Douglas Latimer - cp's OEIS frontend

A225238 Numbers n such that n occurs within its base 2 representation regarded as a fixed necklace, but n is not a substring of the base 2 representation regarded as a string.

1111, 110110, 111011, 1011110, 1101111, 10011001, 10100110, 11001010, 11010011, 110011111, 111010011, 111011101, 1001101111, 11001101111, 11010010100, 100101110100, 101000111011, 101011001010, 101111110111, 110011001100, 110111001101, 111011110011

Offset: 1

Douglas Latimer, May 04 2013

No power of 10 can occur in this sequence.

			111011 (in base 10) = 11011000110100011 (in base 2). Regarding this base 2 representation as a fixed necklace, we can list characters in the order 11110110001101000 by starting with the characters "11" at the end of the base 2 representation. In this listing 111011 occurs (1{111011}0001101000). 111011 however does not occur in the original base 2 representation 11011000110100011. Thus 111011 is in the sequence.

Eric W. Weisstein, MathWorld: Necklace

The union of A038102 and this sequence is A225237.

{inseq(w)=local(bw,mm,texp,btod,bigb,lbb,swsq,ii, hwf);
bw=binary(w); mm=length(bw); texp=0; btod=0;
forstep(i=mm, 1, -1, btod=btod+bw[i]*10^texp; texp++);
bigb=binary(btod); lbb=length(bigb);
for(k=0, lbb - 1 , swsq=1;
for(j=1, mm, ii=(j+k)%lbb; if(ii==0, ii=lbb);
if(bw[j]!=bigb[ii], swsq=-1)); if(swsq==1, hwf=k; break));
if(swsq==1,if(hwf>lbb-mm, swsq=btod, swsq=-1)); return(swsq)}
{ptd=0;for(w=0, 10^9, jj=inseq(w); if(jj>=0, ptd++; print1(jj,", "); if(ptd>23,break)))}

A225237 Numbers n such that n occurs within its base 2 representation regarded as a fixed necklace.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1100, 1101, 1111, 10000, 10001, 10011, 10100, 10101, 10111, 11000, 11001, 11100, 11101, 100000, 100001, 101000, 101010, 101100, 101101, 101111, 110000, 110001, 110101, 110110, 111011, 111100, 111101, 1000000

Offset: 1

Douglas Latimer, May 04 2013

Every power of 10 occurs in this sequence.

			10 (in base 10) = 1010 (in base 2). Regarding this base 2 representation as a fixed necklace, we can list characters in the order 1010 by starting with the first character. In this listing 10 occurs ({10}10). Thus 10 is in the sequence.
111011 (in base 10) = 11011000110100011 (in base 2). Regarding this base 2 representation as a fixed necklace, we can list characters in the order 11110110001101000 by starting with the characters "11" at the end of the base 2 representation. In this listing 111011 occurs (1{111011}0001101000). Thus 111011 is in the sequence.

Eric W. Weisstein, MathWorld: Necklace

A038102 is a subsequence of this. For the terms of the sequence not found in A038102, see A225238.

{inseq(w)=local(bw,mm,texp,btod,bigb,lbb,swsq,ii);
bw=binary(w);
mm=length(bw); texp=0; btod=0;
forstep(i=mm, 1, -1, btod=btod+bw[i]*10^texp; texp++);
bigb=binary(btod); lbb=length(bigb);
for(k=0, lbb - 1 , swsq=1;
for(j=1, mm, ii=(j+k)%lbb; if(ii==0, ii=lbb);
if(bw[j]!=bigb[ii], swsq=-1));
if(swsq==1, break)
); if(swsq==1,swsq=btod);
return(swsq)}
{ptd=0; for(w=0, 10^9, jj=inseq(w); if(jj>=0, ptd++; print1(jj,", "); if(ptd>39,break)))}

A212495 Numbers all of whose base 11 digits are even.

Original entry on oeis.org

0, 2, 4, 6, 8, 10, 22, 24, 26, 28, 30, 32, 44, 46, 48, 50, 52, 54, 66, 68, 70, 72, 74, 76, 88, 90, 92, 94, 96, 98, 110, 112, 114, 116, 118, 120, 242, 244, 246, 248, 250, 252, 264, 266, 268, 270, 272, 274, 286, 288, 290, 292, 294, 296, 308, 310, 312, 314, 316

Offset: 1

Douglas Latimer, May 18 2012

Similar in definition to A033036.

As n increases, it is most likely that A212495(n) < A033036(n), although exceptions to this rule can be found. I conjecture that A212495(n) > A033036(n) for only finitely many values of n.

			30 is represented by "28" in base 11. Both digits in this representation are even, thus 30 belongs to the sequence.

Select[Range[0,350],AllTrue[IntegerDigits[#,11],EvenQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, May 03 2018 *)

{is(c) = local(d);while(c != 0, d=c%11; c=(c-d)/11; if(d%2==1, return(0))) ; 1}
for(i=0, 317, if(is(i), print1(i, ", ")))

A182471 If n = Product (p_j^k_j), j=1 to m, then a(n) = Floor( Sum (p_j + k_j) / 2m ) (a(1) = 0 by convention).

Original entry on oeis.org

0, 1, 2, 2, 3, 1, 4, 2, 2, 2, 6, 2, 7, 2, 2, 3, 9, 2, 10, 2, 3, 3, 12, 2, 3, 4, 3, 3, 15, 2, 16, 3, 4, 5, 3, 2, 19, 5, 4, 2, 21, 2, 22, 4, 2, 6, 24, 2, 4, 2, 5, 4, 27, 2, 4, 3, 6, 8, 30, 2, 31, 8, 3, 4, 5, 3, 34, 5, 7, 2, 36, 2, 37, 10, 2, 6, 5, 3, 40, 3, 3

Offset: 1

Douglas Latimer, May 01 2012

nonn

a(n) = k (a selected constant) for only a finite number of values of n. The maximal value of n for a desired value of a(n) grows rapidly.
Thus for a(n)=2, n=2^1*3^1*5^5 is maximal.
For a(n)=3, n=2^1*3^1*5^1*7^11 is maximal.

			21 = (3^1)(7^1), with 2 distinct prime factors. Floor(((3+1)+(7+1))/(2*2)) = 3, thus a(21) = 3.

A008475

{for(k=2, 175,
M=factor(k); smt = 0; h = 0;
for(i=1, matsize(M)[1], for(j=1, matsize(M)[2], smt=smt+M[i,j]; h++));
print1( floor(smt/h), ", "))} \\\ Douglas Latimer, Apr 30 2012

A182358 Numbers n for which the number of divisors of n is congruent to 2 mod 4.

Original entry on oeis.org

2, 3, 5, 7, 11, 12, 13, 17, 18, 19, 20, 23, 28, 29, 31, 32, 37, 41, 43, 44, 45, 47, 48, 50, 52, 53, 59, 61, 63, 67, 68, 71, 73, 75, 76, 79, 80, 83, 89, 92, 97, 98, 99, 101, 103, 107, 109, 112, 113, 116, 117, 124, 127, 131, 137, 139, 147, 148, 149, 151, 153

Offset: 1

Douglas Latimer, Apr 26 2012

The product of any 2 terms a(i)*a(j) is not a member of the sequence.

tau(n) is congruent to 2 modulo 4 iff only one prime in the prime factorization of n has exponent of the form 4*m + 1, and no prime in the prime factorization of n has exponent of the form 4*k + 3.

			The divisors of 12 are: 1, 2, 3, 4, 6, 12 [6 divisors]. 6 is congruent to 2 modulo 4. Thus 12 is a member of this sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

This is an extension of A000040 (the prime numbers, which each have 2 divisors). The definition of this sequence uses A000005 (the number of divisors of n).

Select[Range[200],Mod[DivisorSigma[0,#],4]==2&] (* Harvey P. Dale, Sep 07 2020 *)

{plnt=1 ;  for(k=1, 10^7,
if(numdiv(k) % 4 == 2, print1(k, ", "); plnt++ ; if(100 <  plnt, break() )))}

is(n)=my(p=core(n));isprime(p)&&valuation(n,p)%4==1 \\ Charles R Greathouse IV, Apr 26 2012

list(lim)=my(v=List(),t);forprime(p=2,lim,for(m=1,sqrtint(lim\p), if(m%p==0, next); t=p*m^2; for(n=1,sqrtint(sqrtint(lim\t)), listput(v, t*n^4)))); vecsort(Vec(v),,8) \\ Charles R Greathouse IV, Apr 26 2012

Terms are of the form p * m^2 * n^4 for any prime p, m coprime to p, and n. - Charles R Greathouse IV, Apr 26 2012

A182120 Numbers for which the canonical prime factorization contains only exponents which are congruent to 2 modulo 3.

Original entry on oeis.org

1, 4, 9, 25, 32, 36, 49, 100, 121, 169, 196, 225, 243, 256, 288, 289, 361, 441, 484, 529, 676, 800, 841, 900, 961, 972, 1089, 1156, 1225, 1369, 1444, 1521, 1568, 1681, 1764, 1849, 2048, 2116, 2209, 2304, 2601, 2809, 3025, 3125, 3249, 3364, 3481, 3721, 3844

Offset: 1

Douglas Latimer, Apr 12 2012

By convention 1 is included as the first term.

			100 is included, as its canonical prime factorization (2^2)*(5^2) contains only exponents which are congruent to 2 modulo 3.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Douglas Latimer)

A062503 is a subsequence.
Subsequence of A001694.
Cf. A002117, A366762.

Join[{1},Select[Range[5000],Union[Mod[Transpose[FactorInteger[#]][[2]],3]] == {2}&]] (* Harvey P. Dale, Aug 18 2014 *)

{plnt=1; k=1; print1(k, ", "); plnt++;
mxind=76 ; mxind++ ; for(k=2, 2*10^6,
M=factor(k);passes=1;
sz = matsize(M)[1];
for(k=1,sz,  if( M[k,2] % 3 != 2, passes=0));
if( passes == 1 ,
print1(k, ", "); plnt++) ; if(mxind ==  plnt, break() ))}

is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 2]%3 != 2, return(0))); 1;} \\ Amiram Eldar, Oct 21 2023

Sum_{n>=1} 1/a(n) = zeta(3) * Product_{p prime} (1 + 1/p^2 - 1/p^3) = 1.56984817927051410948... . - Amiram Eldar, Oct 21 2023

A211485 Numbers for which the canonical prime factorization contains only an odd number of exponents, all of which are congruent to 1 modulo 3.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 16, 17, 19, 23, 29, 30, 31, 37, 41, 42, 43, 47, 53, 59, 61, 66, 67, 70, 71, 73, 78, 79, 81, 83, 89, 97, 101, 102, 103, 105, 107, 109, 110, 113, 114, 127, 128, 130, 131, 137, 138, 139, 149, 151, 154, 157, 163, 165, 167, 170, 173, 174, 179

Offset: 1

Douglas Latimer, Apr 12 2012

By convention 1 is included as the first term, in order to simplify the statement of certain results concerning this sequence.

			3 is included, as its canonical prime factorization 3^1 contains only an odd number of exponents, all of which are congruent to 1 modulo 3.
81 is in the sequence, because its canonical prime factorization is 3^4, and that one exponent, 4, is congruent to 1 modulo 3.

Douglas Latimer, Table of n, a(n) for n = 1..1000

The definition is similar to that for A211484. This sequence includes the prime numbers A000040, and includes A030059.

oneQ[n_]:=Module[{f=FactorInteger[n][[All,2]]},OddQ[Length[f]]&&Union[ Mod[ f,3]]=={1}]; Select[Range[200],oneQ] (* Harvey P. Dale, Jul 03 2019 *)

{plnt=1;k=1; print1(k, ", "); plnt++;
mxind=76 ; mxind++ ; for(k=2, 10^6,
M=factor(k);passes=1;
sz = matsize(M)[1];
for(k=1,sz, if(sz%2 != 1, passes=0;break()); if( M[k,2] % 3 != 1, passes=0));
if( passes == 1 , print1(k, ", "); plnt++) ; if(mxind ==  plnt, break() ))}

A211484 Numbers for which the canonical prime factorization contains only an even number of exponents, all of which are congruent to 1 modulo 3.

Original entry on oeis.org

1, 6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 48, 51, 55, 57, 58, 62, 65, 69, 74, 77, 80, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 112, 115, 118, 119, 122, 123, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 162, 166, 176, 177, 178, 183

Offset: 1

Douglas Latimer, Apr 12 2012

By convention 1 is included as the first term, in order to simplify the statement of certain results involving this sequence.

			6 is included, as its canonical prime factorization (2^1)*(3^1) contains an even number of exponents, all of which are congruent to 1 modulo 3.

Douglas Latimer, Table of n, a(n) for n = 1..1000

This sequence includes A030229 and hence A006881. The definition is similar to that for A211485.

pfQ[n_]:=Module[{f=Transpose[FactorInteger[n]][[2]]},EvenQ[Length[f]] && Union[ Mod[f,3]]=={1}]; Join[{1},Select[Range[200],pfQ]] (* Harvey P. Dale, Mar 24 2016 *)

{plnt=1; k=1; print1(k, ", "); plnt++;
mxind=76 ; mxind++ ; for(k=2, 10^6,
M=factor(k);passes=1;
sz = matsize(M)[1];
for(k=1,sz, if(sz%2 != 0, passes=0;break()); if( M[k,2] % 3 != 1, passes=0));
if( passes == 1 , print1(k, ", "); plnt++) ; if(mxind ==  plnt, break() ))}

is(n,f=factor(n))=omega(f)%2==0 && factorback(f[,2]%3)==1 \\ Charles R Greathouse IV, Sep 07 2017

A211337 Numbers k for which the number of divisors, tau(k), is congruent to 1 modulo 3.

Original entry on oeis.org

1, 6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 46, 48, 51, 55, 57, 58, 62, 64, 65, 69, 74, 77, 80, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 112, 115, 118, 119, 120, 122, 123, 125, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 162

Offset: 1

Douglas Latimer, Apr 07 2012

nonn

Any term a(n) can be expressed as 1 term from A211484 times 1 nonzero term from A000578. - Douglas Latimer, Apr 20 2012

The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 4, 36, 366, 3635, 36499, 365456, 3654240, 36538501, 365382167, 3653804173, ... . Conjecture: the asymptotic density of this sequence exists and equals 3*zeta(3)/Pi^2 = 0.3653814847007... (A346602), so, a(n) ~ k*n with k = Pi^2/(3*zeta(3)) = 2.73686555524... . This conjecture is true if this sequence and A211338 have the same density (see A059269). - Amiram Eldar, Jan 06 2024

			The divisors of 10  are: 1, 2, 5, 10 (4 divisors). 4 is congruent to 1 modulo 3. Thus 10 is a member of this sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Douglas Latimer)

This is an extension of A030513 (numbers with 4 divisors).
The union of A059269 and A211338 is the complementary sequence to this one.
The definition of this sequence uses A000005 (the number of divisors of n).
Cf. A000578, A074794, A211484, A346602.

Select[Range[162], Mod[DivisorSigma[0, #], 3] == 1 &] (* T. D. Noe, Apr 21 2012 *)

{plnt=1 ; mxind=100 ;for(k=1, 10^6,
if(numdiv(k) % 3 == 1, print(k); plnt++; if(mxind+1 ==  plnt, break() )))}

Conjecture: a(n) ~ k*n where k = 2/prod(1 - (p-1)/(p^(3*k))) = 2.7290077... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012

A211338 Numbers k for which the number of divisors, tau(k), is congruent to 2 modulo 3.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 16, 17, 19, 23, 24, 29, 30, 31, 37, 40, 41, 42, 43, 47, 53, 54, 56, 59, 61, 66, 67, 70, 71, 73, 78, 79, 81, 83, 88, 89, 97, 101, 102, 103, 104, 105, 107, 109, 110, 113, 114, 127, 128, 130, 131, 135, 136, 137, 138, 139, 149, 151, 152, 154

Offset: 1

Douglas Latimer, Apr 07 2012

nonn

The product of any 2 terms a(i)*a(j) is not a member of the sequence.
Any term a(n) can be expressed as 1 term (required to be greater than 1) from A211485 times 1 nonzero term from A000578. - Douglas Latimer, Apr 20 2012
The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 4, 37, 368, 3681, 36596, 365336, 3653499, 36537962, 365381169, 3653826361, ... . Conjecture: the asymptotic density of this sequence exists and equals 3*zeta(3)/Pi^2 = 0.3653814847007... (A346602), so, a(n) ~ k*n with k = Pi^2/(3*zeta(3)) = 2.73686555524... . This conjecture is true if this sequence and A211337 have the same density (see A059269). - Amiram Eldar, Jan 06 2024

			The divisors of 16 are: 1, 2, 4, 8, 16 (5 divisors). 5 is congruent to 2 modulo 3. Thus 16 is a member of this sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Douglas Latimer)

This is an extension of A000040 (the prime numbers, which each have 2 divisors).
The union of A059269 and A211337 is the complementary sequence to this one.
The definition of this sequence uses A000005 (the number of divisors of n).
Cf. A000578, A074796, A211485, A346602.

Select[Range[154], Mod[DivisorSigma[0, #], 3] == 2 &] (* T. D. Noe, Apr 21 2012 *)

{plnt=1 ; mxind=100 ;for(k=1, 10^6,
if(numdiv(k) % 3 == 2, print(k); plnt++; if(mxind+1 == plnt, break() )))}

Conjecture: a(n) ~ k*n where k = 2/prod(1 - (p-1)/(p^(3*k))) = 2.7290077... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012

cp's OEIS Frontend

User: Douglas Latimer

A225238 Numbers n such that n occurs within its base 2 representation regarded as a fixed necklace, but n is not a substring of the base 2 representation regarded as a string.

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A225237 Numbers n such that n occurs within its base 2 representation regarded as a fixed necklace.

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A212495 Numbers all of whose base 11 digits are even.

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A182471 If n = Product (p_j^k_j), j=1 to m, then a(n) = Floor( Sum (p_j + k_j) / 2m ) (a(1) = 0 by convention).

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A182358 Numbers n for which the number of divisors of n is congruent to 2 mod 4.

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A182120 Numbers for which the canonical prime factorization contains only exponents which are congruent to 2 modulo 3.

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A211485 Numbers for which the canonical prime factorization contains only an odd number of exponents, all of which are congruent to 1 modulo 3.

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A211484 Numbers for which the canonical prime factorization contains only an even number of exponents, all of which are congruent to 1 modulo 3.

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