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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A266635 Permutation of natural numbers: a(n) = A126760(A264965(A007310(n))).

Original entry on oeis.org

1, 3, 2, 5, 7, 9, 6, 19, 4, 8, 13, 15, 23, 11, 27, 16, 25, 17, 22, 10, 21, 20, 49, 65, 12, 73, 41, 61, 37, 24, 29, 77, 14, 55, 47, 79, 31, 26, 43, 59, 18, 67, 35, 38, 81, 33, 62, 54, 57, 46, 78, 30, 70, 53, 74, 42, 69, 34, 163, 50, 58, 45, 66, 91, 71, 28, 60, 44, 76, 39, 68, 36, 63, 52, 217, 32, 64, 51, 80, 109
Offset: 1

Views

Author

Antti Karttunen, Jan 04 2016

Keywords

Comments

Permutation induced when A264965 is restricted to numbers neither divisible by 2 nor 3 (A007310).

Crossrefs

Programs

Formula

a(n) = A126760(A264965(A007310(n))).

A266636 Permutation of natural numbers: a(n) = A126760(A264966(A007310(n))).

Original entry on oeis.org

1, 3, 2, 9, 4, 7, 5, 10, 6, 20, 14, 25, 11, 33, 12, 16, 18, 41, 8, 22, 21, 19, 13, 30, 17, 38, 15, 66, 31, 52, 37, 76, 46, 58, 43, 72, 29, 44, 70, 82, 27, 56, 39, 68, 62, 50, 35, 84, 23, 60, 78, 74, 54, 48, 34, 85, 49, 61, 40, 67, 28, 47, 73, 77, 24, 63, 42, 71, 57, 53, 65, 83, 26, 55, 81, 69, 32, 51, 36, 79, 45
Offset: 1

Views

Author

Antti Karttunen, Jan 04 2016

Keywords

Comments

Permutation induced when A264966 is restricted to numbers neither divisible by 2 nor 3 (A007310).

Crossrefs

Programs

Formula

a(n) = A126760(A264966(A007310(n))).

A349391 Dirichlet convolution of A126760 with omega.

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 1, 3, 2, 5, 1, 7, 1, 6, 5, 4, 1, 7, 1, 9, 6, 7, 1, 10, 3, 8, 3, 11, 1, 16, 1, 5, 7, 9, 7, 12, 1, 10, 8, 13, 1, 20, 1, 13, 9, 11, 1, 13, 4, 18, 9, 15, 1, 10, 8, 16, 10, 13, 1, 27, 1, 14, 11, 6, 9, 24, 1, 17, 11, 32, 1, 17, 1, 16, 18, 19, 9, 28, 1, 17, 4, 17, 1, 34, 10, 18, 13, 19, 1, 27, 10, 21
Offset: 1

Views

Author

Antti Karttunen, Nov 15 2021

Keywords

Crossrefs

Cf. A347233, A347234, A349390, A349392, A349393, A349395 for other Dirichlet convolutions of A126760. And also A347957.

Programs

  • Mathematica
    f[n_] := 2 * Floor[(m = n/2^IntegerExponent[n, 2]/3^IntegerExponent[n, 3])/6] + Mod[m, 3]; a[n_] := DivisorSum[n, f[#] * PrimeNu[n/#] &]; Array[a, 100] (* Amiram Eldar, Nov 16 2021 *)
  • PARI
    A126760(n) = {n&&n\=3^valuation(n, 3)<A126760
    A349391(n) = sumdiv(n,d,A126760(n/d)*omega(d));

Formula

a(n) = Sum_{d|n} A126760(n/d) * A001221(d).

A254048 a(n) = A126760(A007494(n)).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 4, 1, 3, 2, 6, 1, 2, 3, 8, 1, 5, 1, 10, 2, 1, 4, 12, 1, 7, 5, 14, 3, 4, 2, 16, 1, 9, 6, 18, 1, 3, 7, 20, 2, 11, 3, 22, 4, 6, 8, 24, 1, 13, 9, 26, 5, 2, 1, 28, 3, 15, 10, 30, 2, 8, 11, 32, 1, 17, 4, 34, 6, 5, 12, 36, 1, 19, 13, 38, 7, 10, 5, 40, 2, 21, 14, 42, 3, 1, 15, 44, 4, 23, 2, 46, 8, 12, 16, 48, 1, 25, 17, 50, 9, 7, 6, 52, 5, 27, 18, 54, 1, 14, 19, 56, 3, 29, 7, 58, 10, 4, 20, 60, 2
Offset: 0

Views

Author

Antti Karttunen, Jan 28 2015

Keywords

Crossrefs

Programs

Formula

a(n) = A126760(A007494(n)).
Other identities:
a(4n) = A126760(n).
a(4n+1) = A126760(3n+1).
a(4n+2) = A126760(2n+1) = A253887(n+1).
a(4n+3) = 2n+2.
For all n >= 1, a(n) = A126760(A139391(n)). [Conjecture. The proof should be easy. Holds at least up to n = 2^25 = 33554432.]

A353420 a(n) = A126760(A003961(n)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 1, 9, 3, 5, 2, 6, 4, 12, 1, 7, 9, 8, 3, 19, 5, 10, 2, 17, 6, 42, 4, 11, 12, 13, 1, 22, 7, 26, 9, 14, 8, 29, 3, 15, 19, 16, 5, 59, 10, 18, 2, 41, 17, 32, 6, 20, 42, 31, 4, 39, 11, 21, 12, 23, 13, 92, 1, 40, 22, 24, 7, 49, 26, 25, 9, 27, 14, 82, 8, 48, 29, 28, 3, 209, 15, 30, 19, 45, 16, 52, 5, 33
Offset: 1

Views

Author

Antti Karttunen, Apr 20 2022

Keywords

Crossrefs

Cf. A353335 (Dirichlet inverse), A353336 (sum with it).

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A126760(n) = {n&&n\=3^valuation(n, 3)<A126760
    A353420(n) = A126760(A003961(n));

Formula

a(n) = A353336(4*n) = A353336(n) - A353335(n).
For all n >= 1, a(n) = a(2*n) = a(A000265(n)).
For all n >= 1, A249745(a(n)) = A003602(n).

A353460 Dirichlet convolution of A126760 with A349134 (the Dirichlet inverse of Kimberling's paraphrases).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Apr 20 2022

Keywords

Comments

Taking the Dirichlet convolution between this sequence and A349371 gives A349393, and similarly for many other such analogous pairs.

Crossrefs

Cf. A003602, A126760, A349134, A353461 (Dirichlet inverse), A353462 (sum with it).

Programs

Formula

a(n) = Sum_{d|n} A126760(d) * A349134(n/d).

A353461 Dirichlet convolution of A003602 (Kimberling's paraphrases) with A323881 (the Dirichlet inverse of A126760).

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 0, 3, 0, 2, 0, 2, 0, 3, 0, 3, 0, 3, 0, 4, 0, 4, 0, 2, 0, 9, 0, 5, 0, 5, 0, 7, 0, 1, 0, 6, 0, 8, 0, 7, 0, 7, 0, 9, 0, 8, 0, 5, 0, 11, 0, 9, 0, 1, 0, 12, 0, 10, 0, 10, 0, 12, 0, 2, 0, 11, 0, 15, 0, 12, 0, 12, 0, 10, 0, 3, 0, 13, 0, 27, 0, 14, 0, 2, 0, 19, 0, 15, 0, 4, 0, 20, 0, 3, 0, 16, 0, 21
Offset: 1

Views

Author

Antti Karttunen, Apr 20 2022

Keywords

Comments

Taking the Dirichlet convolution between this sequence and A349393 gives A349371, and similarly for many other such analogous pairs.

Crossrefs

Cf. A003602, A126760, A323881, A353460 (Dirichlet inverse), A353462 (sum with it).
Cf. also A349371, A349393.

Programs

  • PARI
    up_to = 65537;
    A003602(n) = (1+(n>>valuation(n,2)))/2;
    A126760(n) = {n&&n\=3^valuation(n, 3)<A126760
    DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1])*sumdiv(n, d, if(dA126760(n)));
    A323881(n) = v323881[n];
    A353461(n) = sumdiv(n,d,A003602(d)*A323881(n/d));

Formula

a(n) = Sum_{d|n} A003602(d) * A323881(n/d).
a(n) = A353462(n) - A353460(n).

A349339 Odd bisection of the Möbius transform of A126760.

Original entry on oeis.org

1, 0, 1, 2, 0, 3, 4, 0, 5, 6, 0, 7, 7, 0, 9, 10, 0, 8, 12, 0, 13, 14, 0, 15, 14, 0, 17, 14, 0, 19, 20, 0, 16, 22, 0, 23, 24, 0, 20, 26, 0, 27, 22, 0, 29, 24, 0, 24, 32, 0, 33, 34, 0, 35, 36, 0, 37, 30, 0, 32, 37, 0, 33, 42, 0, 43, 36, 0, 45, 46, 0, 40, 38, 0, 49, 50, 0, 40, 52, 0, 44, 54, 0, 55, 52, 0, 57, 40, 0, 59
Offset: 1

Views

Author

Antti Karttunen, Nov 16 2021

Keywords

Crossrefs

Cf. also A072451 (the odd bisection of the Möbius transform of A003602).

Programs

Formula

a(n) = A347233(2*n-1).

A359165 Difference between A126760 and its Möbius transform.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 2, 3, 4, 1, 1, 2, 5, 1, 3, 1, 2, 1, 1, 4, 6, 4, 1, 1, 7, 5, 2, 1, 3, 1, 4, 2, 8, 1, 1, 3, 9, 6, 5, 1, 1, 5, 3, 7, 10, 1, 2, 1, 11, 3, 1, 6, 4, 1, 6, 8, 12, 1, 1, 1, 13, 9, 7, 6, 5, 1, 2, 1, 14, 1, 3, 7, 15, 10, 4, 1, 2, 7, 8, 11, 16, 8, 1, 1, 17
Offset: 1

Views

Author

Antti Karttunen, Dec 22 2022

Keywords

Crossrefs

Programs

Formula

a(n) = A126760(n) - A347233(n).
a(n) = Sum_{d|n, dA347233(d).
a(n) = -Sum_{d|n, dA008683(n/d)*A126760(d).

A007310 Numbers congruent to 1 or 5 mod 6.

Original entry on oeis.org

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175
Offset: 1

Views

Author

C. Christofferson (Magpie56(AT)aol.com)

Keywords

Comments

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1. Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

Examples

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...
		

References

  • K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Crossrefs

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Programs

Formula

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)
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