Gregory Allen - cp's OEIS frontend

User: Gregory Allen

Gregory Allen has authored 5 sequences.

A359664 Prime Maze Room 11, opposite parity of A059459 starting from prime room 11.

11, 43, 41, 2089, 2081, 2083, 2087, 10889035741470030830827987437816582768679, 10889035741470030830827987437816582768647

Offset: 1

Gregory Allen, Jan 10 2023

This is the opposite parity sequence of A059459 and lexicographically least of this sequence.
It is currently not known whether both of these sequences are infinite.
I was able to calculate 40 terms; a(40) is a 3261-digit prime.
a(1) = 11; a(n+1) is obtained by writing a(n) in binary and trying to complement just one bit, starting with the least significant bit, until a new prime is reached. (Terms 2 and 3 are excluded values from the main sequence.)
Conjecture: Room 2 and Room 11 are unlinked, i.e., two separate mazes or branches/trees, as they are of opposite parities.

Gregory Allen, Table of n, a(n) for n = 1..40
William Paulsen, Prime Maze, See Prime room 11.
Carlos Rivera, Problem 25. William Paulsen's Prime Numbers Maze, The Prime Puzzles and Problems Connection.

Cf. A059459.

maxBits = 2^14;
ClearAll[a];
a[1] = 3;
a[2] = 2;
a[3] = 11;
n = 4;
a[n_] :=
 a[n] = If[PrimeQ[a[n - 1]],
   bits = PadLeft[IntegerDigits[a[n - 1], 2], maxBits];
   For[i = 1, i <= maxBits, i++, bits2 = bits;
    bits2[[-i]] = 1 - bits[[-i]];
    If[i == maxBits, Print["maxBits reached"]; Break[],
     If[PrimeQ[an = FromDigits[bits2, 2]] &&
       FreeQ[Table[a[k], {k, 1, n - 1}], an], Return[an]]]],
   0]; Table[a[n], {n, 42}]

A341442 a(n) is the position of the start of the first occurrence of prime(n) after the decimal point in the expansion of e.

Original entry on oeis.org

4, 17, 11, 1, 200, 27, 88, 108, 16, 131, 189, 270, 124, 134, 25, 18, 11, 242, 59, 1, 157, 168, 205, 221, 35, 195, 941, 283, 1748, 355, 370, 4604, 1574, 1998, 223, 413, 201, 483, 232, 599, 2875, 120, 1382, 108, 607, 1067, 426, 2494, 1329, 517, 178, 574, 2133

Offset: 1

Gregory Allen, Feb 11 2021

			The first position at which prime(1)=2 occurs to the right of the decimal point in e=2.71828... is the 4th digit after the decimal point, so a(1)=4.

Cf. A000040, A001113, A037024, A078197, A128050.

en=Characters[ToString@N[E,10000]];
For[x=1,x<=100,x++,Print["x=",x," ",prn=Prime[x]," ",pos=First[SequencePosition[en,Characters[ToString[prn]]]-2]]]

a(n) = A078197(prime(n)). - Rémy Sigrist, Feb 12 2021

A128050 Position of start of first occurrence of prime(n) after the decimal point in expansion of golden ratio phi.

Original entry on oeis.org

19, 5, 22, 10, 34, 55, 35, 188, 131, 174, 137, 98, 90, 27, 102, 111, 166, 1, 150, 217, 479, 44, 25, 13, 81, 458, 1242, 744, 563, 96, 1602, 186, 97, 995, 259, 939, 1999, 1204, 641, 1191, 43, 833, 1682, 2833, 2708, 188, 647, 130, 62, 734, 2337, 1106, 307, 1156, 2532

Offset: 1

Gregory Allen, Feb 13 2007

			Golden ratio phi = 1.6180339887498948482045868343656381177... (see A001622).
First occurrence of prime(1) = 2 is at the 19th digit after the decimal point, hence a(1) = 19.
First occurrence of prime(5) = 11 starts at the 34th digit after the decimal point, hence a(5) = 34.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A001622, A037024, A014777.

k:=3000; R := RealField(k); [ Position(IntegerToString(Round(10^k*(-1 + (Sqrt(elt)+1) / elt))), IntegerToString(NthPrime(n))) : n in [1..55] ]; /* Klaus Brockhaus, Feb 15 2007 */

Module[{p = Rest[First[RealDigits[GoldenRatio, 10, 10^4]]], n = 0, a}, Reap[While[(a = SequencePosition[p, IntegerDigits[Prime[++n]], 1]) != {}, Sow[a[[1, 1]]]]][[2, 1]]] (* Paolo Xausa, Aug 01 2024 *)

Edited, corrected and extended by Klaus Brockhaus, Feb 15 2007

A059458 A binary sequence: a(1) = 10 (2 in decimal) and a(n+1) is obtained by trying to complement just one bit of a(n), starting with the least significant bit, until a new prime is reached.

Original entry on oeis.org

10, 11, 111, 101, 1101, 11101, 11111, 10111, 10011, 10001, 10000000010001, 10000001010001, 10100001010001, 100001010001, 100001010011, 1010011, 1000011, 1000111, 1001111, 10001001111, 10000001111, 10000000111, 10000100111

Offset: 1

Gregory Allen, Feb 02 2001

This is the lexicographically least (in positions of the flipped bits) such sequence.

It is not known if the sequence is infinite.

The decimal sequence is given in A059459. A base-ten analog is in A059471.

Maple
```
See A059459 for Maple program.
```

More terms from David W. Wilson, Feb 05 2001. For many further terms (but written in base 10) see A059459.

A059459 a(1) = 2; a(n+1) is obtained by writing a(n) in binary and trying to complement just one bit, starting with the least significant bit, until a new prime is reached.

Original entry on oeis.org

2, 3, 7, 5, 13, 29, 31, 23, 19, 17, 8209, 8273, 10321, 2129, 2131, 83, 67, 71, 79, 1103, 1039, 1031, 1063, 1061, 1069, 263213, 263209, 263201, 265249, 265313, 264289, 280673, 280681, 280697, 280699, 280703, 280639, 280607, 280603, 280859, 280843, 281867, 265483, 265547, 265579, 265571, 266083, 266081, 266089, 266093, 266029

Offset: 1

Gregory Allen, Feb 02 2001

This is the lexicographically least (in positions of the flipped bits) such sequence.
It is not known if the sequence is infinite.
"The prime maze - consider the prime numbers in base 2, starting with the smallest prime (10)2. One can move to another prime number by either changing only one digit of the number, or adding a 1 to the front of the number. Can we reach 11 = (1011)2.? 3331? The Mersennes?" (See 'Prime Links + +'.) If we start at 11 and exclude terms 2 and 3 we get terms 11, 43, 41, and so on. This is the opposite parity sequence.
a(130), if it exists, is greater than 2^130000. - Charles R Greathouse IV, Jan 02 2014
a(130) is equal to a(129) + 2^400092. - Giovanni Resta, Jul 19 2017

Charles R Greathouse IV, Table of n, a(n) for n = 1..129 (first 104 terms from T. D. Noe)
Chris K. Caldwell, Prime Links + +.
W. Paulsen, The Prime Number Maze, Web Pages.
W. Paulsen, The Prime Number Maze, Fib. Quart., 40 (2002), 272-279.
Carlos Rivera, Problem 25.- William Paulsen's Prime Numbers Maze, The Prime Puzzles and Problems Connection.

Cf. A059458 (for this sequence written in binary), A059471. A strictly ascending analog: A059661, positions of the flipped bits: A059663.

A059459search := proc(a,upto_bit,upto_length) local i,n,t; if(nops(a) >= upto_length) then RETURN(a); fi; t := a[nops(a)]; for i from 0 to upto_bit do n := XORnos(t,(2^i)); if(isprime(n) and (not member(n,a))) then print([op(a), n]); RETURN(A059459search([op(a), n],upto_bit,upto_length)); fi; od; RETURN([op(a),`and no more`]); end;
E.g., call as: A059459search([2],128,200);

maxBits = 2^11; ClearAll[a]; a[1] = 2; a[n_] := a[n] = If[ PrimeQ[ a[n-1] ], bits = PadLeft[ IntegerDigits[ a[n-1], 2], maxBits]; For[i = 1, i <= maxBits, i++, bits2 = bits; bits2[[-i]] = 1 - bits[[-i]]; If[ i == maxBits, Print[ "maxBits reached" ]; Break[], If[ PrimeQ[an = FromDigits[ bits2, 2]] && FreeQ[ Table[ a[k], {k, 1, n-1}], an], Return[an] ] ] ], 0]; Table[ a[n], {n, 129}] (* Jean-François Alcover, Jan 17 2012 *)
f[lst_List] := Block[{db2 = IntegerDigits[lst[[-1]], 2]}, exp = Length@ db2; While[pp = db2; pp[[exp]] = If[OddQ@db2[[exp]], 0, 1]; pp = FromDigits[pp, 2]; !PrimeQ[pp] || MemberQ[lst, pp], exp--; If[exp == 0, exp++; PrependTo[db2, 0]]]; Append[lst, pp]]; Nest[f, {2}, 128] (* Robert G. Wilson v, Jul 17 2017 *)

step(n)=my(k,t); while(vecsearch(v, t=bitxor(n,1<Charles R Greathouse IV, Jan 02 2014

from sympy import isprime
from itertools import islice
def agen():
    seen, cand = set(), 2
    while True:
        an = cand; bit = 1; seen.add(an); yield an
        while cand in seen or not isprime(cand):
            cand = an-bit if an&bit else an+bit
            bit <<= 1
print(list(islice(agen(), 51))) # Michael S. Branicky, Oct 01 2022

More terms and Maple program from Antti Karttunen, Feb 03 2001, who remarks that he was able to extend the sequence to the 104th term 151115727453207491916143 using the bit-flip-limit 128.

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User: Gregory Allen

A359664 Prime Maze Room 11, opposite parity of A059459 starting from prime room 11.

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A341442 a(n) is the position of the start of the first occurrence of prime(n) after the decimal point in the expansion of e.

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A128050 Position of start of first occurrence of prime(n) after the decimal point in expansion of golden ratio phi.

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A059458 A binary sequence: a(1) = 10 (2 in decimal) and a(n+1) is obtained by trying to complement just one bit of a(n), starting with the least significant bit, until a new prime is reached.

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A059459 a(1) = 2; a(n+1) is obtained by writing a(n) in binary and trying to complement just one bit, starting with the least significant bit, until a new prime is reached.

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