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 1-4 of 4 results.

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

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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)

A275947 Number of distinct slopes with multiple nonzero digits in factorial base representation of n: a(n) = A056170(A275734(n)). (See comments for more exact definition).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 15 2016

Keywords

Comments

a(n) gives the number of distinct elements 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 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). Thus there are two slopes with more than one nonzero digit, and a(525) = 2.
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 distinct elements that occur multiple times are 0 and 1, thus a(525) = 2.
		

Crossrefs

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

Programs

Formula

a(n) = A056170(A275734(n)).
Other identities and observations. For all n >= 0.
a(n) = A275949(A225901(n)).
A060502(n) = A275946(n) + a(n).
a(n) <= A275962(n).

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).
Showing 1-4 of 4 results.