Christoph Schreier - cp's OEIS frontend

A334172 Bitwise XNOR of prime(n) and prime(n + 1).

2, 1, 5, 3, 9, 3, 29, 27, 21, 29, 5, 51, 61, 59, 37, 49, 57, 1, 123, 113, 121, 99, 117, 71, 123, 125, 115, 121, 99, 113, 3, 245, 253, 225, 253, 245, 193, 251, 245, 225, 249, 245, 129, 251, 253, 235, 243, 195, 249, 243, 249, 225, 245, 5, 505, 501, 509, 485

Offset: 1

Christoph Schreier, Apr 17 2020

XOR is exclusive OR, meaning that one bit is on and the other bit is off. XNOR is the negation of XOR, meaning that either both bits are on or both bits are off. For example, 4 in binary is 100 and 6 is 110. Then 100 XOR 110 is 010 but 100 XNOR 110 is 101.

From Bertrand's postulate it follows that prime(n + 1) requires only one bit more than prime(n) if they're not the same bit width. In most computer implementations, however, the numbers are placed zero-padded into fixed bit widths using two's complement, making it necessary to make adjustments to avoid unintentionally negative numbers. - Alonso del Arte, Apr 18 2020

			The second prime is 3 (11 in binary) and the third prime is 5 (101 in binary). We see that 011 XNOR 101 = 001. Hence a(2) = 1.
The fourth prime is 7 (111 in binary). We see that 101 XNOR 111 = 101. Hence a(3) = 5.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, XNOR
Wikipedia, Bitwise operation

Cf. A000040, A112591, A175329, A175330, A334143.

a:= n-> (p-> Bits[Not](Bits[Xor](p, ithprime(n+1)),
             bits=1+ilog2(p)))(ithprime(n)):
seq(a(n), n=1..70);  # Alois P. Heinz, Apr 17 2020

Table[BitNot[BitXor[Prime[n], Prime[n + 1]]] + 2^Ceiling[Log[2, Prime[n + 1]]], {n, 50}] (* Alonso del Arte, Apr 17 2020 *)

neg(p) = bitneg(p, #binary(p));
a(n) = my(p=prime(n), q=nextprime(p+1)); bitor(bitand(p, q), bitand(neg(p), neg(q))); \\ Michel Marcus, Apr 17 2020

def XNORprime(n):
    return ~(primes[n] ^ primes[n+1]) + (1 << primes[n+1].bit_length())

val prime: LazyList[Int] = 2 #:: LazyList.from(3).filter(i => prime.takeWhile {
   j => j * j <= i
}.forall {
   k => i % k != 0
})
(0 to 63).map(n => ~(prime(n) ^ prime(n + 1)) + 2 * Integer.highestOneBit(prime(n + 1))) // Alonso del Arte, Apr 18 2020

a(n) = A035327(A112591(n)).

A334122 a(n) is the sum of all primes <= n, mod n.

Original entry on oeis.org

0, 0, 2, 1, 0, 4, 3, 1, 8, 7, 6, 4, 2, 13, 11, 9, 7, 4, 1, 17, 14, 11, 8, 4, 0, 22, 19, 16, 13, 9, 5, 0, 28, 24, 20, 16, 12, 7, 2, 37, 33, 28, 23, 17, 11, 5, 46, 40, 34, 28, 22, 16, 10, 3, 51, 45, 39, 33, 27, 20, 13, 5, 60, 53, 46, 39, 32, 24, 16, 8, 0, 63, 55

Offset: 1

Christoph Schreier, Apr 15 2020

			a(7) = (2+3+5+7) mod 7 = 17 mod 7 = 3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000720, A034387, A090396.

b:= proc(n) b(n):= `if`(n<2, 0, b(n-1)+`if`(isprime(n), n, 0)) end:
a:= n-> irem(b(n), n):
seq(a(n), n=1..80);  # Alois P. Heinz, Apr 15 2020

Mod[Accumulate[(# * Boole @ PrimeQ[#]) & /@ (r = Range[100])], r] (* Amiram Eldar, Apr 15 2020 *)

a(n) = my(np=primepi(n)); vecsum(primes(np)) % n; \\ Michel Marcus, Apr 16 2020

from sympy import primerange
a = lambda n: sum(primerange(n + 1)) % n # David Radcliffe, May 10 2025

a(n) = A034387(n) mod n.

A334143 a(n) = bitwise NOR of prime(n) and prime(n+1).

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 12, 8, 0, 0, 0, 18, 20, 16, 0, 0, 0, 0, 56, 48, 48, 32, 36, 6, 26, 24, 16, 16, 2, 0, 0, 116, 116, 96, 104, 96, 64, 88, 80, 64, 72, 64, 0, 58, 56, 40, 32, 0, 24, 18, 16, 0, 4, 4, 248, 240, 240, 224, 226, 228, 192, 200, 200, 192, 194, 128, 164

Offset: 1

Christoph Schreier, Apr 15 2020

			a(6) = prime(6) NOR prime(7) = 13 NOR 17 = 2.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Wikipedia, Bitwise operation

Cf. A000040, A112591, A145382, A175330.

a:= n-> Bits[Nor](ithprime(n), ithprime(n+1)):
seq(a(n), n=1..70);  # Alois P. Heinz, Apr 15 2020

A334143[n_]:=With[{b=BitOr[Prime[n],Prime[n+1]]},2^BitLength[b]-b-1];Array[A334143,100] (* Paolo Xausa, Oct 13 2023 *)

a(n) = my(x=bitor(prime(n), prime(n+1))); bitneg(x, #binary(x)); \\ Michel Marcus, Apr 16 2020

def NORprime(n):
    s = str(bin(primes[n]))[2:]
    t = str(bin(primes[n-1]))[2:]
    k = (len(s) -  len(t))
    t = k*'0' + t
    r = ''
    for i in range(len(s)):
        if s[i] == t[i] and s[i] == '0':
            r += '1'
        else:
            r += '0'
    return int(r,2)

a(n) = A035327(A175329(n)).

A334085 GCD of n and the product of all primes < n.

Original entry on oeis.org

1, 1, 1, 2, 1, 6, 1, 2, 3, 10, 1, 6, 1, 14, 15, 2, 1, 6, 1, 10, 21, 22, 1, 6, 5, 26, 3, 14, 1, 30, 1, 2, 33, 34, 35, 6, 1, 38, 39, 10, 1, 42, 1, 22, 15, 46, 1, 6, 7, 10, 51, 26, 1, 6, 55, 14, 57, 58, 1, 30, 1, 62, 21, 2, 65, 66, 1, 34, 69, 70, 1, 6, 1, 74, 15

Offset: 1

Christoph Schreier, Apr 14 2020

a(n) = 1 iff n is prime or n = 1.

			For n=8 the product of all primes < 8 = A034386(7) = 210; gcd(8,210) = 2.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Christoph Schreier)

Cf. A000040, A034386, A007947.

a:= n-> `if`(isprime(n), 1, mul(i[1], i=ifactors(n)[2])):
seq(a(n), n=1..80);  # Alois P. Heinz, Apr 17 2020

a[n_] := GCD[n, Times @@ Select[Range[n - 1], PrimeQ]]; Array[a, 60] (* Amiram Eldar, Apr 14 2020 *)

a(n) = gcd(n, prod(k=1, n-1, if (isprime(k), k, 1))); \\ Michel Marcus, Apr 14 2020

a(n,f=factor(n))=if(f[,2]==[1]~,1,factorback(f[,1])) \\ Charles R Greathouse IV, Apr 14 2020

def is_prime(n):
    if n == 2 or n == 3:
        return True
    if n % 2 == 0 or n < 2:
        return False
    for i in range(3,int(n**0.5)+1,2):
        if n % i == 0:
            return False
    return True
def A(n):
    if n < 3:
        raise ValueError('n must be > 2')
    else:
        a = 1
        for i in range(n):
            if is_prime(i):
                a *= i
    return math.gcd(n,a)

a(n) = gcd(n, A034386(n-1)).

a(n) = 1 if n is prime, else a(n) = A007947(n).

A334045 Bitwise NOR of binary representation of n and n-1.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 0, 0, 6, 4, 4, 0, 2, 0, 0, 0, 14, 12, 12, 8, 10, 8, 8, 0, 6, 4, 4, 0, 2, 0, 0, 0, 30, 28, 28, 24, 26, 24, 24, 16, 22, 20, 20, 16, 18, 16, 16, 0, 14, 12, 12, 8, 10, 8, 8, 0, 6, 4, 4, 0, 2, 0, 0, 0, 62, 60, 60, 56, 58, 56, 56, 48, 54, 52, 52

Offset: 1

Christoph Schreier, Apr 13 2020

All terms are even.
a(1) = 0, a(2) = 0, and a(2^n + 1) = 2^n - 2 for n > 0. Are there any other cases where n - a(n) < 4? - Charles R Greathouse IV, Apr 13 2020
The answer to the above question is no. Write n as n = (2m+1)*k, i.e. k = A006519(n) is the highest power of 2 dividing n. If m = 0, a(n) = 0 and n - a(n) = n. If m > 0, then a(n) = 2v*k, where v is the 1's complement of m. Thus n-a(n) = (2(m-v)+1)*k. Since m in binary has a leading 1, m - v >= 1 and thus n - a(n) >= 3 with n - a(n) = 3 when n > 2, k = 1 and m - v = 1, i.e. m is a power of 2 and n is of the form 2^r + 1. - Chai Wah Wu, Apr 13 2020

			a(11) = 11 NOR 10 = bin 1011 NOR 1010 = bin 100 = 4.

Wikipedia, Bitwise operation

Cf. A038712 (n XOR n-1), A086799 (n OR n-1), A129760 (n AND n-1).

a:= n-> Bits[Nor](n, n-1):
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 13 2020

a(n) = my(x=bitor(n-1, n)); bitneg(x, #binary(x)); \\ Michel Marcus, Apr 13 2020

def norbitwise(n):
    a = str(bin(n))[2:]
    b = str(bin(n-1))[2:]
    if len(b) < len(a):
        b = '0' + b
    c = ''
    for i in range(len(a)):
        if a[i] == b[i] and a[i] == '0':
            c += '1'
        else:
            c += '0'
    return int(c,2)

def A334045(n):
    m = n|(n-1)
    return 2**(len(bin(m))-2)-1-m # Chai Wah Wu, Apr 13 2020

A334042 Write n^2 in binary, interchange 0's and 1's, convert back to decimal.

Original entry on oeis.org

1, 0, 3, 6, 15, 6, 27, 14, 63, 46, 27, 6, 111, 86, 59, 30, 255, 222, 187, 150, 111, 70, 27, 494, 447, 398, 347, 294, 239, 182, 123, 62, 1023, 958, 891, 822, 751, 678, 603, 526, 447, 366, 283, 198, 111, 22, 1979, 1886, 1791, 1694, 1595, 1494, 1391, 1286, 1179

Offset: 0

Christoph Schreier, Apr 13 2020

Daniel Starodubtsev, Table of n, a(n) for n = 0..10000

Cf. A000290, A035327.

a:= n-> (l-> add((1-l[i])*2^(i-1), i=1..nops(l)))(convert(n, base, 2)):
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 13 2020

a[n_] := FromDigits[1 - IntegerDigits[n^2, 2], 2]; Array[a, 55, 0] (* Amiram Eldar, Apr 13 2020 *)

a(n)=if(n, my(s=n^2); bitneg(s,exponent(s)+1), 1) \\ Charles R Greathouse IV, Apr 13 2020

def oppsquare(n):
    s = str(bin(n**2))[2:]
    t = ''
    for i in range(len(s)):
        if s[i] == '0':
            t += '1'
        else:
            t += '0'
    return int(t,2)

def A334042(n):
    return 2**(len(bin(n**2))-2)-1-n**2 # Chai Wah Wu, Apr 13 2020

a(n) = A035327(A000290(n)). - Alois P. Heinz, Apr 13 2020

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User: Christoph Schreier

A334172 Bitwise XNOR of prime(n) and prime(n + 1).

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A334122 a(n) is the sum of all primes <= n, mod n.

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A334143 a(n) = bitwise NOR of prime(n) and prime(n+1).

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A334085 GCD of n and the product of all primes < n.

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A334045 Bitwise NOR of binary representation of n and n-1.

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A334042 Write n^2 in binary, interchange 0's and 1's, convert back to decimal.

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Mathematica