A092119 -id:A092119 - cp's OEIS frontend

A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1

Offset: 1

N. J. A. Sloane

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where mBetween every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
  1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
  1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
  1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 2nd ed., 2001-2003; see Dim- and Dim+ on p. 98; Dividing Rulers, on pp. 436-437; The Ruler Game, pp. 469-470; Ruler Fours, Fives, ... Fifteens on p. 470.
L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
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Antonin Callard, Léo Paviet Salomon, and Pascal Vanier, Computability of extender sets in multidimensional subshifts, arXiv:2401.07549 [cs.DM], 2024.
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Eric Weisstein's World of Mathematics, Binary Carry Sequence, Ruler Function, and Towers of Hanoi
Index entries for "core" sequences
Index entries for sequences related to binary expansion of n
Index entries for sequences that are fixed points of mappings
Index entries for sequences related to Towers of Hanoi

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A000005, A000041, A000079, A003188, A003278, A003602, A007949, A018238, A047999, A051731, A054525, A054852, A065176, A082850, A089080, A092119, A117303, A129360, A130093, A173238, A181988, A220466.
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
-- Reinhard Zumkeller, Sep 27 2011

a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
-- Walt Rorie-Baety, Mar 22 2013

nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016

[Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015

A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
# This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018

Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)

a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */

a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */

{a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */

{a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */

def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017

A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020

def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022

[valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015

(define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Name edited following suggestion by David James Sycamore, Oct 05 2023

A173241 Euler transform of A051064, the ruler function sequence for k=3.

Original entry on oeis.org

1, 1, 2, 4, 6, 9, 16, 22, 33, 51, 71, 100, 147, 199, 275, 384, 515, 692, 944, 1242, 1645, 2186, 2847, 3706, 4848, 6231, 8019, 10330, 13153, 16729, 21305, 26864, 33858, 42658, 53366, 66668, 83277, 103378, 128200, 158846, 195895, 241237, 296860, 363796, 445285, 544465, 663520

Offset: 0

Gary W. Adamson, Feb 13 2010

nonn

Let P(x) = polcoeff A000041: (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...) and
A(x) = polcoeff A173241: (1 + x + 2x^2 + 4x^3 + 6x^4 + 9x^5 + ...); then
P(x) = A(x) / A(x^3).
A092119 = Euler transform of the ruler function for k=2: A001511.

			Equals 1/((1-x)*(1-x^2)*(1-x^3)^2*(1-x^4)*(1-x^5)*(1-x^6)^2*(1-x^7)*...); where in (1-x)^k, k = A051064: (1, 1, 2, 1, 1, 2, 1, 1, 3, ...).

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A000041, A001511, A051064, A092119, A173238, A173239.

N=66; x='x+O('x^N); /* that many terms */
gf=1/prod(e=0, ceil(log(N)/log(3)), eta(x^(3^e)));
Vec(gf) /* show terms */ /* Joerg Arndt, Jun 21 2011 */

G.f.: 1/Product_{k>=0} P(x^(3^k)) where P(x)=Product_{k>=1} (1-x^k). - Joerg Arndt, Jun 21 2011
Euler transform of A051064, where A051064 = the ruler function for k=3:
(1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, ...).

More terms from Joerg Arndt, Jun 21 2011

A173238 Triangle by columns, A000041 in every column shifted down twice for columns > 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 5, 2, 1, 7, 3, 1, 11, 5, 2, 1, 15, 7, 3, 1, 22, 11, 5, 2, 1, 30, 15, 7, 3, 1, 42, 22, 11, 5, 2, 1, 56, 30, 15, 7, 3, 1, 77, 42, 22, 11, 5, 2, 1, 101, 56, 30, 15, 7, 3, 1, 135, 77, 42, 22, 11, 5, 2, 1, 176, 101, 56, 30, 15, 7, 3, 1

Offset: 0

Gary W. Adamson, Feb 13 2010

Row sums = A024786 starting with A024786(2): (1, 1, 3, 4, 8, 11, 19, 26, ...) = number of 2's in all partitions of n.
A173238 as an infinite lower triangular matrix * [1, 2, 3, ...] = A103650.
Let the triangle = M. Then Lim_{n->inf} M^n = (1, 1, 3, 4, 10, 13, 26, ...), the left-shifted vector considered as a sequence = A092119, the Euler transform of the ruler sequence, A001511.
Given P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x)/A(x^2), where A(x) = polcoeff A092119: (1 + x + 3x^2 + 4x^3 + ...).
Conjectures: Given the infinite set of triangles with A000041 in every column shifted down 0, 1, 2, ... n times, row sums of n-th triangle (where A173238 = 2nd in the set) = the numbers of n's in all partitions of n. E.g., row sums = of A173238 = A024786, the numbers of 2's in all partitions of n.
Similarly, row sums of triangle A173239 with columns >0 shifted down thrice = numbers of 3's in all partitions of n, and so on. Refer to comments in A000041 regarding the numbers of 1's in partitions of n.
...
Second conjecture: Given the infinite set of analogous triangles with columns shifted down 2, 3, 4, ..., k times, we let such triangles = T(k) and perform lim_{n->inf} T^n(k), obtaining the left-shifted vectors considered as sequences. The conjecture states that the infinite set of such left-shifted vectors = the Euler transform of the infinite set of Ruler functions starting with the ruler function for k=2 = A001511: (1, 2, 1, 3, 1, 2, 1, ...)
To obtain the k-th ruler functions, begin with the natural numbers, 1,...2,...3,...4,...5,...6,...7,...8,...9,...; then for k = 2 we get A001511: 1,...2,...1,...3,...1,...2,...1,...4,...1,...; by finding the highest exponent of k dividing n, then adding 1. Similarly, for k = 2, we obtain A051064: 1,...1,...2,...1,...1,...2,...1,...1,...3,...
Next, we obtain the Euler transforms of the ruler functions (e.g., Euler transform of A001511 = A092119: (1, 1, 3, 4, 10, 13, 26, ...), noting that A092119 is lim_{n->inf} A173238^n, the left-shifted vector.
...
Third conjecture: Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + ...), and A(k)(x) = the Euler transform of k-th ruler sequence, (k=2,3,...). Then P(x) = A(k)(x) / A(k)(x^k).
Examples: for k=2, A(x) = A092119: (1, 1, 3, 4, 10, 13, ...), then P(x) = (1 + x + 2x^2 + 3x^3 + ...) = (1 + x + 3x^2 + 4x^3 + ...) / (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...). For k=3 relating to triangle A173238, the left-shifted vector = the Euler transform of A051064 = A(x) for k=3, then P(x) = A(x) / A(x^3).
The conjecture extends the analogous conclusions to all k.
From Gary W. Adamson, Feb 25 2010: (Start)
Proof of second conjecture received from Helmut Prodinger 02/28/10 with a summary by R. J. Mathar:
Consider Product_{n>=0} z^(t^n)/(1-z^(t^n)) = Sum_{k>=1} (1+v_t(k))z^k where v_t(n) is the number of trailing zeros in the t-ary expansion of n, and its Euler transform A(z) = product_{k >= 1} 1/(1-z^k)^{1+v_t(k)}, then A(z)/A(z^t) = product_{k >= 1} 1/(1-z^k) is the partition generating function.
Here is the proof: A(z)/A(z^t) = Product_{k>=1} (1-z^(tk))^(1+v_t(k))/(1-z^k)^(1+v_t(k))
= Product_{k>=1} (1-z^(tk))^(v_t(tk))/(1-z^k)^(1+v_t(k))
= Product_{k>=1} (1-z^k)^(v_t(k))/(1-z^k)^(1+v_t(k)) (*)
= Product_{k>=1} 1/(1-z^k)
as desired. Notice that for (*), that v_t(n)=0 if n is not divisible by t. [Helmut Prodinger, hproding(AT)sun.ac.za, Feb 28 2010] (End)

			First few rows of the triangle:
    1;
    1;
    2,   1;
    3,   1;
    5,   2,  1;
    7,   3,  1;
   11,   5,  2,  1;
   15,   7,  3,  1;
   22,  11,  5,  2,  1;
   30,  15,  7,  3,  1;
   42,  22, 11,  5,  2, 1;
   56,  30, 15,  7,  3, 1;
   77,  42, 22, 11,  5, 2, 1;
  101,  56, 30, 15,  7, 3, 1;
  135,  77, 42, 22, 11, 5, 2, 1;
  176, 101, 56, 30, 15, 7, 3, 1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..9999 (rows 0 <= n <= 199, flattened)

Cf. A000041, A001511, A092119, A173239.

Table[PartitionsP[n - 2 k], {n, 17}, {k, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 23 2021 *)

T(n,k) = A000041(n-2*k) for k=0..floor(n/2).

a(24), a(25) corrected by Georg Fischer, Nov 23 2021

A327736 Expansion of 1 / (1 - Sum_{i>=1, j>=0} x^(i*2^j)).

Original entry on oeis.org

1, 1, 3, 6, 16, 35, 85, 195, 465, 1081, 2549, 5962, 14016, 32847, 77119, 180866, 424466, 995753, 2336497, 5481712, 12861904, 30176671, 70802913, 166120289, 389761751, 914476925, 2145596677, 5034105820, 11811287658, 27712248159, 65019931641, 152553127471, 357928110743

Offset: 0

Ilya Gutkovskiy, Sep 23 2019

nonn

Invert transform of A001511.

Cf. A000041, A001511, A092119, A129921.

nmax = 32; CoefficientList[Series[1/(1 - Sum[x^(2^k)/(1 - x^(2^k)), {k, 0, Floor[Log[2, nmax]] + 1}]), {x, 0, nmax}], x]
a[0] = 1; a[n_] := a[n] = Sum[IntegerExponent[2 k, 2] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 32}]

G.f.: 1 / (1 - Sum_{k>=0} x^(2^k) / (1 - x^(2^k))).

a(0) = 1; a(n) = Sum_{k=1..n} A001511(k) * a(n-k).

A373295 Expansion of 1/Product_{k>=1} (1 - x^k)^(valuation(k,4) + 1).

Original entry on oeis.org

1, 1, 2, 3, 6, 8, 13, 18, 29, 39, 57, 77, 112, 148, 205, 271, 373, 485, 649, 841, 1116, 1431, 1865, 2379, 3080, 3896, 4979, 6268, 7961, 9953, 12524, 15585, 19505, 24135, 29984, 36943, 45678, 56007, 68841, 84080, 102912, 125164, 152449, 184756, 224184, 270691, 327094, 393675

Offset: 0

Seiichi Manyama, May 31 2024

nonn

Cf. A092119, A173241, A373296, A373297, A373298.

Cf. A000041, A115362, A174065.

my(N=50, x='x+O('x^N)); Vec(1/prod(k=1, N, (1-x^k)^(valuation(k, 4)+1)))

G.f.: A(x) = 1/Product_{i>=1, j>=0} (1 - x^(i * 4^j)).

Let A(x) be the g.f. of this sequence, and P(x) be the g.f. of A000041, then P(x) = A(x)/A(x^4).

A373296 Euler transform of A055457.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 12, 17, 25, 35, 51, 69, 96, 129, 175, 235, 312, 410, 539, 700, 913, 1173, 1508, 1923, 2450, 3106, 3921, 4928, 6180, 7715, 9622, 11935, 14783, 18243, 22470, 27601, 33819, 41327, 50407, 61325, 74494, 90244, 109154, 131732, 158725, 190892, 229171, 274633, 328615

Offset: 0

Seiichi Manyama, May 31 2024

nonn

Cf. A092119, A173241, A373295, A373297, A373298.

Cf. A000041, A055457, A373219.

my(N=50, x='x+O('x^N)); Vec(1/prod(k=1, N, (1-x^k)^(valuation(k, 5)+1)))

G.f.: A(x) = 1/Product_{i>=1, j>=0} (1 - x^(i * 5^j)).

Let A(x) be the g.f. of this sequence, and P(x) be the g.f. of A000041, then P(x) = A(x)/A(x^5).

A373297 Euler transform of A373216.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 12, 16, 24, 33, 47, 63, 90, 118, 161, 212, 283, 367, 487, 624, 812, 1037, 1332, 1685, 2152, 2700, 3409, 4259, 5333, 6617, 8242, 10165, 12568, 15436, 18970, 23178, 28360, 34487, 41970, 50850, 61599, 74322, 89696, 107809, 129572, 155235, 185881, 221936

Offset: 0

Seiichi Manyama, May 31 2024

nonn

Cf. A092119, A173241, A373295, A373296, A373298.

Cf. A000041, A373216, A373220.

my(N=50, x='x+O('x^N)); Vec(1/prod(k=1, N, (1-x^k)^(valuation(k, 6)+1)))

G.f.: A(x) = 1/Product_{i>=1, j>=0} (1 - x^(i * 6^j)).

Let A(x) be the g.f. of this sequence, and P(x) be the g.f. of A000041, then P(x) = A(x)/A(x^6).

A373298 Euler transform of A373217.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 11, 16, 23, 32, 45, 61, 84, 112, 152, 200, 265, 345, 451, 581, 750, 960, 1225, 1552, 1965, 2470, 3101, 3872, 4830, 5990, 7421, 9152, 11270, 13825, 16932, 20672, 25191, 30608, 37129, 44920, 54257, 65376, 78660, 94419, 113172, 135370, 161687, 192752

Offset: 0

Seiichi Manyama, May 31 2024

nonn

Cf. A092119, A173241, A373295, A373296, A373297.

Cf. A000041, A373217, A373221.

my(N=50, x='x+O('x^N)); Vec(1/prod(k=1, N, (1-x^k)^(valuation(k, 7)+1)))

G.f.: A(x) = 1/Product_{i>=1, j>=0} (1 - x^(i * 7^j)).

Let A(x) be the g.f. of this sequence, and P(x) be the g.f. of A000041, then P(x) = A(x)/A(x^7).

A327727 Expansion of Product_{i>=1, j>=0} (1 + x^(i2^j)) / (1 - x^(i2^j)).

Original entry on oeis.org

1, 2, 6, 12, 28, 52, 104, 184, 340, 578, 1004, 1652, 2752, 4404, 7088, 11080, 17362, 26592, 40730, 61284, 92096, 136408, 201608, 294456, 428952, 618658, 889684, 1268624, 1803520, 2545164, 3580784, 5005584, 6976046, 9667164, 13356364, 18360368, 25165732

Offset: 0

Ilya Gutkovskiy, Sep 23 2019

nonn

Convolution of the sequences A000041 and A092119.

Cf. A000041, A001511, A085058, A092119, A170925.

nmax = 36; CoefficientList[Series[Product[1/(1 - x^k)^(IntegerExponent[2 k, 2] + 1), {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d (IntegerExponent[2 d, 2] + 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 36}]

seq(n)={Vec(1/prod(k=1, n, (1 - x^k + O(x*x^n))^(2+valuation(k, 2))))} \\ Andrew Howroyd, Sep 23 2019

G.f.: Product_{k>=1} ((1 + x^k) / (1 - x^k))^A001511(k).

G.f.: Product_{k>=1} 1 / (1 - x^k)^(A001511(k) + 1).

Showing 1-9 of 9 results.

cp's OEIS Frontend

A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

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A173241 Euler transform of A051064, the ruler function sequence for k=3.

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A173238 Triangle by columns, A000041 in every column shifted down twice for columns > 0.

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A327736 Expansion of 1 / (1 - Sum_{i>=1, j>=0} x^(i*2^j)).

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A373295 Expansion of 1/Product_{k>=1} (1 - x^k)^(valuation(k,4) + 1).

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A373296 Euler transform of A055457.

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A373297 Euler transform of A373216.

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A327727 Expansion of Product_{i>=1, j>=0} (1 + x^(i2^j)) / (1 - x^(i2^j)).