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 34 results. Next

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).

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

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 09 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 times a nonzero digit k occurs in the factorial base representation of n. See the examples.

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.
		

Crossrefs

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.

Programs

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

Formula

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)

A260736 a(0) = 0; for n >= 1, a(n) = A000035(n) + a(A257684(n)); in the factorial representation of n the number of digits with maximal possible value allowed in its location.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 27 2015

Keywords

Comments

In the factorial representation of n, given as {d_k, ..., d_3, d_2, d_1}, the maximal allowed digit for any position j is j. This sequence gives the number of digits in the whole representation [A007623(n)] that attain that maximum allowed value.

Examples

			For n=19, which has factorial representation "301", the digits at position 1 and 3, namely "1" and "3" are equal to their one-based position index, in other words, the maximal digits allowed in those positions (while "0" at position 2 is not), thus a(19) = 2.
		

Crossrefs

Cf. also A257511.

Programs

  • Mathematica
    a[n_] := Module[{k = n, m = 2, c = 0, r}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r == m - 1, c++]; m++]; c]; Array[a, 100, 0] (* Amiram Eldar, Jan 23 2024 *)
  • Python
    from sympy import 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 0 if n==0 else n%2 + a(a257684(n)) print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 20 2017

Formula

a(0) = 0; for n >= 1, a(n) = A000035(n) + a(A257684(n)).
Other identities. For all n >= 1:
a(n!-1) = n-1. [n!-1 also gives the first position where n-1 occurs.]

A275733 a(0) = 1; for n >= 1, a(n) = A275732(n) * A003961(a(A257684(n))).

Original entry on oeis.org

1, 2, 3, 6, 3, 6, 5, 10, 15, 30, 15, 30, 5, 10, 15, 30, 15, 30, 5, 10, 15, 30, 15, 30, 7, 14, 21, 42, 21, 42, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 7, 14, 21, 42, 21, 42, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 7, 14, 21, 42, 21, 42
Offset: 0

Views

Author

Antti Karttunen, Aug 08 2016

Keywords

Comments

a(n) = product of primes whose indices are positions of nonzero-digits in factorial base representation of n (see A007623). Here positions are one-based, so that the least significant digit is the position 1, the next least significant the position 2, etc.

Examples

			For n=19, A007623(19) = 301, thus a(19) = prime(3)*prime(1) = 5*2 = 10.
For n=52, A007623(52) = 2020, thus a(52) = prime(2)*prime(4) = 3*7 = 21.
		

Crossrefs

Subsequence of A005117.
Cf. A275727.
Cf. also A275725, A275734, A275735 for other such prime factorization encodings of A060117/A060118-related polynomials.

Formula

a(0) = 1; for n >= 1, a(n) = A275732(n) * A003961(a(A257684(n))).
Other identities and observations. For all n >= 0:
a(A007489(n)) = A002110(n).
A001221(a(n)) = A001222(a(n)) = A060130(n).
A048675(a(n)) = A275727(n).
A061395(a(n)) = A084558(n).

A275727 a(0) = 0, for n >= 1, a(n) = A275736(n) + 2*a(A257684(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 09 2016

Keywords

Comments

a(n) has ones in those positions of its base-2 representation where n has nonzero digits in its factorial base representation.

Examples

			For n=19, A007623(19) = 301, thus a(19) = 5 because A007088(5) = 101.
		

Crossrefs

Formula

a(0) = 0, for n >= 1, a(n) = A275736(n) + 2*a(A257684(n)).
a(n) = A048675(A275733(n)).
Other identities and observations. For all n >= 0:
A000120(a(n)) = A060130(n).

A257694 a(0) = 1; for n >= 1, a(n) = A060130(n) * a(A257684(n)).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 4, 6, 1, 2, 2, 3, 4, 6, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 6, 8, 2, 3, 3, 4, 6, 8, 1, 2, 2, 3, 4, 6, 2, 3, 3, 4, 6, 8, 4, 6, 6, 8, 9, 12, 4, 6, 6, 8, 9, 12, 1, 2, 2, 3, 4, 6, 2, 3, 3, 4, 6, 8, 4, 6, 6, 8, 9, 12, 8, 12, 12, 16, 18, 24, 1, 2, 2, 3, 4, 6, 2, 3, 3, 4, 6, 8, 4, 6, 6, 8, 9, 12, 8, 12, 12, 16, 18, 24, 1
Offset: 0

Views

Author

Antti Karttunen, May 05 2015

Keywords

Crossrefs

Formula

a(0) = 1; for n >= 1, a(n) = A060130(n) * a(A257684(n)).
Other identities:
For all n >= 1, a(A033312(n)) = A000142(n-1).

A275808 a(0) = 0, for n >= 1, a(n) = A275736(n) XOR a(A257684(n)), where XOR is given by A003987.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 09 2016

Keywords

Crossrefs

Cf. A275809 (positions of zeros), A275810 (and their first differences).
Cf. also A275728.

Formula

a(0) = 0, for n >= 1, a(n) = A275736(n) XOR a(A257684(n)), where 2-argument function XOR is given by A003987.
a(n) = A248663(A275734(n)).

A257695 a(0) = 1; for n >= 1, a(n) = lcm(A060130(n), a(A257684(n))).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 6, 1, 2, 2, 3, 2, 6, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 6, 4, 2, 3, 3, 4, 6, 4, 1, 2, 2, 3, 2, 6, 2, 3, 3, 4, 6, 4, 2, 6, 6, 4, 3, 12, 2, 6, 6, 4, 3, 12, 1, 2, 2, 3, 2, 6, 2, 3, 3, 4, 6, 4, 2, 6, 6, 4, 3, 12, 2, 6, 6, 4, 6, 12, 1, 2, 2, 3, 2, 6, 2, 3, 3, 4, 6, 4, 2, 6, 6, 4, 3, 12, 2, 6, 6, 4, 6, 12, 1
Offset: 0

Views

Author

Antti Karttunen, May 05 2015

Keywords

Crossrefs

Formula

a(0) = 1; for n >= 1, a(n) = lcm(A060130(n), a(A257684(n))).

A257696 a(0) = 0; for n >= 1, a(n) = gcd(A060130(n), a(A257684(n))).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, May 05 2015

Keywords

Crossrefs

Formula

a(0) = 0; for n >= 1, a(n) = gcd(A060130(n), a(A257684(n))).

A275728 a(0) = 0, for n >= 1, a(n) = A275736(n) + a(A257684(n)); a(n) = A048675(A275734(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 09 2016

Keywords

Comments

See graph: pine trees on a snowy mountain. - N. J. A. Sloane, Aug 12 2016

Crossrefs

Formula

a(0) = 0, for n >= 1, a(n) = A275736(n) + a(A257684(n)).
a(n) = A048675(A275734(n)).
Showing 1-10 of 34 results. Next