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 1-6 of 6 results.

A254312 Rectangular array A read by upward antidiagonals in which the entry in row n and column k is defined by A(n,k) = (2^a(n)*(6*k - (3 - (-1)^a(n))*(1 - (-1)^n)/2) - 2^n + 4)/6, n,k >= 1, where {a(n)} is the Beatty sequence A117630 defined by a(n) = floor(n*log(3)/log(3/2)).

Original entry on oeis.org

3, 32, 7, 170, 64, 11, 1022, 426, 96, 15, 2726, 2046, 682, 128, 19, 65526, 10918, 3070, 938, 160, 23, 174742, 131062, 19110, 4094, 1194, 192, 27, 2097110, 436886, 196598, 27302, 5118, 1450, 224, 31, 11184726, 4194262, 699030, 262134, 35494, 6142, 1706, 256, 35
Offset: 1

Views

Author

L. Edson Jeffery, May 03 2015

Keywords

Comments

Conjecture: The array A contains without duplication all natural numbers m such that m < S(m), where the function S is as defined in A257480; i.e., the sequence is a permutation of A254311.

Examples

			Array A begins:
.         3       7      11      15       19       23       27       31
.        32      64      96     128      160      192      224      256
.       170     426     682     938     1194     1450     1706     1962
.      1022    2046    3070    4094     5118     6142     7166     8190
.      2726   10918   19110   27302    35494    43686    51878    60070
.     65526  131062  196598  262134   327670   393206   458742   524278
.    174742  436886  699030  961174  1223318  1485462  1747606  2009750
.   2097110 4194262 6291414 8388566 10485718 12582870 14680022 16777174
		

Crossrefs

Cf. A004767, A174312 (rows 1 and 2).

Programs

  • Mathematica
    (* Array antidiagonals flattened: *)
    a[n_] := Floor[n*Log[3/2, 3]]; A254312[n_, k_] := (2^a[n]*(6*k - (3 - (-1)^a[n])*(1 - (-1)^n)/2) - 2^n + 4)/6; Flatten[Table[A254312[n - k + 1, k], {n, 9}, {k, n}]]

A056576 Highest k with 2^k <= 3^n.

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 9, 11, 12, 14, 15, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 68, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 91, 93, 95, 96, 98, 99, 101, 103, 104, 106, 107
Offset: 0

Views

Author

Henry Bottomley, Jun 29 2000

Keywords

Examples

			a(3)=4 because 3^3=27 and 2^4=16 is power of 2 immediately below 27.
		

Crossrefs

Cf. A000079 (powers of 2), A000244 (powers of 3), A020914, A022921.
Cf. A056850, A117630 (complement), A020857 (decimal expansion of log_2(3)), A076227, A100982.

Programs

Formula

a(n) = floor(log_2(3^n)) = log_2(A000244(n)-A056576(n)) = a(n-1)+A022921(n-1).
a(n) = A020914(n) - 1. - L. Edson Jeffery, Dec 12 2014

A052130 a(n) is the number of numbers between 1 and 2^m with m-n prime factors (counted with multiplicity), for m sufficiently large.

Original entry on oeis.org

1, 2, 7, 15, 37, 84, 187, 421, 914, 2001, 4283, 9184, 19611, 41604, 87993, 185387, 389954, 817053, 1709640, 3567978, 7433670, 15460810, 32103728, 66567488, 137840687, 285076323, 588891185, 1215204568, 2505088087, 5159284087
Offset: 0

Views

Author

Bernd-Rainer Lauber (br.lauber(AT)surf1.de), Jan 21 2000

Keywords

Comments

a(n) = number of products of half-odd-primes <= 2^n. E.g., a(2) = 7 since 1, 3/2, (3/2)^2, (3/2)^3, (3/2)*(5/2), 5/2, 7/2 are all <= 2^2. - David W. Wilson, Feb 01 2000
m is sufficiently large precisely when 2^m > 3^(m-n), i.e., when m >= floor(n*log(3)/log(1.5)) = A117630(n+1) = A126281(n) for n > 1. (Robert G. Wilson v asks if this conjecture holds in a comment to A126281.) - David A. Corneth, Apr 09 2015
From Robert G. Wilson v, Apr 13 2020: (Start)
This sequence shows a sufficiently large row of A126279 read backwards or a sufficiently large column of A126279 read vertically.
log(y) ~ a + b*x + c*x^2, where a=1.1422, b=0.7419, and c=-0.00035, with an r^2 of 1.0. (End)
[But what is y? - Editors, Jun 15 2021]

Examples

			Between 1 and 2^m there is just one number with m prime factors, namely 2^m, so a(0) = 1.
For m >= 3, up to 2^m there are 2 numbers with m-1 prime factors, 2^(m-1) and 3*2^(m-2), so a(1) = 2.
		

Crossrefs

Programs

  • Mathematica
    AlmostPrimePi[k_Integer, n_] := Module[{a, i}, a[0] = 1; If[k == 1, PrimePi[n], Sum[PrimePi[n/Times @@ Prime[ Array[a, k - 1]]] - a[k - 1] + 1, Evaluate[ Sequence @@ Table[{a[i], a[i - 1], PrimePi[(n/Times @@ Prime[Array[a, i - 1]])^(1/(k - i + 1))]}, {i, k - 1}]] ]]]; (* Eric W. Weisstein, Feb 07 2006 *)
    Table[ AlmostPrimePi[Floor[n(1 + 1/Sqrt@2)] + 2, 2^(n + Floor[n(1 + 1/Sqrt@2)]) + 2], {n, 2, 30}] (* Robert G. Wilson v, Feb 21 2006 *)
  • Python
    from math import prod, isqrt
    from sympy import primepi, primerange, integer_nthroot
    def A052130(n):
        if n<=1: return n+1
        def g(x,a,b,c,m): yield from (((d,) for d in enumerate(primerange(b,isqrt(x//c)+1),a)) if m==2 else (((a2,b2),)+d for a2,b2 in enumerate(primerange(b,integer_nthroot(x//c,m)[0]+1),a) for d in g(x,a2,b2,c*b2,m-1)))
        k = 1
        while 3**k<(r:=1<Chai Wah Wu, Dec 03 2024

Extensions

More terms from David W. Wilson, Feb 01 2000
a(24)-a(29) from Robert G. Wilson v, Feb 21 2006

A056734 Positive numbers k such that, in base 3, 2^k and 2^(k+1) have the same number of digits and the same number of 0's.

Original entry on oeis.org

2, 5, 8, 10, 18, 21, 27, 29, 35, 40, 62, 67, 83, 92, 138, 146, 165, 184, 298, 346, 428, 487, 666, 750, 785, 929, 937, 1064, 1086, 1156, 1162, 1240, 1614, 1706, 1739, 1788, 2327, 2389, 2533, 2649, 2937, 3240, 3403, 3489, 3549, 3619, 3693, 3817, 3866, 4175
Offset: 1

Views

Author

Russell Harper (rharper(AT)intouchsurvey.com), Aug 13 2000

Keywords

Comments

Using empirical data for 1 <= k <= 10000, it has been found that the distribution of these terms correlates well (R^2 = 0.9513) with f(k) = c*k^(1/2) with c approximately 0.73. In addition, f'(k) approximates the probability that any particular k has this property. Any terms in A056154 must also be in this sequence.

Examples

			First term: 2^2 = 11_3, 2^3 = 22_3, both with 0 zeros and both of length 2.
Second term: 2^5 = 1012_3, 2^6 = 2101_3, both with 1 zero and both of length 4.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[4200],IntegerLength[2^#,3]==IntegerLength[2^(#+1),3] && DigitCount[ 2^#,3,0]==DigitCount[2^(#+1),3,0]&] (* Harvey P. Dale, Dec 10 2021 *)
  • PARI
    isok(k) = my(da=digits(2^k, 3), db=digits(2^(k+1), 3)); (#da == #db) && (#select(x->(x==0), da) == #select(x->(x==0), db)); \\ Michel Marcus, Jul 01 2021

A126281 a(n) is the least m to satisfy the requirements of A052130.

Original entry on oeis.org

1, 2, 5, 8, 10, 13, 16, 18, 21, 24, 27, 29, 32, 35, 37, 40, 43, 46, 48, 51, 54, 56, 59, 62, 65, 67, 70, 73
Offset: 1

Views

Author

Robert G. Wilson v, Dec 24 2006

Keywords

Comments

A052130: For m very large, a(n) = number of numbers between 1 and 2^m with m-n prime factors (counted with multiplicity).
In observing the triangular array of A126279, the array T(k,n) defined as the k-th almost prime count of n-th power of two, it is noticed that the k-th term from the right converges to a fixed value beginning with the n-th power of two.
Will this sequence continue to match A117630: floor(n*log(3)/log(3/2)) ?

Crossrefs

Programs

  • Mathematica
    AlmostPrimePi[k_Integer, n_] := Module[{a, i}, a[0] = 1; If[k == 1, PrimePi[n], Sum[PrimePi[n/Times @@ Prime[ Array[a, k - 1]]] - a[k - 1] + 1, Evaluate[ Sequence @@ Table[{a[i], a[i - 1], PrimePi[(n/Times @@ Prime[Array[a, i - 1]])^(1/(k - i + 1))]}, {i, k - 1}]] ]]]; f[n_] := Block[{a = 0, m = n}, While[ b = AlmostPrimePi[m-n+1, 2^m]; b > a, m++; a = b]; m--; m]; Array[f, 24] (* Eric W. Weisstein, Feb 07 2006 *)

Extensions

a(25)-a(28) from Robert G. Wilson v, Sep 07 2012
Expression in comment corrected by L. Edson Jeffery, Apr 03 2015

A368131 a(n) = floor(n * log(4/3) / log(3/2)).

Original entry on oeis.org

0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 14, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 29, 29, 30, 31, 31, 32, 33, 34, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 46, 46, 47, 48
Offset: 0

Views

Author

Ruud H.G. van Tol, Jan 25 2024

Keywords

Comments

Highest k with 3^(n+k) <= 4^n * 2^k.

Crossrefs

Cf. A054414, A117630, A325913, A369522 (slope).

Programs

  • Mathematica
    Table[Floor[n*Log[4/3]/Log[3/2]],{n,0,68}] (* James C. McMahon, Jan 27 2024 *)
  • PARI
    alist(N) = my(a=-1, b=1, k=0); vector(N, i, a+=2; b*=3; if(logint(b, 2) < a, a++; b*=3; k++); k); \\ note that i is n+1

Formula

a(n) = floor(n * log(3) / log(3/2)) - 2*n.
a(n) = floor(n * arctanh(1/7) / arctanh(1/5)).
a(n) = A325913(n) - n.
a(n) = A117630(n) - 2*n.
a(n) = A054414(n) - 2*n - 1.
Showing 1-6 of 6 results.