A038502 -id:A038502 - cp's OEIS frontend

A266189 Self-inverse permutation of nonnegative integers: a(n) = A263273(A264985(A263273(n))).

0, 1, 3, 2, 4, 10, 6, 9, 12, 7, 5, 11, 8, 13, 37, 24, 28, 31, 21, 19, 57, 18, 27, 30, 15, 36, 39, 22, 16, 34, 23, 17, 35, 69, 29, 32, 25, 14, 38, 26, 40, 118, 78, 109, 112, 75, 46, 100, 72, 82, 91, 51, 85, 94, 66, 64, 192, 20, 73, 219, 60, 171, 138, 63, 55, 165, 54, 81, 84, 33, 90, 111, 48, 58, 174, 45, 108, 93, 42, 117, 120, 67, 49

Offset: 0

Antti Karttunen, Jan 02 2016

Antti Karttunen, Table of n, a(n) for n = 0..6561
Index entries for sequences that are permutations of the natural numbers

Cf. A263273, A264985, A265353, A265354.

Cf. also A265902, A266190.

f[n_] := Block[{g, h}, g[x_] := x/3^IntegerExponent[x, 3]; h[x_] := x/g@ x; If[n == 0, 0, FromDigits[Reverse@ IntegerDigits[#, 3], 3] &@ g[n] h[n]]]; s = Select[f /@ Range@ 5000, OddQ]; t = Table[(s[[n + 1]] - 1)/2, {n, 0, 1000}]; Table[f@ t[[f@ n + 1]], {n, 0, 82}] (* Michael De Vlieger, Jan 04 2016, after Jean-François Alcover at A263273 *)

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a264985(n): return (a263273(2*n + 1) - 1)/2
def a(n): return a263273(a264985(a263273(n))) # Indranil Ghosh, May 22 2017

(define (A266189 n) (A263273 (A264985 (A263273 n))))

a(n) = A263273(A264985(A263273(n))).
As a composition of related permutations:
a(n) = A263273(A265353(n)).
a(n) = A265354(A263273(n)).

A293445 A multiplicative encoding (base-2 compressed) for the exponents of 3 obtained when using Shevelev's algorithm for computing A053446.

Original entry on oeis.org

2, 2, 3, 12, 36, 3, 12, 24, 6, 48, 12, 20736, 82944, 12, 18, 864, 248832, 6, 20, 19906560, 59719680, 80, 8640, 720, 25920, 34560, 5, 80, 103195607040, 240, 480, 622080, 137594142720, 138240, 20, 59440669655040, 138240, 20, 14929920, 29859840, 240, 59719680, 8640, 720, 414720, 8640, 540, 447897600, 960, 46080, 34560, 59719680, 295814814232058265600, 5, 80

Offset: 1

Antti Karttunen, Oct 09 2017

nonn

			A001651(5) = 7 as 7 is the fifth number not divisible by 3. According to the algorithm described in the comment of A053446 we have in the form of a "finite continued fraction"
    1 + 14
    ------ + 7
       3^1
    ---------- + 14
          3^1
    ----------------- + 7
            3^2
    ---------------------- = 1
               3^2
Cumulatively multiplying (with A019565) together the prime-numbers corresponding to 1-bits in the binary expansions of the exponents of 3 in the denominators (that are 1, 1, 2, 2, in binary 1, 1, 10, 10, with 1's in bit-positions 0 and 1), yields prime(0+1) * prime(0+1) * prime(1+1) * prime(1+1) = 2^2 * 3^2 = 36, thus a(5) = 36.
(Adapted from _Vladimir Shevelev_'s explanation in A053446.)
Another example: A001651(19) = 28 as 28 is the 19th number not divisible by 3. (1 + 28) is not a multiple of 3, so we start with (1 + 2*28) = 1+56 = 57 and proceed as:
    1 + 56
    ------ + 56                     [that is, (57/3) + 56 = 75]
       3^1
    ---------- + 56                 [that is, (75/3) + 56 = 81]
          3^1
    -----------------  = 1          [that is, (81/81) = 1]
            3^4
So we obtained exponents 1, 1, 4 (in binary "1", "1" and "100") where the 1-bits are in positions 0, 0 and 2. We form a product prime(0+1) * prime(0+1) * prime(2+1) = 2*2*5, thus a(19) = 20.

Antti Karttunen, Table of n, a(n) for n = 1..1458

Cf. A001651, A007949, A019565, A038502, A053446, A293220.
Cf. A293446 (restricted growth transform of this sequence).
Cf. also A292265.

(define (A293445 n) (define (next_one k m) (if (zero? (modulo (+ k m) 3)) (+ k m) (+ k m m))) (let* ((u (A001651 n)) (x_init (next_one 1 u))) (let loop ((x x_init) (z (A019565 (A007949 x_init)))) (let ((r (A038502 x))) (if (= 1 r) z (let ((x_next (next_one r u))) (loop x_next (* z (A019565 (A007949 x_next))))))))))
(define (A001651 n) (let ((x (- n 1))) (if (even? x) (+ 1 (* 3 (/ x 2))) (- (* 3 (/ (+ x 1) 2)) 1))))
(define (A038500 n) (A000244 (A007949 n)))
(define (A038502 n) (/ n (A038500 n)))

A048675(a(n)) = A053446(n).

A212307 Numerator of n!/3^n.

Original entry on oeis.org

1, 1, 2, 2, 8, 40, 80, 560, 4480, 4480, 44800, 492800, 1971200, 25625600, 358758400, 1793792000, 28700672000, 487911424000, 975822848000, 18540634112000, 370812682240000, 2595688775680000, 57105153064960000, 1313418520494080000, 10507348163952640000

Offset: 0

Jean-François Alcover and Paul Curtz, Oct 30 2013

Also the 3rd column of A152656 (or of A216919).

Cf. A001316, A049606, A125824 (denominators), A152656, A216919.

Cf. A000142, A060828.

Table[Numerator[n!/3^n], {n, 0, 32}]
(* or *) CoefficientList[Series[Exp[3x], {x, 0, 32}], x] // Denominator

a(n) = numerator(n!/3^n); \\ Michel Marcus, Oct 30 2013

a(n) = Product_{i=1..n} A038502(i). - Tom Edgar, Mar 22 2014

a(n) = A000142(n)/A060828(n). - Ridouane Oudra, Sep 23 2024

A348934 Numbers k congruent to 1 or 5 mod 6, for which A348930(k^2) > k^2.

Original entry on oeis.org

5, 11, 17, 23, 25, 29, 41, 47, 49, 53, 55, 59, 71, 83, 85, 89, 101, 107, 113, 115, 121, 125, 131, 137, 145, 149, 167, 169, 173, 179, 187, 191, 197, 205, 227, 233, 235, 239, 245, 251, 253, 257, 263, 265, 269, 275, 281, 289, 293, 295, 311, 317, 319, 343, 347, 353, 355, 359, 361, 383, 389, 391, 401, 415, 419, 425, 431

Offset: 1

Antti Karttunen, Nov 04 2021

nonn

See comments in A348933.

Index entries for sequences related to sigma(n)

Cf. A348930, A348932, A348933, A348936.

s[n_] := n / 3^IntegerExponent[n, 3]; Select[Range[450], MemberQ[{1, 5}, Mod[#, 6]] && s[DivisorSigma[1, #^2]] > #^2 &] (* Amiram Eldar, Nov 04 2021 *)

A038502(n) = (n/3^valuation(n, 3));
A348930(n) = A038502(sigma(n));
isA348934(n) = ((n%2)&&(n%3)&&(A348930(n^2)>(n^2)));

A378578 G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.

Original entry on oeis.org

1, 1, 1, 2, 13, 19, 52, 412, 73, 1405, 11735, 20000, 7300, 388606, 664316, 2325118, 20832709, 11815463, 95438089, 861817318, 1495813613, 5231996647, 47291366710, 3025568936, 199838851432, 1828302724054, 3320026962314, 439614522008, 73390614310810, 131344935434920, 55179693272894, 3321671735661494

Offset: 0

Paul D. Hanna, Jan 03 2025

nonn

Conjecture: a(n) == binomial(3*n,n)/(2*n+1) (mod 2) for n >= 0.

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 13*x^4 + 19*x^5 + 52*x^6 + 412*x^7 + 73*x^8 + 1405*x^9 + 11735*x^10 + 20000*x^11 + 7300*x^12 + ...
The expansion of A(x)^3 begins
A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 57*x^4 + 156*x^5 + 412*x^6 + 1971*x^7 + 4215*x^8 + 11735*x^9 + 60000*x^10 + 197100*x^11 + ...
where g.f. A(x) is obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.
SPECIFIC VALUES.
A(t) = 8/5 at t = 0.257842038645833456558...
A(t) = 3/2 at t = 0.24869526467110689667648213094860932113462559982219...
A(t) = 4/3 at t = 0.21321674572378383093755902318049913517774115880785...
A(t) = 5/4 at t = 0.18151790234803008203317827057063199289923020437324...
A(t) = 6/5 at t = 0.15638236848650043639095127985605468995430265567872...
A(1/4) = 1.5104498750225954401497052152244291483533940069402...
A(1/5) = 1.2948177731384287040434619555644894329636242990640...
A(1/6) = 1.2192544950152148905144159115908573870687883121699...
A(1/8) = 1.1487089332444818621810139499703458742589412625833...
A(1/9) = 1.1286969902151862545955480786537992451685724113531...
PARITY OF TERMS.
The run lengths of the odd terms, which starts
[3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, ...],
appears to equal A282162 (offset 1), the first differences of the upper Wythoff sequence (A001950).
The run lengths of the even terms, which starts
[1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 42, 1, 2, 5, 1, 10, 1, 2, 85, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 170, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 42, 1, 2, 5, 1, 10, 1, 2, 341, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 42, 1, 2, 5, 1, 10, 1, 2, 85, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 682, 1, 2, 5, 1, 10, ...],
appears to equal A085358, the runs of zeros in binomial(3*n,n)/(2*n+1) (mod 2), the records of which are given by A000975 and occur at Fibonacci numbers.

Paul D. Hanna, Table of n, a(n) for n = 0..800

Cf. A378577, A085358, A001764, A282162.

N = 30; A=vector(N+1); A[1]=1; \\ N = number of terms
{a(n) = if(n==0,1, A[n+1] = Vec(1 + x*Ser(A)^3)[n+1]; A[n+1] = A[n+1] / 3^valuation(A[n+1], 3) )}
for(n=0, N, print1(a(n), ", "))

a(n) = A038502( Sum_{k=0..n-1} a(k) * Sum_{j=0..n-1-k} a(j)*a(n-1-k-j) ) for n > 0 with a(0) = 1, where A038502(m) = m/3^A007949(m) and A007949(m) = 3-adic valuation of m.

A099585 Remove all 3s from prime(n) - 1.

Original entry on oeis.org

1, 2, 4, 2, 10, 4, 16, 2, 22, 28, 10, 4, 40, 14, 46, 52, 58, 20, 22, 70, 8, 26, 82, 88, 32, 100, 34, 106, 4, 112, 14, 130, 136, 46, 148, 50, 52, 2, 166, 172, 178, 20, 190, 64, 196, 22, 70, 74, 226, 76, 232, 238, 80, 250, 256, 262, 268, 10, 92, 280, 94, 292, 34, 310

Offset: 1

Ralf Stephan, Oct 24 2004

nonn

Equals A038502( A006093(n) ). prime(n)-1 = a(n) * 3^A099584(n).

Cf. A057023.

a(n) = (prime(n)-1)/3^valuation(prime(n)-1,3)

A273893 Denominator of n/3^n.

Original entry on oeis.org

1, 3, 9, 9, 81, 243, 243, 2187, 6561, 2187, 59049, 177147, 177147, 1594323, 4782969, 4782969, 43046721, 129140163, 43046721, 1162261467, 3486784401, 3486784401, 31381059609, 94143178827, 94143178827, 847288609443, 2541865828329, 282429536481, 22876792454961

Offset: 0

Paul Curtz, Jun 02 2016

The reduced values are Ms(n) = 0, 1/3, 2/9, 1/9, 4/81, 5/243, 2/243, 7/2187, 8/6561, 1/2187, ... .
Numerators: 0, 1, 2, 1, 4, ... = A038502(n).
Ms(-n) = 0, -3, -18, ... = - A036290(n).
Difference table of Ms(n):
0,          1/3,    2/9,    1/9,   4/81, 5/243, 2/243, ...
1/3,       -1/9,   -1/9,  -5/81, -7/243, -1/81, ...
-4/9,         0,   4/81,  8/243,  4/243, ...
4/9,       4/81, -4/243, -4/243, ...
-32/81, -16/243,      0, ...
80/243,  16/243, ...
-64/243, ...
etc.
The difference table of O(n) = n/2^n (Oresme numbers) has its 0's on the main diagonal. Here the 0's appear every two rows. For n/4^n,they appear every three rows. (The denominators of O(n) are 2^A093048(n)).
All terms are powers of 3 (A000244).

Cf. A007949, A000244, A013732, A013733, A036290, A038502, A075101.

Table[Denominator[n/3^n], {n, 0, 28}] (* Michael De Vlieger, Jun 03 2016 *)

a(n) = denominator(n/3^n) \\ Felix Fröhlich, Jun 07 2016

[1] + [3^(n-n.valuation(3)) for n in [1..30]] # Tom Edgar, Jun 02 2016

For n>0, a(n) = 3^(n - valuation(n,3)) = 3^(n - A007949(n)). - Tom Edgar, Jun 02 2016
a(3n+1) = 3^(3n+1), a(3n+2) = 3^(3n+2).
a(3n+6) = 27*(3n+3).
From Peter Bala, Feb 25 2019: (Start)
a(n) = 3^n/gcd(n,3^n).
O.g.f.: 1 + F(3*x) - (2/3)*F((3*x)^3) - (2/9)*F((3*x)^9) - (2/27)*F((3*x)^27) - ..., where F(x) = x/(1 - x).
O.g.f. for reciprocals: Sum_{n >= 0} x^n/a(n) = 1 +  F((x/3)) + 2*( F((x/3)^3) + 3*F((x/3)^9) + 9*F((x/3)^27) + ... ). Cf. A038502. (End)

A347717 Number of states of the minimal deterministic finite automaton that accepts ternary strings that represent numbers that are divisible by n.

Original entry on oeis.org

1, 2, 2, 4, 5, 3, 7, 8, 3, 10, 11, 5, 13, 14, 6, 16, 17, 4, 19, 20, 8, 22, 23, 9, 25, 26, 4, 28, 29, 11, 31, 32, 12, 34, 35, 6, 37, 38, 14, 40, 41, 15, 43, 44, 7, 46, 47, 17, 49, 50, 18, 52, 53, 5, 55, 56, 20, 58, 59, 21, 61, 62, 9, 64, 65, 23, 67, 68, 24, 70, 71, 10, 73

Offset: 1

Liangzhou Chen, Sep 10 2021

a(n) = n for all n coprime to 3.

			The minimal DFA when n=6, giving the form of transition table:
+-------+-----------+
|       | tr. func. |
| state +---+---+---+
|       | 0 | 1 | 2 |
+-------+---+---+---+
|->*A   | A | C | B | (mod 6 = 0)
+-------+---+---+---+
|   B   | A | C | B | (mod 6 = 2,4)
+-------+---+---+---+
|   C   | C | B | C | (mod 6 = 1,3,5)
+-------+---+---+---+

Liangzhou Chen, Table of n, a(n) for n = 1..10000

Cf. A007949, A038502, A119674.

a(n) = n / 3^valuation(n, 3) + valuation(n, 3)

a(1) = 1, a(2) = 2, a(3n) = a(n) + 1, a(3n+1) = 3n+1, a(3n+2) = 3n+2.

A370757 a(n) is the least k > 0 such that 1/n and 1/k have equivalent repeating decimal digits.

Original entry on oeis.org

1, 1, 3, 1, 1, 6, 7, 1, 9, 1, 11, 3, 13, 7, 6, 1, 17, 18, 19, 1, 21, 22, 23, 6, 1, 26, 27, 7, 29, 3, 31, 1, 33, 17, 7, 36, 37, 19, 39, 1, 41, 42, 43, 44, 45, 23, 47, 3, 49, 1, 51, 13, 53, 54, 55, 7, 57, 29, 59, 6, 61, 31, 63, 1, 26, 66, 67, 17, 69, 7, 71, 72

Offset: 1

Rémy Sigrist, Feb 29 2024

In other words, a(n) is the least k > 0 such that the fractional parts of (10^i)/n and (10^j)/k are equal for some integers i, j.

a(n) is not always a divisor of n. For example, a(65) = 26 is not a divisor of 65. - David A. Corneth, Mar 01 2024

			The first terms, alongside the decimal expansion of 1/n with its repeating decimal digits in parentheses, are:
  n   a(n)  1/n
  --  ----  -----------
   1     1  1.(0)
   2     1  0.5(0)
   3     3  0.(3)
   4     1  0.25(0)
   5     1  0.2(0)
   6     6  0.1(6)
   7     7  0.(142857)
   8     1  0.125(0)
   9     9  0.(1)
  10     1  0.1(0)
  11    11  0.(09)
  12     3  0.08(3)
  13    13  0.(076923)
  14     7  0.07(142857)
  15     6  0.0(6)

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program
Index entries for sequences related to decimal expansion of 1/n

Cf. A000265 (base-2 analog), A038502 (base-3 analog), A132739 (base-5 analog), A242603 (base-7 analog).

Cf. A003592, A007732, A132740.

PARI
```
\\ See Links section.
```

from itertools import count
from sympy import multiplicity, n_order
def A370757(n):
    m2, m5 = (~n & n-1).bit_length(), multiplicity(5,n)
    r = max(m2,m5)
    w, m = 10**r, 10**(t:=n_order(10,n2) if (n2:=(n>>m2)//5**m5)>1 else 1)-1
    c = w//n
    s = str(m*w//n-c*m).zfill(t)
    l = len(s)
    for k in count(1):
        m2, m5 = (~k & k-1).bit_length(), multiplicity(5,k)
        r = max(m2,m5)
        w, m = 10**r, 10**(t:=n_order(10,k2) if (k2:=(k>>m2)//5**m5)>1 else 1)-1
        c = w//k
        if any(s[i:]+s[:i] == str(m*w//k-c*m).zfill(t) for i in range(l)):
            return k # Chai Wah Wu, Mar 03 2024

a(n) = 1 iff n belongs to A003592.
a(10*n) = a(n).
A007732(a(n)) = A007732(n).

cp's OEIS Frontend

A266189 Self-inverse permutation of nonnegative integers: a(n) = A263273(A264985(A263273(n))).

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A293445 A multiplicative encoding (base-2 compressed) for the exponents of 3 obtained when using Shevelev's algorithm for computing A053446.

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A212307 Numerator of n!/3^n.

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A348934 Numbers k congruent to 1 or 5 mod 6, for which A348930(k^2) > k^2.

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A378578 G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.

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A099585 Remove all 3s from prime(n) - 1.

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A273893 Denominator of n/3^n.

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A347717 Number of states of the minimal deterministic finite automaton that accepts ternary strings that represent numbers that are divisible by n.

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