A347834 -id:A347834 - cp's OEIS frontend

A072201 a(n) = 4*a(n-1) + 1, a(1) = 15.

15, 61, 245, 981, 3925, 15701, 62805, 251221, 1004885, 4019541, 16078165, 64312661, 257250645, 1029002581, 4116010325, 16464041301, 65856165205, 263424660821, 1053698643285, 4214794573141, 16859178292565

Offset: 1

N. Rathankar (rathankar(AT)yahoo.com), Jul 03 2002

These are the integers N which on application of the Collatz function yield the number 23. The Collatz function: if N is an odd number then (3N+1)/2^r yields a positive odd integer for some value of r (which in this case is 11).

Numbers whose binary representation is 1111 together with n - 1 times 01. For example, a(4) = 981 = 1111010101 (2). - Omar E. Pol, Nov 24 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A002450, A178415, A347834.

[(23*4^n-2)/6: n in [1..30]]; // Vincenzo Librandi, Aug 17 2011

NestList[4#+1&,15,20] (* Harvey P. Dale, Aug 16 2011 *)

a(n) = (23*4^n - 2)/6.
From Colin Barker, Aug 17 2012: (Start)
a(n) = 5*a(n-1) - 4*a(n-2).
G.f.: x*(15-14*x)/((1-x)*(1-4*x)). (End)
a(n) = 46*A002450(n-1) + 15. - Yosu Yurramendi, Jan 24 2017
a(n) = A178415(8, n) = A347834(6, n-1), arrays, for n >= 1. - Wolfdieter Lang, Nov 29 2021

Edited and extended by Henry Bottomley, Aug 05 2002

A347839 An array of the positive integers congruent to 2 modulo 3 (A016789), read by antidiagonals upwards, giving the present triangle.

Original entry on oeis.org

2, 5, 8, 11, 20, 32, 14, 44, 80, 128, 17, 56, 176, 320, 512, 23, 68, 224, 704, 1280, 2048, 26, 92, 272, 896, 2816, 5120, 8192, 29, 104, 368, 1088, 3584, 11264, 20480, 32768, 35, 116, 416, 1472, 4352, 14336, 45056, 81920, 131072, 38, 140, 464, 1664, 5888, 17408, 57344, 180224, 327680, 524288

Offset: 1

Wolfdieter Lang, Oct 21 2021

This array a = (a(k, n))_{k >= 1,n >= 0} is underlying array A of A347834. See the first formula. It has a simple recurrence for the rows k, given the first column a(k, 0) = A347838(k), which lists the positive integers congruent to {2, 5, 11} modulo 12.
In the array one can add the negative of the powers of 4 as row for k = 0, i.e., -A000302(n), for n >= 0.
All positive numbers congruent to 2 modulo 3 (A017617) appear once in this array. Proof from the array A of A347834 of the positive integers congruent to {1,3,5,7} modulo 8, and the present first formula: The members of column n = 0 give all the positive integers congruent to {2, 5, 11} modulo 12 once, and the members of columns n >= 1 give all the positive integers congruent to 8 modulo 12 (A017617) once. These members combined lead to the positive integers congruent to 2 modulo 3.

			The array a(k, n) begins:
k \ n  0   1   2    3    4     5      6      7       8       9       10 ...
---------------------------------------------------------------------------
1:     2   8  32  128  512  2048   8192  32768  131072  524288  2097152 ...
2:     5  20  80  320 1280  5120  20480  81920  327680 1310720  5242880 ...
3:    11  44 176  704 2816 11264  45056 180224  720896 2883584 11534336 ...
4:    14  56 224  896 3584 14336  57344 229376  917504 3670016 14680064 ...
5:    17  68 272 1088 4352 17408  69632 278528 1114112 4456448 17825792 ...
6:    23  92 368 1472 5888 23552  94208 376832 1507328 6029312 24117248 ...
7:    26 104 416 1664 6656 26624 106496 425984 1703936 6815744 27262976 ...
8:    29 116 464 1856 7424 29696 118784 475136 1900544 7602176 30408704 ...
9:    35 140 560 2240 8960 35840 143360 573440 2293760 9175040 36700160 ...
10:   38 152 608 2432 9728 38912 155648 622592 2490368 9961472 39845888 ...
...
----------------------------------------------------------------------------
The triangle t(n,k) begins:
k \ n  0   1   2    3    4     5     6      7      8      9 ...
---------------------------------------------------------------
1:     2
2:     5   8
3:    11  20  32
4:    14  44  80  128
5:    17  56 176  320  512
6:    23  68 224  704 1280  2048
7:    26  92 272  896 2816  5120  8192
8:    29 104 368 1088 3584 11264 20480  32768
9:    35 116 416 1472 4352 14336 45056  81920 131072
10:   38 140 464 1664 5888 17408 57344 180224 327680 524288
...
-----------------------------------------------------------------

Cf. A347834, A016789, A017617.
The rows k are given by -A000302 (for k=0), A004171, A003947(n+1), A002089, 2*A002042, ...
The columns n are given by 4^n*A347838 for n >= 0.

A := (n, k) -> 4^n*(3*(k + iquo(k, 3)) - 1):
for k from 1 to 10 do seq(A(n, k), n = 0..10) od;
# Alternatively:
gf  := n -> (4^n*((z*(z*(7*z + 3) + 3) - 1)))/((z - 1)^2*(1 + z + z^2)):
ser := n -> series(gf(n), z, 12):
col := (n, len) -> seq(coeff(ser(n), z, k), k = 1..len):
seq(print(col(n, 10)), n = 0..10); # Peter Luschny, Oct 26 2021

A[n_, k_] := 4^n (3(k + Quotient[k, 3]) - 1);
Table[A[n-k, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 07 2021, from Maple code *)

Array a:
a(k, n) = (3*A(k, n) + 1)/2, with the array A from A347834, for k >= 1, and n >= 0.
a(k, n) = 4^n*A347838(k) = 4^n*(2 + 3*k + 3*floor((k + 1)/3)).
Recurrence for rows k: a(k, n) = 4*a(k, n-1), for n >= 1, with a(k, 0) = A347838(k).
O.g.f.: expansion in z gives the o.g.f.s for rows k, also for k = 0: -A000302; expansion in x gives the o.g.f.s for columns n.
G(z, x) = (-1 + 3*z + 3*z^2 + 7*z^3)/((1 - z)*(1 - z^3)*(1 - 4*x)).
Triangle t:
t(k, n) = a(k-n, n), for k >= 1, and n = 0, 1, ..., k-1.

A350053 a(n) = (2^(3*n + 3 + (-1)^n) - (6 + (-1)^n))/9, for n >= 1.

Original entry on oeis.org

3, 113, 227, 7281, 14563, 466033, 932067, 29826161, 59652323, 1908874353, 3817748707, 122167958641, 244335917283, 7818749353073, 15637498706147, 500399958596721, 1000799917193443, 32025597350190193, 64051194700380387

Offset: 1

Wolfdieter Lang, Jan 20 2022

Labels of nodes at level L = 1 of the Collatz tree with only odd numbers congruent to 1, 3, and 7 modulo 8, named here CToddr.
a(n) is given by the successor of the non-leaf node labels of the (reduced) Collatz tree with odd numbers (named here CTodd) at level 1 given by A198586(n), for n >= 1. See a comment in A347834 for the construction of CTodd. (For all labels of CTodd at level 1 see {A002450(k)}_{k>=2}.) The present sequence gives the labels of the (further) reduced rooted tree CToddr, at level L = 1. Level L = 0 has the root labeled 1, and this node has a directed 1-cycle.
The successor of a node label u of the tree CTodd is given by (4*u - 1)/3 if u == 1 (mod 6), (2*u - 1)/3 if u == 5 (mod 6), and there is no successor if the label u == 3 (mod 6) (a leaf).
This sequence is motivated by a draft of Immo O. Kerner (see A347834 and the link).
Sorted set of all A385109(A198584(i)), i>0 (conjectured but easy to see). - Ralf Stephan, Jun 18 2025

Winston de Greef, Table of n, a(n) for n = 1..1100
Index entries for linear recurrences with constant coefficients, signature (0,65,0,-64).

Cf. A002450, A198586, A228871, A347834, A350054.

a[n_] := (2^(3*n + 3 + (-1)^n) - (6 + (-1)^n))/9; Array[a, 20] (* Amiram Eldar, Jan 21 2022 *) (* or *)
LinearRecurrence[{0, 65, 0, -64}, {3, 113, 227, 7281}, 20] (* Georg Fischer, Sep 30 2022 *)

a(n) = (2^(3*n + 3 + (-1)^n))\9 \\ Winston de Greef, Jan 28 2024

Bisection: a(2*k-1) = (2^(6*k-1) - 5)/9 = A228871(k), a(2*k) = (4^(3*k+2) - 7)/9 = A350054(k), for k >= 1.
a(n) = (2^(3*n+ 2 + b(n)) - (5 + b(n)))/9, with b(n) = 1 + (-1)^n = A010673(n-1), for n >= 1. See the name.
G.f.: Bisection: x*(3 + 32*x)/((1 - x)*(1 - 64*x)) and x*(113 - 64*x)/((1 - x)*(1 - 64*x)).
G.f.: x*(3 + 113*x + 32*x^2 - 64*x^3)/((1 - x^2)*(1 - 64*x^2)).

A350668 Numbers congruent to 2, 4, and 6 modulo 9: positions of 2 in A159955.

Original entry on oeis.org

2, 4, 6, 11, 13, 15, 20, 22, 24, 29, 31, 33, 38, 40, 42, 47, 49, 51, 56, 58, 60, 65, 67, 69, 74, 76, 78, 83, 85, 87, 92, 94, 96, 101, 103, 105, 110, 112, 114, 119, 121, 123, 128, 130, 132, 137, 139, 141, 146, 148

Offset: 0

Wolfdieter Lang, Jan 29 2022

This sequence, together with A350666 and A350667, gives a 3-set partition of the nonnegative integers.
This sequence {a(n)}_{n>=0} gives the indices of the row sequences of array A = A347834, that are modulo 6 periodic with period length 3, namely
{A347834(a(n), m) mod 6}_{m >= 0} = {repeat(3, 1, 5)}.

			Rows of array {A347834(a(n), m)}_{m >= 0}, with modulo 6 congruence:
n = 0: row 2: {3, 13, 53, 213, 853, 3413, 13653, ...} mod 6 = {repeat(3, 1, 5)},
n = 1: row 4: {9, 37, 149, 597, 2389, 9557, ...} (mod 6) = {repeat(3, 1, 5)},
...

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A049310, A159955, A347834, A350666, A350667.

Select[Range[0, 150], MemberQ[{2, 4, 6}, Mod[#, 9]] &] (* Amiram Eldar, Jan 29 2022 *)
LinearRecurrence[{1,0,1,-1},{2,4,6,11},80] (* Harvey P. Dale, Jul 12 2024 *)

A159955(a(n)) = 2.
Trisection: a(3*k) = 2 + 9*k, a(3*k + 1) = 4 + 9*k, and a(3*k + 3) = 6 + 9*k, for k >= 0.
G.f.: (2 + 2*x + 2*x^2 + 3*x^3)/((1 - x)*(1 - x^3)).
a(n) = 1 + 3*n + cos(2*n*Pi/3) + sin(2*n*Pi/3)/sqrt(3). - Stefano Spezia, Jan 30 2022
a(n) = 1 + 3*n + S(2*n, 1) = 1+3*n+A057078(n), with the Chebyshev S polynomials from A049310, using the partial fraction decomposition of the g.f., or the previous formula.

A323824 a(0) = 6; thereafter a(n) = 4*a(n-1) + 1.

Original entry on oeis.org

6, 25, 101, 405, 1621, 6485, 25941, 103765, 415061, 1660245, 6640981, 26563925, 106255701, 425022805, 1700091221, 6800364885, 27201459541, 108805838165, 435223352661, 1740893410645, 6963573642581, 27854294570325, 111417178281301, 445668713125205

Offset: 0

N. J. A. Sloane, Feb 01 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Mike Warburton, Ulam-Warburton Automaton - Counting Cells with Quadratics, arXiv:1901.10565 [math.CO], 2019.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A178415, A347834.

A323824[n_]:=(19*4^n-1)/3;Array[A323824,30,0] (* or *)
LinearRecurrence[{5,-4},{6,25},30] (* Paolo Xausa, Nov 14 2023 *)
NestList[4#+1&,6,30] (* Harvey P. Dale, Nov 26 2024 *)

Vec((6 - 5*x) / ((1 - x)*(1 - 4*x)) + O(x^25)) \\ Colin Barker, Feb 01 2019

a(n) = (19*4^n - 1) / 3 \\ Colin Barker, Feb 01 2019

G.f.: (6 - 5*x) / ((1 - x)*(1 - 4*x)).
a(n) = (19*4^n - 1) / 3. - Colin Barker, Feb 01 2019
a(n) = A178415(7, n) = A347834(10, n-1), arrays, for n >= 1. - Wolfdieter Lang, Nov 29 2021

A330246 a(n) = 4^(n+1) + (4^n-1)/3.

Original entry on oeis.org

4, 17, 69, 277, 1109, 4437, 17749, 70997, 283989, 1135957, 4543829, 18175317, 72701269, 290805077, 1163220309, 4652881237, 18611524949, 74446099797, 297784399189, 1191137596757, 4764550387029, 19058201548117, 76232806192469, 304931224769877, 1219724899079509

Offset: 0

Vincenzo Librandi, Jan 09 2020

After 4, these numbers are the third column of the rectangular array in A238475.

Similar to A272743.
Cf. A000302, A002450, A052909, A178415, A237930, A238475, A347834.
Together with 1: first bisection of A136326.

Magma
```
[4^(n+1)+(4^n-1)/3: n in [0..30]];
```

Table[(4^(n + 1) + (4^n - 1) / 3), {n, 0, 30}]

G.f.: (4 - 3*x) / ((1 - x)*(1 - 4*x)).
a(n) = 5*a(n-1) - 4*a(n-2) for n > 1.
a(n) = 4*a(n-1) + 1 for n > 0.
a(n) = (13*4^n -1)/3, for n >= 0. - Wolfdieter Lang, Sep 16 2021
a(n) = A178415(5, n) = A347834(7, n-1), arrays, for n >= 1. - Wolfdieter Lang, Nov 29 2021

A350666 Numbers congruent to 0, 5, and 7 modulo 9: positions of 0 in A159955.

Original entry on oeis.org

0, 5, 7, 9, 14, 16, 18, 23, 25, 27, 32, 34, 36, 41, 43, 45, 50, 52, 54, 59, 61, 63, 68, 70, 72, 77, 79, 81, 86, 88, 90, 95, 97, 99, 104, 106, 108, 113, 115, 117, 122, 124, 126, 131, 133, 135, 140, 142, 144, 149

Offset: 0

Wolfdieter Lang, Jan 29 2022

This sequence, together with A350667 and A350668, gives a 3-set partition of the nonnegative integers.

This sequence {a(n)}, for n >= 1, gives the indices of the row sequences of array A = A347834, that are modulo 6 periodic with period length 3, namely: {A347834(a(n), m) mod 6}_{m >= 0} = {repeat(0, 3, 1)}.

			Rows of array {A347834(a(n), m)}_{m>=0}, with modulo 6 congruence:
n = 1: row 5: {11, 45, 181, 725, 2901, 11605,...} mod 6 = {5, 3, 1, 5, 3, 1, ...},
n = 2: row 7: {17, 69, 277, 1109, 4437, 17749, ...} mod 6 = {repeat(5, 3, 1)},
...

Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A159955, A347834, A350667, A350668.

Select[Range[0, 150], MemberQ[{0, 5, 7}, Mod[#, 9]] &] (* Amiram Eldar, Jan 29 2022 *)
Table[1 + 3n - ChebyshevU[n,-1/2],{n,0,49}] (* Stefano Spezia, Jan 30 2022 *)

A159955(a(n)) = 0.
Trisection: a(3*k) = 9*k, a(3*k+1) = 5 + 9*k, and a(3*k+2) = 7 + 9*k, for k >= 0.
G.f.: x*(5 + 2*x + 2*x^2)/((1 - x)*(1 - x^3)).
a(n) = 1 + 3*n - U(n, -1/2) = 1+3*n-A049347(n), where U(n, x) is a Chebyshev U-polynomial. - Stefano Spezia, Jan 30 2022
a(n) = 1 + 3*n - (2/sqrt(3))*sin(2*(n+1)*Pi/3) (from the previous formula).

A350667 Numbers congruent to 1, 3, and 8 modulo 9: positions of 1 in A159955.

Original entry on oeis.org

1, 3, 8, 10, 12, 17, 19, 21, 26, 28, 30, 35, 37, 39, 44, 46, 48, 53, 55, 57, 62, 64, 66, 71, 73, 75, 80, 82, 84, 89, 91, 93, 98, 100, 102, 107, 109, 111, 116, 118, 120, 125, 127, 129, 134, 136, 138, 143, 145, 147

Offset: 0

Wolfdieter Lang, Jan 29 2022

This sequence, together with A350666 and A350668, gives a 3-set partition of the nonnegative integers.

This sequence {a(n)}A347834,%20that%20are%20modulo%206%20periodic%20with%20period%20length%203,%20namely%20%7BA347834(a(n),%20m)%20mod%206%7D">{n>=0}, gives the indices of the row sequences of array A = A347834, that are modulo 6 periodic with period length 3, namely {A347834(a(n), m) mod 6}{m>=0} = {repeat(1, 5, 3)}.

			Rows of array {A347834(a(n), m)}_{m>=0}, with modulo 6 congruence:
n = 0: row 1: {1, 5, 21, 85, 341, 1365, 5461, ...} mod 6 = {repeat(1, 5, 3)},
n = 1: row 3: {7, 29, 117, 469, 1877, 7509, ...} mod 6 = {repeat(1, 5, 3)},
...

Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A049310, A055264, A159955, A347834, A350666, A350668.

Select[Range[0, 150], MemberQ[{1, 3, 8}, Mod[#, 9]] &] (* Amiram Eldar, Jan 29 2022 *)

A159955(a(n)) = 1.
Trisection: a(3*k) = 1 + 9*k, a(3*k+1) = 3 + 9*k, and a(3*k+3) = 8 + 9*k, for k >= 0.
G.f.: (1 + 2*x + 5*x^2 + x^3)/((1 - x)*(1 - x^3)).
a(n) = 1 + 3*n - 2*sin(2*n*Pi/3)/sqrt(3). - Stefano Spezia, Jan 30 2022
a(n) = 1 + 3*n - S(n-1,-1), with S(-1, x) = 0, with the Chebyshev S polynomials from A049310. From the g.f., or from the previous formula (see also Spezia's formula in A350666).

A347838 Positive numbers that are congruent to 2, 5, or 11 modulo 12.

Original entry on oeis.org

2, 5, 11, 14, 17, 23, 26, 29, 35, 38, 41, 47, 50, 53, 59, 62, 65, 71, 74, 77, 83, 86, 89, 95, 98, 101, 107, 110, 113, 119, 122, 125, 131, 134, 137, 143, 146, 149, 155, 158, 161, 167, 170, 173, 179, 182, 185, 191, 194, 197, 203, 206, 209, 215, 218, 221, 227, 230, 233, 239

Offset: 1

Wolfdieter Lang, Oct 21 2021

This sequence follows from the first column sequence of the array A347834, namely A047529 ({1,3,7} (mod 8)), as given in the formula below.

Together with A017617, the positive integers congruent to 8 modulo 12, one obtains A016789, the positive integers congruent to 2 modulo 3. See the array A347839.

Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A002264, A016921, A016789, A017617, A017653, A040117(k+1), A047261, A047529, A102283, A347834, A347839.

Map[(3 # + 1)/2 &, LinearRecurrence[{1, 0, 1, -1}, {1, 3, 7, 9}, 60]] (* Michael De Vlieger, Oct 21 2021 *)

a(n) = (3*A047529(n) + 1)/2.
Trisection: a(3*k+1) = 2 + 12*k, a(3*k+2) = 5 + 12*k, a(3*k+3) = 11 + 12*k, or with a(3*k) = -1 + 12*k for k >= 0.
O.g.f. with a(0) =-1: G(x) = (-1 + 3*x + 3*x^2 + 7*x^3)/((1 - x)*(1 - x^3)) = -6/(1-x) + 4/(1-x)^2 + (1 + x)/(1 + x + x^2). Note that (1 - x)*(1 - x^3) = (1-x)^2*(1 + x + x^2) =  1 - x - x^3 + x^4.
a(n) = a(n-1) + a(n-3) - a(n-4), for n >= 4, given a(n) for 0..3, with a(0) = -1.
a(n) = 2*b(n) + 3*b(n-1) + 6*b(n-2) + b(n-3), with b(n) = floor((n+2)/3) = A002264(n+2).
a(n) = -1 + 3*n + 3*floor(n/3) (from the partial fraction decomposition of G).
E.g.f.: 1 + 2*exp(x)*(2*x - 1) + exp(-x/2)*(3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))/3. - Stefano Spezia, Dec 08 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = ((sqrt(2)+1)*Pi + sqrt(3)*log(sqrt(3)+2) + sqrt(6)*log(5-2*sqrt(6)))/12. - Amiram Eldar, Dec 30 2021

cp's OEIS Frontend

A072201 a(n) = 4*a(n-1) + 1, a(1) = 15.

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A347839 An array of the positive integers congruent to 2 modulo 3 (A016789), read by antidiagonals upwards, giving the present triangle.

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A350053 a(n) = (2^(3*n + 3 + (-1)^n) - (6 + (-1)^n))/9, for n >= 1.

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A350668 Numbers congruent to 2, 4, and 6 modulo 9: positions of 2 in A159955.

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A323824 a(0) = 6; thereafter a(n) = 4*a(n-1) + 1.

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A330246 a(n) = 4^(n+1) + (4^n-1)/3.

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A350666 Numbers congruent to 0, 5, and 7 modulo 9: positions of 0 in A159955.

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A350667 Numbers congruent to 1, 3, and 8 modulo 9: positions of 1 in A159955.