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 1-5 of 5 results.

A080585 Partial sums of A080584.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37
Offset: 0

Views

Author

N. J. A. Sloane, Feb 23 2003

Keywords

Programs

  • Mathematica
    Accumulate[Flatten[Table[{PadRight[{},3*2^n,0],PadRight[{},3*2^n, 1]},{n,0,4}]]] (* Harvey P. Dale, Jun 01 2012 *)

Formula

Also: a(n) = 3*2^A000523(A002264(n+6)/2)*(1-3*A080584(n))+A080584(n)*(n+7)-3. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

A079000 a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is odd".

Original entry on oeis.org

1, 4, 6, 7, 8, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 23, 25, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 95, 97
Offset: 1

Views

Author

Matthew Vandermast, Feb 01 2003

Keywords

Comments

a(a(n)) = 2n + 3 for n>1.

Examples

			a(2) cannot be 2 because 2 is even; it cannot be 3 because that would require 2 to be a member of the sequence. Hence a(2)=4 and the next odd member of the sequence is the fourth member.

References

  • Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
  • N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

Links

Crossrefs

Cf. A079250-A079259, A079313, A079325, A064437, A003605, A079352, A079358.
Cf. also A080596, A080731, A080752, A121053, A080032.
Partial sums give A080566. Differences give A079948.

Programs

  • Maple
    Digits := 50; A079000 := proc(n) local k,j; if n<=2 then n^2; else k := floor(evalf(log( (n+3)/6 )/log(2)) ); j := n-(9*2^k-3); 12*2^k-3+3*j/2 +abs(j)/2; fi; end;
A002264 := n->floor(n/3): A079944 := n->floor(log[2](4*(n+2)/3))-floor(log[2](n+2)): A000523 := n->floor(log[2](n)): f := n->A079944(A002264(n-4)): g := n->A000523(A002264(n+2)/2): A079000 := proc(n) if n>3 then RETURN(simplify(3*n+3-3*2^g(n)+(-1)^f(n)*(9*2^g(n)-n-3))/2) else if n>0 then RETURN([1,4,6][n]) else RETURN(0) fi fi: end;
  • Mathematica
    a[1] = 1; a[n_] := (k = Floor[Log[2, (n+3)/6]]; j = n-(9*2^k - 3); 12*2^k-3 + 3*j/2 + Abs[j]/2); Table[a[n], {n, 1, 71}] (* Jean-François Alcover, May 21 2012, after Maple *)

Formula

a(1) = 1, a(2) = 4, then a(9*2^k-3+j) = 12*2^k-3+3*j/2+|j|/2 for k>=0, -3*2^k <= j <= 3*2^k. Also a(3n) = 3*b(n/3), a(3n+1) = 2*b(n)+b(n+1), a(3n+2) = b(n)+2*b(n+1) for n>=2, where b = A079905. - N. J. A. Sloane and Benoit Cloitre, Feb 20 2003
a(n+1) - 2*a(n) + a(n-1) = 1 for n = 9*2^k - 3, k>=0, = -1 for n = 2 and 3*2^k-3, k>=1 and = 0 otherwise.
a(n) = (3*n + 3 - 3*2^g(n) + (-1)^f(n)*(9*2^g(n) - n - 3))/2 for n>3, f(n) = A079944(A002264(n-4)) and g(n) = A000523(A002264(n+2)/2). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003
Also a(n) = n + 3*2^A000523(A002264(n+2)/2)*(1 - 3*A080584(n-4)) + A080584(n-4)*(n+3) for n>3, where A080584(n)=A079944(A002264(n)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

A079882 A run of 2^n 1's followed by a run of 2^n 2's, for n=0, 1, 2, ...

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Feb 21 2003

Keywords

Comments

In the sequence of nonnegative integers (cf. A001477) substitute all n by 2^floor(n/2) occurrences of (1 + n mod 2); a(n)=A173920(n+2,3) for n>0. [From Reinhard Zumkeller, Mar 04 2010]

Links

Crossrefs

Partial sums give A079945. Equals 1 + A079944. Cf. A080584.
First differences of A080637.

Programs

  • Maple
    f1 := n->[seq(1,i=1..2^n)]; f2 := n->[seq(2,i=1..2^n)]; s := []; for i from 0 to 10 do s := [op(s), op(f1(i)), op(f2(i))]; od: s;
  • Mathematica
    Table[{PadRight[{},2^n,1],PadRight[{},2^n,2]},{n,0,5}]//Flatten (* Harvey P. Dale, Jul 22 2016 *)

Formula

a(n) = floor(log[2](8*(n+2)/3)) - floor(log[2](n+2)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003

A080586 A run of 3*2^n 1's followed by a run of 3*2^n 2's, for n=0, 1, 2, ...

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Feb 23 2003

Keywords

Crossrefs

Cf. A079948, A080587.
Equals 1 + A080584.

Programs

  • Maple
    f := (c,n)->seq(c,i=1..3*2^n); [f(1,0),f(2,0),f(1,1),f(2,1),f(1,2),f(2,2),f(1,3),f(2,3)]; f;
  • Mathematica
    Flatten[Table[{PadRight[{},3*2^n,1],PadRight[{},3*2^n,2]},{n,0,4}]] (* Harvey P. Dale, May 04 2014 *)

Formula

a(n) = ( 3 - (-1)^A079944(A002264(n)) )/2, A079944(A002264(n))=floor(log[2](4*(floor((n+6)/3))/3)) - floor(log[2](floor((n+6)/3))) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003
Also a(n) = 1+A079944(A002264(n))=floor(log[2](8*(floor((n+6)/3))/3)) - floor(log[2](floor((n+6)/3))) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

A080587 Partial sums of A080586.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 17, 19, 21, 23, 25, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 89, 91, 93, 95, 97, 99, 101
Offset: 0

Views

Author

N. J. A. Sloane, Feb 23 2003

Keywords

Links

Programs

  • Mathematica
    Accumulate[Flatten[Table[{PadRight[{},3*2^n,1],PadRight[{},3*2^n,2]},{n,0,4}]]] (* Harvey P. Dale, May 04 2014 *)

Formula

Also: a(n) = n-2+3*2^A000523(A002264(n+6)/2)*(1-3*A080584(n))+A080584(n)*(n+7). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003
Also: a(n) = 3*2^(A000523(A002264(n+6))-1)*(4-3*A080586(n))+A080586(n)*(n+7)-9. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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