This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
The triangle begins 2 4 16 12, 32 8, 52, 128 40, 56, 84, 136, 160, 180, 256 60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024
Segments of A022328 starting with 0, deleted maxima in brackets: . 1: 0 [1] . 2: 0 2 1 [3] . 3: 0 2 [4] 1 3 . 4: 0 5 2 4 1 [6] 3 . 5: 0 5 2 [7] 4 1 6 3 . 6: 0 8 5 2 7 4 1 [9] 6 3 . 7: 0 8 5 2 10 7 4 1 9 6 3 [11] . 8: 0 8 5 2 10 7 4 [12] 1 9 6 3 11 . 9: 0 8 5 13 2 10 7 4 12 1 9 6 [14] 3 11 . 10: 0 8 5 13 2 10 7 [15] 4 12 1 9 6 14 3 11 . 11: 0 8 16 5 13 2 10 7 15 4 12 1 9 [17] 6 14 3 11 . 12: 0 8 16 5 13 2 10 18 7 15 4 12 1 9 17 6 14 3 11 [19] . 13: 0 8 16 5 13 2 10 18 7 15 4 12 [20] 1 9 17 6 14 3 11 19 . 14: 0 8 16 5 13 21 2 10 18 7 15 4 12 20 1 9 17 6 14 [22] 3 11 19 . 15: 0 8 16 5 13 21 2 10 18 7 15 [23] 4 12 20 1 9 17 6 14 22 3 11 19
import Data.List (delete)
a258033 n = a258033_list !! (n-1)
a258033_list = 0 : f (tail a022328_list) where
f xs = (0 : (delete (maximum ys) ys)) ++ f zs
where (ys, (_ : zs)) = span (> 0) xs
Every odd number greater than 1 yields a msa value. a(1) is 4 and it corresponds to A260590(1). a(2) is 2 since A260590(2) is 2. a(3) is 7 since A260590(3) is 7. a(4) is 5 since A260590(5) is 5, A260590(4) is 2 but it already appears as a(2).
msa[n_] := If[ OddQ@ n, (3n + 1)/2, n/2]; f[n_] := Block[{k = 2n + 1}, Length@ NestWhileList[ msa@# &, k, # >= k &] - 1]; k = 1; lst = {}; While[k < 10000001, a = f@ k; If[ !MemberQ[lst, a], AppendTo[lst, a]]; k++]; lst
import math
numberOfTerms = 20
L6 = [1,0]
def c(n):
return math.floor(n/(math.log2(3)-1))
def p(a,b):
return math.factorial(a)/(math.factorial(a-b)*math.factorial(b))
def anotherTerm(newTermCount):
global L6
for a in range(newTermCount+1-len(L6)):
y = len(L6)
newElement = 0
for k in range(y):
newElement -= int(L6[k]*p(c(y)+y-k-2, c(y)-2))
L6.append(newElement)
anotherTerm(numberOfTerms)
print("A325904")
for a in range(numberOfTerms+1):
print(a, "|", L6[a])
@cached_function
def a(n):
if n < 2: return 0^n
A = floor(n/(log(3, 2) - 1)) - 2
return -sum(a(k)*binomial(A + n - k, A) for k in (0..n-1))
[a(n) for n in range(100)] # Peter Luschny, Sep 10 2019
Example with 4x+b (0 <= b < 4): 4x is even, thus gives 2x, 2 < 4 (first case). 4x+1, is odd thus 3(4x+1)+1 = 12x+4 is even, thus (12x+4)/2/2=3x+1 3 < 4, first case. 4x+2 is even, (4x+2)/2=2x+1, 2 < 4, first case. 4x+3 with same way gives 9x+8. 9 is odd and 9 > 4, second case. That explains why the second (n=2) term in sequence is 3.
a[n_] := Module[{b, p0, p1, minimized = 0}, For[b = 1, b <= 2^n, b++, {p0, p1} = {b, 2^n}; While[Mod[p1, 2] == 0 && p1 >= 2^n, {p0, p1} = If[Mod[p0, 2] == 0, {p0/2, p1/2}, {3*p0+1, 3*p1}]; If[p1<2^n, minimized += 1]]]; minimized]; Table[Print[an = a[n]]; an, {n, 1, 40}] (* Jean-François Alcover, Feb 12 2014, translated from D. S. McNeil's Sage code *)
upto(P=18)= my(r=Vec([1, 1], P)); forstep(x=3,2^P,4, my(s=x, p=0); until(s<=x, s= if(s%2, 3*s+1, s)/2; if(p++ > P, next(2))); if((2^p>x), r[p]++)); for(i=2, #r, r[i]+= 2*r[i-1]); print(r); \\ Ruud H.G. van Tol, Mar 13 2023
def A182137(n): minimized = 0 for b in range(2**n): p = [b, 2**n] while p[1] % 2 == 0 and p[1] >= 2**n: p[0],p[1] = [p[0]/2, p[1]/2] if p[0] % 2 == 0 else [3*p[0]+1, 3*p[1]] if p[1] < 2**n: minimized += 1 return minimized # D. S. McNeil, Apr 14 2012
for n from 0 to 120 do : for k from 0 to 100 do: x:=floor(2^k /3^n):if x=1 then printf(`%d, `,k):else fi:od:od:
a(n)=ceil(n*log(3)/log(2)) \\ Charles R Greathouse IV, Jan 15 2012
a(8) = 2 = 4 - 2 for binary 1000 and ternary 22. a(64) = 3 = 7 - 4 for binary 1000000 and ternary 2101.
a:= n-> ilog[2](n)-ilog[3](n): seq(a(n), n=1..100); # Alois P. Heinz, Mar 15 2020
a[n_]: = Floor @ Log[2, n] - Floor @ Log[3, n]; Array[a, 100] (* Amiram Eldar, Mar 16 2020 *)
a(n) = logint(n,2) - logint(n,3); \\ Kevin Ryde, May 15 2020
L = 1 ; M = 1 ; B = 2 ; T = 3 ; S = 0 do N = 2 while length( S ) < 258 if B = N then do ; B = B * 2 ; L = L + 1 ; end if T = N then do ; T = T * 3 ; M = M + 1 ; end S = S || ',' L - M end N say S ; return S
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