cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A083346 Denominator of r(n) = Sum(e/p: n=Product(p^e)).

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 3, 10, 11, 3, 13, 14, 15, 1, 17, 6, 19, 5, 21, 22, 23, 6, 5, 26, 1, 7, 29, 30, 31, 2, 33, 34, 35, 3, 37, 38, 39, 10, 41, 42, 43, 11, 15, 46, 47, 3, 7, 10, 51, 13, 53, 2, 55, 14, 57, 58, 59, 15, 61, 62, 21, 1, 65, 66, 67, 17, 69, 70, 71, 6, 73, 74, 15, 19, 77, 78
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 25 2003

Keywords

Comments

Multiplicative with a(p^e) = 1 iff p|e, p otherwise. For f(n) = A083345(n)/A083346(n), f(p^i*q^j*...) = f(p^i)+f(q^j)+ ... The denominator of each term is 1 or the prime, thus the denominator of the sum is the product of the denominators of the components. - Christian G. Bower, May 16 2005
n divided by the greatest common divisor of n and its arithmetic derivative, i.e., a(n) = n/gcd(n,n') = A000027(n)/A085731(n). - Giorgio Balzarotti, Apr 14 2011

Examples

			n=12 = 2*2*3 = 2^2 * 3^1 -> r(12) = 2/2 + 1/3 = (6+2)/6, therefore a(12)=3, A083345(12)=4;
n=18 = 2*3*3 = 2^1 * 3^2 -> r(18) = 1/2 + 2/3 = (3+4)/6, therefore a(18)=6, A083345(18)=7.
		

Crossrefs

Cf. A083345 (numerator).
Cf. A035263 (parity of terms), A003159 (positions of odd terms), A036554 (of even terms).
Cf. A065463, A072873, A083347, A083348, A359588 (Dirichlet inverse).

Programs

  • Mathematica
    a[n_] := Product[Module[{p, e}, {p, e} = pe; If[Divisible[e, p], 1, p]], {pe, FactorInteger[n]}];
    Array[a, 100] (* Jean-François Alcover, Oct 06 2021 *)
  • PARI
    A083346(n) = { my(f=factor(n)); denominator(vecsum(vector(#f~,i,f[i,2]/f[i,1]))); }; \\ Antti Karttunen, Mar 01 2018

Formula

Sum_{k=1..n} a(k) ~ c * n^2, where c = A065463 * Product_{p prime} (p^(2*p)*(p^2+p-1)-p^3)/((p^2+p-1)*(p^(2*p)-1)) = 0.3374565531... . - Amiram Eldar, Nov 18 2022

Extensions

Incorrect formula removed by Antti Karttunen, Jan 09 2023

A056832 All a(n) = 1 or 2; a(1) = 1; get next 2^k terms by repeating first 2^k terms and changing last element so sum of first 2^(k+1) terms is odd.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1
Offset: 1

Views

Author

Jonas Wallgren, Aug 30 2000

Keywords

Comments

Dekking (2016) calls this the Toeplitz sequence or period-doubling sequence. - N. J. A. Sloane, Nov 08 2016
Fixed point of the morphism 1->12 and 2->11 (1 -> 12 -> 1211 -> 12111212 -> ...). - Benoit Cloitre, May 31 2004
a(n) is multiplicative. - Christian G. Bower, Jun 03 2005
a(n) is the least k such that A010060(n-1+k) = 1 - A010060(n-1); the sequence {a(n+1)-1} is the characteristic sequence for A079523. - Vladimir Shevelev, Jun 22 2009
The squarefree part of the even part of n. - Peter Munn, Dec 03 2020

Examples

			1 -> 1,2 -> 1,2,1,1 -> 1,2,1,1,1,2,1,2 -> 1,2,1,1,1,2,1,2,1,2,1,1,1,2,1,1.
Here we have 1 element, then 2 elements, then 4, 8, 16, etc.
		

References

  • Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991; pp. 277-279.

Crossrefs

Cf. A197911 (partial sums).
Essentially same as first differences of Thue-Morse, A010060. - N. J. A. Sloane, Jul 02 2015
See A035263 for an equivalent version.
Limit of A317956(n) for large n.
Row/column 2 of A059895.
Positions of 1s: A003159.
Positions of 2s: A036554.
A002425, A006519, A079523, A096268, A214682, A234957 are used in a formula defining this sequence.
A059897 is used to express relationship between terms of this sequence.

Programs

  • Haskell
    a056832 n = a056832_list !! (n-1)
    a056832_list = 1 : f [1] where
       f xs = y : f (y : xs) where
              y = 1 + sum (zipWith (*) xs $ reverse xs) `mod` 2
    -- Reinhard Zumkeller, Jul 29 2014
    
  • Mathematica
    Nest[ Function[l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {1, 1}})]}], {1}, 7] (* Robert G. Wilson v, Mar 03 2005 *)
    Table[Mod[-(-1)^(n + 1) (-1)^n Numerator[EulerE[2 n + 1, 1]], 3] , {n, 0, 120}] (* Michael De Vlieger, Aug 15 2016, after Jean-François Alcover at A002425 *)
  • PARI
    a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%3
    
  • PARI
    a(n)=if(n<1, 0, valuation(n,2)%2+1) /* Michael Somos, Jun 18 2005 */
    
  • Python
    def A056832(n): return 1+((~n&n-1).bit_length()&1) # Chai Wah Wu, Jan 09 2023

Formula

a(n) = ((-1)^(n+1)*A002425(n)) modulo 3. - Benoit Cloitre, Dec 30 2003
a(1)=1, a(n) = 1 + ((Sum_{i=1..n-1} a(i)*a(n-i)) mod 2). - Benoit Cloitre, Mar 16 2004
a(n) is multiplicative with a(2^e) = 1 + (1-(-1)^e)/2, a(p^e)=1 if p > 2. - Michael Somos, Jun 18 2005
[a(2^n+1) .. a(2^(n+1)-1)] = [a(1) .. a(2^n-1)]; a(2^(n+1)) = 3 - a(2^n).
For n > 0, a(n) = 2 - A035263(n). - Benoit Cloitre, Nov 24 2002
a(n)=2 if n-1 is in A079523; a(n)=1 otherwise. - Vladimir Shevelev, Jun 22 2009
a(n) = A096268(n-1) + 1. - Reinhard Zumkeller, Jul 29 2014
From Peter Munn, Dec 03 2020: (Start)
a(n) = A007913(A006519(n)) = A006519(n)/A234957(n).
a(n) = A059895(n, 2) = n/A214682(n).
a(n*k) = (a(n) * a(k)) mod 3.
a(A059897(n, k)) = A059897(a(n), a(k)).
(End)
Asymptotic mean: lim_{m->oo} (1/m) * Sum__{k=1..m} a(k) = 4/3. - Amiram Eldar, Mar 09 2021

A161579 Positions n such that A010060(n) = A010060(n+3).

Original entry on oeis.org

0, 1, 3, 4, 6, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 38, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 72, 73, 75, 76, 77, 79, 80, 81, 83, 84, 86, 88, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 105, 107
Offset: 1

Views

Author

Vladimir Shevelev, Jun 14 2009

Keywords

Comments

Or: union of A131323 with the sequence of terms of the form A131323(n)-2, and with the sequence of terms of the form A036554(n)-2.
Conjecture: In every sequence of numbers n such that A010060(n)=A010060(n+k), for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. - Vladimir Shevelev, Jul 31 2009

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] == tm[n+3], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
  • PARI
    is(n)=hammingweight(n)%2==hammingweight(n+3)%2 \\ Charles R Greathouse IV, Aug 20 2013

Formula

Equals {A001477} \ {A161580}.

Extensions

More terms from R. J. Mathar, Aug 17 2009

A161580 Positions n such that A010060(n) + A010060(n+3) = 1.

Original entry on oeis.org

2, 5, 7, 10, 14, 18, 21, 23, 26, 29, 31, 34, 37, 39, 42, 46, 50, 53, 55, 58, 62, 66, 69, 71, 74, 78, 82, 85, 87, 90, 93, 95, 98, 101, 103, 106, 110, 114, 117, 119, 122, 125, 127, 130, 133, 135, 138, 142, 146, 149, 151, 154, 157, 159, 162, 165, 167, 170, 174, 178, 181, 183, 186
Offset: 1

Views

Author

Vladimir Shevelev, Jun 14 2009

Keywords

Comments

Conjecture: In every sequence of numbers n such that A010060(n) + A010060(n+k) = 1, for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [From Vladimir Shevelev, Jul 31 2009]

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] + tm[n+3] == 1, Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
  • PARI
    is(n)=hammingweight(n)%2+hammingweight(n+3)%2==1 \\ Charles R Greathouse IV, Mar 22 2013

Formula

Extensions

More terms from R. J. Mathar, Aug 17 2009

A161627 Positions n such that A010060(n)=A010060(n+4).

Original entry on oeis.org

4, 5, 6, 7, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, 52, 53, 54, 55, 68, 69, 70, 71, 84, 85, 86, 87, 92, 93, 94, 95, 100, 101, 102, 103, 116, 117, 118, 119, 124, 125, 126, 127, 132, 133, 134, 135, 148, 149, 150, 151, 156, 157, 158, 159, 164, 165, 166, 167, 180, 181, 182
Offset: 1

Views

Author

Vladimir Shevelev, Jun 15 2009

Keywords

Comments

Or: union of the numbers of the form 4*A079523(n)+k, k=0, 1, 2, or 3.
Locates patterns of the form 1xxx1 or 0xxx0 in the Thue-Morse sequence.

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] == tm[n+4], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
    SequencePosition[ThueMorse[Range[200]],{x_,,,_,x_}][[All,1]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Apr 16 2017 *)
  • PARI
    is(n)=hammingweight(n)%2==hammingweight(n+4)%2 \\ Charles R Greathouse IV, Aug 20 2013

Extensions

Extended by R. J. Mathar, Aug 28 2009

A161639 Positions n such that A010060(n) = A010060(n+8).

Original entry on oeis.org

8, 9, 10, 11, 12, 13, 14, 15, 40, 41, 42, 43, 44, 45, 46, 47, 56, 57, 58, 59, 60, 61, 62, 63, 72, 73, 74, 75, 76, 77, 78, 79, 104, 105, 106, 107, 108, 109, 110, 111, 136, 137, 138, 139, 140, 141, 142, 143, 168, 169, 170, 171, 172, 173, 174, 175, 184, 185, 186, 187, 188, 189
Offset: 1

Views

Author

Vladimir Shevelev, Jun 15 2009

Keywords

Comments

Locates correlations of the form 1xxxxxxx1 or 0xxxxxxx0 in the Thue-Morse sequence.
Or: union of numbers 8*A079523(n)+k, k=0, 1, 2, 3, 4, 5, 6, or 7.
Generalization: the numbers n such that A010060(n) = A010060(n+2^m) constitute the union of sequences {2^m*A079523(n)+k}, k=0,1,...,2^m-1.

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] == tm[n+8], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
    SequencePosition[ThueMorse[Range[0,200]],{x_,,,_,,,_,,x}][[All,1]]-1 (* Harvey P. Dale, Jul 23 2021 *)
  • PARI
    is(n)=hammingweight(n)%2==hammingweight(n+8)%2 \\ Charles R Greathouse IV, Aug 20 2013

Extensions

Duplicate of 174 removed by R. J. Mathar, Aug 28 2009

A029883 First differences of Thue-Morse sequence A001285.

Original entry on oeis.org

1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, 0, -1
Offset: 1

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Also first differences of {0,1} Thue-Morse sequence A010060.- N. J. A. Sloane, Jan 05 2021
Fixed point of the morphism a->abc, b->ac, c->b, with a = 1, b = 0, c = -1, starting with a(1) = 1. - Philippe Deléham
From Thomas Anton, Sep 22 2020: (Start)
This sequence, interpreted as an infinite word, is squarefree.
Let & represent concatenation. For a word w of integers, let -w be the same word with each symbol negated. Then, starting with the empty word, this sequence can be obtained by iteratively applying the transformation T(w) = w & 1 & -w & 0 & -w & -1 & w. (End)

Crossrefs

Apart from signs, same as A035263. Cf. A001285, A010060, A036554, A091785, A091855.
a(n+1) = A036577(n) - 1 = A036585(n) - 2.

Programs

  • Mathematica
    Nest[ Function[ l, {Flatten[(l /. {0 -> {1, -1}, 1 -> {1, 0, -1}, -1 -> {0}})]}], {1}, 7] (* Robert G. Wilson v, Feb 26 2005 *)
    ThueMorse /@ Range[0, 105] // Differences (* Jean-François Alcover, Oct 15 2019 *)
  • PARI
    a(n)=if(n<1||valuation(n,2)%2,0,-(-1)^subst(Pol(binary(n)),x,1)) /* Michael Somos, Jul 08 2004 */
    
  • PARI
    a(n)=hammingweight(n)%2-hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 26 2013
    
  • Python
    def A029883(n): return (bin(n).count('1')&1)-(bin(n-1).count('1')&1) # Chai Wah Wu, Mar 03 2023

Formula

Recurrence: a(4*n) = a(n), a(4*n+1) = a(2*n+1), a(4*n+2) = 0, a(4*n+3) = -a(2*n+1), starting a(1) = 1.
a(n) = 2 - A007413(n). a(A036554(n)) = 0; a(A091785(n)) = -1; a(A091855(n)) = 1. - Philippe Deléham, Mar 20 2004
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = -v+w+u^2-v^2+2*w^2-2*u*w. - Michael Somos, Jul 08 2004

Extensions

Edited by Ralf Stephan, Dec 09 2004

A325424 Complement of A036668: numbers not of the form 2^i*3^j*k, i + j even, (k,6) = 1.

Original entry on oeis.org

2, 3, 8, 10, 12, 14, 15, 18, 21, 22, 26, 27, 32, 33, 34, 38, 39, 40, 46, 48, 50, 51, 56, 57, 58, 60, 62, 69, 70, 72, 74, 75, 82, 84, 86, 87, 88, 90, 93, 94, 98, 104, 105, 106, 108, 110, 111, 118, 122, 123, 126, 128, 129, 130, 132, 134, 135, 136, 141, 142
Offset: 1

Views

Author

Clark Kimberling, Apr 26 2019

Keywords

Comments

These are the numbers 2x and 3x as x ranges through the numbers in A036668.
Numbers whose squarefree part is divisible by exactly one of {2, 3}. - Peter Munn, Aug 24 2020
The asymptotic density of this sequence is 5/12. - Amiram Eldar, Sep 20 2020

Crossrefs

Symmetric difference of: A003159 and A007417; A036554 and A145204\{0}.

Programs

  • Mathematica
    a = {1}; Do[AppendTo[a, NestWhile[# + 1 &, Last[a] + 1, Apply[Or,
    Map[MemberQ[a, #] &, Select[Flatten[{#/3, #/2}],
    IntegerQ]]] &]], {150}]; a     (* A036668 *)
    Complement[Range[Last[a]], a]  (* A325424 *)
    (* Peter J. C. Moses, Apr 23 2019 *)
  • Python
    from itertools import count
    def A325424(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c = n
            for i in range(x.bit_length()+1):
                i2 = 1<x:
                        break
                    m = x//k
                    c += (m-1)//6+(m-5)//6+2
            return c
        return bisection(f,n,n) # Chai Wah Wu, Jan 28 2025

Formula

(2 * {A036668}) union (3 * {A036668}). - Sean A. Irvine, May 19 2019

A108269 Numbers of the form (2*m - 1)*4^k where m >= 1, k >= 1.

Original entry on oeis.org

4, 12, 16, 20, 28, 36, 44, 48, 52, 60, 64, 68, 76, 80, 84, 92, 100, 108, 112, 116, 124, 132, 140, 144, 148, 156, 164, 172, 176, 180, 188, 192, 196, 204, 208, 212, 220, 228, 236, 240, 244, 252, 256, 260, 268, 272, 276, 284, 292, 300, 304, 308, 316, 320, 324, 332
Offset: 1

Author

Andras Erszegi (erszegi.andras(AT)chello.hu), May 30 2005

Keywords

Comments

Numbers of terms in nonnegative integer sequences the sum of which is never a square.
The sum of a sequence of consecutive nonnegative integers starting with k is never a square for any k, if and only if the number of the terms in the sequence can be expressed as (2*m - 1) * 2^(2*n), m and n being any positive integers. (Proved by Alfred Vella, Jun 14 2005.)
Odious and evil terms alternate. - Vladimir Shevelev, Jun 22 2009
Even numbers whose binary representation ends in an even number of zeros. - Amiram Eldar, Jan 12 2021
From Antti Karttunen, Jan 28 2023: (Start)
Numbers k for which the parity of k is equal to that of A048675(k).
A multiplicative semigroup; if m and n are in the sequence then so is m*n. (End)

Examples

			a( 1, 1 ) = 4, a( 2, 1) = 12, etc.
For a( 1, 1 ): the sum of 4 consecutive nonnegative integers (4k+6, if the first term is k) is never a square.
		

Crossrefs

Intersection of A005843 and A003159.
Cf. A000069, A001969, A017113 (primitive terms), A036554, A328981 (characteristic function), A359794 (complement).

Programs

  • Mathematica
    Select[2 * Range[200], EvenQ @ IntegerExponent[#, 2] &] (* Amiram Eldar, Jan 12 2021 *)
  • PARI
    is(n)=my(e=valuation(n,2)); e>1 && e%2==0 \\ Charles R Greathouse IV, Nov 03 2016
    
  • Python
    def A108269(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c, s = n+(x+1>>1), bin(x)[2:]
            l = len(s)
            for i in range(l&1,l,2):
                c += int(s[i])+int('0'+s[:i],2)
            return c
        return bisection(f,n,n) # Chai Wah Wu, Jan 29 2025

Formula

a(n) = 6*n + O(log n). - Charles R Greathouse IV, Nov 03 2016 [Corrected by Amiram Eldar, Jan 12 2021]
a(n) = 2 * A036554(n) = 4 * A003159(n). - Amiram Eldar, Jan 12 2021

Extensions

Entry revised by N. J. A. Sloane, Jun 26 2005
More terms from Amiram Eldar, Jan 12 2021

A232745 Numbers k for which the largest m such that m! divides k is even.

Original entry on oeis.org

2, 4, 8, 10, 14, 16, 20, 22, 24, 26, 28, 32, 34, 38, 40, 44, 46, 48, 50, 52, 56, 58, 62, 64, 68, 70, 72, 74, 76, 80, 82, 86, 88, 92, 94, 96, 98, 100, 104, 106, 110, 112, 116, 118, 122, 124, 128, 130, 134, 136, 140, 142, 144, 146, 148, 152, 154, 158, 160, 164
Offset: 1

Author

Antti Karttunen, Dec 01 2013

Keywords

Comments

Numbers k for which A055881(k) is even.
Equally: Numbers k which have an odd number of the trailing zeros in their factorial base representation A007623(k).
The sequence can be described in the following manner: Sequence includes all multiples of 2! (even numbers), except that it excludes from those the multiples of 3! (6), except that it includes the multiples of 4! (24), except that it excludes the multiples of 5! (120), except that it includes the multiples of 6! (720), except that it excludes the multiples of 7! (5040), except that it includes the multiples of 8! (40320), except that it excludes the multiples of 9! (362880), except that it includes the multiples of 10! (3628800), except that ..., ad infinitum.
The number of terms not exceeding m! for m>=1 is A000166(m). The asymptotic density of this sequence is 1/e (A068985). - Amiram Eldar, Feb 26 2021

Crossrefs

Complement: A232744.
b(n) = A153880(A232744(n)) gives a subset of this sequence.
Analogous sequences for binary system: A003159 & A036554.

Programs

  • Mathematica
    seq[max_] := Select[Range[max!], OddQ @ LengthWhile[Reverse @ IntegerDigits[#, MixedRadix[Range[max, 2, -1]]], #1 == 0 &] &]; seq[5] (* Amiram Eldar, Feb 26 2021 *)
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