Barry Carter - cp's OEIS frontend

User: Barry Carter

Barry Carter has authored 4 sequences.

A293161 a(n) = ceiling(A293160(n)/2).

1, 1, 2, 3, 4, 7, 10, 16, 24, 39, 59, 96, 150, 233, 367, 588, 925, 1463, 2299, 3648, 5776, 9139, 14432, 22916, 36178, 57371

Offset: 1

N. J. A. Sloane, Oct 12 2017, following a suggestion from Barry Carter

Cf. A293160.

A293160 Number of distinct terms in row n of Stern's diatomic array, A049456.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 13, 20, 31, 48, 78, 118, 191, 300, 465, 734, 1175, 1850, 2926, 4597, 7296, 11552, 18278, 28863, 45832, 72356, 114742, 181721, 287926, 455748, 722458, 1144370, 1813975, 2873751, 4553643, 7213620, 11432169, 18120733, 28716294, 45491133

Offset: 0

N. J. A. Sloane, Oct 12 2017, answering a question raised by Barry Carter in an email message. Barry Carter worked out the first 26 terms

Equivalently, a(n) is the number of distinct terms in row n of the Stern-Brocot sequence (A002487) when that sequence is divided into blocks of lengths 1, 2, 4, 8, 16, 32, ...
It would be nice to have a formula or recurrence, or even some bounds. Empirically, a(n) seems to be roughly 2^(2n/3) for the known values. Note that the first half of row n has about 2^(n-2) terms, and the maximal multiplicity is given by A293957(n), so 2^(n-2)/A293957(n) is a lower bound on a(n), which seems not too bad for the known values. - N. J. A. Sloane, Nov 04 2017
The multiset of terms in row n of Stern's diatomic array, with unique elements counted by a(n), is the same as the multiset of numerators of fractions in row n of Kepler's tree. Indeed, a fraction p/q is in row n-1 of Kepler's tree if and only if p/q and q/p are in row n of Calkin-Wilf tree. To form row n of Stern's diatomic array, one should take either numerators and denominators of fractions less than 1 or all numerators from Calkin-Wilf tree row n, in either case p/q and q/p contribute p and q. In Kepler's tree, a fraction p/q contributes p and q as numerators to the next row, i.e. row n. See A294442 for Kepler's tree and A294444 for the number of distinct denominators in it. - Andrey Zabolotskiy, Dec 05 2024

			Row 4 of A294442 contains eight fractions: 1/5, 4/5, 3/7, 4/7, 2/7, 2/7, 3/8, 5/8.
There are five distinct numerators, so a(4) = 5.

R. J. Mathar, Java program to compute the sequence
R. J. Mathar, The Kepler binary tree of reduced fractions
Don Reble, C++ program for A135510 and A293160

Cf. A002487, A049456, A070878, A293161, A293165, A293957.
See A135510 for the smallest positive missing number in each row.
Cf. A294442, A294444, A295783 (first differences).

A049456 := proc(n, k)
    option remember;
    if n =1 then
        if k >= 0 and k <=1 then
            1;
        else
            0 ;
        end if;
    elif type(k, 'even') then
        procname(n-1, k/2) ;
    else
        procname(n-1, (k+1)/2)+procname(n-1, (k-1)/2) ;
    end if;
end proc: # R. J. Mathar, Dec 12 2014
# A293160. This is not especially fast, but it will easily calculate the first 26 terms and confirm Barry Carter's values.
rho:=n->[seq(A049456(n,k),k=0..2^(n-1))];
w:=n->nops(convert(rho(n),set));
[seq(w(n),n=1..26)];
# Alternative program:
# S[n] is the list of fractions, written as pairs [i, j], in row n of Kepler's triangle; nc is the number of distinct numerators, and dc the number of distinct denominators
S[0]:=[[1, 1]]; S[1]:=[[1, 2]];
nc:=[1, 1]; dc:=[1, 1];
for n from 2 to 18 do
S[n]:=[];
for k from 1 to nops(S[n-1]) do
t1:=S[n-1][k];
a:=[t1[1], t1[1]+t1[2]];
b:=[t1[2], t1[1]+t1[2]];
S[n]:=[op(S[n]), a, b];
od:
listn:={};
for k from 1 to nops(S[n]) do listn:={op(listn), S[n][k][1]}; od:
c:=nops(listn);  nc:=[op(nc), c];
listd:={};
for k from 1 to nops(S[n]) do listd:={op(listd), S[n][k][2]}; od:
c:=nops(listd);  dc:=[op(dc), c];
od:
nc; # this sequence
dc; # A294444
# N. J. A. Sloane, Nov 20 2017

Length[Union[#]]& /@ NestList[Riffle[#, Total /@ Partition[#, 2, 1]]&, {1, 1}, 26] (* Jean-François Alcover, Mar 25 2020, after Harvey P. Dale in A049456 *)
Map[Length@ Union@ Numerator@ # &, #] &@ Nest[Append[#, Flatten@ Map[{#1/(#1 + #2), #2/(#1 + #2)} & @@ {Numerator@ #, Denominator@ #} &, Last@ #]] &, {{1/1}, {1/2}}, 21] (* Michael De Vlieger, Apr 18 2018 *)

from itertools import chain, product
from functools import reduce
def A293160(n): return n if n <= 1 else len({1}|set(sum(reduce(lambda x,y:(x[0],x[0]+x[1]) if y else (x[0]+x[1],x[1]),chain(k,(1,)),(1,0))) for k in product((False,True),repeat=n-2))) # Chai Wah Wu, Jun 20 2022

a(28)-a(39) from Don Reble, Oct 16 2017

a(0) prepended and content related to Kepler's tree added from a duplicate entry (where the terms up to a(28) have been independently obtained by Michael De Vlieger) by Andrey Zabolotskiy, Dec 06 2024

A276707 Number of terms of A069090 with exactly n digits.

Original entry on oeis.org

4, 12, 60, 381, 2522, 19094, 151286, 1237792, 10354144, 88407746, 766869330

Offset: 1

Barry Carter, Sep 15 2016

Barry Carter, Mathematica work on first n for which randomly rolled digits will yield a prime

Cf. A069090.

A276707 := proc(n)
    local a,k;
    a := 0 ;
    k := nextprime(10^(n-1)) ;
    while k < 10^n do
        if isA069090(k) then
            a := a+1 ;
        end if;
        k := nextprime(k) ;
    end do:
    a ;
end proc: # R. J. Mathar, Dec 15 2016

from sympy import primerange, isprime
def ok(p):
    s = str(p)
    if len(s) == 1: return True
    return all(not isprime(int(s[:i])) for i in range(1, len(s)))
def a(n): return sum(ok(p) for p in primerange(10**(n-1), 10**n))
print([a(n) for n in range(1, 7)]) # Michael S. Branicky, Jul 03 2021

# faster version skipping bad prefixes
from sympy import isprime, nextprime
def a(n):
    if n == 1: return 4
    p, c = nextprime(10**(n-1)), 0
    while p < 10**n:
        s, fail = str(p), False
        for i in range(1, n):
            ti = int(s[:i])
            if isprime(ti): fail = i; break
        if fail: p = nextprime((ti+1)*10**(n-i))
        else: p, c = nextprime(p), c+1
    return c
print([a(n) for n in range(1, 8)]) # Michael S. Branicky, Jul 03 2021

a(9)-a(11) from Michael S. Branicky, Jul 03 2021

A276473 Number of terms of A202259 with precisely n digits.

Original entry on oeis.org

6, 38, 320, 2819, 25668, 237586, 2224574, 21007948, 199725336, 1908845614, 18321586810, 176478166845

Offset: 1

Barry Carter, Sep 12 2016

When generating n random digits in order, number of ways to fail to generate a prime at any step.

a(n+1) >= 6*a(n), for n > 1, since any term of A202259 counted in a(n) may be extended with 0, 2, 4, 5, 6, or 8. - Michael S. Branicky, Nov 18 2024

Barry Carter, Mathematica work on probability of generating member of A202259 after generating n random numbers in order
Michael S. Branicky, Python program for OEIS A276473

Cf. A202259.

a(9)-a(11) from Michael S. Branicky, Nov 13 2024

a(12) from Michael S. Branicky, Nov 18 2024

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User: Barry Carter

A293161 a(n) = ceiling(A293160(n)/2).

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A293160 Number of distinct terms in row n of Stern's diatomic array, A049456.

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Maple

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A276707 Number of terms of A069090 with exactly n digits.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Python

Python

Extensions

A276473 Number of terms of A202259 with precisely n digits.

Views

Author

Keywords

Comments

Links

Crossrefs

Extensions