A265747 -id:A265747 - cp's OEIS frontend

A364379 Greedy Jacobsthal-Niven numbers: numbers that are divisible by the sum of the digits in their representation in Jacobsthal greedy base (A265747).

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 20, 21, 22, 24, 26, 27, 28, 32, 33, 36, 40, 42, 43, 44, 45, 46, 48, 51, 52, 54, 56, 57, 60, 64, 68, 69, 72, 75, 76, 80, 84, 85, 86, 87, 88, 90, 92, 93, 96, 99, 100, 104, 105, 106, 108, 111, 112, 115, 116, 117, 120

Offset: 1

Amiram Eldar, Jul 21 2023

Numbers k such that A265745(k) | k.

The positive Jacobsthal numbers, A001045(n) for n >= 1, are terms since their representation in Jacobsthal greedy base is one 1 followed by n-1 0's, so A265745(A001045(n)) = 1 divides A001045(n).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A265745, A265747.
Subsequences: A364380, A364381, A364382, A364383.
Similar sequences: A005349, A049445, A064150, A064438, A064481, A118363, A328208, A328212, A331085, A333426, A342726, A334308, A331728, A342426, A344341, A351714, A351719, A352089, A352107, A352320, A352342, A352508, A364216.

greedyJacobNivenQ[n_] := Divisible[n, A265745[n]]; Select[Range[120], greedyJacobNivenQ] (* using A265745[n] *)

isA364379(n) = !(n % A265745(n)); \\ using A265745(n)

A265746 Jacobsthal greedy base (A265747) interpreted as base-3 numbers, then shown in decimal.

Original entry on oeis.org

0, 1, 2, 3, 4, 9, 10, 11, 12, 13, 18, 27, 28, 29, 30, 31, 36, 37, 38, 39, 40, 81, 82, 83, 84, 85, 90, 91, 92, 93, 94, 99, 108, 109, 110, 111, 112, 117, 118, 119, 120, 121, 162, 243, 244, 245, 246, 247, 252, 253, 254, 255, 256, 261, 270, 271, 272, 273

Offset: 0

Antti Karttunen, Dec 17 2015

Analogously to "Fibbinary numbers" (A003714) and "Catquaternary numbers" (A244161), this sequence could be called "Jacoternary numbers".

Antti Karttunen, Table of n, a(n) for n = 0..10923

Cf. A000244, A001045, A130249.
Cf. A265745, A265747.
Cf. also A003714, A244161.

a[n_] := FromDigits[IntegerDigits[A265747[n]], 3]; Array[a, 100, 0] (* Amiram Eldar, Jul 21 2023 using A265747[n] *)

a(n) = fromdigits(digits(A265747(n)), 3); \\ Amiram Eldar, Jul 21 2023, using A265747(n)

a(0) = 0; for n >= 1, a(n) = 3^(A130249(n)-2) + a(n - A001045(A130249(n))).

A324477 k appears t+1 times, where t = A364377(k) is the number of trailing zeros in the greedy Jacobsthal representation of k, A265747(k).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 8, 9, 10, 10, 10, 11, 11, 11, 11, 12, 13, 14, 14, 15, 16, 16, 16, 17, 18, 19, 19, 20, 21, 21, 21, 21, 21, 22, 23, 24, 24, 25, 26, 26, 26, 27, 28, 29, 29, 30, 31, 31, 31, 32, 32, 32, 32, 33, 34, 35, 35, 36, 37, 37, 37, 38, 39, 40, 40

Offset: 1

Nathan Fox and N. J. A. Sloane, Mar 09 2019

Amiram Eldar, Table of n, a(n) for n = 1..10000
Nathan Fox, Trees, Fibonacci Numbers, and Nested Recurrences, Rutgers University Experimental Math Seminar, Mar 07, 2019.

Cf. A001045, A265747, A364377.

A046699, A316628, A324473 and A324475 have similar definitions.

Table[Table[k, {IntegerExponent[A265747[k], 10] + 1}], {k, 1, 40}] // Flatten (* Amiram Eldar, Jul 21 2023 using A265747[n] *)

More terms from Amiram Eldar, Jul 21 2023

A364377 The number of trailing 0's in the representation of n in Jacobsthal greedy base (A265747).

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 0, 1, 0, 2, 3, 0, 0, 1, 0, 2, 0, 0, 1, 0, 4, 0, 0, 1, 0, 2, 0, 0, 1, 0, 2, 3, 0, 0, 1, 0, 2, 0, 0, 1, 0, 4, 5, 0, 0, 1, 0, 2, 0, 0, 1, 0, 2, 3, 0, 0, 1, 0, 2, 0, 0, 1, 0, 4, 0, 0, 1, 0, 2, 0, 0, 1, 0, 2, 3, 0, 0, 1, 0, 2, 0, 0, 1, 0, 6, 0, 0

Offset: 1

Amiram Eldar, Jul 21 2023

The first position of k, for k = 0, 1, 2, ..., is A001045(k+2).
The asymptotic density of the occurrences of 2*k is 9/4^(k+2), and of 2*k+1 is 3/4^(k+2), both for k >= 0.
The asymptotic mean of this sequence is 11/12, and its asymptotic standard deviation is sqrt(283)/12.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001045, A265747, A324477.

a[n_] := IntegerExponent[A265747[n], 10]; Array[a, 100] (* using A265747[n] *)

a(n) = valuation(A265747(n), 10); \\ using A265747(n)

A364378 Numbers whose representation in Jacobsthal greedy base (A265747) is palindromic.

Original entry on oeis.org

0, 1, 2, 4, 6, 9, 12, 20, 22, 27, 36, 41, 44, 60, 68, 84, 86, 97, 112, 123, 132, 143, 158, 169, 172, 204, 220, 252, 260, 292, 308, 340, 342, 363, 396, 417, 432, 453, 486, 507, 516, 537, 570, 591, 606, 627, 660, 681, 684, 748, 780, 844, 860, 924, 956, 1020, 1028

Offset: 1

Amiram Eldar, Jul 21 2023

A128209(n) = A001045(n) + 1 is a term for n >= 3, since its representation is two 1's with n-3 0's between them.
A084639(n) is a term for n >= 1 since its representation is n 1's.
A014825(n) is a term for n >= 1 since its representation is n-1 0's interleaved with n 1's.

			The first 10 terms are:
   n  a(n)  A265747(a(n))
  --  ----  -------------
   1     0              0
   2     1              1
   3     2              2
   4     4             11
   5     6            101
   6     9            111
   7    12           1001
   8    20           1111
   9    22          10001
  10    27          10101

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001045, A014825, A084639, A128209, A265747.

Similar sequences: A002113, A006995, A014190, A094202, A331191, A351712, A351717, A352087, A352105, A352319, A352341, A364214.

palJacobQ[n_] := PalindromeQ[A265747[n]]; Select[Range[0, 1000], palJacobQ] (* using A265747[n] *)

is(n) = {my(dig = digits(A265747(n))); dig == Vecrev(dig);} \\ using A265747(n)

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 5, 4, 5, 4, 5, 6, 5, 6, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2

Offset: 0

Antti Karttunen, Dec 17 2015

Sum of digits in "Jacobsthal greedy base", A265747.
It would be nice to know for sure whether this sequence gives also the least number of Jacobsthal numbers that add to n, i.e., that there cannot be even better nongreedy solutions.
The integer 63=21+21+21 has 3 for its 'non-greedy' solution, and a(63) = 5 for its greedy solution 63=43+11+5+3+1. - Yuriko Suwa, Jul 11 2021
Positions where a(n) is different from A372555(n) are n=63, 84, 148, 169, 191, 212, 234, 255, etc. See A372557. - Antti Karttunen, May 07 2024

			a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
For n=1 we need just A001045(2) = 1, thus a(1) = 1.
For n=2 we need A001045(2) + A001045(2) = 1 + 1, thus a(2) = 2.
For n=4 we need A001045(3) + A001045(2) = 3 + 1, thus a(4) = 2.
For n=6 we form the greedy sum as A001045(4) + A001045(2) = 5 + 1, thus a(6) = 2. Alternatively, we could form the sum as A001045(3) + A001045(3) = 3 + 3, but the number of summands in that case is no less.
For n=7 we need A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = 3.
For n=8 we need A001045(4) + A001045(3) = 5 + 3, thus a(8) = 2.
For n=10 we need A001045(4) + A001045(4) = 5 + 5, thus a(10) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..10923

Cf. A001045, A130249, A197911, A003158, A265746, A265747, A364377, A364379, A364380, A372288, A372555, A372557.
Cf. A054111 (apparently the positions of the first occurrence of each n > 0).
Cf. also A007895, A014420, A053610, A265404, A265743, A265744.

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265745[n_] := A265745[n] = 1 + A265745[n - jacob[maxInd[n]]]; A265745[0] = 0; Array[A265745, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

A130249(n) = floor(log(3*n + 1)/log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = {if(n == 0, 0, my(d = n - A001045(A130249(n))); if(d == 0, 1, 1 + A265745(d)));} \\ Amiram Eldar, Jul 21 2023

def greedyJ(n): n1 = (3*n+1).bit_length() - 1; return (2**n1 - (-1)**n1)//3
def a(n): return 0 if n == 0 else 1 + a(n - greedyJ(n))
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 11 2021

a(0) = 0; for n >= 1, a(n) = 1 + a(n - A001045(A130249(n))). [This formula uses a simple greedy algorithm.]

A084639 Expansion of x(1+2x)/((1+x)(1-x)(1-2*x)).

Original entry on oeis.org

0, 1, 4, 9, 20, 41, 84, 169, 340, 681, 1364, 2729, 5460, 10921, 21844, 43689, 87380, 174761, 349524, 699049, 1398100, 2796201, 5592404, 11184809, 22369620, 44739241, 89478484, 178956969, 357913940, 715827881, 1431655764, 2863311529, 5726623060, 11453246121

Offset: 0

Paul Barry, Jun 06 2003

Original name was: Generalized Jacobsthal numbers.
This is the sequence A(0,1;1,2;3) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
Entries correspond to value bound adjustment for an N-bit string having M bits set and a(n+1) bit transitions. Wolfram Alpha can easily generate an entry. a(5)=41 stems from input as 1111110_2 - 1010101_2. The subtraction pattern alternates (begins at 1), and bit count is ptr+2 both terms, with the lead term having only its LSB clear. - Bill McEachen, Jul 15 2011
Also a(n) = 2*A000975(n) if n even, a(n) = 2*A000975(n) - 1 if n odd. - Michel Lagneau, Jan 11 2012
In the above comment by Bill McEachen the binary pattern (in an obvious notation) is for even n 1^(n+1)0 - (10)^((n+2)/2) and for odd n 1^(n+1)0 - (10)^((n+1)/2)1. That is for even n a(n) = sum(2^k, k=1..(n+1)) - sum(2^(2*k-1), k=1..(n+2)/2)  = (2^(n+2) - 4)/3, and for odd n a(n) = sum(2^k , k=1..(n+1)) - sum(2^(2*k), k=0..(n+1)/2) = (2^(n+2) - 5)/3. This checks with the formula a(n) = (2^(n+3) + (-1)^n - 9)/6 given below. After a correspondence with Bill McEachen. - Wolfdieter Lang, Jan 24 2014
Michel Lagneau's comment above is equal to the fact that a(n) = A000975(n)-1, or in other words, this sequence gives the partial sums of Jacobsthal sequence, starting from its second 1, A001045(2). From this also follows that this sequence gives the positions of repunits in "Jacobsthal greedy base", A265747. - Antti Karttunen, Dec 17 2015
From Kensuke Matsuoka, Aug 11 2020: (Start)
This sequence is the sum of diagonally arranged powers of 2 repeated in an L shape. For example, a(1)=1, a(2) = 4, a(3)=9, a(4)= 20, a(5)=41, a(6)=84 are obtained from the figure below.
  32
  16  8
   8  4  2
   4  2  1  2
   2  1  2  4  8
   1  2  4  8 16 32
From this figure, a(n) = a(n-2) + 2^n is obtained. (End)
For n > 0, also the total distance that the disks travel from the leftmost peg to the middle peg in the Tower of Hanoi puzzle, in the unique solution with 2^n - 1 moves (see links). - Sela Fried, Dec 17 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Sela Fried, Economically solving the Tower of Hanoi puzzle.
A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart. 34, 40-54, 1996.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
El-Mehdi Mehiri, Saad Mneimneh, and Hacène Belbachir, The Towers of Fibonacci, Lucas, Pell, and Jacobsthal, arXiv:2502.11045 [math.CO], 2025. See p. 12.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A000975, A000225, A001045.

Cf. A265747.

[2^(n+2)/3+(-1)^n/6-3/2: n in [0..35]]; // Vincenzo Librandi, Aug 08 2011

a:=proc(n) (2^(n+3) + (-1)^n - 9)/6 end proc: [seq(a(n), n=0..33)]; # Wolfdieter Lang, Jan 24 2014

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + 2 a[n - 2] + 3; Array[a, 32, 0] (* Or *)
a[0] = 0; a[1] = 1; a[n_] := a[n] = 3 a[n - 1] - 2 a[n - 2] + (-1)^n; Array[a, 32, 0]
CoefficientList[Series[x*(1+2*x)/((1+x)*(1-x)*(1-2*x)),{x,0,40}],x] (* or *) LinearRecurrence[{2,1,-2},{0,1,4},40]  (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)

a(n)=2^(n+2)/3-if(n%2,5,4)/3 \\ Charles R Greathouse IV, Aug 08 2011

concat(0, Vec(x*(1+2*x)/((1+x)*(1-x)*(1-2*x)) + O(x^100))) \\ Altug Alkan, Dec 17 2015

def A084639(n): return (4<Chai Wah Wu, Apr 25 2025

G.f.: x*(1+2*x)/((1+x)*(1-x)*(1-2*x)).
E.g.f.: 4*exp(2*x)/3-3*exp(x)/2+exp(-x)/6.
a(n) = a(n-1)+2*a(n-2)+3, a(0)=0, a(1)=1.
a(n) = 2^(n+2)/3+(-1)^n/6-3/2.
a(n) = A001045(n+2) - A000034(n).
a(n) = 5*a(n-2)-4*a(n-4). Cf. A084640, A101622. - Paul Curtz, Apr 03 2008
a(n) = 2*a(n-1) + a(n-2) -2*a(n-3). - R. J. Mathar, Jun 28 2010
a(n) = a(n-1)+2*a(n-2)+3, n>1. - Gary Detlefs, Dec 19 2010
a(n) = 3*a(n-1)-2*a(n-2) +(-1)^n, n>1. - Gary Detlefs, Dec 19 2010
a(n) = a(n-2) + 2^n for n >= 2. - Kensuke Matsuoka, Aug 11 2020

Replaced duplicate of a formula by another recurrence - R. J. Mathar, Jun 28 2010

A276326 Numbers expressed in greedy A001563-base.

Original entry on oeis.org

0, 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 40, 41, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 140, 141, 200, 201, 202, 203, 210, 211, 212, 213, 220, 221, 222, 223, 230, 231, 232, 233, 240, 241, 300, 301, 302, 303, 310, 311, 312, 313, 320, 321, 322, 323, 330, 331, 332, 333, 340, 341, 400

Offset: 0

Antti Karttunen, Aug 30 2016

Terms A001563(1) = 1, A001563(2) = 4, A001563(3) = 18, ... give the base values for the digit positions from 1 onward. Digit places are filled by always trying to find the largest possible term of A001563 that still fits into the sum.

A130744(8) = 3225600 = 10*A001563(8) is the first number which yields an ambiguous representation when expressed in decimal, because in this base it is actually "A0000000" (where digit "A" stands for ten).

			To recover n from a(n) the digits in positions i = 1, 2, 3, ... (starting indexing from the least significant digit at right) are multiplied by A001563(i) and added together:
  ----------------
   n         a(n)
  ----------------
   0           0
   1           1
   2           2
   3           3
   4          10
   5          11
   6          12
   7          13
   8          20
   9          21
  10          22
  11          23
  12          30
  13          31
  14          32
  15          33
  16          40
  17          41 (as 4*A001563(2) + 1*A001563(1) = 17)
  18         100 (as 1*A001563(3) + 0*A001563(2) + 0*A001563(1) = 18)
and:
3225599 99111111 (as 3225599 = 9*b(8) + 9*b(7) + b(6) + b(5) + b(4) + b(3) + b(2) + b(1)), where b(n) = A001563(n).

Antti Karttunen, Table of n, a(n) for n = 0..4320

Cf. A001563, A130744, A258198, A276335.
Cf. A276327 (the least significant nonzero digit).
Cf. A276328 (the sum of digits).
Cf. A276333 (the most significant digit).
Cf. A276336 (a largest digit).
Cf. A276337 (number of nonzero digits).
Cf. A033312 (repunits).
Cf. A276091 (no digits larger than one).
Differs from A007090 for the first time at n=16 and from A055655 at n=18.
Cf. also A007623, A007961, A000433, A014418, A265747.

f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], (# #!) &[# - i]]], {i, 0, # - 1}] &@ NestWhile[# + 1 &, 0, (# #!) &[# + 1] <= n &]; Rest[a][[All, 1]]]; Table[FromDigits@ f@ n, {n, 72}] (* Michael De Vlieger, Aug 31 2016 *)

(define (A276326 n) (let loop ((n n) (s 0)) (if (zero? n) s (let ((dig (A276333 n))) (if (> dig 9) (error "A276326: ambiguous representation of n, digit > 9 would be needed: " n dig) (loop (A276335 n) (+ s (* dig (expt 10 (- (A258198 n) 1))))))))))

A372288 Array read by upward antidiagonals: A(n, k) = A265745(A372282(n, k)), n,k >= 1, where A265745(n) is the sum of digits in "Jacobsthal greedy base".

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 3, 3, 1, 1, 1, 1, 3, 3, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 5, 3, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 5, 3

Offset: 1

Antti Karttunen, Apr 28 2024

Collatz conjecture is equal to the claim that each column will eventually settle to constant 1's, somewhere under its topmost row. This works as only the bisection A002450 of Jacobsthal numbers (A001045) contains numbers of the form 4k+1, while the other bisection contains only numbers of the form 4k+3, which do not occur among the range of A372351. See also the comments in A371094.

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13     14 15    16 17 18 19 20    21 22
---+----------------------------------------------------------------------------
1  | 1, 1, 1, 3, 3, 1, 3, 3, 3, 3, 1, 3, 3,     3, 3,    3, 3, 3, 3, 5,    5, 1,
2  | 1, 1, 1, 3, 3, 3, 1, 3, 3, 3, 1, 5, 5,     5, 3,    5, 3, 3, 3, 5,    5, 3,
3  | 1, 1, 1, 3, 3, 3, 1, 5, 1, 3, 1, 3, 3,     5, 3,    5, 5, 1, 3, 3,    5, 3,
4  | 1, 1, 1, 3, 3, 1, 1, 3, 1, 3, 1, 1, 3,     5, 3,    3, 3, 1, 3, 5,    5, 3,
5  | 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3,     5, 1,    5, 3, 1, 3, 3,    3, 3,
6  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3,     3, 1,    5, 3, 1, 1, 5,    5, 3,
7  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    3, 3, 1, 1, 3,    5, 3,
8  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 3,    3, 3,
9  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     3, 1,    5, 1, 1, 1, 3,    5, 1,
10 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 3,    5, 1,
11 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1, 2155, 1, 1, 1, 1,    5, 1,
12 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 1, 6251, 1,
13 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 10347, 1,    5, 1, 1, 1, 1,    5, 1,
14 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 1,    5, 1,
15 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    7, 1, 1, 1, 1,    5, 1,
16 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 1,    7, 1,

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A001045, A002450, A265745, A265747, A371094, A372351, A372282, A372283, A372287.

Cf. also array A372561 (formed by columns whose indices in this array are given by A372443).

up_to = 105;
A130249(n) = (#binary(3*n+1)-1);
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = { my(s=0); while(n,s++; n -= A001045(A130249(n))); (s); };
A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };
A372282sq(n,k) = if(1==n,2*k-1,A371094(A372282sq(n-1,k)));
A372288sq(n,k) = A265745(A372282sq(n,k));
A372288list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372288sq((a-(col-1)),col))); (v); };
v372288 = A372288list(up_to);
A372288(n) = v372288[n];

A364380 Numbers k such that k and k+1 are both greedy Jacobsthal-Niven numbers (A364379).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 10, 11, 14, 15, 20, 21, 26, 27, 32, 42, 43, 44, 45, 51, 56, 68, 75, 84, 85, 86, 87, 92, 99, 104, 105, 111, 115, 116, 125, 128, 135, 144, 155, 170, 171, 176, 182, 183, 195, 204, 213, 219, 224, 260, 264, 267, 275, 304, 305, 324, 329, 341, 344

Offset: 1

Amiram Eldar, Jul 21 2023

The positive Jacobsthal numbers, A001045(n) for n >= 1, are terms since their representation in Jacobsthal greedy base is one 1 followed by n-1 0's, so A265745(A001045(n)) = 1 divides A001045(n), and the representation of A001045(n) + 1 is 2 if n <= 2 and otherwise n-3 0's between two 1's, so A265745(A001045(n) + 1) = 2 which divides A001045(n) + 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A265745, A265747.
Subsequence of A364379.
Subsequences: A364381, A364382, A364383.
Similar sequences: A330927, A328205, A328209, A328213, A330931, A331086, A333427, A334309, A331820, A342427, A344342, A351715, A351720, A352090, A352108, A352321, A352343, A352509, A364217.

consecGreedyJN[kmax_, len_] := Module[{m = 1, c = Table[False, {len}], s = {}}, Do[c = Join[Rest[c], {greedyJacobNivenQ[k]}]; If[And @@ c, AppendTo[s, k - len + 1]], {k, 1, kmax}]; s]; consecGreedyJN[350, 2] (* using the function greedyJacobNivenQ[n] from A364379 *)

lista(kmax, len) = {my(c = vector(len)); for(k = 1, kmax, c = concat(vecextract(c, "^1"), isA364379(k)); if(vecsum(c) == len, print1(k-len+1, ", ")));} \\ using the function isA364379(n) from A364379
lista(350, 2)

cp's OEIS Frontend

A364379 Greedy Jacobsthal-Niven numbers: numbers that are divisible by the sum of the digits in their representation in Jacobsthal greedy base (A265747).

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A265746 Jacobsthal greedy base (A265747) interpreted as base-3 numbers, then shown in decimal.

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A324477 k appears t+1 times, where t = A364377(k) is the number of trailing zeros in the greedy Jacobsthal representation of k, A265747(k).

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A364377 The number of trailing 0's in the representation of n in Jacobsthal greedy base (A265747).

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A364378 Numbers whose representation in Jacobsthal greedy base (A265747) is palindromic.

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A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

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A084639 Expansion of x*(1+2*x)/((1+x)*(1-x)*(1-2*x)).

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A084639 Expansion of x(1+2x)/((1+x)(1-x)(1-2*x)).