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 20 results. Next

A275948 Number of nonzero digits that occur only once in factorial base representation of n: a(n) = A056169(A275735(n)).

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 0, 0, 0, 2, 1, 1, 2, 2, 1, 0, 1, 1, 2, 2, 1, 2, 3, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 1, 0, 2, 1, 1, 1, 3, 2, 1, 2, 2, 1, 0, 1, 2, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 2, 3, 3, 2, 1, 2, 1, 2, 2, 1, 2, 3, 2, 1, 1, 1, 3, 2, 2, 3, 3, 2, 1, 2, 0, 1, 1, 0, 1, 2, 1, 2, 2, 1, 2, 3, 2, 1, 1, 1, 3, 2, 2, 3, 3, 2, 1, 2, 2, 3, 3, 2, 3, 4, 1
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 is one nonzero digit, thus a(2) = 1.
For n=3 (= "11") there is no nonzero digit which would occur just once, thus a(3) = 0.
For n=23 (= "321") there are three nonzero digits and each of those digits occurs just once, thus a(23) = 3.
For n=44 (= "1310") there are two distinct nonzero digits ("1" and "3"), but only the other (3) occurs just once, thus a(44) = 1.
		

Crossrefs

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++]; Count[Tally[Select[s, # > 0 &]][[;;,2]], 1]]; Array[a, 100, 0] (* Amiram Eldar, Feb 07 2024 *)
  • Python
    from sympy import prime, factorint
    from operator import mul
    from functools import reduce
    import collections
    def a056169(n):
        f=factorint(n)
        return 0 if n==1 else sum([1 for i in f if f[i]==1])
    def a007623(n, p=2): return n if n
  • Scheme
    (define (A275948 n) (A056169 (A275735 n)))
    

Formula

a(n) = A056169(A275735(n)).
Other identities. For all n >= 0.
a(n) = A275946(A225901(n)).
A275806(n) = a(n) + A275949(n).
A060130(n) = a(n) + A275964(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).

A278235 Filter-sequence for factorial base (digit levels): Least number with the same prime signature as A275735(n).

Original entry on oeis.org

1, 2, 2, 4, 2, 6, 2, 4, 4, 8, 6, 12, 2, 6, 6, 12, 4, 12, 2, 6, 6, 12, 6, 30, 2, 4, 4, 8, 6, 12, 4, 8, 8, 16, 12, 24, 6, 12, 12, 24, 12, 36, 6, 12, 12, 24, 30, 60, 2, 6, 6, 12, 4, 12, 6, 12, 12, 24, 12, 36, 4, 12, 12, 36, 8, 24, 6, 30, 30, 60, 12, 60, 2, 6, 6, 12, 6, 30, 6, 12, 12, 24, 30, 60, 6, 30, 30, 60, 12, 60, 4, 12, 12, 36, 12, 60, 2, 6, 6, 12, 6, 30, 6
Offset: 0

Views

Author

Antti Karttunen, Nov 16 2016

Keywords

Comments

This sequence can be used for filtering certain factorial base (A007623) related sequences, because it matches only with any such sequence b that can be computed as b(n) = f(A275735(n)), where f(n) is any function that depends only on the prime signature of n (some of these are listed under the index entry for "sequences computed from exponents in ...").
Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.

Crossrefs

Other factorial base related filter-sequences: A278225, A278234, A278236.
Sequences that partition N into same or coarser equivalence classes: A060130, A257696 (?), A264990, A275806, A275948, A275964 (this is a proper a subset of the sequences that match with A278236).

Programs

Formula

a(n) = A046523(A275735(n)).
a(n) = A278234(A225901(n)).

A275949 Number of distinct nonzero digits that occur multiple times in factorial base representation of n: a(n) = A056170(A275735(n)).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 2, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 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 distinct nonzero digit which occurs more than once, thus a(3) = 1.
For n=41 ("1221") there are two distinct nonzero digits ("1" and "2"), and both occur more than once, thus a(41) = 2.
For n=44 ( "1310") there are two distinct nonzero digits ("1" and "3"), but only the other (1) occurs more than once, thus a(44) = 1.
		

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++]; Count[Tally[Select[s, # > 0 &]][[;;, 2]], ?(# > 1 &)]]; Array[a, 100, 0] (* _Amiram Eldar, Feb 14 2024 *)
  • Python
    from sympy import prime, factorint
    from operator import mul
    from functools import reduce
    import collections
    def a056170(n):
        f = factorint(n)
        return sum([1 for i in f if f[i]!=1])
    def a007623(n, p=2): return n if n
  • Scheme
    (define (A275949 n) (A056170 (A275735 n)))
    

Formula

a(n) = A056170(A275735(n)).
Other identities and observations. For all n >= 0.
a(n) = A275947(A225901(n)).
A275806(n) = A275948(n) + a(n).
a(n) <= A275964(n).

A275729 A275735-polynomials evaluated at x=2: a(n) = A048675(A275735(n)).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 2, 3, 3, 4, 4, 5, 4, 5, 5, 6, 6, 7, 1, 2, 2, 3, 3, 4, 2, 3, 3, 4, 4, 5, 3, 4, 4, 5, 5, 6, 5, 6, 6, 7, 7, 8, 2, 3, 3, 4, 4, 5, 3, 4, 4, 5, 5, 6, 4, 5, 5, 6, 6, 7, 6, 7, 7, 8, 8, 9, 4, 5, 5, 6, 6, 7, 5, 6, 6, 7, 7, 8, 6, 7, 7, 8, 8, 9, 8, 9, 9, 10, 10, 11, 8, 9, 9, 10, 10, 11, 9, 10, 10, 11, 11, 12, 10, 11, 11, 12, 12, 13, 12, 13, 13, 14, 14, 15, 1
Offset: 0

Views

Author

Antti Karttunen, Aug 09 2016

Keywords

Crossrefs

Differs from A034968 for the first time at n=18, where a(18) = 4, while A034968(18) = 3.

Programs

Formula

a(n) = A048675(A275735(n)).

A003961 Completely multiplicative with a(prime(k)) = prime(k+1).

Original entry on oeis.org

1, 3, 5, 9, 7, 15, 11, 27, 25, 21, 13, 45, 17, 33, 35, 81, 19, 75, 23, 63, 55, 39, 29, 135, 49, 51, 125, 99, 31, 105, 37, 243, 65, 57, 77, 225, 41, 69, 85, 189, 43, 165, 47, 117, 175, 87, 53, 405, 121, 147, 95, 153, 59, 375, 91, 297, 115, 93, 61, 315, 67, 111, 275, 729, 119
Offset: 1

Views

Author

Keywords

Comments

Meyers (see Guy reference) conjectures that for all r >= 1, the least odd number not in the set {a(i): i < prime(r)} is prime(r+1). - N. J. A. Sloane, Jan 08 2021
Meyers' conjecture would be refuted if and only if for some r there were such a large gap between prime(r) and prime(r+1) that there existed a composite c for which prime(r) < c < a(c) < prime(r+1), in which case (by Bertrand's postulate) c would necessarily be a term of A246281. - Antti Karttunen, Mar 29 2021
a(n) is odd for all n and for each odd m there exists a k with a(k) = m (see A064216). a(n) > n for n > 1: bijection between the odd and all numbers. - Reinhard Zumkeller, Sep 26 2001
a(n) and n have the same number of distinct primes with (A001222) and without multiplicity (A001221). - Michel Marcus, Jun 13 2014
From Antti Karttunen, Nov 01 2019: (Start)
More generally, a(n) has the same prime signature as n, A046523(a(n)) = A046523(n). Also A246277(a(n)) = A246277(n) and A287170(a(n)) = A287170(n).
Many permutations and other sequences that employ prime factorization of n to encode either polynomials, partitions (via Heinz numbers) or multisets in general can be easily defined by using this sequence as one of their constituent functions. See the last line in the Crossrefs section for examples.
(End)

Examples

			a(12) = a(2^2 * 3) = a(prime(1)^2 * prime(2)) = prime(2)^2 * prime(3) = 3^2 * 5 = 45.
a(A002110(n)) = A002110(n + 1) / 2.
		

References

  • Richard K. Guy, editor, Problems From Western Number Theory Conferences, Labor Day, 1983, Problem 367 (Proposed by Leroy F. Meyers, The Ohio State U.).

Crossrefs

See A045965 for another version.
Row 1 of table A242378 (which gives the "k-th powers" of this sequence), row 3 of A297845 and of A306697. See also arrays A066117, A246278, A255483, A308503, A329050.
Cf. A064989 (a left inverse), A064216, A000040, A002110, A000265, A027746, A046523, A048673 (= (a(n)+1)/2), A108228 (= (a(n)-1)/2), A191002 (= a(n)*n), A252748 (= a(n)-2n), A286385 (= a(n)-sigma(n)), A283980 (= a(n)*A006519(n)), A341529 (= a(n)*sigma(n)), A326042, A049084, A001221, A001222, A122111, A225546, A260443, A245606, A244319, A246269 (= A065338(a(n))), A322361 (= gcd(n, a(n))), A305293.
Cf. A249734, A249735 (bisections).
Cf. A246261 (a(n) is of the form 4k+1), A246263 (of the form 4k+3), A246271, A246272, A246259, A246281 (n such that a(n) < 2n), A246282 (n such that a(n) > 2n), A252742.
Cf. A275717 (a(n) > a(n-1)), A275718 (a(n) < a(n-1)).
Cf. A003972 (Möbius transform), A003973 (Inverse Möbius transform), A318321.
Cf. A300841, A305421, A322991, A250469, A269379 for analogous shift-operators in other factorization and quasi-factorization systems.
Cf. also following permutations and other sequences that can be defined with the help of this sequence: A005940, A163511, A122111, A260443, A206296, A265408, A265750, A275733, A275735, A297845, A091202 & A091203, A250245 & A250246, A302023 & A302024, A302025 & A302026.
A version for partition numbers is A003964, strict A357853.
A permutation of A005408.
Applying the same transformation again gives A357852.
Other multiplicative sequences: A064988, A357977, A357978, A357980, A357983.
A056239 adds up prime indices, row-sums of A112798.

Programs

  • Haskell
    a003961 1 = 1
    a003961 n = product $ map (a000040 . (+ 1) . a049084) $ a027746_row n
    -- Reinhard Zumkeller, Apr 09 2012, Oct 09 2011
    (MIT/GNU Scheme, with Aubrey Jaffer's SLIB Scheme library)
    (require 'factor)
    (define (A003961 n) (apply * (map A000040 (map 1+ (map A049084 (factor n))))))
    ;; Antti Karttunen, May 20 2014
    
  • Maple
    a:= n-> mul(nextprime(i[1])^i[2], i=ifactors(n)[2]):
    seq(a(n), n=1..80);  # Alois P. Heinz, Sep 13 2017
  • Mathematica
    a[p_?PrimeQ] := a[p] = Prime[ PrimePi[p] + 1]; a[1] = 1; a[n_] := a[n] = Times @@ (a[#1]^#2& @@@ FactorInteger[n]); Table[a[n], {n, 1, 65}] (* Jean-François Alcover, Dec 01 2011, updated Sep 20 2019 *)
    Table[Times @@ Map[#1^#2 & @@ # &, FactorInteger[n] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[n == 1], {n, 65}] (* Michael De Vlieger, Mar 24 2017 *)
  • PARI
    a(n)=local(f); if(n<1,0,f=factor(n); prod(k=1,matsize(f)[1],nextprime(1+f[k,1])^f[k,2]))
    
  • PARI
    a(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ Michel Marcus, May 17 2014
    
  • Perl
    use ntheory ":all";  sub a003961 { vecprod(map { next_prime($) } factor(shift)); }  # _Dana Jacobsen, Mar 06 2016
    
  • Python
    from sympy import factorint, prime, primepi, prod
    def a(n):
        f=factorint(n)
        return 1 if n==1 else prod(prime(primepi(i) + 1)**f[i] for i in f)
    [a(n) for n in range(1, 11)] # Indranil Ghosh, May 13 2017

Formula

If n = Product p(k)^e(k) then a(n) = Product p(k+1)^e(k).
Multiplicative with a(p^e) = A000040(A000720(p)+1)^e. - David W. Wilson, Aug 01 2001
a(n) = Product_{k=1..A001221(n)} A000040(A049084(A027748(n,k))+1)^A124010(n,k). - Reinhard Zumkeller, Oct 09 2011 [Corrected by Peter Munn, Nov 11 2019]
A064989(a(n)) = n and a(A064989(n)) = A000265(n). - Antti Karttunen, May 20 2014 & Nov 01 2019
A001221(a(n)) = A001221(n) and A001222(a(n)) = A001222(n). - Michel Marcus, Jun 13 2014
From Peter Munn, Oct 31 2019: (Start)
a(n) = A225546((A225546(n))^2).
a(A225546(n)) = A225546(n^2).
(End)
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((p^2-p)/(p^2-nextprime(p))) = 2.06399637... . - Amiram Eldar, Nov 18 2022

A181819 Prime shadow of n: a(1) = 1; for n>1, if n = Product prime(i)^e(i), then a(n) = Product prime(e(i)).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 10, 3, 4, 5, 6, 2, 8, 2, 11, 4, 4, 4, 9, 2, 4, 4, 10, 2, 8, 2, 6, 6, 4, 2, 14, 3, 6, 4, 6, 2, 10, 4, 10, 4, 4, 2, 12, 2, 4, 6, 13, 4, 8, 2, 6, 4, 8, 2, 15, 2, 4, 6, 6, 4, 8, 2, 14, 7, 4, 2, 12, 4, 4, 4, 10, 2, 12, 4, 6, 4, 4, 4, 22, 2, 6, 6, 9, 2, 8, 2, 10, 8
Offset: 1

Views

Author

Matthew Vandermast, Dec 07 2010

Keywords

Comments

a(n) depends only on prime signature of n (cf. A025487). a(m) = a(n) iff m and n have the same prime signature, i.e., iff A046523(m) = A046523(n).
Because A046523 (the smallest representative of prime signature of n) and this sequence are functions of each other as A046523(n) = A181821(a(n)) and a(n) = a(A046523(n)), it implies that for all i, j: a(i) = a(j) <=> A046523(i) = A046523(j) <=> A101296(i) = A101296(j), i.e., that equivalence-class-wise this is equal to A101296, and furthermore, applying any function f on this sequence gives us a sequence b(n) = f(a(n)) whose equivalence class partitioning is equal to or coarser than that of A101296, i.e., b is then a sequence that depends only on the prime signature of n (the multiset of exponents of its prime factors), although not necessarily in a very intuitive way. - Antti Karttunen, Apr 28 2022

Examples

			20 = 2^2*5 has the exponents (2,1) in its prime factorization. Accordingly, a(20) = prime(2)*prime(1) = A000040(2)*A000040(1) = 3*2 = 6.
		

Crossrefs

Programs

Formula

From Antti Karttunen, Feb 07 2016: (Start)
a(1) = 1; for n > 1, a(n) = A000040(A067029(n)) * a(A028234(n)).
a(1) = 1; for n > 1, a(n) = A008578(A001511(n)) * a(A064989(n)).
Other identities. For all n >= 1:
a(A124859(n)) = A122111(a(n)) = A238745(n). - from Matthew Vandermast's formulas for the latter sequence.
(End)
a(n) = A246029(A156552(n)). - Antti Karttunen, Oct 15 2016
From Antti Karttunen, Apr 28 & Apr 30 2022: (Start)
A181821(a(n)) = A046523(n) and a(A046523(n)) = a(n). [See comments]
a(n) = A329900(A124859(n)) = A319626(A124859(n)).
a(n) = A246029(A156552(n)).
a(a(n)) = A328830(n).
a(A304660(n)) = n.
a(A108951(n)) = A122111(n).
a(A185633(n)) = A322312(n).
a(A025487(n)) = A181820(n).
a(A276076(n)) = A275735(n) and a(A276086(n)) = A328835(n).
As the sequence converts prime exponents to prime indices, it effects the following mappings:
A001221(a(n)) = A071625(n). [Number of distinct indices --> Number of distinct exponents]
A001222(a(n)) = A001221(n). [Number of indices (i.e., the number of prime factors with multiplicity) --> Number of exponents (i.e., the number of distinct prime factors)]
A056239(a(n)) = A001222(n). [Sum of indices --> Sum of exponents]
A066328(a(n)) = A136565(n). [Sum of distinct indices --> Sum of distinct exponents]
A003963(a(n)) = A005361(n). [Product of indices --> Product of exponents]
A290103(a(n)) = A072411(n). [LCM of indices --> LCM of exponents]
A156061(a(n)) = A290107(n). [Product of distinct indices --> Product of distinct exponents]
A257993(a(n)) = A134193(n). [Index of the least prime not dividing n --> The least number not among the exponents]
A055396(a(n)) = A051904(n). [Index of the least prime dividing n --> Minimal exponent]
A061395(a(n)) = A051903(n). [Index of the greatest prime dividing n --> Maximal exponent]
A008966(a(n)) = A351564(n). [All indices are distinct (i.e., n is squarefree) --> All exponents are distinct]
A007814(a(n)) = A056169(n). [Number of occurrences of index 1 (i.e., the 2-adic valuation of n) --> Number of occurrences of exponent 1]
A056169(a(n)) = A136567(n). [Number of unitary prime divisors --> Number of exponents occurring only once]
A064989(a(n)) = a(A003557(n)) = A295879(n). [Indices decremented after <--> Exponents decremented before]
Other mappings:
A007947(a(n)) = a(A328400(n)) = A329601(n).
A181821(A007947(a(n))) = A328400(n).
A064553(a(n)) = A000005(n) and A000005(a(n)) = A182860(n).
A051903(a(n)) = A351946(n).
A003557(a(n)) = A351944(n).
A258851(a(n)) = A353379(n).
A008480(a(n)) = A309004(n).
a(A325501(n)) = A325507(n) and a(A325502(n)) = A038754(n+1).
a(n!) = A325508(n).
(End)

Extensions

Name "Prime shadow" (coined by Gus Wiseman in A325755) prefixed to the definition by Antti Karttunen, Apr 27 2022

A225901 Write n in factorial base, then replace each nonzero digit d of radix k with k-d.

Original entry on oeis.org

0, 1, 4, 5, 2, 3, 18, 19, 22, 23, 20, 21, 12, 13, 16, 17, 14, 15, 6, 7, 10, 11, 8, 9, 96, 97, 100, 101, 98, 99, 114, 115, 118, 119, 116, 117, 108, 109, 112, 113, 110, 111, 102, 103, 106, 107, 104, 105, 72, 73, 76, 77, 74, 75, 90, 91, 94, 95, 92, 93, 84, 85, 88, 89, 86, 87, 78, 79, 82, 83, 80, 81, 48, 49, 52, 53, 50, 51, 66, 67, 70, 71, 68
Offset: 0

Views

Author

Paul Tek, May 20 2013

Keywords

Comments

Analogous to A004488 or A048647 for the factorial base.
A self-inverse permutation of the natural numbers.
From Antti Karttunen, Aug 16-29 2016: (Start)
Consider the following way to view a factorial base representation of nonnegative integer n. For each nonzero digit d_i present in the factorial base representation of n (where i is the radix = 2.. = one more than 1-based position from the right), we place a pebble to the level (height) d_i at the corresponding column i of the triangular diagram like below, while for any zeros the corresponding columns are left empty:
.
Level
6 o
─ ─
5 . .
─ ─ ─
4 . . .
─ ─ ─ ─
3 . . . .
─ ─ ─ ─ ─
2 . . o . .
─ ─ ─ ─ ─ ─
1 . o . . o o
─ ─ ─ ─ ─ ─ ─
Radix: 7 6 5 4 3 2
Digits: 6 1 2 0 1 1 = A007623(4491)
Instead of levels, we can observe on which "slope" each pebble (nonzero digit) is located at. Formally, the slope of nonzero digit d_i with radix i is (i - d_i). Thus in above example, both the most significant digit (6) and the least significant 1 are on slope 1 (called "maximal slope", because it contains digits that are maximal allowed in those positions), while the second 1 from the right is on slope 2 ("submaximal slope").
This involution (A225901) sends each nonzero digit at level k to the slope k (and vice versa) by flipping such a diagram by the shallow diagonal axis that originates from the bottom right corner. Thus, from above diagram we obtain:
Slope (= digit's radix - digit's value)
1
2 .
3 . .╲
4 . .╲o╲
5 . .╲.╲.╲
6 . .╲.╲o╲.╲
. .╲.╲.╲.╲o╲
o╲.╲.╲.╲.╲o╲
-----------------
1 5 3 0 2 1 = A007623(1397)
and indeed, a(4491) = 1397 and a(1397) = 4491.
Thus this permutation maps between polynomial encodings A275734 & A275735 and all the respective sequences obtained from them, where the former set of sequences are concerned with the "slopes" and the latter set with the "levels" of the factorial base representation. See the Crossrefs section.
Sequences A231716 and A275956 are closed with respect to this sequence, in other words, for all n, a(A231716(n)) is a term of A231716 and a(A275956(n)) is a term of A275956.
(End)

Examples

			a(1000) = a(1*6! + 2*5! + 1*4! + 2*3! + 2*2!) = (7-1)*6! + (6-2)*5! + (5-1)*4! + (4-2)*3! + (3-2)*2! = 4910.
a(1397) = a(1*6! + 5*5! + 3*4! + 0*3! + 2*2! + 1*1!) = (7-1)*6! + (6-5)*5! + (5-3)*4! + (3-2)*2! + (2-1)*1! = 4491.
		

Crossrefs

Cf. A275959 (fixed points), A231716, A275956.
This involution maps between the following sequences related to "levels" and "slopes" (see comments): A275806 <--> A060502, A257511 <--> A260736, A264990 <--> A275811, A275729 <--> A275728, A275948 <--> A275946, A275949 <--> A275947, A275964 <--> A275962, A059590 <--> A276091.

Programs

  • Mathematica
    b = MixedRadix[Reverse@ Range[2, 12]]; Table[FromDigits[Map[Boole[# > 0] &, #] (Reverse@ Range[2, Length@ # + 1] - #), b] &@ IntegerDigits[n, b], {n, 0, 82}] (* Version 10.2, or *)
    f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], Times @@ Range[# - i]]], {i, 0, #}] &@ NestWhile[# + 1 &, 0, Times @@ Range[# + 1] <= n &]; Most@ Rest[a][[All, 1]] /. {} -> {0}]; g[w_List] := Total[Times @@@ Transpose@ {Map[Times @@ # &, Range@ Range[0, Length@ w]], Reverse@ Append[w, 0]}]; Table[g[Map[Boole[# > 0] &, #] (Reverse@ Range[2, Length@ # + 1] - #)] &@ f@ n, {n, 0, 82}] (* Michael De Vlieger, Aug 29 2016 *)
  • PARI
    a(n)=my(s=0,d,k=2);while(n,d=n%k;n=n\k;if(d,s=s+(k-d)*(k-1)!);k=k+1);return(s)
    
  • Python
    from sympy import factorial as f
    def a(n):
        s=0
        k=2
        while(n):
            d=n%k
            n=(n//k)
            if d: s=s+(k - d)*f(k - 1)
            k+=1
        return s
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017
  • Scheme
    (define (A225901 n) (let loop ((n n) (z 0) (m 2) (f 1)) (cond ((zero? n) z) (else (loop (quotient n m) (if (zero? (modulo n m)) z (+ z (* f (- m (modulo n m))))) (+ 1 m) (* f m))))))
    ;; One implementing the first recurrence, with memoization-macro definec:
    (definec (A225901 n) (if (zero? n) n (+ (A276091 (A275736 n)) (A153880 (A225901 (A257684 n))))))
    ;; Antti Karttunen, Aug 29 2016
    

Formula

From Antti Karttunen, Aug 29 2016: (Start)
a(0) = 0; for n >= 1, a(n) = A276091(A275736(n)) + A153880(a(A257684(n))).
or, for n >= 1, a(n) = A276149(n) + a(A257687(n)).
(End)
Other identities. For n >= 0:
a(n!) = A001563(n).
a(n!-1) = A007489(n-1).
From Antti Karttunen, Aug 16 2016: (Start)
A275734(a(n)) = A275735(n) and vice versa, A275735(a(n)) = A275734(n).
A060130(a(n)) = A060130(n). [The flip preserves the number of nonzero digits.]
A153880(n) = a(A255411(a(n))) and A255411(n) = a(A153880(a(n))). [This involution conjugates between the two fundamental factorial base shifts.]
a(n) = A257684(a(A153880(n))) = A266193(a(A255411(n))). [Follows from above.]
A276011(n) = A273662(a(A273670(n))).
A276012(n) = A273663(a(A256450(n))).
(End)

A060130 Number of nonzero digits in factorial base representation (A007623) of n; minimum number of transpositions needed to compose each permutation in the lists A060117 & A060118.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Mar 02 2001

Keywords

Examples

			19 = 3*(3!) + 0*(2!) + 1*(1!), thus it is written as "301" in factorial base (A007623). The count of nonzero digits in that representation is 2, so a(19) = 2.
		

Crossrefs

Cf. A227130 (positions of even terms), A227132 (of odd terms).
The topmost row and the leftmost column in array A230415, the left edge of triangle A230417.
Differs from similar A267263 for the first time at n=30.

Programs

  • Maple
    A060130(n) = count_nonfixed(convert(PermUnrank3R(n), 'disjcyc'))-nops(convert(PermUnrank3R(n), 'disjcyc')) or nops(fac_base(n))-nops(positions(0, fac_base(n)))
    fac_base := n -> fac_base_aux(n, 2); fac_base_aux := proc(n, i) if(0 = n) then RETURN([]); else RETURN([op(fac_base_aux(floor(n/i), i+1)), (n mod i)]); fi; end;
    count_nonfixed := l -> convert(map(nops, l), `+`);
    positions := proc(e, ll) local a, k, l, m; l := ll; m := 1; a := []; while(member(e, l[m..nops(l)], 'k')) do a := [op(a), (k+m-1)]; m := k+m; od; RETURN(a); end;
    # For procedure PermUnrank3R see A060117
  • Mathematica
    Block[{nn = 105, r}, r = MixedRadix[Reverse@ Range[2, -1 + SelectFirst[Range@ 12, #! > nn &]]]; Array[Count[IntegerDigits[#, r], k_ /; k > 0] &, nn, 0]] (* Michael De Vlieger, Dec 30 2017 *)
  • Scheme
    (define (A060130 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (quotient n i) (+ 1 i) (+ s (if (zero? (remainder n i)) 0 1)))))))
    ;; Two other implementations, that use memoization-macro definec:
    (definec (A060130 n) (if (zero? n) n (+ 1 (A060130 (A257687 n)))))
    (definec (A060130 n) (if (zero? n) n (+ (A257511 n) (A060130 (A257684 n)))))
    ;; Antti Karttunen, Dec 30 2017

Formula

a(0) = 0; for n > 0, a(n) = 1 + a(A257687(n)).
a(0) = 0; for n > 0, a(n) = A257511(n) + a(A257684(n)).
a(n) = A060129(n) - A060128(n).
a(n) = A084558(n) - A257510(n).
a(n) = A275946(n) + A275962(n).
a(n) = A275948(n) + A275964(n).
a(n) = A055091(A060119(n)).
a(n) = A069010(A277012(n)) = A000120(A275727(n)).
a(n) = A001221(A275733(n)) = A001222(A275733(n)).
a(n) = A001222(A275734(n)) = A001222(A275735(n)) = A001221(A276076(n)).
a(n) = A046660(A275725(n)).
a(A225901(n)) = a(n).
A257511(n) <= a(n) <= A034968(n).
A275806(n) <= a(n).
a(A275804(n)) = A060502(A275804(n)). [A275804 gives all the positions where this coincides with A060502.]
a(A276091(n)) = A260736(A276091(n)). [A276091 gives all the positions where this coincides with A260736.]

Extensions

Example-section added, name edited, the old Maple-code moved away from the formula-section, and replaced with all the new formulas by Antti Karttunen, Dec 30 2017

A276076 Factorial base exp-function: digits in factorial base representation of n become the exponents of successive prime factors whose product a(n) is.

Original entry on oeis.org

1, 2, 3, 6, 9, 18, 5, 10, 15, 30, 45, 90, 25, 50, 75, 150, 225, 450, 125, 250, 375, 750, 1125, 2250, 7, 14, 21, 42, 63, 126, 35, 70, 105, 210, 315, 630, 175, 350, 525, 1050, 1575, 3150, 875, 1750, 2625, 5250, 7875, 15750, 49, 98, 147, 294, 441, 882, 245, 490, 735, 1470, 2205, 4410, 1225, 2450, 3675, 7350, 11025, 22050, 6125, 12250, 18375, 36750, 55125, 110250, 343
Offset: 0

Views

Author

Antti Karttunen, Aug 18 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 digit in one-based position k of the factorial base representation of n. See the examples.

Examples

			   n  A007623   polynomial     encoded as             a(n)
   -------------------------------------------------------
   0       0    0-polynomial   (empty product)        = 1
   1       1    1*x^0          prime(1)^1             = 2
   2      10    1*x^1          prime(2)^1             = 3
   3      11    1*x^1 + 1*x^0  prime(2) * prime(1)    = 6
   4      20    2*x^1          prime(2)^2             = 9
   5      21    2*x^1 + 1*x^0  prime(2)^2 * prime(1)  = 18
   6     100    1*x^2          prime(3)^1             = 5
   7     101    1*x^2 + 1*x^0  prime(3) * prime(1)    = 10
and:
  23     321  3*x^2 + 2*x + 1  prime(3)^3 * prime(2)^2 * prime(1)
                                      = 5^3 * 3^2 * 2 = 2250.
		

Crossrefs

Cf. A276075 (a left inverse).
Cf. A276078 (same terms in ascending order).
Cf. also A275733, A275734, A275735, A275725 for other such encodings of factorial base related polynomials, and A276086 for a primorial base analog.

Programs

  • Mathematica
    a[n_] := Module[{k = n, m = 2, r, p = 2, q = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, q *= p^r; p = NextPrime[p]; m++]; q]; Array[a, 100, 0] (* Amiram Eldar, Feb 07 2024 *)

Formula

a(0) = 1, for n >= 1, a(n) = A275733(n) * a(A276009(n)).
Or: for n >= 1, a(n) = a(A257687(n)) * A000040(A084558(n))^A099563(n).
Other identities.
For all n >= 0:
A276075(a(n)) = n.
A001221(a(n)) = A060130(n).
A001222(a(n)) = A034968(n).
A051903(a(n)) = A246359(n).
A048675(a(n)) = A276073(n).
A248663(a(n)) = A276074(n).
a(A007489(n)) = A002110(n).
a(A059590(n)) = A019565(n).
For all n >= 1:
a(A000142(n)) = A000040(n).
a(A033312(n)) = A076954(n-1).
From Antti Karttunen, Apr 18 2022: (Start)
a(n) = A276086(A351576(n)).
A276085(a(n)) = A351576(n)
A003557(a(n)) = A351577(n).
A003415(a(n)) = A351950(n).
A069359(a(n)) = A351951(n).
A083345(a(n)) = A342001(a(n)) = A351952(n).
A351945(a(n)) = A351954(n).
A181819(a(n)) = A275735(n).
(End)
lambda(a(n)) = A262725(n+1), where lambda is Liouville's function, A008836. - Antti Karttunen and Peter Munn, Aug 09 2024

Extensions

Name changed by Antti Karttunen, Apr 18 2022
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