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.

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A265354 Permutation of nonnegative integers: a(n) = A263273(A264985(n)).

Original entry on oeis.org

0, 1, 3, 2, 4, 9, 6, 10, 12, 7, 5, 19, 8, 13, 27, 18, 28, 36, 21, 11, 57, 24, 37, 30, 15, 31, 39, 22, 16, 64, 23, 14, 55, 20, 46, 58, 25, 17, 73, 26, 40, 81, 54, 82, 108, 63, 29, 171, 72, 109, 90, 45, 85, 117, 66, 34, 192, 69, 38, 165, 60, 100, 174, 75, 35, 219, 78, 118, 84, 33, 91, 93, 48, 32, 138, 51, 112, 111, 42, 94, 120, 67
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Composition of A263273 with the permutation obtained from its odd bisection.

Crossrefs

Inverse: A265353.
Cf. also A265352, A265355, A265356.

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a264985(n): return (a263273(2*n + 1) - 1)/2
    def a(n): return a263273(a264985(n)) # Indranil Ghosh, May 22 2017
  • Scheme
    (define (A265354 n) (A263273 (A264985 n)))
    

Formula

a(n) = A263273(A264985(n)).

A242603 Largest divisor of n not divisible by 7. Remove factors 7 from n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 1, 8, 9, 10, 11, 12, 13, 2, 15, 16, 17, 18, 19, 20, 3, 22, 23, 24, 25, 26, 27, 4, 29, 30, 31, 32, 33, 34, 5, 36, 37, 38, 39, 40, 41, 6, 43, 44, 45, 46, 47, 48, 1, 50, 51, 52, 53, 54, 55, 8, 57, 58, 59, 60, 61, 62, 9, 64, 65, 66, 67, 68, 69, 10, 71, 72, 73, 74, 75, 76, 11
Offset: 1

Views

Author

Wolfdieter Lang, Jun 18 2014

Keywords

Comments

This is member p = 7 in the p-family of sequences (p a prime).
See A000265, A038502 and A132739 for primes 2, 3 and 5, also for formulas, programs and references.
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 21 2019

Examples

			From _Indranil Ghosh_, Jan 31 2017: (Start)
For n = 12, the divisors of 12 are 1,2,3,4,6 and 12. The largest divisor not divisible by 7 is 12. So, a(12) = 12.
For n = 14, the divisors of 14 are 1,2,7 and 14. The largest divisor not divisible by 7 is 2. So, a(14) = 2. (End)
From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - (6*7)*G(x^7) - (6*49)*G(x^49) - (6*343)*G(x^343) - ..., where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (6/7)*H(x^7) - (6/49)*H(x^49) - (6/343)*H(x^343) - ..., where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (6/7^2)*L(x^7) - (6/49^2)*L(x^49) - (6/343^2)*L(x^343) - ..., where L(x) = Log(1/(1 - x)).
Also, Sum_{n >= 1} (1/a(n))*x^n = L(x) + (6/7)*L(x^7) + (6/7)*L(x^49) + (6/7)*L(x^343) ... . (End)
		

Crossrefs

Programs

  • Mathematica
    Table[n/7^IntegerExponent[n, 7], {n, 80}] (* Alonso del Arte, Jun 18 2014 *)
  • PARI
    a(n) = f = factor(n);  for (i=1, #f~, if (f[i,1]==7, f[i, 1]=1)); factorback(f); \\ Michel Marcus, Jun 18 2014
    
  • PARI
    a(n) = n \ 7^valuation(n, 7) \\ David A. Corneth, Feb 21 2019
    
  • Python
    def A242603(n):
        for i in range(n,0,-1):
            if n%i==0 and i%7!=0:
                return i # Indranil Ghosh, Jan 31 2017

Formula

Multiplicative with a(p^e) = 1 if p = 7, else p^e.
Dirichlet g.f.: zeta(s-1)*7*(7^(s-1) - 1)/(7^s - 1).
a(n) = n/A268354(n).
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,7^n).
O.g.f.: F(x) - 6*F(x^7) - 6*F(x^49) - 6*F(x^243) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers. More generally, for m >= 1,
Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) - (7^m - 1)( F(m,x^7) + F(m,x^49) + F(m,x^243) + ...), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m_th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence a(n) produces generating functions for the sequences (n^m*a(n))n>=1, m in Z. Some examples are given below. (End)
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

A265353 Permutation of nonnegative integers: a(n) = A264985(A263273(n)).

Original entry on oeis.org

0, 1, 3, 2, 4, 10, 6, 9, 12, 5, 7, 19, 8, 13, 31, 24, 28, 37, 15, 11, 33, 18, 27, 30, 21, 36, 39, 14, 16, 46, 23, 25, 73, 69, 55, 64, 17, 22, 58, 26, 40, 94, 78, 85, 112, 51, 34, 100, 72, 82, 91, 75, 109, 118, 42, 32, 96, 20, 35, 105, 60, 99, 102, 45, 29, 87, 54, 81, 84, 57, 90, 93, 48, 38, 114, 63, 108, 111, 66, 117, 120, 41
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Composition of A263273 with the permutation obtained from its odd bisection.

Crossrefs

Inverse: A265354.

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a264985(n): return (a263273(2*n + 1) - 1)/2
    def a(n): return a264985(a263273(n)) # Indranil Ghosh, May 22 2017
  • Scheme
    (define (A265353 n) (A264985 (A263273 n)))
    

Formula

a(n) = A264985(A263273(n)).

A343430 Part of n composed of prime factors of the form 3k-1.

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 1, 8, 1, 10, 11, 4, 1, 2, 5, 16, 17, 2, 1, 20, 1, 22, 23, 8, 25, 2, 1, 4, 29, 10, 1, 32, 11, 34, 5, 4, 1, 2, 1, 40, 41, 2, 1, 44, 5, 46, 47, 16, 1, 50, 17, 4, 53, 2, 55, 8, 1, 58, 59, 20, 1, 2, 1, 64, 5, 22, 1, 68, 23, 10, 71, 8, 1, 2, 25, 4, 11, 2, 1, 80, 1, 82, 83, 4, 85
Offset: 1

Views

Author

Peter Munn, Jun 08 2021

Keywords

Comments

Largest term of A004612 that divides n.
Modulo 6, the prime numbers are partitioned into 4 nonempty sets: {2}, {3}, primes of the form 6k-1 (A007528) and primes of the form 6k+1 (A002476). The modulo 3 partition is nearly the same, but unites the only even prime, 2, with primes of the form 6k-1 in the set of primes we use here.
A positive integer m is a Loeschian number (a term of A003136) if and only if a(A007913(m)) = 1, that is the squarefree part of m has no prime factors of the form 3k-1.

Examples

			n = 60 has prime factorization 60 = 2 * 2 * 3 * 5. Factors 2 = 3*1 - 1 and 5 = 3*2 - 1 have form 3k-1, whereas 3 does not (having form 3k). Multiplying the factors of form 3k-1, we get 2 * 2 * 5 = 20. So a(60) = 20.
		

Crossrefs

Equivalent sequences for prime factors of other forms: A006519 (2 only), A000265 (2k+1), A038500 (3 only), A038502 (3k+/-1), A170818 (4k+1), A097706 (4k-1), A248909 (6k+1), A343431 (6k-1).
Range of terms: A004612 (closure under multiplication of A003627).
Cf. A002476, A007528, squarefree part (A007913) of terms of A003136.
First 28 terms are the same as A247503.

Programs

  • Mathematica
    f[p_, e_] := If[Mod[p, 3] == 2, p^e, 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jun 11 2021 *)
  • PARI
    a(n) = {my(f = factor(n)); for (i=1, #f~, if ((f[i, 1] + 1) % 3, f[i, 1] = 1); ); factorback(f); } \\ after Michel Marcus at A248909
    
  • Python
    from math import prod
    from sympy import factorint
    def A343430(n): return prod(p**e for p, e in factorint(n).items() if p%3==2) # Chai Wah Wu, Dec 23 2022

Formula

Completely multiplicative with a(p) = p if p is of the form 3k-1, otherwise a(p) = 1.
For k >= 1, a(n) = a(k*n) / gcd(k, a(k*n)).
a(n) = A006519(n) * A343431(n).
a(n) = (n / A038500(n)) / A248909(n) = A038502(n) / A248909(n).
A006519(a(n)) = a(A006519(n)) = A006519(n).
A343431(a(n)) = a(A343431(n)) = A343431(n).
A038500(a(n)) = a(A038500(n)) = 1.
A248909(a(n)) = a(A248909(n)) = 1.

A067622 Consider the power series (x + 1)^(1/3) = 1 + x/3-x^2/9 + 5x^3/81 + ...; sequence gives numerators of coefficients.

Original entry on oeis.org

1, 1, -1, 5, -10, 22, -154, 374, -935, 21505, -55913, 147407, -1179256, 3174920, -8617640, 70664648, -194327782, 537259162, -13431479050, 37466757350, -104906920580, 884215473460, -2491879970660, 7042269482300, -59859290599550
Offset: 0

Views

Author

Benoit Cloitre, Feb 02 2002

Keywords

Comments

a(n) is also the numerator of the binomial coefficient C(k,n) evaluated at k=1/3, e.g. a(4) = (1/24)k(k-1)(k-2)(k-3), plug in k=1/3 and take numerator. - James R. Buddenhagen, Aug 16 2014

Crossrefs

Denominators are A067623.

Programs

  • Maple
    s := convert(taylor((x+1)^(1/3), x, 50), polynom): for n from 0 to 50 do printf(`%a,`, abs(numer(coeff(s, x, n)))) od;
    seq(numer(subs(k=1/3,expand(binomial(k,n)))),n=0..50) # James R. Buddenhagen, Aug 16 2014

Formula

a(n) =(-1)^n*A004990(n)*A067623(n)/A000244(n); ignoring signs, a(n) =A038502(A004990(n)) =A038502(A034164(n-2)). a(n)'s sign is (-1)^(n+1) if n>0.

Extensions

Edited by Henry Bottomley and James Sellers, Feb 11 2002

A265904 Self-inverse permutation of nonnegative integers: a(n) = A263272(A263273(A263272(n))).

Original entry on oeis.org

0, 1, 2, 3, 4, 11, 6, 29, 8, 9, 10, 5, 12, 13, 38, 33, 92, 17, 18, 83, 20, 87, 110, 35, 24, 89, 26, 27, 28, 7, 30, 37, 32, 15, 86, 23, 36, 31, 14, 39, 40, 119, 114, 281, 44, 99, 254, 65, 276, 335, 98, 51, 260, 71, 54, 245, 56, 249, 326, 101, 60, 263, 74, 261, 272, 47, 330, 353, 116, 105, 278, 53, 72, 251, 62, 267, 332, 107, 78, 269, 80, 81, 82, 19
Offset: 0

Views

Author

Antti Karttunen, Jan 02 2016

Keywords

Comments

A263273 conjugated with the permutation obtained from its even bisection.

Crossrefs

Cf. also A265902.
Cf. A265369, A266190, A266401, A266403 (other conjugates or similar derivations of A263273).

Programs

  • Mathematica
    f[n_] := Block[{g, h}, g[x_] := x/3^IntegerExponent[x, 3]; h[x_] := x/g@ x; If[n == 0, 0, FromDigits[Reverse@ IntegerDigits[#, 3], 3] &@g[n] h[n]]]; t = Table[f[2 n]/2, {n, 0, 1000}]; Table[t[[f[t[[n + 1]]] + 1]], {n, 0, 83}] (* Michael De Vlieger, Jan 04 2016, after Jean-François Alcover at A263273 *)
  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a263272(n): return a263273(2*n)/2
    def a(n): return a263272(a263273(a263272(n))) # Indranil Ghosh, May 25 2017
  • Scheme
    (define (A265904 n) (A263272 (A263273 (A263272 n))))
    

Formula

a(n) = A263272(A263273(A263272(n))).
As a composition of related permutations:
a(n) = A263272(A265352(n)).
a(n) = A265351(A263272(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]

A244414 Remove highest power of 6 from n.

Original entry on oeis.org

1, 2, 3, 4, 5, 1, 7, 8, 9, 10, 11, 2, 13, 14, 15, 16, 17, 3, 19, 20, 21, 22, 23, 4, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 1, 37, 38, 39, 40, 41, 7, 43, 44, 45, 46, 47, 8, 49, 50, 51, 52, 53, 9, 55, 56, 57, 58, 59, 10, 61, 62, 63, 64, 65, 11
Offset: 1

Views

Author

Wolfdieter Lang, Jun 27 2014

Keywords

Comments

This is instance g = 6 of the g-family of sequences, call it r(g,n), where for g >= 2 the highest power of g is removed from n. See the crossrefs.
The present sequence is not multiplicative: a(6) = 1 not a(2)*a(3) = 6. In the prime factor decomposition one has to consider a(2^e2*3^e^3) as one entity, also for e2 >= 0, e3 >= 0 with a(1) = 1, and apply the rule given in the formula section. With this rule the sequence will be multiplicative in an unusual sense. - Wolfdieter Lang, Feb 12 2018

Examples

			a(1) = 1 = 1/6^A122841(1) = 1/6^0.
a(9) = a(2^0*3^2), min(0,2) = 0, a(9) = 2^(0-0)*3^(2-0) = 1*9 = 9.
a(12) = a(2^2*3^1), m = min(2,1) = 1, a(12) = 2^(2-1)*3^(1-1) = 2^1*1 = 2.
a(30) = a(2*3*5) = a(2^1*3^1)*a(5) = 1*a(5) = 5.
		

Crossrefs

A007310, A007913, A008833 are used to express relationship between terms of this sequence.

Programs

  • Mathematica
    a[n_] := n/6^IntegerExponent[n, 6]; Array[a, 66] (* Robert G. Wilson v, Feb 12 2018 *)
  • PARI
    a(n) = n/6^valuation(n,6); \\ Joerg Arndt, Jun 28 2014

Formula

a(n) = n/6^A122841(n), n >= 1.
For n >= 2, a(n) is sort of multiplicative if a(2^e2*3^e3) = 2^(e2 - m)*3^(e3 - m) with m = m(e2, e3) = min(e2, e3), for e2, e3 >= 0, a(1) = 1, and a(p^e) = p^e for primes p >= 5.
From Peter Munn, Jun 04 2020: (Start)
Proximity to being multiplicative may be expressed as follows:
a(n * A007310(k)) = a(n) * a(A007310(k));
a(n^2) = a(n)^2;
a(n) = a(A007913(n)) * a(A008833(n)).
(End)
Sum_{k=1..n} a(k) ~ (3/7) * n^2. - Amiram Eldar, Nov 20 2022

Extensions

Incorrect multiplicity claim corrected by Wolfdieter Lang, Feb 12 2018

A264984 Even bisection of A263273; terms of A263262 doubled.

Original entry on oeis.org

0, 2, 4, 6, 8, 10, 12, 22, 16, 18, 20, 14, 24, 26, 28, 30, 64, 46, 36, 58, 40, 66, 76, 34, 48, 70, 52, 54, 56, 38, 60, 74, 32, 42, 68, 50, 72, 62, 44, 78, 80, 82, 84, 190, 136, 90, 172, 118, 192, 226, 100, 138, 208, 154, 108, 166, 112, 174, 220, 94, 120, 202, 148, 198, 184, 130
Offset: 0

Views

Author

Antti Karttunen, Dec 05 2015

Keywords

Crossrefs

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a(n): return a263273(2*n) # Indranil Ghosh, May 22 2017
  • Scheme
    (define (A264984 n) (A263273 (+ n n)))
    

Formula

a(n) = 2 * A263272(n).
a(n) = A263273(2*n).
Other identities. For all n >= 0:
A010873(a(n)) = 2 * A000035(n) = A010673(n).

A265342 Permutation of even numbers: a(n) = 2 * A265351(n).

Original entry on oeis.org

0, 2, 4, 6, 8, 22, 12, 10, 16, 18, 20, 58, 24, 26, 76, 66, 64, 70, 36, 14, 40, 30, 28, 34, 48, 46, 52, 54, 56, 166, 60, 62, 184, 174, 172, 178, 72, 74, 220, 78, 80, 238, 228, 226, 232, 198, 68, 202, 192, 190, 196, 210, 208, 214, 108, 38, 112, 42, 44, 130, 120, 118, 124, 90, 32, 94, 84, 82, 88, 102, 100, 106, 144
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Iterating this sequence as 1, a(1), a(a(1)), a(a(a(1))), ... yields A264980.

Crossrefs

Cf. A265351.
Cf. also A265341, A263273, A264980.

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a263272(n): return a263273(2*n)/2
    def a(n): return 2*a263272(a263273(n)) # Indranil Ghosh, May 25 2017
  • Scheme
    (define (A265342 n) (* 2 (A265351 n)))
    

Formula

a(n) = 2 * A265351(n).

A348931 Numbers k congruent to 1 or 5 mod 6, for which A348930(k) < k.

Original entry on oeis.org

5, 11, 17, 23, 29, 35, 41, 47, 49, 53, 55, 59, 65, 71, 77, 83, 85, 89, 95, 101, 107, 113, 115, 119, 125, 131, 137, 143, 145, 149, 155, 161, 167, 169, 173, 179, 185, 187, 191, 197, 203, 205, 209, 215, 221, 227, 233, 235, 239, 245, 251, 253, 257, 263, 265, 269, 275, 281, 287, 293, 295, 299, 305, 311, 317, 319, 323
Offset: 1

Views

Author

Antti Karttunen, Nov 04 2021

Keywords

Comments

See comments in A348930.

Crossrefs

Programs

  • Mathematica
    s[n_] := n / 3^IntegerExponent[n, 3]; Select[Range[350], MemberQ[{1, 5}, Mod[#, 6]] && s[DivisorSigma[1, #]] < # &] (* Amiram Eldar, Nov 04 2021 *)
  • PARI
    A038502(n) = (n/3^valuation(n, 3));
    A348930(n) = A038502(sigma(n));
    isA348931(n) = ((n%2)&&(n%3)&&(A348930(n)
    				
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