A349407
The Farkas map: a(n) = x/3 if x mod 3 = 0; a(n) = (3x+1)/2 if x mod 3 <> 0 and x mod 4 = 3; a(n) = (x+1)/2 if x mod 3 <> 0 and x mod 4 = 1, where x = 2*n-1.
Original entry on oeis.org
1, 1, 3, 11, 3, 17, 7, 5, 9, 29, 7, 35, 13, 9, 15, 47, 11, 53, 19, 13, 21, 65, 15, 71, 25, 17, 27, 83, 19, 89, 31, 21, 33, 101, 23, 107, 37, 25, 39, 119, 27, 125, 43, 29, 45, 137, 31, 143, 49, 33, 51, 155, 35, 161, 55, 37, 57, 173, 39, 179, 61, 41, 63, 191, 43
Offset: 1
From _Omar E. Pol_, Jan 01 2022: (Start)
Written as a rectangular array with six columns read by rows the sequence begins:
1, 1, 3, 11, 3, 17;
7, 5, 9, 29, 7, 35;
13, 9, 15, 47, 11, 53;
19, 13, 21, 65, 15, 71;
25, 17, 27, 83, 19, 89;
31, 21, 33, 101, 23, 107;
37, 25, 39, 119, 27, 125;
43, 29, 45, 137, 31, 143;
49, 33, 51, 155, 35, 161;
55, 37, 57, 173, 39, 179;
...
(End)
- H. M. Farkas, "Variants of the 3N+1 Conjecture and Multiplicative Semigroups", in Entov, Pinchover and Sageev, "Geometry, Spectral Theory, Groups, and Dynamics", Contemporary Mathematics, vol. 387, American Mathematical Society, 2005, p. 121.
- J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, American Mathematical Society, 2010, p. 74.
- Paolo Xausa, Table of n, a(n) for n = 1..10000
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).
- Index entries for sequences related to 3x+1 (or Collatz) problem
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LinearRecurrence[{0,0,0,0,0,2,0,0,0,0,0,-1},{1,1,3,11,3,17,7,5,9,29,7,35},100]
Table[Which[Mod[n,3]==0,n/3,Mod[n,4]==3,(3n+1)/2,True,(n+1)/2],{n,1,200,2}] (* Harvey P. Dale, May 15 2022 *)
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a(n)=if (n%3==2, 2*n\3, if (n%2, n, 3*n-1)) \\ Charles R Greathouse IV, Nov 16 2021
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def a(n):
x = 2*n - 1
return x//3 if x%3 == 0 else ((3*x+1)//2 if x%4 == 3 else (x+1)//2)
print([a(n) for n in range(1, 66)]) # Michael S. Branicky, Nov 16 2021
A239127
Rectangular companion array to M(n,k), given in A239126, showing the end numbers N(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.
Original entry on oeis.org
5, 11, 17, 17, 35, 53, 23, 53, 107, 161, 29, 71, 161, 323, 485, 35, 89, 215, 485, 971, 1457, 41, 107, 269, 647, 1457, 2915, 4373, 47, 125, 323, 809, 1943, 4373, 8747, 13121, 53, 143, 377, 971, 2429, 5831, 13121, 26243, 39365, 59, 161, 431, 1133, 2915, 7289, 17495, 39365, 78731, 118097
Offset: 1
The rectangular array N(n, k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 ...
1: 5 11 17 23 29 35 41 47 53 59
2: 17 35 53 71 89 107 125 143 161 179
3: 53 107 161 215 269 323 377 431 485 539
4: 161 323 485 647 809 971 1133 1295 1457 1619
5: 485 971 1457 1943 2429 2915 3401 3887 4373 4859
6: 1457 2915 4373 5831 7289 8747 10205 11663 13121 14579
7: 4373 8747 13121 17495 21869 26243 30617 34991 39365 43739
8: 13121 26243 39365 52487 65609 78731 91853 104975 118097 131219
9: 39365 78731 118097 157463 196829 236195 275561 314927 354293 393659
10: 118097 236195 354293 472391 590489 708587 826685 944783 1062881 1180979
...
-------------------------------------------------------------------------------
The triangle TN(m, n) begins (zeros are not shown):
m\n 1 2 3 4 5 6 7 8 9 10 ...
1: 5
2: 11 17
3: 17 35 53
4: 23 53 107 161
5: 29 71 161 323 485
6: 35 89 215 485 971 1457
7: 41 107 269 647 1457 2915 4373
8: 47 125 323 809 1943 4373 8747 13121
9: 53 143 377 971 2429 5831 13121 26243 39365
10: 59 161 431 1133 2915 7289 17495 39365 78731 118097
...
n=1, ud, k=1: M(1, 1) = 3 = TM(1, 1), N(1,1) = 5 with the Collatz sequence [3, 10, 5] of length 3.
n=1, ud, k=2: M(1, 2) = 7 = TM(2, 1), N(1,2) = 11 with the Collatz sequence [7, 22, 11] of length 3.
n=4, (ud)^4, k=2: M(4, 2) = 63 = TM(5, 4), N(4,2) = 323 with the Collatz sequence [63, 190, 95, 286, 143, 430, 215, 646, 323] of length 9.
n=5, (ud)^5, k=1: M(5, 1) = 63 = TM(5, 5), N(5,1) = 485 with the Collatz sequence [63, 190, 95, 286, 143, 430, 215, 646, 323, 970, 485] of length 11.
- Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
- Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
A350522
a(n) = 18*n + 16.
Original entry on oeis.org
16, 34, 52, 70, 88, 106, 124, 142, 160, 178, 196, 214, 232, 250, 268, 286, 304, 322, 340, 358, 376, 394, 412, 430, 448, 466, 484, 502, 520, 538, 556, 574, 592, 610, 628, 646, 664, 682, 700, 718, 736, 754, 772, 790, 808, 826, 844, 862, 880, 898, 916, 934, 952, 970
Offset: 0
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List([0..53], n-> 18*n+16)
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[18*n+16: n in [0..53]];
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seq(18*n+16, n=0..53);
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Table[18n+16, {n, 0, 53}]
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makelist(18*n+16, n, 0, 53);
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a(n)=18*n+16
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[18*n+16 for n in range(53)]
A251731
Least k such that k^3 + q is divisible by 3^n where q is the n-th number congruent to 1 or -1 (mod 18).
Original entry on oeis.org
2, 1, 2, 16, 32, 145, 62, 1363, 3458, 19492, 58928, 89308, 70028, 1594318, 1890551, 189871, 31401806, 47918575, 190704887, 163454602, 502048577, 9481323661, 11627845304, 34656488290, 115450061084, 286130228125, 2303721331049, 1569269836240, 22013516320412
Offset: 1
a(1) = 2 because the first number of the form +-1 (mod 18) is 1, and 2^3 + 1 = 9 = 3*3^1;
a(2) = 1 because the second number of the form +-1 (mod 18) is 17, and 1^3 + 17 = 18 = 2*3^2;
a(3) = 2 because the third number of the form +-1 (mod 18) is 19, and 2^3 + 19 = 27 = 3^3;
a(4)= 16 because the fourth number of the form +-1 (mod 18) is 35, and 16^3 + 35 = 4131 = 51*3^4.
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f:= proc(n) local q,R,k;
if n::odd then q:= 9*n-8 else q:= 9*n-1 fi;
min(map(subs,[msolve(k^3+q,3^n)],k))
end proc:
map(f, [$1..30]); # Robert Israel, Dec 23 2018
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lst1={1};Do[lst1=Union[lst1,Union[{18*n+1},{18*n-1}]],{n,1,10}];lst={};Do[k=1;While[Mod[k^3+lst1[[n]],3^n]!=0,k++];Print[n," ",k],{n,1,10}];lst
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a(n) = {if (n % 2, q = 9*(n-1)+1, q = 9*n-1); m = 3^n; k = 1; while ((k^3+q) % m, k++); k;} \\ Michel Marcus, Jan 07 2015
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