A259663 -id:A259663 - cp's OEIS frontend

A316625 Terms in A259663, in ascending order.

1, 3, 5, 7, 11, 13, 15, 19, 21, 23, 31, 35, 47, 53, 55, 63, 79, 85, 87, 95, 99, 127, 143, 151, 191, 213, 223, 227, 255, 271, 319, 341, 351, 383, 407, 483, 511, 575, 663, 739, 767, 783, 853, 863, 895, 1023, 1175, 1251, 1279, 1365, 1407, 1535, 1599, 1807, 1887, 2047

Offset: 1

Bob Selcoe, Jul 08 2018

nonn

See A259663 for discussion of these terms in relation to Collatz sequences.
There are k terms in the interval [2^k, 2^(k+1)], k >= 1; terms in each interval are of the form 2^k + a(n) for some n.
The sequence is a permutation (without repeating terms) of the following numbers:
2^i-1 and 7*2^i-1 when i is odd, i >= 1;
3^2^i-1 and 5^2^i-1 when i is even, i >= 2;
For fixed k >= 4: least residues of 3^j*(2^(2^(k-3) + i*2^(k-2) - j)) - 1 mod 2^(2^(k-3) + i*2^(k-2) + k-j), i >= 0, 0 <= j < 2^(k-3) + i*2^(k-2) .  (See example).

			k=5, i=1 -- terms are least residues of 3^j*2^(12-j)-1 mod 2^(17-j), 0 <= j < 12:
j=0: 4096-1 mod 131072 = 4095;
j=1: 3*2048-1 mod 65536 = 6143;
j=2: 9*1024-1 mod 32768 = 9215;
j=3: 27*512-1 mod 16384 = 13823;
j=4: 81*256-1 mod 8192 = 20735 mod 8192 == 4351;
j=5: 243*128-1 mod 4096 = 31103 mod 4096 == 2431;
j=6: 729*64-1 mod 2048 = 46655 mod 2048 == 1599;
j=7: 2187*32-1 mod 1024 = 69983 mod 1024 == 351;
j=8: 6561*16-1 mod 512 = 104975 mod 512 == 15;
j=9: 19683*8-1 mod 256 = 157463 mod 256 == 23;
j=10: 59049*4-1 mod 128 = 236195 mod 128 == 35;
j=11: 177147*2-1 mod 64 = 354293 mod 64 == 53.
Note: k=5, i=0 is equivalent to starting with j=0: 15 mod 512.

Cf. A259663.

More terms from Michel Marcus, Jul 10 2018

A328037 Irregular triangle T(n,k) read by rows: "quotient trajectories" in reduced Collatz sequences; i.e., T(n,k) = q-value(A256598(n,k)) where q-value(z) = (z - A259663(m,j))/2^(m+j) and (m,j) is the unique pair such that z == A259663(m,j) (mod 2^(m+j)). (See Comments for definitions.)

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 1, 0, 2, 0, 0, 0, 1, 5, 0, 0, 2, 6, 20, 15, 5, 17, 3, 9, 29, 2, 8, 24, 74, 27, 5, 15, 47, 17, 53

Offset: 0

Bob Selcoe, Oct 03 2019

Coefficients T(m,j) in the array A259663 are least residues in congruence classes T(m,j) mod 2^(m+j). T"(m,j) denotes all members of that class.
Reduced Collatz sequences (i.e., reduced sequences) are standard Collatz sequences excluding even terms. Row n in triangle A256598 shows the reduced sequence starting with 2n+1.
Let every positive odd number z = T"(m,j)_q, where q is the quotient of z in T"(m,j). For example, T(2,3) = 7 in A259663, so T"(2,3) contains all numbers == 7 (mod 32). So z = T"(2,3)_0 = 7, z = T"(2,3)_1 = 39, z = T"(2,3)_2 = 71, etc.
T(n,k) is the q-value of A256598(n,k) (see Example below). Thus, each row n is defined here as a "quotient trajectory" for the reduced sequence with starting term 2n+1.

			Triangle starts:
  0;
  0, 0, 0;
  0, 0;
  0, 0, 2, 0, 0, 0;
  1, 0, 0, 2, 0, 0, 0;
  0, 2, 0, 0, 0;
  0, 0, 0;
  0, 0, 0, 0, 0, 0;
  2, 0, 0, 0;
  0, 1, 0, 2, 0, 0, 0;
  0, 0;
  0, 0, 0, 0, 0;
  3, 0, 1, 0, 2, 0, 0, 0;
  ...
n=13 starts with 27 = T"(2,2)_1 and takes 41 steps: 1, 5, 0, 0, 2, 6, 20, 15, 5, ..., 0, 0, 0.
Row n=12 maps to the reduced sequence n=12 in A256598: 25 -> 19 -> 29 -> 11 -> 17 -> 13 -> 5 -> 1, which is T"(2,1)_3 -> T"(3,2)_0 -> T"(3,1)_1 -> T"(2,2)_0 -> T"(2,1)_2 -> T"(3,1)_0 -> T"(4,1)_0 -> T"(2,1)_0.

Cf. A256598, A259663.

Tdt(n, k) = if (n==2, if (k%2, 2^k-1, 3*2^k-1), if (n==3, if (k%2, 7*2^k-1, 5*2^k-1), mj = 2^(n-3) % 2^(n-2); mk = k % 2^(n-2); (2^k*3^(mj-mk) - 1) % 2^(n+k))); \\ A259663
qvalue(m) = {my(line = 2, i, md); while (1, i = line; for (j=1, line-1, md = Tdt(i, j); if (m % (2^(i+j)) == md % (2^(i+j)), return((m-md)/2^(i+j))); i--;); line ++;);}
row(n) = {my(oddn = 2*n+1, vl = List(oddn), x); while (oddn != 1, x = 3*oddn+1; oddn = x >> valuation(x, 2); listput(vl, oddn)); my(v = Vec(vl)); for (i=1, #v, v[i] = qvalue(v[i]);); v;} \\ A256598
tabf(nn) = {for (n=0, nn, my(rown = row(n)); for (k=1, #rown, print1(rown[k], ", ")); print;);} \\ Michel Marcus, Oct 04 2019

A087445 Numbers that are congruent to 1 or 5 mod 12.

Original entry on oeis.org

1, 5, 13, 17, 25, 29, 37, 41, 49, 53, 61, 65, 73, 77, 85, 89, 97, 101, 109, 113, 121, 125, 133, 137, 145, 149, 157, 161, 169, 173, 181, 185, 193, 197, 205, 209, 217, 221, 229, 233, 241, 245, 253, 257, 265, 269, 277, 281, 289, 293, 301, 305, 313, 317, 325, 329

Offset: 1

Paul Barry, Sep 04 2003

From Bob Selcoe, Jun 03 2015: (Start)
For k >= 1: all numbers congruent to A002450(k) mod 2^(2k+1) and A072197(k) mod 4^(k+1) not congruent to 0 mod 3. Equivalently, for k >= 3: all numbers congruent to A096773(k) mod 2^k not congruent to 0 mod 3.
Conjecture: at least one number in this sequence must appear in all Collatz sequences.
(End)
The sequence is composed of all numbers in congruence classes T(n,1) mod 2^(n+k) in A259663 (i.e., T"(1) in array T259663(n,k)) not congruent to 0 mod 3. Therefore the conjecture above is true (see A259663 for additional explanation). - Bob Selcoe, Jul 15 2017
Closure of {1} under the map (x,y)->2x+3y [Klarner-Rado, see Lagarias (2016), p. 755]. - N. J. A. Sloane, Oct 06 2016
The above conjecture is true: this is because even numbers and odd numbers divisible by 3 will lead to the set of odd numbers not divisible by 3. Odd numbers of the form 4k - 1 can also be ignored, as this consists of odd numbers that grow between themselves and the next odd term through Collatz iteration. No infinite sequence of growth between consecutive odd terms is possible, so all numbers of the form 4k - 1 will lead to an odd number that shrinks between itself and the next odd number. All numbers 4k - 1 will lead to a number in 4k - 3, the odd numbers that shrink between themselves and the following odd term. What we are left after that elimination is this sequence. - Aidan Simmons, Feb 25 2019

Jeffrey C. Lagarias, Erdős, Klarner and the 3x+ 1 Problem, Amer. Math. Monthly, Vol. 123, No. 8 (2016), pp. 753-776.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001651, A002450, A047241, A087444, A087446, A096773, A259663.

[k:k in [1..330]| k mod 12 in [1,5]]; // Marius A. Burtea, Feb 08 2020

seq(6*(n-1)-(-1)^n,n=1..100); # Robert Israel, Jun 10 2015

LinearRecurrence[{1,1,-1},{1,5,13},70] (* or *) Rest[CoefficientList[ Series[x (1+4x+7x^2)/((1+x)(1-x)^2),{x,0,70}],x]]  (* Harvey P. Dale, Jun 13 2011 *)

a(n)=(n-1)\2*12 + [5,1][n%2+1] \\ Charles R Greathouse IV, Jun 03 2015

G.f.: x*(1+4*x+7*x^2)/((1+x)*(1-x)^2).
E.g.f.: 6*(x-1)*exp(x) + 7 - exp(-x). - corrected by Robert Israel, Jun 10 2015
a(n) = 6*(n-1) - (-1)^n. - Rolf Pleisch, Aug 04 2009
a(n) = 12*n - a(n-1) - 18 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
a(n) = a(n-1) + a(n-2) - a(n-3), with a(0)=1, a(1)=5, a(2)=13. - Harvey P. Dale, Jun 13 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/12 + log(2 + sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 28 2021

A259614 Numbers congruent to {17,29} mod 36.

Original entry on oeis.org

17, 29, 53, 65, 89, 101, 125, 137, 161, 173, 197, 209, 233, 245, 269, 281, 305, 317, 341, 353, 377, 389, 413, 425, 449, 461, 485, 497, 521, 533, 557, 569, 593, 605, 629, 641, 665, 677, 701, 713, 737, 749, 773, 785, 809, 821, 845, 857, 881, 893, 917, 929, 953

Offset: 1

Bob Selcoe, Jun 30 2015

Subsequence of A087445.
Let terms in this sequence be T:
Collatz sequences (C) that contain no T must terminate at 1.
Define C containing at least one T as C(T), and let T(i) {i=1..z} be T in order of appearance in C(T).
All T(i) i>=2 have odd preimages congruent to either {1,5} mod 12 or {11,19} mod 24. Preimages of the second type (P2) are congruent to B mod 2^m (m>=4), where B is a set of numbers with a predictable recurrence pattern (a bit cumbersome to describe here) starting with A259663(n,2), i.e., {11, 19, 3, 35, 99, 483, ...}. All P2 lead to T(i) == A002450((m-2)/2) mod 2^(m-1) when m is even, and T(i) == A072197((m-3)/2) mod 2^(m-1) when m is odd. So, for example, T(i) == 1 mod 8 when P2 == 11 mod 16; T(i) == 13 mod 16 when P2 == 19 mod 32; T(i) == 5 mod 32 when P2 == 3 mod 64; T(i) == 53 mod 64 when P2 == 35 mod 128; etc.
If the Collatz conjecture is true (i.e., all C terminate at 1), then all C(T) contain T(z) after which all subsequent odd terms decrease and are congruent to {1,5} mod 12 that are not congruent to {17,29} mod 36. The first few T(z) are {17, 53, 341, 1109, 1205, ...}. So, for example, the trajectory of odd terms in C with initial term 950 is [475, 713, 535, 803, 1205, 113, 85, 1], where T(1) = 713 and T(2) = T(z) = 1205. In this example, P2 = 803 because 803 == 11 mod 24.

G. C. Greubel, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A002450, A072197, A087445, A259663.

[ n : n in [1..1000] | n mod 36 in [17, 29] ] // Vincenzo Librandi Jul 01 2015

Select[Range[1000], MemberQ[{17, 29}, Mod[#, 36]] &] (* Vincenzo Librandi, Jul 01 2015 *)

G.f.: x*(17+12*x+7*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 26 2015

E.g.f.: 7 + (18*x - 4)*exp(x) - 3*exp(-x). - David Lovler, Sep 10 2022

A327283 Irregular triangle T(n,k) read by rows: "residual summands" in reduced Collatz sequences (see Comments for definition and explanation).

Original entry on oeis.org

1, 1, 5, 1, 5, 19, 73, 347, 1, 7, 29, 103, 373, 1631, 1, 5, 23, 133, 1, 11, 1, 5, 19, 65, 451, 1, 7, 53, 1, 5, 31, 125, 503, 2533, 1, 1, 5, 19, 185, 1, 7, 29, 151, 581, 2255, 10861, 1, 5, 23, 85, 287, 925

Offset: 1

Bob Selcoe, Sep 15 2019

Let R_s be the reduced Collatz sequence (cf. A259663) starting with s and let R_s(k), k >= 0 be the k-th term in R_s. Then R_(2n-1)(k) = (3^k*(2n-1) + T(n,k))/2^j, where j is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k). T(n,k) is defined here as the "residual summand".

The sequence without duplicates is a permutation of A116641.

			Triangle starts:
  1;
  1, 5;
  1;
  1, 5, 19, 73,  347;
  1, 7, 29, 103, 373, 1631;
  1, 5, 23, 133;
  1, 11;
  1, 5, 19, 65,  451;
  1, 7, 53;
  1, 5, 31, 125, 503, 2533;
  1;
  1, 5, 19, 185;
  1, 7, 29, 151, 581, 2255, 10861;
  ...
T(5,4)=103 because R_9(4) = 13; the number of halving steps from R_9(0) to R_9(4) is 6, and 13 = (81*9 + 103)/64.

Cf. A116623, A116641, A259663.

T(n,k) = 2^j*R_(2n-1)(k) - 3^k*(2n-1), as defined in Comments.

T(n,1) = 1; for k>1: T(n,k) = 3*T(n,k-1) + 2^i, where i is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k-1).

cp's OEIS Frontend

A316625 Terms in A259663, in ascending order.

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A328037 Irregular triangle T(n,k) read by rows: "quotient trajectories" in reduced Collatz sequences; i.e., T(n,k) = q-value(A256598(n,k)) where q-value(z) = (z - A259663(m,j))/2^(m+j) and (m,j) is the unique pair such that z == A259663(m,j) (mod 2^(m+j)). (See Comments for definitions.)

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A087445 Numbers that are congruent to 1 or 5 mod 12.

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Magma

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A259614 Numbers congruent to {17,29} mod 36.

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Magma

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A327283 Irregular triangle T(n,k) read by rows: "residual summands" in reduced Collatz sequences (see Comments for definition and explanation).

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Author

Keywords

Comments

Examples

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Formula