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.

Showing 1-10 of 17 results. Next

A275962 Total number of nonzero digits that occur on the multiply occupied slopes of the factorial base representation of n: a(n) = A275812(A275734(n)). (See comments for more exact definition).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 15 2016

Keywords

Comments

a(n) gives the total number of elements (counted with multiplicity) that have multiplicity > 1 in a multiset [(i_x - d_x) | where d_x ranges over each nonzero digit present and i_x is its position from the right].

Examples

			For n=525, in factorial base "41311", there are three occupied slopes. The maximal slope contains the nonzero digits "3.1", the sub-maximal the digits "4..1.", and the sub-sub-sub-maximal just "1..." (the 1 in the position 4 from right is the sole occupier of its own slope). There are two slopes with more than one nonzero digit, each having two such digits, and thus a(525) = 2+2 = 4.
Equally, when we form a multiset of (digit-position - digit-value) differences for all nonzero digits present in "41311", we obtain a multiset [0, 0, 1, 1, 3], in which the elements that occur multiple times are [0, 0, 1, 1], thus a(525) = 4.
		

Crossrefs

Cf. A275804 (indices of zeros), A275805 (of nonzeros).

Programs

Formula

a(n) = A275812(A275734(n)).
Other identities and observations. For all n >= 0.
a(n) = A275964(A225901(n)).
a(n) = A060130(n) - A275946(n).
a(n) >= A275947(n).

A275964 Total number of nonzero digits with multiple occurrences in factorial base representation of n (counted with multiplicity): a(n) = A275812(A275735(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 15 2016

Keywords

Examples

			For n=0, with factorial base representation (A007623) also 0, there are no nonzero digits, thus a(0) = 0.
For n=2, with factorial base representation "10", there are no nonzero digits that are present multiple times, thus a(2) = 0.
For n=3 ("11") there is one nonzero digit which occurs more than once, and it occurs two times in total, thus a(3) = 2.
For n=41 ("1221") there are two distinct nonzero digits ("1" and "2"), and both occur more than once, namely twice each, thus a(41) = 2+2 = 4.
For n=44 ("1310") there are two distinct nonzero digits ("1" and "3"), but only the other (1) occurs more than once (two times), thus a(44) = 2.
For n=279 ("21211") there are two distinct nonzero digits present that occur more than once, digit 2 twice, and digit 1 for three times, thus a(279) = 2+3 = 5.
		

Crossrefs

Cf. A265349 (indices of zeros), A265350 (of terms > 0).

Programs

  • Mathematica
    a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Total[Select[Tally[Select[s, # > 0 &]][[;;,2]], # > 1 &]]]; Array[a, 100, 0] (* Amiram Eldar, Feb 07 2024 *)
  • Scheme
    (define (A275964 n) (A275812 (A275735 n)))

Formula

a(n) = A275812(A275735(n)).
Other identities and observations. For all n >= 0.
a(n) = A275962(A225901(n)).
a(n) = A060130(n) - A275948(n).
a(n) >= A275949(n).

A284268 Sum of coefficients > 1 in the Stern polynomial B(2n+1,x): a(n) = A275812(A277324(n)).

Original entry on oeis.org

0, 0, 2, 0, 2, 3, 4, 0, 2, 6, 7, 5, 5, 6, 6, 0, 2, 8, 9, 9, 10, 11, 11, 7, 7, 11, 12, 9, 8, 9, 8, 0, 2, 10, 12, 11, 13, 17, 16, 12, 13, 18, 20, 16, 15, 17, 15, 9, 9, 15, 17, 16, 17, 19, 18, 12, 11, 16, 17, 13, 11, 12, 10, 0, 2, 12, 15, 14, 17, 22, 21, 15, 17, 25, 27, 24, 23, 26, 22, 15, 16, 24, 29, 26, 28, 32, 30, 21, 20, 28, 30, 24, 21, 23, 19, 11, 11, 19
Offset: 0

Views

Author

Antti Karttunen, Mar 25 2017

Keywords

Comments

Sum of terms larger than one on row 2n+1 of table A125184.

Crossrefs

Odd bisection of A284272.

Programs

Formula

a(n) = A284272((2*n)+1).
a(n) = A275812(A277324(n)).
Other identities. For all n >= 0:
A007306(1+n) = A284267(n) + a(n).

A284272 Sum of coefficients > 1 in the Stern polynomial B(n,x): a(n) = A275812(A260443(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 4, 0, 0, 0, 2, 2, 6, 2, 7, 3, 5, 0, 5, 4, 6, 0, 6, 0, 0, 0, 2, 2, 8, 2, 9, 6, 9, 2, 10, 7, 11, 3, 11, 5, 7, 0, 7, 5, 11, 4, 12, 6, 9, 0, 8, 6, 9, 0, 8, 0, 0, 0, 2, 2, 10, 2, 12, 8, 11, 2, 13, 9, 17, 6, 16, 9, 12, 2, 13, 10, 18, 7, 20, 11, 16, 3, 15, 11, 17, 5, 15, 7, 9, 0, 9, 7, 15, 5, 17, 11, 16, 4, 17, 12, 19, 6, 18, 9
Offset: 0

Views

Author

Antti Karttunen, Mar 25 2017

Keywords

Comments

Sum of terms larger than one on row n of table A125184.

Crossrefs

Cf. A002487, A125184, A260443, A275812, A284264, A284271, A284268 (odd bisection).

Programs

Formula

a(n) = A275812(A260443(n)).
Other identities and observations. For all n >= 0:
A002487(n) = A284271(n) + a(n).
a(n) >= 2*A284264(n).

A046660 Excess of n = number of prime divisors (with multiplicity) - number of prime divisors (without multiplicity).

Original entry on oeis.org

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

Views

Author

Keywords

Comments

a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3 * 3 and 375 = 3 * 5^3 both have prime signature (3, 1).
a(n) = 0 for squarefree n.
A162511(n) = (-1)^a(n). - Reinhard Zumkeller, Jul 08 2009
a(n) = the number of divisors of n that are each a composite power of a prime. - Leroy Quet, Dec 02 2009
a(A005117(n)) = 0; a(A060687(n)) = 1; a(A195086(n)) = 2; a(A195087(n)) = 3; a(A195088(n)) = 4; a(A195089(n)) = 5; a(A195090(n)) = 6; a(A195091(n)) = 7; a(A195092(n)) = 8; a(A195093(n)) = 9; a(A195069(n)) = 10. - Reinhard Zumkeller, Nov 29 2015

References

  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, pp. 51-52.

Crossrefs

Programs

  • Haskell
    import Math.NumberTheory.Primes.Factorisation (factorise)
    a046660 n = sum es - length es where es = snd $ unzip $ factorise n
    -- Reinhard Zumkeller, Nov 28 2015, Jan 09 2013
    
  • Maple
    with(numtheory); A046660 := n -> bigomega(n)-nops(factorset(n)):
    seq(A046660(k), k=1..100); # Wesley Ivan Hurt, Oct 27 2013
    # Or:
    with(NumberTheory): A046660 := n -> NumberOfPrimeFactors(n) - NumberOfPrimeFactors(n, 'distinct'):  # Peter Luschny, Jul 14 2023
  • Mathematica
    Table[PrimeOmega[n] - PrimeNu[n], {n, 50}] (* or *) muf[n_] := Module[{fi = FactorInteger[n]}, Total[Transpose[fi][[2]]] - Length[fi]]; Array[muf, 50] (* Harvey P. Dale, Sep 07 2011. The second program is several times faster than the first program for generating large numbers of terms. *)
  • PARI
    a(n)=bigomega(n)-omega(n) \\ Charles R Greathouse IV, Nov 14 2012
    
  • PARI
    a(n)=my(f=factor(n)[,2]); vecsum(f)-#f \\ Charles R Greathouse IV, Aug 01 2016
    
  • Python
    from sympy import factorint
    def A046660(n): return sum(e-1 for e in factorint(n).values()) # Chai Wah Wu, Jul 18 2023

Formula

a(n) = Omega(n) - omega(n) = A001222(n) - A001221(n).
Additive with a(p^e) = e - 1.
a(n) = Sum_{k = 1..A001221(n)} (A124010(n,k) - 1). - Reinhard Zumkeller, Jan 09 2013
G.f.: Sum_{p prime, k>=2} x^(p^k)/(1 - x^(p^k)). - Ilya Gutkovskiy, Jan 06 2017
Asymptotic mean: lim_{m->oo} (1/m) Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p*(p-1)) = 0.773156... (A136141). - Amiram Eldar, Jul 28 2020
a(n) = Sum_{p|n} A286563(n/p,p), where p is prime. - Ridouane Oudra, Sep 13 2023
a(n) = A275812(n) - A056170(n). - Amiram Eldar, Jan 09 2024
a(n) = A001222(A003557(n)). - Peter Munn, Feb 06 2024

Extensions

More terms from David W. Wilson

A056170 Number of non-unitary prime divisors of n.

Original entry on oeis.org

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

Views

Author

Labos Elemer, Jul 27 2000

Keywords

Comments

A prime factor of n is unitary iff its exponent is 1 in the prime factorization of n. (Of course for any prime p, GCD(p, n/p) is either 1 or p. For a unitary prime factor it must be 1.)
Number of squared primes dividing n. - Reinhard Zumkeller, May 18 2002
a(A005117(n)) = 0; a(A013929(n)) > 0; a(A190641(n)) = 1. - Reinhard Zumkeller, Dec 29 2012
First differences of A013940. - Jason Kimberley, Feb 01 2017
Number of exponents larger than 1 in the prime factorization of n. - Antti Karttunen, Nov 28 2017

Crossrefs

Programs

Formula

Additive with a(p^e) = 0 if e = 1, 1 otherwise.
G.f.: Sum_{k>=1} x^(prime(k)^2)/(1 - x^(prime(k)^2)). - Ilya Gutkovskiy, Jan 01 2017
a(n) = log_2(A000005(A071773(n))). - observed by Velin Yanev, Aug 20 2017, confirmed by Antti Karttunen, Nov 28 2017
From Antti Karttunen, Nov 28 2017: (Start)
a(n) = A001221(n) - A056169(n).
a(n) = omega(A000188(n)) = omega(A003557(n)) = omega(A057521(n)) = omega(A295666(n)), where omega = A001221.
For all n >= 1 it holds that:
a(A003557(n)) = A295659(n).
a(n) >= A162641(n).
(End)
Dirichlet g.f.: primezeta(2s)*zeta(s). - Benedict W. J. Irwin, Jul 11 2018
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^2 = 0.452247... (A085548). - Amiram Eldar, Nov 01 2020
a(n) = A275812(n) - A046660(n). - Amiram Eldar, Jan 09 2024

Extensions

Minor edits by Franklin T. Adams-Watters, Mar 23 2011

A056169 Number of unitary prime divisors of n.

Original entry on oeis.org

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

Views

Author

Labos Elemer, Jul 27 2000

Keywords

Comments

The zeros of this sequences are the powerful numbers (A001694). There are no arbitrarily long subsequences with a given upper bound; for example, every sequence of 4 values includes one divisible by 2 but not 4, so there are no more than 3 consecutive zeros. Similarly, there can be no more than 23 consecutive values with none divisible by both 2 and 3 but neither 4 nor 9 (so a(n) >= 2), etc. In general, this gives an upper bound that is a (relatively) small multiple of the k-th primorial number (prime(k)#). One suspects that the actual upper bounds for such subsequences are quite a bit lower; e.g., Erdős conjectured that there are no three consecutive powerful numbers. - Franklin T. Adams-Watters, Aug 08 2006
In particular, for every A048670(k)*A002110(k) consecutive terms, at least one is greater than or equal to k. - Charlie Neder, Jan 03 2019
Following Catalan's conjecture (which became Mihăilescu's theorem in 2002), the first case of two consecutive zeros in this sequence is for a(8) and a(9), because 8 = 2^3 and 9 = 3^2, and there are no other consecutive zeros for consecutive powers. However, there are other pairs of consecutive zeros at powerful numbers (A001694, A060355). The next example is a(288) = a(289) = 0, because 288 = 2^5 * 3^2 and 289 = 17^2, then also a(675) and a(676). - Bernard Schott, Jan 06 2019
a(2k-1) is the number of primes p such that p || x + y and p^2 || x^(2k-1) + y^(2k-1) for some positive integers x and y. For any positive integers x, y and k > 1, there is no prime p such that p || x + y and p^2 || x^(2k) + y^(2k). - Jinyuan Wang, Apr 08 2020

Examples

			9 = 3^2 so a(9) = 0; 10 = 2 * 5 so a(10) = 2; 11 = 11^1 so a(11) = 1.
		

Crossrefs

Programs

  • Haskell
    a056169 = length . filter (== 1) . a124010_row
    -- Reinhard Zumkeller, Sep 10 2013
    
  • Maple
    a:= n-> nops(select(i-> i[2]=1, ifactors(n)[2])):
    seq(a(n), n=1..120);  # Alois P. Heinz, Mar 27 2017
  • Mathematica
    Join[{0},Table[Count[Transpose[FactorInteger[n]][[2]],1],{n,2,110}]] (* Harvey P. Dale, Mar 15 2012 *)
    Table[DivisorSum[n, 1 &, And[PrimeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Nov 28 2017 *)
  • PARI
    a(n)=my(f=factor(n)[,2]); sum(i=1,#f,f[i]==1) \\ Charles R Greathouse IV, Apr 29 2015
    
  • Python
    from sympy import factorint
    def a(n):
        f=factorint(n)
        return 0 if n==1 else sum(1 for i in f if f[i]==1)
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 19 2017
    
  • Scheme
    ;; With memoization-macro definec.
    (definec (A056169 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 1 0) (A056169 (A028234 n))))) ;; Antti Karttunen, Nov 28 2017

Formula

A prime factor of n is unitary iff its exponent is 1 in prime factorization of n. In general, gcd(p, n/p) = 1 or = p.
Additive with a(p^e) = 1 if e = 1, 0 otherwise.
a(n) = #{k: A124010(n,k) = 1, k = 1..A001221}. - Reinhard Zumkeller, Sep 10 2013
From Antti Karttunen, Nov 28 2017: (Start)
a(1) = 0; for n > 1, a(n) = A063524(A067029(n)) + a(A028234(n)).
a(n) = A001221(A055231(n)) = A001222(A055231(n)).
a(n) = A001221(n) - A056170(n) = A001221(n) - A001221(A000188(n)).
a(n) = A001222(n) - A275812(n).
a(n) = A162642(n) - A295662(n).
a(n) <= A162642(n) <= a(n) + A295659(n).
a(n) <= A295664(n).
(End)
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = Sum_{p prime} (1/p^2) = 0.452247... (A085548). - Amiram Eldar, Sep 28 2023

A275734 Prime-factorization representations of "factorial base slope polynomials": a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).

Original entry on oeis.org

1, 2, 3, 6, 2, 4, 5, 10, 15, 30, 10, 20, 3, 6, 9, 18, 6, 12, 2, 4, 6, 12, 4, 8, 7, 14, 21, 42, 14, 28, 35, 70, 105, 210, 70, 140, 21, 42, 63, 126, 42, 84, 14, 28, 42, 84, 28, 56, 5, 10, 15, 30, 10, 20, 25, 50, 75, 150, 50, 100, 15, 30, 45, 90, 30, 60, 10, 20, 30, 60, 20, 40, 3, 6, 9, 18, 6, 12, 15, 30, 45, 90, 30, 60, 9, 18, 27
Offset: 0

Views

Author

Antti Karttunen, Aug 08 2016

Keywords

Comments

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of nonzero digits that occur on the slope (k-1) levels below the "maximal slope" in the factorial base representation of n. See A275811 for the definition of the "digit slopes" in this context.

Examples

			For n=23 ("321" in factorial base representation, A007623), all three nonzero digits are maximal for their positions (they all occur on "maximal slope"), thus a(23) = prime(1)^3 = 2^3 = 8.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the "maximal slope", while the most significant 1 is on the "sub-sub-sub-maximal", thus a(29) = prime(1)^2 * prime(4)^1 = 2*7 = 28.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the sub-maximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus a(37) = prime(1) * prime(2) * prime(4) = 2*3*7 = 42.
For n=55 ("2101"), the least significant 1 is on the maximal slope, and the digits "21" at the beginning are together on the sub-sub-maximal slope (as they are both two less than the maximal digit values 4 and 3 allowed in those positions), thus a(55) = prime(1)^1 * prime(3)^2 = 2*25 = 50.
		

Crossrefs

Cf. A275811.
Cf. A275804 (indices of squarefree terms), A275805 (of terms not squarefree).
Cf. also A275725, A275733, A275735, A276076 for other such prime factorization encodings of A060117/A060118-related polynomials.

Programs

  • Python
    from operator import mul
    from sympy import prime, factorial as f
    def a007623(n, p=2): return n if n

    0 else '0' for i in x)[::-1] return 0 if n==1 else sum(int(y[i])*f(i + 1) for i in range(len(y))) def a(n): return 1 if n==0 else a275732(n)*a(a257684(n)) print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017

Formula

a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).
Other identities and observations. For all n >= 0:
a(n) = A275735(A225901(n)).
a(A007489(n)) = A002110(n).
A001221(a(n)) = A060502(n).
A001222(a(n)) = A060130(n).
A007814(a(n)) = A260736(n).
A051903(a(n)) = A275811(n).
A048675(a(n)) = A275728(n).
A248663(a(n)) = A275808(n).
A056169(a(n)) = A275946(n).
A056170(a(n)) = A275947(n).
A275812(a(n)) = A275962(n).

A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).

Original entry on oeis.org

2, 6, 18, 30, 90, 270, 450, 210, 630, 6750, 20250, 9450, 15750, 47250, 22050, 2310, 6930, 330750, 3543750, 1653750, 4961250, 53156250, 24806250, 727650, 1212750, 57881250, 173643750, 18191250, 8489250, 25467750, 2668050, 30030, 90090, 40020750, 1910081250, 891371250, 9550406250, 455814843750, 212713593750
Offset: 0

Views

Author

Antti Karttunen, Oct 10 2016

Keywords

Comments

From David A. Corneth, Oct 22 2016: (Start)
The exponents of the prime factorization of a(n) are first nondecreasing, then nonincreasing.
The exponent of 2 in the prime factorization of a(n) is 1. (End)

Examples

			A method to find terms of this sequence, explained by an example to find a(7). To find k = a(7), we find k such that A048675(k) = 2*7+1 = 15. 7 has the binary partitions: {[7, 0, 0], [5, 1, 0], [3, 2, 0], [1, 3, 0], [3, 0, 1], [1, 1, 1]}. To each of those, we prepend a 1. This gives the binary partitions of 15 starting with a 1. For example, for the first we get [1, 7, 0, 0]. We see that only [1, 5, 1, 0], [1, 3, 2, 0] and [1, 1, 1, 1] start nondecreasing, then nonincreasing, so we only check those. These numbers will be the exponents in a prime factorization. [1, 5, 1, 0] corresponds to prime(1)^1 * prime(2)^5 * prime(3)^1 * prime(4)^0 = 2430. We find that [1, 1, 1, 1] gives k = 210 for which A048675(k) = 15 so a(7) = 210. - _David A. Corneth_, Oct 22 2016
		

Crossrefs

Cf. A277200 (same sequence sorted into ascending order).

Programs

  • Mathematica
    a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[a[2 n + 1], {n, 0, 38}] (* Michael De Vlieger, Apr 05 2017 *)
  • Python
    from sympy import factorint, prime, primepi
    from operator import mul
    def a003961(n):
        F=factorint(n)
        return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F])
    def a260443(n): return n + 1 if n<2 else a003961(a260443(n//2)) if n%2==0 else a260443((n - 1)//2)*a260443((n + 1)//2)
    def a(n): return a260443(2*n + 1)
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 21 2017

Formula

a(n) = A260443((2*n)+1).
a(0) = 2; for n >= 1, a(n) = A260443(n) * A260443(n+1).
Other identities. For all n >= 0:
A007949(a(n)) = A005811(n). [See comments in A125184.]
A156552(a(n)) = A277189(n), a(n) = A005940(1+A277189(n)).
A048675(a(n)) = 2n + 1. - David A. Corneth, Oct 22 2016
A001222(a(n)) = A007306(1+n).
A056169(a(n)) = A284267(n).
A275812(a(n)) = A284268(n).
A248663(a(n)) = A283975(n).
A000188(a(n)) = A283484(n).
A247503(a(n)) = A284563(n).
A248101(a(n)) = A284564(n).
A046523(a(n)) = A284573(n).
a(n) = A277198(n) * A284008(n).
a(n) = A284576(n) * A284578(n) = A284577(n) * A000290(A284578(n)).

Extensions

More linking formulas added by Antti Karttunen, Apr 16 2017

A275725 a(n) = A275723(A002110(1+A084558(n)), n); prime factorization encodings of cycle-polynomials computed for finite permutations listed in the order that is used in tables A060117 / A060118.

Original entry on oeis.org

2, 4, 18, 8, 12, 8, 150, 100, 54, 16, 24, 16, 90, 40, 54, 16, 36, 16, 60, 40, 36, 16, 24, 16, 1470, 980, 882, 392, 588, 392, 750, 500, 162, 32, 48, 32, 270, 80, 162, 32, 108, 32, 120, 80, 72, 32, 48, 32, 1050, 700, 378, 112, 168, 112, 750, 500, 162, 32, 48, 32, 450, 200, 162, 32, 72, 32, 300, 200, 108, 32, 48, 32, 630, 280, 378, 112, 252, 112, 450, 200
Offset: 0

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Author

Antti Karttunen, Aug 09 2016

Keywords

Comments

In this context "cycle-polynomials" are single-variable polynomials where the coefficients (encoded with the exponents of prime factorization of n) are equal to the lengths of cycles in the permutation listed with index n in tables A060117 or A060118. See the examples.

Examples

			Consider the first eight permutations (indices 0-7) listed in A060117:
  1 [Only the first 1-cycle explicitly listed thus a(0) = 2^1 = 2]
  2,1 [One transposition (2-cycle) in beginning, thus a(1) = 2^2 = 4]
  1,3,2 [One fixed element in beginning, then transposition, thus a(2) = 2^1 * 3^2 = 18]
  3,1,2 [One 3-cycle, thus a(3) = 2^3 = 8]
  3,2,1 [One transposition jumping over a fixed element, a(4) = 2^2 * 3^1 = 12]
  2,3,1 [One 3-cycle, thus a(5) = 2^3 = 8]
  1,2,4,3 [Two 1-cycles, then a 2-cycle, thus a(6) = 2^1 * 3^1 * 5^2 = 150].
  2,1,4,3 [Two 2-cycles, not crossed, thus a(7) = 2^2 * 5^2 = 100]
and also the seventeenth one at n=16 [A007623(16)=220] where we have:
  3,4,1,2 [Two 2-cycles crossed, thus a(16) = 2^2 * 3^2 = 36].
		

Crossrefs

Cf. A275807 (terms divided by 2).
Cf. also A275733, A275734, A275735 for other such prime factorization encodings of A060117/A060118-related polynomials.

Programs

Formula

a(n) = A275723(A002110(1+A084558(n)), n).
Other identities:
A001221(a(n)) = 1+A257510(n) (for all n >= 1).
A001222(a(n)) = 1+A084558(n).
A007814(a(n)) = A275832(n).
A048675(a(n)) = A275726(n).
A051903(a(n)) = A275803(n).
A056169(a(n)) = A275851(n).
A046660(a(n)) = A060130(n).
A072411(a(n)) = A060131(n).
A056170(a(n)) = A060128(n).
A275812(a(n)) = A060129(n).
a(n!) = 2 * A243054(n) = A000040(n)*A002110(n) for all n >= 1.
Showing 1-10 of 17 results. Next