A055120 -id:A055120 - cp's OEIS frontend

A243589 Numbers returned when each binary digit of n is replaced by the sum modulo 2 of the digits to its (wrapped) left and (wrapped) right.

0, 0, 0, 3, 5, 6, 0, 5, 15, 0, 10, 15, 5, 10, 0, 9, 27, 12, 30, 3, 17, 6, 20, 29, 15, 24, 10, 23, 5, 18, 0, 17, 51, 20, 54, 27, 57, 30, 60, 5, 39, 0, 34, 15, 45, 10, 40, 57, 27, 60, 30, 51, 17, 54, 20, 45, 15, 40, 10, 39, 5, 34, 0, 33, 99, 36, 102, 43, 105, 46

Offset: 1

Anthony Sand, Jun 07 2014

a(n) = ror(n) XOR rol(n), where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 24 2016

Numbers returned by the following function: take the t binary digits of n, d(1)..d(t), and replace each with the sum d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) maps to t and (i+1 > t) maps to 1.

			For 1, the function returns d(1) = (d(1) + d(1)) mod 2 = (1 + 1) mod 2 = 0.
For 5, the initial digits are (1,0,1).
d(1) = (d(3) + d(2)) mod 2 = (1 + 0) mod 2 = 1; d(2) = (d(1) + d(3)) mod 2 = (1 + 1) mod 2 = 0; d(3) = (d(2) + d(1)) mod 2 = (0 + 1) mod 2 = 1.
The function returns (1,0,1) = 101 = 5 in base 10.

Anthony Sand, Table of n, a(n) for n = 1..1000

Cf. A006257, A035327, A038572, A055120.

for n in range(1, 100):
    BL = len(bin(n))-2
    x = (n>>1) + ((n&1) << (BL-1))   # A038572(n)
    x^= (n*2) - (1<A006257(n)  for n>0
    print(x, end=', ')

for digits d(1)..d(t), d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) -> t, (i+1 > t) -> 1.

A243590 Numbers returned when each digit of n is replaced by the sum modulo 10 of the digits to its (wrapped) left and (wrapped) right.

Original entry on oeis.org

2, 4, 6, 8, 0, 2, 4, 6, 8, 2, 22, 42, 62, 82, 2, 22, 42, 62, 82, 4, 24, 44, 64, 84, 4, 24, 44, 64, 84, 6, 26, 46, 66, 86, 6, 26, 46, 66, 86, 8, 28, 48, 68, 88, 8, 28, 48, 68, 88, 0, 20, 40, 60, 80, 0, 20, 40, 60, 80, 2, 22, 42, 62, 82, 2, 22, 42, 62, 82, 4, 24

Offset: 1

Anthony Sand, Jun 07 2014

Numbers returned by the following function: take the t digits of n, d(1)..d(t), and replace each with the sum d(i) = (d(i-1) + d(i+1)) mod 10, where (i-1 = 0) maps to t and (i+1 > t) maps to 1.

			For 1, the function returns (1 + 1) mod 10 = 2.
For 5, the function returns (5 + 5) mod 10 = 0.
For 125, the initial digits are (1,2,5).
d(1) <- (d(3) + d(2)) mod 10 = (5 + 2) mod 10 = 7; d(2) <- (d(1) + d(3)) mod 10 = (1 + 5) mod 10 = 6; d(3) <- (d(2) + d(1)) mod 10 = (2 + 1) mod 10 = 3.
The function returns (7,6,3) = 763.

Anthony Sand, Table of n, a(n) for n = 1..1000

Cf. A055120, A035327.

for digits d(1)..d(t), d(i) = (d(i-1) + d(i+1)) mod 10, where (i-1 = 0) -> t, (i+1 > t) -> 1.

A345343 Numbers that yield a prime when any single digit is replaced by its digital complement.

Original entry on oeis.org

3, 5, 7, 8, 17, 33, 39, 47, 63, 71, 77, 93, 107, 171, 177, 221, 223, 287, 333, 339, 401, 441, 447, 699, 823, 827, 857, 883, 999, 1421, 1781, 2087, 2089, 2171, 2233, 2539, 3253, 3829, 3963, 4007, 4173, 4977, 5051, 5059, 5503, 5507, 6363, 7217, 7491, 7541, 8447, 10247

Offset: 1

Lamine Ngom, Jun 14 2021

Digital complement of a digit d is 10-d if d > 0, 0 otherwise.

			3829 is a term since 7829, 3229, 3889 and 3821 are all primes.

Cf. A055120, A089740, A318785.

q[n_] := Module[{d = IntegerDigits[n]}, And @@ PrimeQ[Table[FromDigits[ReplacePart [d, i -> If[d[[i]] == 0, d[[i]], 10 - d[[i]]]]], {i, 1, Length[d]}]]]; Select[Range[10^4], q] (* Amiram Eldar, Jun 15 2021 *)

f(x) = if (x, 10-x);
isok(m) = {my(d=digits(m)); for (i=1, #d, d[i] = f(d[i]); if (!isprime(fromdigits(d)), return (0)); d[i] = f(d[i]);); return (1);} \\ Michel Marcus, Jun 15 2021

from sympy import isprime
def comp(d, i): return d[:i] + str((10-int(d[i]))%10) + d[i+1:]
def ok(n):
    d = str(n)
    return all(isprime(int(comp(d, i))) for i in range(len(d)))
print(list(filter(ok, range(1, 11000)))) # Michael S. Branicky, Jun 14 2021

A300487 Numbers k whose 10's complement mod 10 of their digits is equal to phi(k), the Euler totient function of k.

Original entry on oeis.org

74, 834, 80940, 809400, 833334, 7414114, 7422694, 7539694, 8094000, 80940000, 809400000, 8094000000, 80940000000, 83335786566, 809400000000, 7539682539694, 8094000000000, 80940000000000

Offset: 1

Paolo P. Lava, Mar 07 2018

Any number of the form 8094*10^j, with j>0, is part of the sequence because its Euler totient function is 2016*10^j.

Contains subsequence 834, 833334, 833333333333334, ... formed by numbers (10^k/4 + 2)/3 for k in A296059. - Max Alekseyev, Mar 09 2024

			phi(74) = 36 that is the 10's complement of the digits of 74.

Cf. A000010, A055120, A296059.

with(numtheory): P:=proc(q) local a,b,k,n;
for n from 1 to q do a:=convert(phi(n),base,10);
for k from 1 to nops(a) do a[k]:=(10-a[k]) mod 10; od; b:=0;
for k from 1 to nops(a) do b:=b*10+a[nops(a)-k+1]; od;
if b=n then print(n); fi; od; end: P(10^9);

isok(x) = {my(dx = digits(x), dy = vector(#dx, k, (10-dx[k]) % 10)); fromdigits(dy) == eulerphi(x); } \\ Michel Marcus, Mar 12 2018

a(11)-a(15) from Giovanni Resta, Mar 09 2018

a(16)-a(18) from Max Alekseyev, Mar 09 2024

A300796 Numbers x whose 10's complements y have the same sum of divisors of x, with x <> y.

Original entry on oeis.org

3762, 4125, 4865, 5135, 5875, 6238, 37620, 41250, 42825, 44571, 48650, 48839, 49496, 50504, 51161, 51350, 55429, 57175, 58750, 62380, 376200, 389232, 397584, 399441, 412500, 417864, 428250, 434355, 436185, 445710, 446369, 472535, 481325, 483662, 483792, 486500

Offset: 1

Paolo P. Lava, Mar 13 2018

Many patterns can be found, e.g. 3762*10^j, 4125*10^j, 4865*10^j, 5135*10^j, 5875*10^j, 6238*10^j, etc.

			3762 is in the sequence because sigma(3762) = sigma(10^4-3762) = 9360.
5875 is in the sequence because sigma(5875) = sigma(10^4-5875) = 7488.

Cf. A000203, A055120, A300447.

with(numtheory): P:=proc(q) local a,n;
for n from 1 to q do a:=10^(ilog10(n)+1)-n;
if n<>a and sigma(n)=sigma(a) then print(n); fi; od; end: P(10^6);

c10Q[n_]:=Module[{c=10^IntegerLength[n]-n},c!=n&&DivisorSigma[1,n] == DivisorSigma[1,c]]; Select[Range[500000],c10Q] (* Harvey P. Dale, Sep 24 2021 *)

A345529 Primes that yield a prime when any single digit is replaced by its 10's complement.

Original entry on oeis.org

3, 5, 7, 17, 47, 71, 107, 223, 401, 823, 827, 857, 883, 2087, 2089, 2539, 3253, 4007, 5051, 5059, 5503, 5507, 7541, 8447, 10247, 12401, 18041, 25303, 33529, 33589, 35533, 40427, 44171, 45557, 53503, 53653, 53899, 54401, 55001, 55009, 55333, 55817, 57077, 71147, 81017, 82003, 93553

Offset: 1

Lamine Ngom, Jun 20 2021

Digital complement of a digit d is 10-d if d > 0, 0 otherwise.

Primes in A345343.

			71147 is a term since 31147, 79147, 71947, 71167 and 71143 are all primes.

Cf. A345343, A055120, A089740, A318785.

q[n_] := PrimeQ[n] && Module[{d = IntegerDigits[n]}, And @@ PrimeQ[Table[ FromDigits[ReplacePart[d, i -> If[d[[i]] == 0, d[[i]], 10 - d[[i]]]]], {i, 1, Length[d]}]]]; Select[Range[10^5], q] (* Amiram Eldar, Jul 06 2021 *)

from sympy import isprime, primerange
def comp(d, i): return d[:i] + str((10-int(d[i]))%10) + d[i+1:]
def ok(p):
    d = str(p)
    return all(isprime(int(comp(d, i))) for i in range(len(d)))
print(list(filter(ok, primerange(1, 95000)))) # Michael S. Branicky, Jun 20 2021

cp's OEIS Frontend

A243589 Numbers returned when each binary digit of n is replaced by the sum modulo 2 of the digits to its (wrapped) left and (wrapped) right.

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A243590 Numbers returned when each digit of n is replaced by the sum modulo 10 of the digits to its (wrapped) left and (wrapped) right.

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A345343 Numbers that yield a prime when any single digit is replaced by its digital complement.

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Python

A300487 Numbers k whose 10's complement mod 10 of their digits is equal to phi(k), the Euler totient function of k.

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Maple

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A300796 Numbers x whose 10's complements y have the same sum of divisors of x, with x <> y.

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Comments

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Programs

Maple

Mathematica

A345529 Primes that yield a prime when any single digit is replaced by its 10's complement.

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Examples

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Programs

Mathematica

Python