Ian Band - cp's OEIS frontend

User: Ian Band

Ian Band has authored 2 sequences.

A327562 a(0) = a(1) = 1; for n > 1, a(n) = (a(n-1) + a(n-2)) / gcd(a(n-1), a(n-2)) if a(n-1) and a(n-2) are not coprime, otherwise a(n) = a(n-1) + a(n-2) + 1.

1, 1, 3, 5, 9, 15, 8, 24, 4, 7, 12, 20, 8, 7, 16, 24, 5, 30, 7, 38, 46, 42, 44, 43, 88, 132, 5, 138, 144, 47, 192, 240, 9, 83, 93, 177, 90, 89, 180, 270, 5, 55, 12, 68, 20, 22, 21, 44, 66, 5, 72, 78, 25, 104, 130, 9, 140, 150, 29, 180, 210, 13, 224, 238, 33, 272, 306

Offset: 0

Ian Band, Sep 16 2019

nonn

The sequence increases rapidly after n = 92.
From roughly n = 480 to n = 600, the sequence increases relatively fast and is almost a perfect exponential function.
Redefining a(0) and a(1) can result in drastically different sequences.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A133058.

a:=[1,1]; for n in [3..67] do  if Gcd(a[n-1],a[n-2]) ne 1 then Append(~a,(a[n-1]+a[n-2])/Gcd(a[n-1],a[n-2])); else Append(~a,a[n-1]+a[n-2]+1); end if; end for; a; // Marius A. Burtea, Sep 19 2019

a[0]=a[1]=1; a[n_] := a[n] = Block[{g = GCD[a[n-1], a[n-2]]}, If[g==1,
a[n-1] + a[n-2] + 1, (a[n-1] + a[n-2])/g]]; Array[a, 67, 0] (* Giovanni Resta, Sep 19 2019 *)
nxt[{a_,b_}]:={b,If[!CoprimeQ[a,b],(a+b)/GCD[a,b],a+b+1]}; NestList[nxt,{1,1},70][[;;,1]] (* Harvey P. Dale, Feb 14 2024 *)

import math
def a(n): # Iteratively generates an array containing the first n terms of a(n), n should be greater than 2
    a1 = 1 # this will hold a(n-1), its initial value is a(1)
    a2 = 1 # this will hold a(n-2), its initial value is a(0)
    terms = [None] * n
    terms[0] = a2
    terms[1] = a1
    for i in range(2, n):
        gcdPrev2 = math.gcd(a1, a2)
        if(gcdPrev2 > 1):
            terms[i] = int((a1 + a2) / gcdPrev2)
        else:
            terms[i] = a1 + a2 + 1
        a2 = a1
        a1 = terms[i]
    return terms

a(0) = a(1) = 1; for n > 1, a(n) = (a(n-1) + a(n-2)) / gcd(a(n-1), a(n-2)) if gcd(a(n-1), a(n-2)) > 1, otherwise a(n) = a(n-1) + a(n-2) + 1.

A327296 a(n) is the decimal value of the largest binary palindrome that can be obtained from the digits of the binary representation of n.

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 5, 7, 1, 9, 9, 7, 9, 7, 7, 15, 1, 17, 17, 21, 17, 21, 21, 27, 17, 21, 21, 27, 21, 27, 27, 31, 1, 33, 33, 21, 33, 21, 21, 51, 33, 21, 21, 51, 21, 51, 51, 31, 33, 21, 21, 51, 21, 51, 51, 31, 21, 51, 51, 31, 51, 31, 31, 63, 1, 65, 65, 73, 65, 73

Offset: 0

Ian Band, Sep 16 2019

a(n) is the largest-valued binary palindrome that can be constructed from the binary representation of n. The palindrome may not have a leading 0.
When constructing a palindrome from a binary string, digits may only be permuted and deleted. I.e., the constructed palindrome will contain the same number or fewer 1's and 0's as the original binary string.
The algorithm is as follows:
If there is an odd number of 1's in the original binary string, start by placing a 1 in the middle of the output palindrome. Note that the remaining number of 1's is now even. If the remaining number of 1's is zero, then the constructed palindrome is simply "1" because the constructed palindrome cannot have leading zeros. If the remaining number of 1's is greater than zero, then place the remaining 0's evenly on either side of the previously placed "1". Because the output must be a palindrome, if the original string contains an odd number of 0's, then one of the 0's will be discarded. Lastly, place the remaining 1's on either end of the constructed palindrome.
If there is an even number of 1's in the original binary string, start by placing all of the 0's from the original string in the center of the palindrome, then place the 1's evenly on either side of the 0's.

			The largest palindrome that can be constructed:
For 101  it is  101, thus a(5)  = 5.
For 1000 it is    1, thus a(8)  = 1.
For 1010 it is 1001, thus a(10) = 9.
For 1011 it is  111, thus a(11) = 7.

Ian Band, Table of n, a(n) for n = 0..4094

Cf. A006995 (binary palindromes), A057148.

a[n_] := Block[{o,z}, {o,z} = DigitCount[n, 2]; If[o==1, 1, If[OddQ@ o, {o,z} = Floor[{o,z} /2]; 2^(z + o) + (2^o - 1) (1 + 2^(o + 1 + 2 z)), o = o/2; (2^o - 1) (1 + 2^(o + z))]]]; a /@ Range[0, 70] (* Giovanni Resta, Sep 16 2019 *)

def a(n):
    #convert n to binary string
    bin_str = str(bin(n))[2:]
    #count 1's and 0's in the string
    zeros = bin_str.count('0')
    ones  = len(bin_str) - zeros
    if(ones == 1 or ones == 0):
        return ones
    pal = ""
    if(ones % 2 == 1):
        #start by placing a '1' in the middle of the string
        pal = '1'
        #place as many 0's around the central '1' as possible
        for i in range(0, zeros >> 1):
                pal = '0' + pal + '0'
    else:
        #place all 0's in the center
        for i in range(0, zeros):
            pal += '0'
    #place the remaining 1's (guaranteed to be an even number, because one '1' was placed in the middle) on either side of the palindrome
    for i in range(0, ones >> 1):
        pal = '1' + pal + '1'
    #return integer value of the constructed palindrome
    return int(pal, 2)

More terms from Giovanni Resta, Sep 16 2019

cp's OEIS Frontend

User: Ian Band

A327562 a(0) = a(1) = 1; for n > 1, a(n) = (a(n-1) + a(n-2)) / gcd(a(n-1), a(n-2)) if a(n-1) and a(n-2) are not coprime, otherwise a(n) = a(n-1) + a(n-2) + 1.

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