A265350 -id:A265350 - cp's OEIS frontend

A275735 Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

1, 2, 2, 4, 3, 6, 2, 4, 4, 8, 6, 12, 3, 6, 6, 12, 9, 18, 5, 10, 10, 20, 15, 30, 2, 4, 4, 8, 6, 12, 4, 8, 8, 16, 12, 24, 6, 12, 12, 24, 18, 36, 10, 20, 20, 40, 30, 60, 3, 6, 6, 12, 9, 18, 6, 12, 12, 24, 18, 36, 9, 18, 18, 36, 27, 54, 15, 30, 30, 60, 45, 90, 5, 10, 10, 20, 15, 30, 10, 20, 20, 40, 30, 60, 15, 30, 30, 60, 45, 90, 25, 50, 50, 100, 75

Offset: 0

Antti Karttunen, Aug 09 2016

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 times a nonzero digit k occurs in the factorial base representation of n. See the examples.

			For n = 0 whose factorial base representation (A007623) is also 0, there are no nonzero digits at all, thus there cannot be any prime present in the encoding, and a(0) = 1.
For n = 1 there is just one 1, thus a(1) = prime(1) = 2.
For n = 2 ("10"), there is just one 1-digit, thus a(2) = prime(1) = 2.
For n = 3 ("11") there are two 1-digits, thus a(3) = prime(1)^2 = 4.
For n = 18 ("300") there is just one 3, thus a(18) = prime(3) = 5.
For n = 19 ("301") there is one 1 and one 3, thus a(19) = prime(1)*prime(3) = 2*5 = 10.
For n = 141 ("10311") there are three 1's and one 3, thus a(141) = prime(1)^3 * prime(3) = 2^3 * 5^1 = 40.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A000079, A001221, A001222, A003961, A007623, A008683, A181819, A225901, A257511, A257684, A265349, A265350, A264990, A275729, A275806, A351954.
Cf. also A275725, A275733, A275734 for other such prime factorization encodings of A060117/A060118-related polynomials, and also A276076.
Differs from A227154 for the first time at n=18, where a(18) = 5, while A227154(18) = 4.

A276076(n) = { my(i=0,m=1,f=1,nextf); while((n>0),i=i+1; nextf = (i+1)*f; if((n%nextf),m*=(prime(i)^((n%nextf)/f));n-=(n%nextf));f=nextf); m; };
A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2])));
A275735(n) = A181819(A276076(n)); \\ Antti Karttunen, Apr 03 2022

from sympy import prime
from operator import mul
import collections
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 19 2017

a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).
Other identities and observations. For all n >= 0:
a(n) = A275734(A225901(n)).
A001221(a(n)) = A275806(n).
A001222(a(n)) = A060130(n).
A048675(a(n)) = A275729(n).
A051903(a(n)) = A264990(n).
A008683(a(A265349(n))) = -1 or +1 for all n >= 0.
A008683(a(A265350(n))) = 0 for all n >= 1.
From Antti Karttunen, Apr 03 2022: (Start)
A342001(a(n)) = A351954(n).
a(n) = A181819(A276076(n)). (End)

A264990 a(n) = number of occurrences of a most frequent nonzero digit in factorial base representation (A007623) of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 2, 3, 1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2

Offset: 0

Antti Karttunen, Dec 22 2015

			   n  A007623(n)   a(n) [highest number of times any nonzero digit occurs].
   0 =   0           0 (because no nonzero digits present)
   1 =   1           1
   2 =  10           1
   3 =  11           2
   4 =  20           1
   5 =  21           1
   6 = 100           1
   7 = 101           2
   8 = 110           2
   9 = 111           3
  10 = 120           1
  11 = 121           2
  12 = 200           1
  13 = 201           1
  14 = 210           1
  15 = 211           2
  16 = 220           2
  17 = 221           2
  18 = 300           1
and for n=63 we have:
  63 = 2211          2.

Antti Karttunen, Table of n, a(n) for n = 0..10080
Index entries for sequences related to factorial base representation.

Cf. A007623, A051903, A225901, A257511, A257684, A275735, A275811.
Cf. A265349 (positions of terms <= 1), A265350 (positions of term > 1).
Cf. also A266117, A266118.

a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Max[Tally[Select[s, # > 0 &]][[;;,2]]]]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jan 24 2024 *)

from sympy import prime, factorint
from operator import mul
import collections
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 20 2017

a(0) = 0; for n >= 1, a(n) = max(A257511(n), a(A257684(n))).
Other identities. For all n >= 0:
From Antti Karttunen, Aug 15 2016: (Start)
a(n) = A275811(A225901(n)).
a(n) = A051903(A275735(n)).
(End)

Name changed by Antti Karttunen, Aug 15 2016

A265349 Numbers in whose factorial base representation (A007623) no digit > 0 occurs more than once.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 10, 12, 13, 14, 18, 19, 20, 22, 23, 24, 28, 36, 42, 46, 48, 49, 50, 54, 66, 67, 68, 72, 73, 74, 76, 77, 78, 82, 84, 85, 86, 96, 97, 98, 100, 101, 102, 106, 108, 109, 110, 114, 115, 116, 118, 119, 120, 124, 132, 138, 142, 168, 186, 192, 196, 204, 216, 220, 228, 234, 238, 240, 241, 242, 246, 258, 259, 260

Offset: 0

Antti Karttunen, Dec 22 2015

Zero is a special case in this sequence, thus a(0) = 0, and the indexing of natural numbers >= 1 present starts from a(1) = 1.

After a(0), positions of ones in A264990.

			23 (in factorial base "321") is present, because none of the digits (which all are nonzero) occurs twice.
48 (in factorial base "2000") is present, because the only nonzero digit, "2", occurs only once.
259 (in factorial base "20301") is present, because none of the nonzero digits occurs more than once.

Antti Karttunen, Table of n, a(n) for n = 0..10080
Indranil Ghosh, Python program for computing this sequence.
Index entries for sequences related to factorial base representation.

Cf. A007623, A264990.
Cf. A265350 (complement).
Cf. A000142 (a subsequence).
Cf. also A266117.

q[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Max[Tally[Select[s, # > 0 &]][[;;,2]]] == 1]; q[0] = True; Select[Range[0, 260], q] (* Amiram Eldar, Jan 24 2024 *)

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

Antti Karttunen, Aug 15 2016

			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.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Indranil Ghosh, Python program for computing this sequence.
Index entries for sequences related to factorial base representation.

Cf. A275735, A275812.
Cf. A265349 (indices of zeros), A265350 (of terms > 0).
Cf. also A060130, A225901, A275948, A275949.

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 *)

(define (A275964 n) (A275812 (A275735 n)))

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).

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

Antti Karttunen, Aug 15 2016

			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.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation.

Cf. A056170, A275735.
Cf. A265349 (indices of zeros), A265350 (of terms > 0).
Cf. also A007623, A225901, A275806, A275947, A275948, A275964.

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 *)

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 nIndranil Ghosh, Jun 20 2017

(define (A275949 n) (A056170 (A275735 n)))

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).

cp's OEIS Frontend

A275735 Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

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A264990 a(n) = number of occurrences of a most frequent nonzero digit in factorial base representation (A007623) of n.

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A265349 Numbers in whose factorial base representation (A007623) no digit > 0 occurs more than once.

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

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

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