111)    (4211)
                                     (21111)   (211111)   (22211)
                                     (111111)  (1111111)  (32111)
                                                          (221111)
                                                          (2111111)
                                                          (11111111)

For subsets of {1..n} we appear to have A101925, A364671, A364672.
The strict case (no differences of 0) appears to be A154402.
Starting with the distinct parts gives A342337.
For disjoint multisets: A363260, subsets A364463, strict A364464.
For overlapping multisets: A364467, ranks A364537, strict A364536.
For subsets instead of submultisets we have A364673.
A000041 counts integer partitions, strict A000009.
A008284 counts partitions by length, strict A008289.
A236912 counts sum-free partitions, complement A237113.
A325325 counts partitions with distinct first differences.
Cf. A002865, A007862, A108917, A229816, A237667, A237668, A320347, A363225, A364272, A364345, A364466.

submultQ[cap_,fat_] := And@@Function[i,Count[fat,i] >= Count[cap,i]] /@ Union[List@@cap];
Table[Length[Select[IntegerPartitions[n], submultQ[Differences[Union[#]],#]&]], {n,0,30}]

A113474 a(n) = a(floor(n/2)) + floor(n/2) with a(1) = 1.

Original entry on oeis.org

1, 2, 2, 4, 4, 5, 5, 8, 8, 9, 9, 11, 11, 12, 12, 16, 16, 17, 17, 19, 19, 20, 20, 23, 23, 24, 24, 26, 26, 27, 27, 32, 32, 33, 33, 35, 35, 36, 36, 39, 39, 40, 40, 42, 42, 43, 43, 47, 47, 48, 48, 50, 50, 51, 51, 54, 54, 55, 55, 57, 57, 58, 58, 64, 64, 65, 65, 67, 67, 68, 68, 71, 71

Offset: 1

Zak Seidov, Jan 08 2006

a(2^n) = 2^n, in other cases a(n) < n. Except for the initial 1 all entries are repeated. Apparently no simple formula is known for a(n).
Taking every other term seems to give A101925. - Dominick Cancilla, Aug 03 2010
1/a(n) is the probability that a randomly chosen divisor of n! is odd. This is because the product n! contains the prime factor 2 a total of a(n) - 1 times (cf. A011371) and thus the prime factor 2 can occur in a divisor of n! a total of a(n) times, ranging between 0 and a(n) - 1 times. - Martin Renner, Dec 28 2022

G. C. Greubel, Table of n, a(n) for n = 1..2500
Tanya Khovanova, There are no coincidences, arXiv:1410.2193 [math.CO], 2014.

Cf. A005187, A011371, A101925.

A113474:= func< n | n+1-Multiplicity(Intseq(n, 2), 1) >;
[A113474(n): n in [1..90]]; // G. C. Greubel, Sep 28 2024

a[1]=1;a[n_]:=a[n]=a[Floor[n/2]]+Floor[n/2]; Table[a[n], {n, 100}]

for(n=1, 90, print1(1 - n + sum(k=0,n, n\2^k), ", ")) \\ G. C. Greubel, Mar 11 2017

a(n)=sum(k=1,logint(n,2), n>>k)+1 \\ Charles R Greathouse IV, Mar 12 2017

a(n)=n+1-hammingweight(n) \\ Charles R Greathouse IV, Dec 29 2022

def A113474(n): return n+1 - sum((n+0).digits(2))
[A113474(n) for n in range(1,90)] # G. C. Greubel, Sep 28 2024

From Paul Barry, Aug 27 2006: (Start)
a(n) = ( Sum_{k=0..n} floor(n/2^k) ) - n + 1.
a(n) = 2 + Sum_{k=0..n} ( floor(n/2^k)-1 ).
a(n) = A005187(n) - n + 1. (End)
a(n) = n + O(log n). - Charles R Greathouse IV, Mar 12 2017
a(n) = A011371(n) + 1 for n > 0. - Martin Renner, Dec 28 2022

A077070 Triangle read by rows: T(n,k) is the power of 2 in denominator of coefficients of Legendre polynomials, where n >= 0 and 0 <= k <= n.

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 4, 4, 4, 4, 7, 5, 6, 5, 7, 8, 8, 7, 7, 8, 8, 10, 9, 10, 8, 10, 9, 10, 11, 11, 11, 11, 11, 11, 11, 11, 15, 12, 13, 12, 14, 12, 13, 12, 15, 16, 16, 14, 14, 15, 15, 14, 14, 16, 16, 18, 17, 18, 15, 17, 16, 17, 15, 18, 17, 18, 19, 19, 19, 19, 18, 18, 18, 18, 19, 19, 19, 19, 22, 20, 21, 20, 22, 19, 20, 19, 22, 20, 21, 20, 22

Offset: 0

Michael Somos, Oct 25 2002

			Triangle T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
   0;
   1,  1;
   3,  2,  3;
   4,  4,  4,  4;
   7,  5,  6,  5,  7;
   8,  8,  7,  7,  8,  8;
  10,  9, 10,  8, 10,  9, 10;
  ...

Alois P. Heinz, Rows n = 0..200, flattened

Cf. A005187 (column k=0), A101925 (column k=1), A077071 (row sums), A144816 (denominators).

Cf. A000120, A007814.

T:= n-> (p-> seq(padic[ordp](denom(coeff(p, x, i)), 2)
             , i=0..2*n, 2))(orthopoly[P](2*n, x)):
seq(T(n), n=0..12);  # Alois P. Heinz, Jan 25 2022

T[n_, k_] := IntegerExponent[Denominator[Coefficient[LegendreP[2n, x], x, 2k]], 2]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 28 2017 *)

{T(n, k) = if( k<0 || k>n, 0, -valuation( polcoeff( pollegendre(2*n), 2*k), 2))}

T(n,k) = 2*n - hammingweight(n-k) - hammingweight(k); \\ Kevin Ryde, Jan 29 2022

T(n, k) = A007814(A144816(n, k)). - Michel Marcus, Jan 29 2022

T(n, k) = 2*n - wt(n-k) - wt(k) where wt = A000120 is the binary weight. - Kevin Ryde, Jan 29 2022

A118319 a(n) = (highest power of 2 dividing n)th integer among those positive integers not occurring in {a(1),a(2),a(3),...,a(n-1)}.

Original entry on oeis.org

1, 3, 2, 7, 4, 6, 5, 15, 8, 10, 9, 14, 11, 13, 12, 31, 16, 18, 17, 22, 19, 21, 20, 30, 23, 25, 24, 29, 26, 28, 27, 63, 32, 34, 33, 38, 35, 37, 36, 46, 39, 41, 40, 45, 42, 44, 43, 62, 47, 49, 48, 53, 50, 52, 51, 61, 54, 56, 55, 60, 57, 59, 58, 127, 64, 66, 65, 70, 67, 69, 68, 78

Offset: 1

Leroy Quet, Apr 23 2006

Sequence is a permutation of the positive integers. a(2n-1) is the smallest positive integer not occurring earlier in the sequence.

A101925 is the odd bisection, and it seems that A045412 is the sorted even bisection: a(2*n) = A045412(a(n)). - Andrey Zabolotskiy, Oct 09 2019

			4 is the highest power of 2 dividing 12. Those positive integers not occurring among the first 11 terms of the sequence form the sequence 11, 12, 13, 14, 16,... Now 14 is the 4th of these integers, so a(12) = 14.

Andrey Zabolotskiy, Table of n, a(n) for n = 1..8192
Index entries for sequences that are permutations of the natural numbers

Cf. A108918 (inverse permutation), A000120, A006519, A045412, A101925.

A118319 := proc(nmin) local a,anxt,i,n ; a := [1] ; while nops(a) < nmin do n := nops(a)+1 ; i := 2^A007814(n); anxt := 0 ; while i > 0 do anxt := anxt+1 ; while anxt in a do anxt := anxt+1 ; od ; i := i-1; od ; a := [op(a),anxt] ; od; a ; end: A118319(80) ; # R. J. Mathar, Sep 06 2007
a := n -> n + 2^padic[ordp](n, 2) - add(convert(n, base, 2)): seq(a(n), n = 1..72); # Peter Luschny, Mar 08 2025

a[1] := 1; a[n_] := a[n] =  Part[ Complement[ Range[2 n], Table[a[i], {i, n - 1}]],  2^IntegerExponent[n, 2]]; Array[a, 100] (* Birkas Gyorgy, Jul 09 2012 *)

PARI
```
a(n) = n + 1<Kevin Ryde, Mar 02 2025
```

a(2^m) = 2^(m+1) - 1; a(2^m+k) = a(k) + 2^m - 1 for 0 < k < 2^m. - Andrey Zabolotskiy, Oct 10 2019

a(n) = n + A006519(n) - A000120(n). - Kevin Ryde, Mar 08 2025

More terms from R. J. Mathar, Sep 06 2007

A144816 Denominators of triangle T(n,k), n>=0, 0<=k<=n, read by rows: T(n,k) is the coefficient of x^(2*k+1) in polynomial t_n(x), used to define continuous and n times differentiable sigmoidal transfer functions.

Original entry on oeis.org

1, 2, 2, 8, 4, 8, 16, 16, 16, 16, 128, 32, 64, 32, 128, 256, 256, 128, 128, 256, 256, 1024, 512, 1024, 256, 1024, 512, 1024, 2048, 2048, 2048, 2048, 2048, 2048, 2048, 2048, 32768, 4096, 8192, 4096, 16384, 4096, 8192, 4096, 32768, 65536, 65536, 16384, 16384, 32768, 32768, 16384, 16384, 65536, 65536

Offset: 0

Alois P. Heinz, Sep 21 2008

			Triangle begins:
    1;
    2,  2;
    8,  4,  8;
   16, 16, 16, 16;
  128, 32, 64, 32, 128;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

See A144815 for more information on T(n,k).
Main diagonal and column k=0 gives A046161.
Column k=1 gives A101926(n-1) = 2^A101925(n-1) = 2^(A005187(n-1)+1).
Cf. A077070.

# Function T(n,k) defined in A144815.
seq(seq(denom(T(n,k)), k=0..n), n=0..10);

row[n_] := Module[{f, a, eq}, f = Function[x, Sum[a[2*k+1]*x^(2*k+1), {k, 0, n}]]; eq = Table[Derivative[k][f][1] == If[k == 0, 1, 0], {k, 0, n}]; Table[a[2*k+1], {k, 0, n}] /. Solve[eq] // First]; Table[row[n] // Denominator, {n, 0, 10}] // Flatten (* Jean-François Alcover, Feb 03 2014 *)

A273926 Given G(x) such that G( G(x)^2 - G(x)^3 ) = x^2, then G(x) = Sum_{n>=1} A273925(n)*x^n / 2^a(n).

Original entry on oeis.org

0, 1, 3, 2, 7, 4, 10, 5, 15, 8, 18, 9, 22, 11, 25, 12, 31, 16, 34, 17, 38, 19, 41, 20, 46, 23, 49, 24, 53, 26, 56, 27, 63, 32, 66, 33, 70, 35, 73, 36, 78, 39, 81, 40, 85, 42, 88, 43, 94, 47, 97, 48, 101, 50, 104, 51, 109, 54, 112, 55, 116, 57, 119, 58, 127, 64, 130, 65, 134, 67, 137, 68, 142, 71, 145, 72, 149, 74, 152, 75, 158, 79, 161, 80, 165, 82, 168, 83, 173, 86, 176, 87, 180, 89, 183, 90, 190, 95, 193, 96, 197, 98, 200, 99, 205, 102, 208, 103, 212, 105, 215, 106, 221, 110, 224, 111, 228, 113, 231, 114

Offset: 1

Paul D. Hanna, Jun 04 2016

nonn

Terms appear to occur only once in the sequence.

Both bisections of this sequence appear to be monotonically increasing.

			G.f.: G(x) = x + 1/2*x^2 + 3/8*x^3 + 3/4*x^4 + 175/128*x^5 + 41/16*x^6 + 4947/1024*x^7 + 321/32*x^8 + 687611/32768*x^9 + 11403/256*x^10 + 25132181/262144*x^11 + 107305/512*x^12 + 1941554203/4194304*x^13 + 2111325/2048*x^14 + 77643067507/33554432*x^15 + 21427329/4096*x^16 + 25549683166419/2147483648*x^17 + 1782548851/65536*x^18 + 1073363084982753/17179869184*x^19 + 18891311061/131072*x^20 + 91744420207896017/274877906944*x^21 + 406630578535/524288*x^22 + 3975787925128277349/2199023255552*x^23 + 4432136534071/1048576*x^24 +...+ A273925(n)*x^n / 2^a(n) +...
such that G( G(x)^2 - G(x)^3 ) = x^2.
The bisections of this sequence begin:
odd bisection (cf. A120738): [0, 3, 7, 10, 15, 18, 22, 25, 31, 34, 38, 41, 46, 49, 53, 56, 63, 66, 70, 73, 78, 81, 85, 88, 94, 97, 101, 104, 109, 112, 116, 119, 127, 130, 134, 137, 142, 145, 149, 152, 158, 161, 165, 168, 173, 176, 180, 183, 190, 193, 197, 200, 205, 208, 212, 215, 221, 224, 228, 231, 236, 239, 243, 246, 255, ...].
even bisection (cf. A101925): [1, 2, 4, 5, 8, 9, 11, 12, 16, 17, 19, 20, 23, 24, 26, 27, 32, 33, 35, 36, 39, 40, 42, 43, 47, 48, 50, 51, 54, 55, 57, 58, 64, 65, 67, 68, 71, 72, 74, 75, 79, 80, 82, 83, 86, 87, 89, 90, 95, 96, 98, 99, 102, 103, 105, 106, 110, 111, 113, 114, 117, 118, 120, 121, 128, ...].

Paul D. Hanna, Table of n, a(n) for n = 1..261

Cf. A273925, A120738, A101925, A000120, A005187.

{a(n) = my(A=x); for(i=0,n, A = serreverse( sqrt(subst(A,x,x^2 - x^3 +x^2*O(x^n) )) )); valuation(denominator(polcoeff(A,n)),2)}
for(n=1,60,print1(a(n),", "))

a(2*n-1) = A120738(n-1) = 4*(n-1) - A000120(n-1), for n>=0 (conjecture).

a(2*n) = A101925(n-1) = A005187(n-1) + 1, for n>=0 (conjecture).

cp's OEIS Frontend

A101926 a(n) = 2^A101925(n).

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A048881 a(n) = A000120(n+1) - 1 = wt(n+1) - 1.

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A046699 a(1) = a(2) = 1, a(n) = a(n - a(n-1)) + a(n-1 - a(n-2)) if n > 2.

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A082850 Let S(0) = {}, S(n) = {S(n-1), S(n-1), n}; sequence gives S(infinity).

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A364675 Number of integer partitions of n whose nonzero first differences are a submultiset of the parts.

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A113474 a(n) = a(floor(n/2)) + floor(n/2) with a(1) = 1.

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