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.

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A212171 Prime signature of n (nonincreasing version): row n of table lists positive exponents in canonical prime factorization of n, in nonincreasing order.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 4, 1, 2, 2, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 3, 1
Offset: 2

Views

Author

Matthew Vandermast, Jun 03 2012

Keywords

Comments

Length of row n equals A001221(n).
The multiset of positive exponents in n's prime factorization completely determines a(n) for a host of OEIS sequences, including several "core" sequences. Of those not cross-referenced here or in A212172, many can be found by searching the database for A025487.
(Note: Differing opinions may exist about whether the prime signature of n should be defined as this multiset itself, or as a symbol or collection of symbols that identify or "signify" this multiset. The definition of this sequence is designed to be compatible with either view, as are the original comments. When n >= 2, the customary ways to signify the multiset of exponents in n's prime factorization are to list the constituent exponents in either nonincreasing or nondecreasing order; this table gives the nonincreasing version.)
Table lists exponents in the order in which they appear in the prime factorization of a member of A025487. This ordering is common in database comments (e.g., A008966).
Each possible multiset of an integer's positive prime factorization exponents corresponds to a unique partition that contains the same elements (cf. A000041). This includes the multiset of 1's positive exponents, { } (the empty multiset), which corresponds to the partition of 0.
Differs from A124010 from a(23) on, corresponding to the factorization of 18 = 2^1*3^2 which is here listed as row 18 = [2, 1], but as [1, 2] (in the order of the prime factors) in A124010 and also in A118914 which lists the prime signatures in nondecreasing order (so that row 12 = 2^2*3^1 is also [1, 2]). - M. F. Hasler, Apr 08 2022

Examples

			First rows of table read:
  1;
  1;
  2;
  1;
  1,1;
  1;
  3;
  2;
  1,1;
  1;
  2,1;
  ...
The multiset of positive exponents in the prime factorization of 6 = 2*3 is {1,1} (1s are often left implicit as exponents). The prime signature of 6 is therefore {1,1}.
12 = 2^2*3 has positive exponents 2 and 1 in its prime factorization, as does 18 = 2*3^2. Rows 12 and 18 of the table both read {2,1}.
		

Crossrefs

Cf. A025487, A001221 (row lengths), A001222 (row sums). A118914 gives the nondecreasing version. A124010 lists exponents in n's prime factorization in natural order, with A124010(1) = 0.
A212172 cross-references over 20 sequences that depend solely on n's prime exponents >= 2, including the "core" sequence A000688. Other sequences determined by the exponents in the prime factorization of n include:
Additive: A001221, A001222, A056169.
A highly incomplete selection of sequences, each definable by the set of prime signatures possessed by its members: A000040, A000290, A000578, A000583, A000961, A001248, A001358, A001597, A001694, A002808, A004709, A005117, A006881, A013929, A030059, A030229, A052486.

Programs

  • Magma
    &cat[Reverse(Sort([pe[2]:pe in Factorisation(n)])):n in[1..76]]; // Jason Kimberley, Jun 13 2012
    
  • PARI
    apply( {A212171_row(n)=vecsort(factor(n)[,2]~,,4)}, [1..40])\\ M. F. Hasler, Apr 19 2022

Formula

Row n of A118914, reversed.
Row n of A124010 for n > 1, with exponents sorted in nonincreasing order. Equivalently, row A046523(n) of A124010 for n > 1.

A257511 Number of 1's in factorial base representation of n (A007623).

Original entry on oeis.org

0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1
Offset: 0

Views

Author

Antti Karttunen, Apr 27 2015

Keywords

Crossrefs

Cf. A255411 (numbers n such that a(n) = 0), A255341 (such that a(n) = 1), A255342 (such that a(n) = 2), A255343 (such that a(n) = 3).
Positions of records: A007489.
Cf. also A257510.

Programs

  • Mathematica
    factBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > i!, i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= j!, m = m - j!; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; dList]; s = Table[FromDigits[factBaseIntDs@ n], {n, 0, 120}];
    First@ DigitCount[#] & /@ s (* Michael De Vlieger, Apr 27 2015, after Alonso del Arte at A007623 *)
    nn = 120; b = Module[{m = 1}, While[Factorial@ m < nn, m++]; MixedRadix[Reverse@ Range[2, m]]]; Table[Count[IntegerDigits[n, b], 1], {n, 0, nn}] (* Michael De Vlieger, Aug 29 2016, Version 10.2 *)
  • Scheme
    (define (A257511 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (floor->exact (/ n i)) (+ 1 i) (+ s (if (= 1 (modulo n i)) 1 0)))))))

Formula

a(0) = 0; for n >= 1, a(n) = A265333(n) + a(A257687(n)). - Antti Karttunen, Aug 29 2016
Other identities and observations. For all n >= 0:
a(n) = A260736(A225901(n)).
a(n) = A001221(A275732(n)) = A001222(A275732(n)).
a(n) = A007814(A275735(n)).
a(n) = A056169(A276076(n)).
a(A007489(n)) = n. [Particularly, A007489(n) gives the position where n first appears.]
a(n) <= A060130(n) <= A034968(n).

A295664 Exponent of the highest power of 2 dividing number of divisors of n: a(n) = A007814(A000005(n)); 2-adic valuation of tau(n).

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 2, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 3, 0, 2, 2, 1, 1, 3, 1, 1, 2, 2, 2, 0, 1, 2, 2, 3, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 3, 2, 3, 2, 2, 1, 2, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 1, 2, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 3, 1, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 0, 1, 3, 1, 3, 3, 2, 1, 2, 1, 3, 2, 1, 1, 3, 2, 1, 1, 2, 2, 4, 0
Offset: 1

Views

Author

Antti Karttunen, Nov 28 2017

Keywords

Comments

In the prime factorization of n = p1^e1 * ... pk^ek, add together the number of trailing 1-bits in each exponent e when they are written in binary.

Crossrefs

Cf. A000290 (positions of zeros).

Programs

  • Mathematica
    Table[IntegerExponent[DivisorSigma[0, n], 2], {n, 120}] (* Michael De Vlieger, Nov 28 2017 *)
  • PARI
    a(n) = valuation(numdiv(n), 2); \\ Michel Marcus, Nov 30 2017
    
  • Python
    from sympy import divisor_count
    def A295664(n): return (~(m:=int(divisor_count(n))) & m-1).bit_length() # Chai Wah Wu, Jul 05 2022

Formula

Additive with a(p^e) = A007814(1+e).
a(1) = 0; for n > 1, a(n) = A007814(1+A067029(n)) + a(A028234(n)).
a(n) = A007814(A000005(n)).
a(n) >= A162642(n) >= A056169(n).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) =-0.223720656976344505701..., where f(x) = -x + (1-x) * Sum_{k>=1} x^(2^k-1)/(1-x^(2^k)). - Amiram Eldar, Sep 28 2023

A295659 Number of exponents larger than 2 in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 28 2017

Keywords

Examples

			For n = 120 = 2^3 * 3^1 * 5^1 there is only one exponent larger than 2, thus a(120) = 1.
For n = 216 = 2^3 * 3^3 there are two exponents larger than 2, thus a(216) = 2.
		

Crossrefs

Cf. A004709 (positions of zeros), A046099 (of nonzeros), A212793.

Programs

Formula

Additive with a(p^e) = 1 if e>2, 0 otherwise.
a(n) = 0 iff A212793(n) = 1.
a(n) = A001221(A053150(n)).
a(n) = A056170(A003557(n)).
a(n) >= A295662(n) = A162642(n) - A056169(n).
a(n) = A295883(n) + A295884(n).
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^3 = 0.174762... (A085541). - Amiram Eldar, Nov 01 2020

A295662 Number of odd exponents larger than one in the canonical prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 28 2017

Keywords

Examples

			For n = 24 = 2^3 * 3^1 there are two odd exponents, but only the other is larger than 1, thus a(24) = 1.
For n = 216 = 2^3 * 3^3 there are two odd exponents larger than 1, thus a(216) = 2.
		

Crossrefs

Cf. A295661 (positions of nonzero terms).

Programs

Formula

Additive with a(p) = 0, a(p^e) = A000035(e) if e > 1.
a(1) = 0; and for n > 1, if A067029(n) = 1, a(n) = a(A028234(n)), otherwise A000035(A067029(n)) + a(A028234(n)).
a(n) = A162642(n) - A056169(n).
a(n) <= A295659(n).
a(n) = 0 iff A295663(n) = 0, and when A295663(n) > 0, a(n) <= A295663(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p^2*(p+1)) = 0.122017493776862257491... . - Amiram Eldar, Sep 28 2023

A369427 The number of unitary divisors of n that are squares of primes.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0
Offset: 1

Views

Author

Amiram Eldar, Jan 23 2024

Keywords

Comments

The number of exponents in the prime factorization of n that are equal to 2.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 2, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecsum(apply(x -> if(x == 2, 1, 0), factor(n)[, 2]));

Formula

Additive with a(p^e) = 1 if e = 2, and 0 otherwise.
a(n) > 0 if and only if n is in A038109.
a(A061742(n)) = n, and a(k) < n for all k < A061742(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (1/p^2 - 1/p^3) = A085548 - A085541 = 0.27748478074162196208... .

A081387 Number of non-unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of prime factors in C(2n,n) whose exponent is greater than one.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 2, 2, 2, 2, 1, 1, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 3, 3, 3, 2, 4, 2, 2, 3, 3, 3, 3, 4, 4, 5, 4, 4, 2, 3, 3, 2, 2, 2, 4, 3, 3, 4, 3, 4, 5, 4, 2, 2, 2, 3, 5, 5, 5, 5, 3, 2, 3, 2, 3, 3, 3
Offset: 1

Views

Author

Labos Elemer, Mar 27 2003

Keywords

Examples

			For n = 14: binomial(28,14) = 40116600 = 2*2*2*3*3*3*5*5*17*19*23; unitary prime divisors: {17,19,23}; non-unitary prime divisors: {2,3,5}, so a(14) = 3.
		

Crossrefs

Programs

Formula

a(n) = A056170(A000984(n)) = A001221(A000984(n)) - A081386(n) = A067434(n) - A081386(n).

A081386 Number of unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of those prime factors in C(2n,n), whose exponent equals one.

Original entry on oeis.org

1, 2, 1, 3, 1, 3, 3, 4, 4, 4, 5, 5, 4, 3, 5, 7, 6, 7, 7, 8, 9, 9, 6, 7, 7, 7, 8, 11, 12, 11, 11, 11, 12, 12, 12, 13, 13, 13, 11, 13, 12, 14, 13, 13, 15, 14, 14, 14, 15, 16, 16, 16, 17, 19, 18, 17, 18, 19, 18, 19, 18, 18, 18, 20, 18, 21, 22, 20, 20, 20, 20, 20, 20, 19, 21, 21, 24, 23
Offset: 1

Views

Author

Labos Elemer, Mar 27 2003

Keywords

Examples

			n=10: C(20,10) = 184756 = 2*2*11*13*17*19; unitary-p-divisors = {11,13,17,19}, so a(10)=4.
		

Crossrefs

Programs

  • Mathematica
    Table[Function[m, Count[Divisors@ m, k_ /; And[PrimeQ@ k, GCD[k, m/k] == 1]]]@ Binomial[2 n, n], {n, 50}] (* Michael De Vlieger, Dec 17 2016 *)
  • PARI
    a(n) = my(f=factor(binomial(2*n, n))); sum(k=1, #f~, f[k,2] == 1); \\ Michel Marcus, Dec 18 2016

Formula

a(n) = A056169(A000984(n)).

A275851 a(n) = number of elements in range [1..(1+A084558(n))] fixed by the permutation with rank n of permutation list A060117 (or A060118).

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 1, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 3, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 1, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 3, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 3, 2, 2, 1, 1, 1, 1, 0, 1, 0, 0, 0, 2, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 3, 2, 1, 1, 2, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 0, 4
Offset: 0

Views

Author

Antti Karttunen, Aug 11 2016

Keywords

Crossrefs

Cf. A275852 (indices of zeros).

Formula

a(n) = A056169(A275725(n)).
a(n) = 1 + A084558(n) - A060129(n).

A063956 Sum of unitary prime divisors of n.

Original entry on oeis.org

0, 2, 3, 0, 5, 5, 7, 0, 0, 7, 11, 3, 13, 9, 8, 0, 17, 2, 19, 5, 10, 13, 23, 3, 0, 15, 0, 7, 29, 10, 31, 0, 14, 19, 12, 0, 37, 21, 16, 5, 41, 12, 43, 11, 5, 25, 47, 3, 0, 2, 20, 13, 53, 2, 16, 7, 22, 31, 59, 8, 61, 33, 7, 0, 18, 16, 67, 17, 26, 14, 71, 0, 73, 39, 3, 19, 18, 18, 79, 5, 0
Offset: 1

Views

Author

Labos Elemer, Sep 04 2001

Keywords

Examples

			The prime divisors of 420 = 2^2 * 3 * 5 * 7. Among them, those that have exponent 1 (i.e., unitary prime divisors) are {3, 5, 7}, so a(420) = 3 + 5 + 7 = 15.
		

Crossrefs

Programs

  • Mathematica
    Table[DivisorSum[n, # &, And[PrimeQ@ #, GCD[#, n/#] == 1] &], {n, 81}] (* Michael De Vlieger, Feb 17 2019 *)
    f[p_, e_] := If[e == 1, p, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 24 2024 *)
    Join[{0},Table[Total[Select[FactorInteger[n],#[[2]]==1&][[;;,1]]],{n,2,100}]] (* Harvey P. Dale, Jan 26 2025 *)
  • PARI
    { for (n=1, 1000, f=factor(n)~; a=0; for (i=1, length(f), if (f[2, i]==1, a+=f[1, i])); write("b063956.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 04 2009

Formula

a(n*m) = a(n) + a(m) - a(gcd(n^2, m)) - a(gcd(n, m^2)) for all n and m > 0 (conjecture). - Velin Yanev, Feb 17 2019
From Amiram Eldar, Jul 24 2024: (Start)
a(n) = A008472(n) - A063958(n).
Additive with a(p^e) = p is e = 1, and 0 otherwise. (End)

Extensions

Example clarified by Harvey P. Dale, Jan 26 2025
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