cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-15 of 15 results.

A282391 Numbers j such that d(j) = d(j + 3*d(j)), where d(j) is the number of divisors of j.

Original entry on oeis.org

5, 7, 10, 11, 13, 14, 15, 17, 21, 22, 23, 26, 27, 30, 31, 32, 34, 37, 39, 41, 42, 45, 46, 47, 50, 53, 54, 57, 60, 61, 62, 65, 67, 72, 73, 74, 78, 82, 83, 90, 94, 96, 97, 98, 99, 101, 103, 104, 106, 107, 111, 114, 120, 122, 128, 129, 130, 131, 133, 134, 143
Offset: 1

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Author

Vladimir Shevelev, Feb 14 2017

Keywords

Comments

The sequence contains the smaller member of every pair of sexy primes (A023201).
The sequence contains no perfect squares. Indeed, let a(m) = k^2 for some m. Then, by the definition, d(k^2 + 3*d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (cf. A046522) that d(k^2) < 2*k. Hence (k+3)^2 - k^2 = 6*k + 1 > 3*d(k^2). Thus k^2 < k^2 + 3*d(k^2) < (k+3)^2. Note that, evidently, k^2 + 3*d(k^2) cannot be (k+2)^2. Let us also show that k^2 + 3*d(k^2) cannot be (k+1)^2, or, equivalently, 3*d(k^2) cannot be equal to 2*k + 1. Indeed, let 3*d(k^2) = 2*k + 1. For some prime p, let p^a || k (that is, p^a | k, but p^(a+1) !| k), a > 0, so 2*k + 1 == 1 (mod p). But now we have 3*p^(a+1) | 3*d(k^2) and thus 3*p^(a+1)|2*k + 1, so 2*k + 1 == 0 (mod p). Contradiction. Therefore, we conclude that k^2 + 3*d(k^2) cannot be a square. Hence, d(k^2 + 3*d(k^2)) is even, which is a contradiction.

Links

Crossrefs

Cf. A000005, A025487, A037916, A175304, A282354.

Programs

Extensions

More terms from Peter J. C. Moses, Feb 14 2017

A372960 Concatenation of the exponents in the prime factorization of 2*n-1.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 11, 1, 1, 11, 1, 2, 3, 1, 1, 11, 11, 1, 11, 1, 1, 21, 1, 2, 11, 1, 11, 11, 1, 1, 21, 11, 1, 11, 1, 1, 12, 11, 1, 4, 1, 11, 11, 1, 11, 11, 11, 1, 21, 1, 1, 111, 1, 1, 11, 1, 11, 21, 11, 2, 11, 3, 1, 11, 1, 11, 31, 1, 1, 11, 11, 11, 12, 1, 1, 21
Offset: 1

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Author

Jean-Marc Rebert, Aug 02 2024

Keywords

Examples

			a(8) = 11, because 2*8 - 1 = 15 = 3^1 * 5^1 and the concatenation of the exponents of the prime factorization is 11.

Links

Crossrefs

Cf. A005408, A037916, A375039.

Programs

  • Mathematica
    Join[{0},Table[FromDigits[Flatten[IntegerDigits/@FactorInteger[2n-1][[;;,2]]]],{n,2,80}]] (* Harvey P. Dale, Jun 04 2025 *)

Formula

a(n) = A037916(2*n-1).

A079617 Occurrences of prime factorization templates, unordered.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 5, 1, 3, 3, 6, 1, 5, 1, 5, 3, 3, 1, 7, 2, 3, 4, 5, 1, 8, 1, 9, 3, 3, 3, 10, 1, 3, 3, 7, 1, 8, 1, 5, 5, 3, 1, 11, 2, 5, 3, 5, 1, 7, 3, 7, 3, 3, 1, 12, 1, 3, 5, 13, 3, 8, 1, 5, 3, 8, 1, 14, 1, 3, 5, 5, 3, 8, 1, 11, 6, 3, 1, 12, 3, 3, 3, 7, 1, 12, 3, 5, 3, 3, 3, 15, 1, 5, 5, 10, 1, 8
Offset: 2

Views

Author

Jon Perry, Jan 29 2003

Keywords

Comments

1=p, 2=p^2, 3=p.q, 4=p^3, 5=p^2.q, 6=p^4 7=p^3.q, 8=p.q.r, 9=p^5, 10=p^2.q^2, 11=p^4.q

Examples

			Primes are given 1. The next prime factorization pattern is 4=p^2, so a(4)=2 and similarly a(6)=3. a(12)=a(18), etc...

Crossrefs

Cf. A037916, A079616.

Programs

  • PARI
    primetemplate2(n)=local(f,fl,fs,res,eres); f=factor(n); fl=length(f[,1]); fs=f[,2]; fs=vecsort(fs); res=""; for (i=1,fl,res=concat(res,fs[i])); eres=eval(res); if (v[eres]==0,v[eres]=vc; vc++); eres vc=1; v=vector(10000); for (j=2,50,print1(v[primetemplate2(j)]","))

Formula

a(n) = A101296(n)-1. - David Wasserman, Dec 27 2004

Extensions

More terms from David Wasserman, Dec 27 2004

A097463 Let P(i) = i-th prime. To get a(n), factor P(n)-1 as a product of primes, then concatenate the exponents.

Original entry on oeis.org

0, 1, 2, 11, 101, 21, 4, 12, 10001, 2001, 111, 22, 301, 1101, 100000001, 200001, 1000000001, 211, 11001, 1011, 32, 110001, 1000000000001, 30001, 51, 202, 1100001, 1000000000000001, 23, 4001, 1201, 101001, 3000001, 110000001, 200000000001
Offset: 1

Views

Author

Pierre CAMI, Aug 23 2004

Keywords

Comments

If P(n)-1 = P(1)^a * P(2)^b *....* P(j)^k then a(n) = ab...k.

Examples

			3-1=2^1, so a(2)=1.
5-1=2^2, so a(3)=2.
7-1=2^1*3^1, so a(4)=11.
23=(2^1)*(11^1)+1. So a(9) = 10001.
37 = 36 + 1 = 2^2*3^2 + 1, so 37 becomes 22 (a=2,b=2).

Crossrefs

Cf. A037916.

Programs

  • PARI
    {forprime(p=2,150,f=factor(p-1);j=1;q=2;s="0";while(j<=matsize(f)[1], if(q==f[j,1],s=concat(s,f[j,2]);j++,s=concat(s,0));q=nextprime(q+1));print1(eval(s),","))} \\ Klaus Brockhaus, Apr 25 2005

Extensions

More terms from Klaus Brockhaus, Apr 25 2005
a(9) corrected by Dennis (tuesdayist(AT)juno.com), Mar 30 2006

A334027 "Look and say" the concatenated exponents in the prime factorization of n.

Original entry on oeis.org

10, 11, 11, 12, 11, 21, 11, 13, 12, 21, 11, 1211, 11, 21, 21, 14, 11, 1112, 11, 1211, 21, 21, 11, 1311, 12, 21, 13, 1211, 11, 31, 11, 15, 21, 21, 21, 22, 11, 21, 21, 1311, 11, 31, 11, 1211, 1211, 21, 11, 1411, 12, 1112, 21, 1211, 11, 1113, 21, 1311, 21, 21, 11, 1221, 11
Offset: 1

Views

Author

Wesley Ivan Hurt, Apr 12 2020

Keywords

Examples

			2 = 2^1 and there is one 1, so a(2) = 11.
3 = 3^1 and there is one 1, so a(3) = 11.
4 = 2^2 and there is one 2, so a(4) = 12.
5 = 5^1 and there is one 1, so a(5) = 11.
6 = 2^1*3^1 and there are two 1's, so a(6) = 21.
7 = 7^1 and there is one 1, so a(7) = 11.
8 = 2^3 and there is one 3, so a(8) = 13.
9 = 3^2 and there is one 2, so a(9) = 12.
10 = 2^1*5^1 and there are two 1's, so a(10) = 21.

Links

Crossrefs

Cf. A037916, A045918.

Formula

a(n) = A045918(A037916(n)).
Previous Showing 11-15 of 15 results.

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