A007583 -id:A007583 - cp's OEIS frontend

A073724 a(n) = (4^(n+1) + 6n + 5)/9.

1, 3, 9, 31, 117, 459, 1825, 7287, 29133, 116515, 466041, 1864143, 7456549, 29826171, 119304657, 477218599, 1908874365, 7635497427, 30541989673, 122167958655, 488671834581, 1954687338283, 7818749353089, 31274997412311

Offset: 0

Wouter Meeussen, Sep 01 2002

a(n) is the number of times a disk is moved from peg 1 to peg 2 during a move of a tower of 2n or (2n-1) disks from peg 1 to peg 2 ("Tower of Hanoi" problem). Binomial transform of A025579.

An approximation to A091841.

			Moving a tower of 4 disks = 2^4 - 1 moves, coded {1,0,5,1,2,3,1,0,5,4,2,5,1,0,5}. The move from peg 1 to peg 2 has code "0" and this occurs 3 times. For 3 disks we also find 3 zeros in {0,1,3,0,4,5,0}. Hence a(2)=3. The coding corresponds to the rank of the permutation {'from peg' 1, 'to peg' 2, 'by peg' 3} or {1,2,3} with rank 0.

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Hacène Belbachir and El-Mehdi Mehiri, Enumerating moves in the optimal solution of the Tower of Hanoi, arXiv:2210.08657 [math.CO], 2022.
F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence [pdf, ps].
Index entries for sequences related to Gijswijt's sequence
Index entries for linear recurrences with constant coefficients, signature (6,-9,4).
Index entries for sequences related to Towers of Hanoi

Cf. A001045, A002450, A007583, A020988, A025579, A047849 (first differences), A090822, A091841.

[(4^(n+1)+6*n+5)/9: n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

Mathematica
```
Table[(4^(n+1)+6n+5)/9, {n, 0, 24}]
```
PARI
```
a(n)=(4*4^n+6*n+5)/9
```

a(n)=polcoeff((1-3*x)/(1-4*x)/(1-x)^2+x*O(x^n),n)

G.f.: (1-3*x)/((1-4*x)*(1-x)^2).
a(n) = Sum_{k=0..n} A047849(k). - L. Edson Jeffery, May 01 2021
From Elmo R. Oliveira, Dec 11 2023: (Start)
a(n) = 6*a(n-1) - 9*a(n-2) + 4*a(n-3) for n>2.
E.g.f.: (1/9)*(4*(exp(4*x)) + 6*x*exp(x) + 5*exp(x)). (End)

A191669 Dispersion of A004767 (4k+3, k>=0), by antidiagonals.

Original entry on oeis.org

1, 3, 2, 11, 7, 4, 43, 27, 15, 5, 171, 107, 59, 19, 6, 683, 427, 235, 75, 23, 8, 2731, 1707, 939, 299, 91, 31, 9, 10923, 6827, 3755, 1195, 363, 123, 35, 10, 43691, 27307, 15019, 4779, 1451, 491, 139, 39, 12, 174763, 109227, 60075, 19115, 5803, 1963, 555, 155

Offset: 1

Clark Kimberling, Jun 11 2011

For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The six sequences and dispersions are listed here:
...
A191452=dispersion of A008586 (4k, k>=1)
A191667=dispersion of A016813 (4k+1, k>=1)
A191668=dispersion of A016825 (4k+2, k>=0)
A191669=dispersion of A004767 (4k+3, k>=0)
A191670=dispersion of A042968 (1 or 2 or 3 mod 4 and >=2)
A191671=dispersion of A004772 (0 or 1 or 3 mod 4 and >=2)
A191672=dispersion of A004773 (0 or 1 or 2 mod 4 and >=2)
A191673=dispersion of A004773 (0 or 2 or 3 mod 4 and >=2)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191452 has 1st col A042968, all else A008486
A191667 has 1st col A004772, all else A016813
A191668 has 1st col A042965, all else A016825
A191669 has 1st col A004773, all else A004767
A191670 has 1st col A008486, all else A042968
A191671 has 1st col A016813, all else A004772
A191672 has 1st col A016825, all else A042965
A191673 has 1st col A004767, all else A004773
...
Regarding the dispersions A191670-A191673, there is a formula for sequences of the type "(a or b or c mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 3), then (a,b,c,a,b,c,a,b,c,...) is given by
   a*f(n+2)+b*f(n+1)+c*f(n), so that "(a or b or c mod m)" is given by
   a*f(n+2)+b*f(n+1)+c*f(n)+m*floor((n-1)/3)), for n>=1.

			Northwest corner:
1...3....11....43....171
2...7....27....107...427
4...15...59....235...939
5...19...75....299...1195
6...23...91....363...1451

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Row 1: A007583, Row 2: A136412.

Cf. A004773, A004767, A191673, A191426.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12; c = 40; c1 = 12;
f[n_] := 4*n-1
Table[f[n], {n, 1, 30}] (* A004767 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191669 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191669 *)

A303448 Numbers m such that both m and (m-1)/2 are Fermat pseudoprimes base 2 (A001567).

Original entry on oeis.org

19781763, 46912496118443, 192153584101141163

Offset: 1

Max Alekseyev, Apr 24 2018

No other terms below 2^65.
Terms a(2) and a(3) are of the form (2^(2k+1)+1)/3 = A007583(k).
Terms A007583(k) belong to this sequence for k in A303009. Correspondingly, a(4) <= A007583(A303009(3)) = (2^83+1)/3 = 3223802185639011132549803.
If a(n) is not divisible by 3, then it also belongs to A175625.

Numbers (a(n)-1)/2 are listed in A303447.
Subsequence of A006970 and A300193.
Cf. A175625, A175942, A298758.

a(n) = 2*A303447(n) + 1.

a(1) from Amiram Eldar, Jan 26 2018

A256265 Sums of two successive terms of A256264.

Original entry on oeis.org

1, 3, 7, 11, 15, 23, 35, 43, 47, 55, 67, 79, 95, 123, 155, 171, 175, 183, 195, 207, 223, 251, 283, 303, 319, 347, 387, 439, 503, 579, 651, 683, 687, 695, 707, 719, 735, 763, 795, 815, 831, 859, 899, 951, 1015, 1091, 1163, 1199, 1215, 1243, 1283, 1335, 1399, 1475, 1563, 1663, 1775, 1899, 2035, 2183, 2343, 2515, 2667, 2731

Offset: 1

Omar E. Pol, Apr 10 2015

nonn

First differs from A139250 (the toothpick sequence) at a(27).
It appears that both sequences share infinitely many terms, for example: a(1)..a(26), a(31)..a(42), a(47)..a(50), a(63)..a(74), a(79)..a(82), etc.
The main entry for this sequence is A256263.

Cf. A007583, A139250, A255747, A256260, A256261, A256263, A256264.

a(n) = A256264(n) + A256264(n-1).

A292849 a(n) is the least positive k such that the Hamming weight of k equals the Hamming weight of k*n.

Original entry on oeis.org

1, 1, 3, 1, 7, 3, 7, 1, 15, 7, 3, 3, 5, 7, 15, 1, 31, 15, 7, 7, 13, 3, 7, 3, 31, 5, 31, 7, 31, 15, 31, 1, 63, 31, 11, 15, 7, 7, 7, 7, 57, 13, 3, 3, 23, 7, 11, 3, 21, 31, 43, 5, 39, 31, 7, 7, 9, 31, 35, 15, 21, 31, 63, 1, 127, 63, 23, 31, 15, 11, 15, 15, 29, 7

Offset: 1

Rémy Sigrist and Altug Alkan, Sep 25 2017

The Hamming weight of a number n is given by A000120(n).
All terms are odd.
Numbers n such that a(n) is not squarefree are 33, 57, 63, 66, 83, 114, 115, 126, 132, 153, 155, ...
Numbers n such that a(n) > n are 5, 9, 17, 25, 27, 29, 33, 41, 65, 97, 101, 109, 113, ...
a(n) = 1 iff n = 2^i for some i >= 0.
a(n) = 3 iff n = A007583(i) * 2^j for some i > 0 and j >= 0.
Apparently:
- if n < 2^k then a(n) < 2^k,
- a(n) = n iff n = A000225(i) for some i > 0.
Proof that a(n) < 2^k if n < 2^k (see preceding comment): We can assume that n is not a power of two and take k such that 2^(k-1) < n < 2^k (so that k is the number of binary digits of n). Now, n - 1 and 2^k - n have complementary binary digits, so the binary digits of (2^k - 1)*n = 2^k*(n - 1) + (2^k - n) consist of the k digits of n - 1 followed by the complementary digits. This implies that the number of binary 1's is k, so that (2^k - 1)*n and 2^k - 1 have the same number of 1's and a(n) <= 2^k - 1. - Pontus von Brömssen, Jan 01 2021
See also A180012 for the base 10 equivalent sequence.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Altug Alkan, Line graph of n - a(n) for n <= 2^20 + 1
Rémy Sigrist, Scatterplot of n XOR a(n) for n <= 2^16

Cf. A000120, A000225, A007583, A115873, A180012.

Table[SelectFirst[Range[1, 2^8 + 1, 2], Equal @@ Thread[DigitCount[{#, # n}, 2, 1]] &], {n, 74}] (* Michael De Vlieger, Sep 25 2017 *)

a(n) = forstep(k=1, oo, 2, if (hammingweight(k) == hammingweight(k*n), return (k)))

a(n) = my(k=1); while ((hammingweight(k)) != hammingweight(k*n), k++); k;

from itertools import count
def A292849(n): return next(k for k in count(1) if k.bit_count()==(k*n).bit_count()) # Chai Wah Wu, Mar 11 2025

a((2^m)*n) = a(n) for all m >= 0 and n >= 1.
a(2^m + 1) = 2^(m + 1) - 1 for all m >= 0.
a(2^m - 1) = 2^m - 1 for all m >= 1.
a(2^m) = 1 for all m >= 0.

A364216 Jacobsthal-Niven numbers: numbers that are divisible by the sum of the digits in their Jacobsthal representation (A280049).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 11, 12, 14, 15, 16, 20, 22, 24, 27, 28, 32, 33, 36, 40, 42, 43, 44, 45, 46, 48, 51, 52, 54, 56, 57, 60, 68, 72, 75, 76, 84, 86, 87, 88, 92, 93, 95, 96, 99, 100, 104, 105, 108, 112, 115, 117, 120, 125, 126, 128, 129, 132, 135, 136, 138, 140

Offset: 1

Amiram Eldar, Jul 14 2023

Numbers k such that A364215(k) | k.

A007583 is a subsequence since A364215(A007583(n)) = 1 for n >= 0.

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

Cf. A280049, A364215.
Subsequences: A007583, A364217, A364218, A364219, A364220, A364221.
Similar sequences: A005349, A049445, A064150, A064438, A064481, A118363, A328208, A328212, A331085, A333426, A342726, A334308, A331728, A342426, A344341, A351714, A351719, A352089, A352107, A352320, A352342, A352508.

seq[kmax_] := Module[{m = 1, s = {}}, Do[If[Divisible[k, DigitCount[m, 2, 1]], AppendTo[s, k]]; While[m++; OddQ[IntegerExponent[m, 2]]], {k, 1, kmax}]; s]; seq[140]

lista(kmax) = {my(m = 1); for(k = 1, kmax, if( !(k % sumdigits(m, 2)), print1(k,", ")); until(valuation(m, 2)%2 == 0, m++));}

A364217 Numbers k such that k and k+1 are both Jacobsthal-Niven numbers (A364216).

Original entry on oeis.org

1, 2, 3, 8, 11, 14, 15, 27, 32, 42, 43, 44, 45, 51, 56, 75, 86, 87, 92, 95, 99, 104, 125, 128, 135, 144, 155, 171, 176, 182, 183, 195, 204, 264, 267, 275, 287, 305, 344, 363, 375, 387, 428, 444, 455, 474, 497, 512, 524, 535, 544, 545, 552, 555, 581, 605, 623, 639

Offset: 1

Amiram Eldar, Jul 14 2023

A001045(2*n+1) = A007583(n) = (2^(2*n+1) + 1)/3 is a term for n >= 0, since its representation is 2*n 1's, so A364215(A001045(2*n+1)) = 1 divides A001045(2*n+1), and the representation of A001045(2*n+1) + 1 = (2^(2*n+1) + 4)/3 is max(2*n-1, 0) 0's between 2 1's, so A364215(A001045(2*n+1) + 1) = 2 which divides (2^(2*n+1) + 4)/3.

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

Cf. A364215.
Subsequence of A364216.
Subsequences: A364218, A364219, A364220, A364221.
Similar sequences: A330927, A328205, A328209, A328213, A330931, A331086, A333427, A334309, A331820, A342427, A344342, A351715, A351720, A352090, A352108, A352321, A352343, A352509.

consecJacobsthalNiven[kmax_, len_] := Module[{m = 1, c = Table[False, {len}], s = {}}, Do[c = Join[Rest[c], {Divisible[k, DigitCount[m, 2, 1]]}]; While[m++; OddQ[IntegerExponent[m, 2]]]; If[And @@ c, AppendTo[s, k - len + 1]], {k, 1, kmax}]; s]; consecJacobsthalNiven[640, 2]

lista(kmax, len) = {my(m = 1, c = vector(len)); for(k = 1, kmax, c = concat(vecextract(c, "^1"), !(k % sumdigits(m, 2))); until(valuation(m, 2)%2 == 0, m++); if(vecsum(c) == len, print1(k-len+1, ", ")));}
lista(640, 2)

A082685 (2*5^n + 2^n)/3.

Original entry on oeis.org

1, 4, 18, 86, 422, 2094, 10438, 52126, 260502, 1302254, 6510758, 32552766, 162761782, 813804814, 4069015878, 20345063006, 101725282262, 508626345774, 2543131597798, 12715657726846, 63578288109942, 317891439501134

Offset: 0

Paul Barry, Apr 11 2003

Binomial transform of A007583

Row sums of A114195. - Paul Barry, Nov 16 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-10)

Cf. A001045.

Table[(2*5^n+2^n)/3,{n,0,30}] (* or *) LinearRecurrence[{7,-10},{1,4},30] (* Harvey P. Dale, Apr 09 2014 *)
CoefficientList[Series[(1 - 3 x)/((1 - 5 x) (1 - 2 x)), {x, 0, 50}], x] (* Vincenzo Librandi, Apr 10 2014 *)

a(n)=(2*5^n+2^n)/3 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1-3x)/((1-5x)(1-2x))
a(n)=sum{k=0..n, sum{j=0..n, C(n, j)C(j+k, 2k)2^(j-k)}}. - Paul Barry, Nov 16 2005
a(0)=1, a(1)=4, a(n)=7*a(n-1)-10*a(n-2). - Harvey P. Dale, Apr 09 2014

A156760 5*4^n-1.

Original entry on oeis.org

4, 19, 79, 319, 1279, 5119, 20479, 81919, 327679, 1310719, 5242879, 20971519, 83886079, 335544319, 1342177279, 5368709119, 21474836479, 85899345919, 343597383679, 1374389534719, 5497558138879, 21990232555519, 87960930222079, 351843720888319

Offset: 0

Paul Curtz, Feb 15 2009

Second column of the array A132207, or, if this array is flattened, a(n)=A132207(A007583(n)).

			Binary.......................................Decimal
100................................................4
10011.............................................19
1001111...........................................79
100111111........................................319
10011111111.....................................1279
1001111111111...................................5119
100111111111111................................20479
10011111111111111..............................81919
1001111111111111111...........................327679
100111111111111111111........................1310719
10011111111111111111111......................5242879
1001111111111111111111111...................20971519
100111111111111111111111111.................83886079
10011111111111111111111111111..............335544319
1001111111111111111111111111111...........1342177279
... - _Philippe Deléham_, Feb 23 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..500

[5*4^n-1: n in [0..30]]; // Vincenzo Librandi, Jul 02 2011

Table[5*4^n - 1, {n, 0, 18}] (* Michael De Vlieger, Apr 20 2015 *)

a(n) mod 9 = A070403(n+2).
a(n+1) = 10*A083420(n)+9 .
a(n) = 5*A000302(n)-1.
a(n) = ( A024036(n+1)+A140529(n) )/2.
a(n) = 4a(n-1)+3, a(0)=4.
a(n) = A003947(n+1)-1 = 5*a(n-1)-4*a(n-2). G.f.: (4-x)/((1-x)(1-4x)). - R. J. Mathar, Feb 23 2009
a(n) = A198693(n) + 2^(2n+1). - Bob Selcoe, Apr 20 2015

Edited and extended by R. J. Mathar, Feb 23 2009

A377627 Cogrowth sequence of the 12-element group C6 X C2 = .

Original entry on oeis.org

1, 1, 1, 2, 29, 211, 926, 3095, 9829, 37130, 164921, 728575, 2973350, 11450531, 43942081, 174174002, 708653429, 2884834891, 11582386286, 46006694735, 182670807229, 729520967450, 2926800830801, 11743814559415, 47006639297270, 187791199242011, 750176293590361

Offset: 0

Sean A. Irvine, Nov 02 2024

Gives the even terms, all the odd terms are 0.

			a(2)=1 corresponds to the word TTTT.
a(3)=2 corresponds to the words SSSSSS and TTTTTT.

Cf. A007583 (D6), A377626 (A4), A377656 (Dic12), A377714 (C4 X C2), A377840 (C8 X C2).

G.f.: (6*x^5+5*x^4+11*x^3-10*x^2+5*x-1) / ((4*x-1) * (x^2+x+1) * (9*x^2-3*x+1)).

cp's OEIS Frontend

A073724 a(n) = (4^(n+1) + 6n + 5)/9.

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A191669 Dispersion of A004767 (4k+3, k>=0), by antidiagonals.

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A303448 Numbers m such that both m and (m-1)/2 are Fermat pseudoprimes base 2 (A001567).

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A256265 Sums of two successive terms of A256264.

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A292849 a(n) is the least positive k such that the Hamming weight of k equals the Hamming weight of k*n.

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A364216 Jacobsthal-Niven numbers: numbers that are divisible by the sum of the digits in their Jacobsthal representation (A280049).

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A364217 Numbers k such that k and k+1 are both Jacobsthal-Niven numbers (A364216).

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A082685 (2*5^n + 2^n)/3.

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