A087445 -id:A087445 - cp's OEIS frontend

A381583 Starts of runs of 3 consecutive integers that are all terms in A381581.

1, 2, 20, 55, 56, 110, 304, 364, 398, 846, 1024, 1084, 1744, 1854, 2044, 2104, 2105, 2527, 2824, 2862, 3870, 4374, 5222, 5223, 5243, 5718, 5928, 6488, 6784, 6844, 6894, 6978, 7142, 7924, 10590, 11240, 11889, 11975, 12248, 14284, 14915, 16638, 17710, 17714, 17824

Offset: 1

Amiram Eldar, Feb 28 2025

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, and A291711(3) = 1 divides 3.
20 is a term since A291711(20) = 4 divides 20, A291711(21) = 1 divides 21, and A291711(22) = 2 divides 22.

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

Cf. A000045, A001906, A087445, A291711.
Subsequence of A381581 and A381582.
Subsequences: A381584, A381585.
Similar sequences: A154701, A328210, A330932, A351721.

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[45, 3]

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, lim, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}

A381584 Starts of runs of 4 consecutive integers that are all terms in A381581.

Original entry on oeis.org

1, 55, 2104, 5222, 24784, 63510, 64264, 69487, 95463, 121393, 184327, 327303, 374589, 463110, 468168, 561069, 572550, 596868, 671407, 740310, 759030, 819948, 902670, 956680, 1023009, 1036230, 1065030, 1259817, 1274910, 1359552, 1683154, 1714470, 1731750, 2182023

Offset: 1

Amiram Eldar, Feb 28 2025

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, A291711(3) = 1 divides 3, and A291711(4) = 2 divides 4.
55 is a term since A291711(55) = 1 divides 55, A291711(56) = 2 divides 56, A291711(57) = 3 divides 57, and A291711(58) = 2 divides 58.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A000045, A001906, A087445, A291711.
Subsequence of A381581, A381582 and A381583.
A381585 is a subsequence.
Similar sequences: A141769, A328211, A328215, A330933.

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[12, 4]

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3 = is1(3), q4); for(k = 4, lim, q4 = is1(k); if(q1 && q2 && q3 && q4, print1(k-3, ", ")); q1 = q2; q2 = q3; q3 = q4);}

A087444 Numbers that are congruent to {1, 4} mod 9.

Original entry on oeis.org

1, 4, 10, 13, 19, 22, 28, 31, 37, 40, 46, 49, 55, 58, 64, 67, 73, 76, 82, 85, 91, 94, 100, 103, 109, 112, 118, 121, 127, 130, 136, 139, 145, 148, 154, 157, 163, 166, 172, 175, 181, 184, 190, 193, 199, 202, 208, 211, 217, 220, 226, 229, 235, 238, 244, 247, 253

Offset: 1

Paul Barry, Sep 04 2003

3*a(n) is conjectured to be the total number of sides (straight double bonds (long side) and single bond (short side)) of a certain equilateral triangle expansion shown in one of the links. The pattern is supposed to become the planar Archimedean net 3.3.3.3.6 when n -> infinity. 3*a(n) is also conjectured to be the total number of sided (bonds) in another version of an equilateral triangle expansion that is supposed to become the planar Archimedean net 3.6.3.6. See the illustrations in the links. - Kival Ngaokrajang, Nov 30 2014

David Lovler, Table of n, a(n) for n = 1..10000
Kival Ngaokrajang, Illustration of initial terms (3.3.3.3.6), (3.6.3.6)
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001651, A047241, A087445, A087446.

Select[Range[300],MemberQ[{1,4},Mod[#,9]]&] (* or *) LinearRecurrence[ {1,1,-1},{1,4,10},60] (* Harvey P. Dale, Jan 22 2019 *)

a(n) = (18*n - 17 - 3*(-1)^n)/4 \\ David Lovler, Aug 20 2022

G.f.: x*(1+3*x+5*x^2)/((1+x)*(1-x)^2).
E.g.f.: 5 + ((9*x - 17/2)*exp(x) - (3/2)*exp(-x))/2.
a(n) = (18*n - 17 - 3*(-1)^n)/4.
a(n) = 9*n - a(n-1) - 13 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010

Kival Ngaokrajang's comment reworded by Wolfdieter Lang, Dec 05 2014

E.g.f. and formula adapted to offset by David Lovler, Aug 20 2022

A087446 Numbers that are congruent to {1, 6} mod 15.

Original entry on oeis.org

1, 6, 16, 21, 31, 36, 46, 51, 61, 66, 76, 81, 91, 96, 106, 111, 121, 126, 136, 141, 151, 156, 166, 171, 181, 186, 196, 201, 211, 216, 226, 231, 241, 246, 256, 261, 271, 276, 286, 291, 301, 306, 316, 321, 331, 336, 346, 351, 361, 366, 376, 381, 391, 396, 406

Offset: 1

Paul Barry, Sep 04 2003

3*a(n) is conjectured to be the number of edges (bonds) visited when walking around the boundary of a certain equilateral triangle construction at the n-th iteration. See the illustration in the link. Note that isthmus edges (bridges) are counted twice. The pattern is supposed to become the planar Archimedean net 3.12.12 when n -> infinity. - Kival Ngaokrajang, Nov 30 2014

Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,1,-1)

Cf. A001651, A047241, A087444, A087445.

#+{1,6}&/@(15*Range[0,30])//Flatten (* or *) LinearRecurrence[{1,1,-1},{1,6,16},60] (* Harvey P. Dale, Dec 05 2018 *)

G.f.: x*(1 + 5*x + 9*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (30*x-1)*exp(x)/4 + 5*exp(-x)/4.
a(n) = (18*n-1)/4 + 5*(-1)^n/4.
a(n) = 15*n - a(n-1) - 23, with a(1)=1. - Vincenzo Librandi, Aug 08 2010

Editing: rewording of Kival Ngaokrajang's comment. - Wolfdieter Lang, Dec 06 2014

A228137 Numbers that are congruent to {1, 4} mod 12.

Original entry on oeis.org

1, 4, 13, 16, 25, 28, 37, 40, 49, 52, 61, 64, 73, 76, 85, 88, 97, 100, 109, 112, 121, 124, 133, 136, 145, 148, 157, 160, 169, 172, 181, 184, 193, 196, 205, 208, 217, 220, 229, 232, 241, 244, 253, 256, 265, 268, 277, 280, 289, 292, 301, 304, 313, 316, 325

Offset: 1

Colin Barker, Aug 12 2013

The squares of the terms of A001651 are the squares of this sequence. - Bruno Berselli, Aug 12 2013

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001651, A087445, A146512, A228138.

Select[Range[300], MemberQ[{1, 4}, Mod[#, 12]] &] (* Amiram Eldar, Dec 28 2021 *)

Vec(x*(8*x^2+3*x+1)/((x-1)^2*(x+1)) + O(x^99))

a(n) = -13/2 - 3*(-1)^n/2 + 6*n.
a(n) = a(n-1) + a(n-2) - a(n-3).
G.f.: x*(8*x^2+3*x+1) / ((x-1)^2*(x+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(3)+3)*Pi/36 + log(2)/4 - sqrt(3)*log(26-15*sqrt(3))/36. - Amiram Eldar, Dec 28 2021
E.g.f.: 8 + ((12*x - 13)*exp(x) - 3*exp(-x))/2. - David Lovler, Sep 04 2022

A174398 Numbers that are congruent to {1, 4, 5, 8} mod 12.

Original entry on oeis.org

1, 4, 5, 8, 13, 16, 17, 20, 25, 28, 29, 32, 37, 40, 41, 44, 49, 52, 53, 56, 61, 64, 65, 68, 73, 76, 77, 80, 85, 88, 89, 92, 97, 100, 101, 104, 109, 112, 113, 116, 121, 124, 125, 128, 133, 136, 137, 140, 145, 148, 149, 152, 157, 160, 161, 164, 169, 172, 173

Offset: 1

Gary Detlefs, Mar 18 2010

Numbers k such that k*(k + 3)/4 + (k + 1)*(k + 2)/6 or k*(5*k + 3)/12 + 1/3 is a nonnegative integer. - Bruno Berselli, Feb 14 2017

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Cf. A087445, A092259.

[n : n in [0..200] | n mod 12 in [1, 4, 5, 8]]; // Wesley Ivan Hurt, Jun 07 2016

Maple
```
seq(3*n +(-1)^floor(n/2), n=0..50);
```

Table[(1+I)*(3*(n-n*I+I-1)+I^(1-n)-I^n)/2, {n, 60}] (* Wesley Ivan Hurt, Jun 07 2016 *)
Select[Range[200],MemberQ[{1,4,5,8},Mod[#,12]]&] (* or *) LinearRecurrence[ {2,-2,2,-1},{1,4,5,8},60] (* Harvey P. Dale, Aug 02 2020 *)

a(n) = 3*n - 3 + (-1)^floor((n-1)/2).
From Wesley Ivan Hurt, Jun 07 2016: (Start)
G.f.: x*(1 + 2*x - x^2 + 4*x^3)/((1 - x)^2*(1 + x^2)).
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) for n>4.
a(n) = (1 + i)*(3*(n - n*i + i - 1) + i^(1-n) - i^n)/2, where i=sqrt(-1).
a(2*k) = A092259(k), a(2*k-1) = A087445(k). (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/6 + log(2)/2. - Amiram Eldar, Dec 31 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

A364602 Triangle T(n,k) with rows of length 2n-1, generated by T(1,1)=0, T(n,1)=T(n-1,1)+2, T(n,2)=4(n-1)-1, and for k>=3, T(n,k)=4*T(n-1,k-2)+1.

Original entry on oeis.org

0, 2, 3, 1, 4, 7, 9, 13, 5, 6, 11, 17, 29, 37, 53, 21, 8, 15, 25, 45, 69, 117, 149, 213, 85, 10, 19, 33, 61, 101, 181, 277, 469, 597, 853, 341, 12, 23, 41, 77, 133, 245, 405, 725, 1109, 1877, 2389, 3413, 1365, 14, 27, 49, 93, 165, 309, 533, 981, 1621, 2901

Offset: 1

Ruud H.G. van Tol, Jul 29 2023

The sequence is a permutation of all integers >= 0.
Each row of T contains n*2-1 terms; the terms in column k increase by 2^k.
T(1,1) = 0; T(2,2) = 3.
T(2,1) = T(1,1)+2 = 2; T(2,3) = 4*T(1,1)+1 = 1 ("knight jump").
In the context of the 3x+1 problem, when a term x is used to represent the odd 4*x+1, its successor is 3*x+1, and k-1 is the 2-adic valuation of 3*x+1.
Right diagonal is A002450.
The terms at the top of the columns are A096773(k), or (2^(k-1)*(3 + 2*(-1)^k) - 1)/3.
When the table is analytically continued upwards by subtracting 2^k, the first layer of values are -A255138(k), or -(2^k*(3 + 2*(-1)^k) + 1)/3.

			Triangle T(n,k) begins:
n/k 1| 2| 3| 4|  5|  6|  7|  8|  9| 10| 11|
1|  0
2|  2  3  1
3|  4  7  9 13   5
4|  6 11 17 29  37  53  21
5|  8 15 25 45  69 117 149 213  85
6| 10 19 33 61 101 181 277 469 597 853 341
7| 12 ...

Cf. A002450, A007310, A071797, A087445, A096773, A255138, A257480.

my(N=8, v=Vec([0, 2, 3, 1], N^2), p=4); for(n=3, N, my(K=2*n-1); for(k=1, K, v[p+k]=if(k<=2, v[p-K+k+2]+2^k, 4*v[p-K+k]+1)); p+=K); v

T(n, k) = 2^k*(n-(6*k+3-(-1)^k)/12)-1/3;

n_of_x(x) = my(n=0); while(1==x%4, x>>=2; n++); n + if(x%2,(x+1)/4,  x/2) + 1;

PARI
```
k_of_x(x) = valuation(3*x+1,2) + 1;
```

For n>1, T(n,k) = T(n-1,k) + 2^k, so T(n,1) = 2*(n-1).
T(n,2) = 4*(n-1)-1 = 2*T(n,1)-1, so T(2,2) = 3.
For n>1 and k>2, T(n,k) = 4*T(n-1,k-2)+1, so T(2,3) = 1.
For i>=0, a(i^2+1) = T(i+1,1).
T(n, k) = 2^k * (n - (6*k + 3 - (-1)^k)/12) - 1/3.
T(n,1) == 0 (mod 2); T(n,2) == 3 (mod 4); T(n,k>=3) == 1 (mod 4).
k = v2(3*T(n,k)+1) + 1, where v2(x) = A007814(x) is the 2-adic valuation of x.

cp's OEIS Frontend

A381583 Starts of runs of 3 consecutive integers that are all terms in A381581.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A381584 Starts of runs of 4 consecutive integers that are all terms in A381581.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A087444 Numbers that are congruent to {1, 4} mod 9.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A087446 Numbers that are congruent to {1, 6} mod 15.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A228137 Numbers that are congruent to {1, 4} mod 12.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A174398 Numbers that are congruent to {1, 4, 5, 8} mod 12.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A364602 Triangle T(n,k) with rows of length 2n-1, generated by T(1,1)=0, T(n,1)=T(n-1,1)+2, T(n,2)=4(n-1)-1, and for k>=3, T(n,k)=4*T(n-1,k-2)+1.