A275968 Smaller of two consecutive primes p and q such that c(p) = c(q), where c(n) = A008908(n) is the length of x, f(x), f(f(x)), ... , 1 in the Collatz conjecture.
173, 409, 419, 421, 439, 487, 521, 557, 571, 617, 761, 887, 919, 1009, 1039, 1117, 1153, 1171, 1217, 1327, 1373, 1549, 1559, 1571, 1657, 1693, 1709, 1721, 1733, 1783, 1831, 1861, 1901, 1993, 1997, 2053, 2089, 2339, 2393, 2521, 2539, 2647, 2657, 2677, 2693, 2777
Offset: 1
Examples
a(1) = p = 173; q = 179 c(p) = c(q) = 32
Links
- Abhiram R Devesh, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
t = Table[Length@ NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # != 1 &] - 1, {n, 10^4}]; Prime@ Flatten@ Position[#, k_ /; Length@ k == 1] &@ Map[Union@ Part[t, #] &, #] &@ Partition[#, 2, 1] &@ Prime@ Range@ 410 (* Michael De Vlieger, Sep 01 2016 *)
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PARI
A008908(n)=my(c=1); while(n>1, n=if(n%2, 3*n+1, n/2); c++); c t=A008908(p=2); forprime(q=3,1e4, tt=A008908(q); if(t==tt, print1(p", ")); p=q; t=tt) \\ Charles R Greathouse IV, Sep 01 2016
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Python
import sympy def lcs(n): a=1 while n>1: if n%2==0: n=n//2 else: n=(3*n)+1 a=a+1 return(a) m=2 while m>0: n=sympy.nextprime(m) if lcs(m)==lcs(n): print(m,) m=n # Abhiram R Devesh, Sep 02 2016
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