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 1-4 of 4 results.

A052006 Numbers k for which Fibonacci(k) is the first member of a 1,1 pair (A052005).

Original entry on oeis.org

17, 30, 43, 53, 66, 79, 89, 102, 115, 125, 138, 151, 161, 174, 187, 200, 210, 223, 236, 246, 259, 272, 282, 295, 308, 321, 331, 344, 357, 367, 380, 393, 403, 416, 429, 442, 452, 465, 478, 488, 501, 514, 524, 537, 550, 560, 573, 586, 599, 609, 622, 635, 645
Offset: 0

Views

Author

Antti Karttunen and Patrick De Geest, Nov 15 1999

Keywords

Comments

Keep adding the terms of sequence A052005 up to the first member of the next 1,1 pair to yield the terms of this sequence. - Patrick De Geest
Those k for which F(k-1) < 2^(floor(log_2(F(k)))) and F(k+1) >= 2^(floor(log_2(F(k)))+1) and F(k+2) >= 2^(floor(log_2(F(k)))+2).

Links

Crossrefs

The first differences are A051392.
Cf. A052005, A000045, A050815, A036284, A037093.

Programs

  • Mathematica
    With[{F = Fibonacci}, Reap[For[n=0, n<1000, n++, If[F[n-1] < 2^Floor[Log[2, F[n]]] && F[n+1] >= 2^(Floor[Log[2, F[n]]]+1) && F[n+2] >= 2^(Floor[Log[ 2, F[n]]]+2), Print[n]; Sow[n]]]][[2, 1]]] (* Jean-François Alcover, Feb 27 2016 *)

A051392 First differences of A052006.

Original entry on oeis.org

13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10
Offset: 1

Views

Author

Antti Karttunen and Patrick De Geest, Nov 15 1999

Keywords

Comments

Does this sequence only contain 10's and 13's?
Yes. Since all blocks of terms of A052005 are of the form 1(12)n, a(n) must be congruent to 1 modulo 3. [1, 2] blocks give an average growth rate of 3/2 powers of phi for every power of two, but since phi^3 > 4, singleton 1's are required to slow growth when errors get too large. Since singleton 1's reduce the growth rate by 1/2 power of phi per power of two, they should occur roughly once every (1/2) / log_2(phi^1.5 / 2) ~ 12.088 powers of phi. Therefore, a(n) will be 13 most of the time, with 10 occurring when needed to maintain this ratio. - Charlie Neder, Oct 24 2018

Links

Programs

  • Mathematica
    With[{F = Fibonacci}, Reap[For[n = 0, n < 1000, n++, If[F[n - 1] < 2^Floor[Log[2, F[n]]] && F[n + 1] >= 2^(Floor[Log[2, F[n]]] + 1) && F[n + 2] >= 2^(Floor[Log[2, F[n]]] + 2), Sow[n]]]][[2, 1]]] // Differences (* Jean-François Alcover, Feb 27 2016 *)

A224702 Number of partition sums between powers of 2 where the partition sums b(k) are A000070 and 2^n <= b(k) < 2^(n+1).

Original entry on oeis.org

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

Views

Author

Frank M Jackson, Apr 16 2013

Keywords

Comments

The sequence of partition sums A000070 is a complete sequence.

Examples

			a(11) = 3 as between 2048 and 4096 there are 3 partition sums namely 2087, 2714, 3506.

Links

Crossrefs

Cf. A000070, A036378, A052005

Programs

  • Mathematica
    getterm[n0_] := Sum[PartitionsP[m0], {m0, 0, n0}]; termcount[n1_] := (m1=0; While[getterm[m1]<2^n1, m1++]; m1); Table[termcount[n+1]-termcount[n], {n, 0, 100}]

A225152 Let b(k) be A036378, then a(n) is the number of b(k) terms such that 2^n < b(k) <= 2^(n+1).

Original entry on oeis.org

2, 0, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 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

Frank M Jackson, Apr 30 2013

Keywords

Comments

A036378 is a complete sequence.

Examples

			a(7) = 2 as between 128 and 256 there are 2 terms (A036378) namely 137 and 255.

Crossrefs

Cf. A036378, A052005.

Programs

  • Mathematica
    getterm[n2_] := PrimePi[2^(n2+1)]-PrimePi[2^n2];
termcount[n3_] := (m1=0; While[getterm[m1]<=2^n3, m1++]; m1);
Table[termcount[p+1]-termcount[p], {p, 0, 39}]
Showing 1-4 of 4 results.

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

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