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 1-3 of 3 results.

A171182 Period 6: repeat [0, 1, 1, 1, 0, 2].

Original entry on oeis.org

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

Views

Author

Juri-Stepan Gerasimov, Dec 04 2009, Dec 07 2009

Keywords

Comments

The number of divisors d of n of the form d=2 or 3. - Vladimir Shevelev, May 21 2010
a(n) = s(n+6), where s(k) is the number of partitions of k into distinct parts such that max(p) = 2 + min(p) for k >= 1, and (s(0)..s(6)) = (0,0,0,0,1,0,2). - Clark Kimberling, Apr 15 2014
Number of r X s integer-sided rectangles such that r < s, r + s = 2n, r | s and (s - r)/2 | s. - Wesley Ivan Hurt, Apr 24 2020
Number of positive integer solutions, (r,s,t), of the equation r^2 + t*s^2 = (n + 6)^2, where r + s = n + 6 and t < r <= s. For example, when n=6 we have the two solutions (4,8,2) and (6,6,3) since 4^2 + 2*8^2 = 12^2 and 6^2 + 3*6^2 = 12^2. - Wesley Ivan Hurt, Oct 04 2020

Links

Crossrefs

Cf. A178142. - Vladimir Shevelev, May 21 2010
Cf. A115357.
Cf. A001221, A059841, A065331, A079978, A089128, A169611.
Number of distinct prime factors <= p: this sequence (p=3), A178146 (p=5), A210679 (p=7).

Programs

Formula

a(n) = A115357(n-2) for n>1. - R. J. Mathar, Dec 09 2009
a(2) = 1, a(3) = 1, a(5) = 0, otherwise a(n) = a(n-2) + a(n-3) - a(n-5), where we put a(n) = 0, if n<0. - Vladimir Shevelev, May 21 2010
a(n) = floor(((n+1) mod 6)/3) + 2*floor(((n+5) mod 6)/5). - Gary Detlefs, Feb 15 2014
From Wesley Ivan Hurt, Aug 27 2014: (Start)
G.f.: (2+2*x+x^2)/(1+x-x^3-x^4).
a(n) + a(n-1) = a(n-3) + a(n-4) for n>4.
a(n) = (1 + floor((n-3)^2/2)) mod 3. (End)
a(n) = (5 + 3*cos(n*Pi) + 4*cos(2*n*Pi/3))/6. - Wesley Ivan Hurt, Jun 19 2016
From Amiram Eldar, Sep 16 2023: (Start)
Additive with a(p^e) = 1 if p <= 3, and 0 otherwise.
a(n) = A059841(n) + A079978(n).
a(n) = A001221(A089128(n)).
a(n) = A001221(A065331(n)). (End)

Extensions

Edited by Charles R Greathouse IV, Mar 23 2010

A210679 Number of distinct prime factors <= 7 of n.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Apr 01 2012

Keywords

Comments

Periodic with period length 210. - Amiram Eldar, Sep 16 2023

Links

Crossrefs

Cf. A001221, A002473, A008364, A027748, A080672, A165743.
Number of distinct prime factors <= p: A171182 (p=3), A178146 (p=5), this sequence (p=7).

Programs

Formula

a(n) <= 4.
a(A008364(n)) = 0; a(A080672(n)) > 0.
a(n) = A001221(n) iff n is 7-smooth: a(A002473(n)) = A001221(A002473(n)). [corrected by Amiram Eldar, Sep 16 2023]
From Amiram Eldar, Sep 16 2023: (Start)
Additive with a(p^e) = 1 if p <= 7, and 0 otherwise.
a(n) = A001221(A165743(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 247/210. (End)

A178147 Sum of squares d^2 of distinct divisors of n, d in {2, 3, 5}.

Original entry on oeis.org

0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0, 29, 9, 4, 0, 13, 25, 4, 9, 4, 0, 38, 0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0, 29, 9, 4, 0, 13, 25, 4, 9, 4, 0, 38, 0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0
Offset: 1

Views

Author

Vladimir Shevelev, May 21 2010, May 23 2010

Keywords

Comments

The sequence is periodic with period {0 4 9 4 25 13 0 4 9 29 0 13 0 4 34 4 0 13 0 29 9 4 0 13 25 4 9 4 0 38} of length 30.
A generalization: let B={b_1,...,b_t} be a set of t positive (not necessarily distinct) integers and m>=0 an integer.
For m>=0, let A(n)=Sum d^m over divisors d of n which are elements of B (with the multiplicities as in B). Calculating directly values of
A(b_i),A(b_i+b_j),A(b_i+b_j+b_k),...,
A(b_1+...+b_t), for the other values of A(n) we have the recursion:
A(n)=Sum{1<=i<=t}A(n-b_i)- Sum{1<=i

Links

Crossrefs

Cf. A005063, A098002, A178142-A178144, A178146.

Formula

a(n)= a(n-2) +a(n-3) -a(n-7)- a(n-8) +a(n-10), n>10.
By the comment, up to 10 it is sufficient to
calculate directly only values a(2)=4, a(3)=9, a(5)=25, a(7)=0, a(8)=4, a(10)=29.
For other n's we can use the recursion, accepting formally a(n)=0 for n<0. So a(1)=0; a(4)=a(2)+a(1)=4;a(6)=a(4)+a(3)=4+9=13,
a(9)=a(7)+a(6)-a(2)-a(1)=0+13-4+0=9.
a(n) = -2*a(n-1) -2*a(n-2) -a(n-3) +a(n-5) +2*a(n-6) +2*a(n-7) +a(n-8). - R. J. Mathar, Jul 13 2010
G.f. -x^2*(4+17*x+30*x^2+55*x^3+80*x^4+38*x^6+76*x^5) / ( (x-1)*(1+x)*(1+x+x^2)*(x^4+x^3+x^2+x+1) ). - R. J. Mathar, Dec 17 2012
Showing 1-3 of 3 results.

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