A135510 -id:A135510 - cp's OEIS frontend

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

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

A178031 Consider the Farey tree A049455/A049456; a(n) = row at which the denominator n first appears (assumes first row is labeled row 1).

Original entry on oeis.org

1, 2, 3, 4, 4, 6, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 7, 8, 8, 8, 8, 8, 8, 9, 8, 8, 8, 9, 9, 8, 9, 9, 9, 10, 9, 9, 9, 10, 9, 9, 9, 9, 9, 10, 9, 9, 10, 10, 10, 11, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 1

N. J. A. Sloane, Dec 16 2010

nonn

Computed by Alan Wechsler, Dec 16 2010.
Richard C. Schroeppel also asked about the analogous sequence giving the last occurrence of denominator n.
The first occurrence of k in this sequence is apparently at n = A135510(k-1), except for k=5. The last occurrence of k is at n = Fibonacci(k). - Andrey Zabolotskiy, Dec 01 2024

			Start with a pair of fractions 0/1, 1/1 and repeatedly insert the "Farey sum" (p+r)/(q+s) in between every pair of adjacent fractions p/q, r/s. The first few iterations are:
1:   0/1                                     1/1
2:   0/1                 1/2                 1/1
3:   0/1       1/3       1/2       2/3       1/1
4:   0/1  1/4  1/3  2/5  1/2  3/5  2/3  3/4  1/1
We only look at the denominators in this table (which form the sequence A049456, or A002487 if the rightmost column is removed).
1 first appears in row 1, so a(1) = 1.
2 first appears in row 2, so a(2) = 2.
3 first appears in row 3, so a(3) = 3.
4 and 5 first appear in row 4, so a(4) = a(5) = 4.

Based on a posting by Richard C. Schroeppel to the Math Fun Mailing List, Dec 15 2010.

Bo Gyu Jeong, Table of n, a(n) for n = 1..10000
Richard J. Mathar, The Kepler binary tree of reduced fractions, 2017.

See A178047 for another version. Cf. A002487, A006842, A006843, A177903, A178042, A135510.

More terms from Bo Gyu Jeong, Oct 20 2012

A293957 When A002487 is written as a triangle the n-th row has length 2^(n-1); a(n) is the maximal multiplicity of any entry in that row, considering the entries strictly between the initial 1 and the central 2.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 4, 5, 6, 8, 12, 16, 22, 29, 36, 48, 67, 84, 118, 151, 203, 270, 362, 472, 636, 846, 1142, 1526, 2024, 2736, 3666, 4918, 6550, 8776, 11796, 15824

Offset: 0

N. J. A. Sloane, Nov 03 2017

The maximal entry is row n is Fibonacci(n+1), and the smallest missing number is A135510(n). The number of distinct numbers in each row is given by A293160.

It would be nice to have a formula for this sequence, or at least some bounds.

			Rows 0 through 6 of A002487 are:
0,
1,
1, 2,
1, 3, 2, 3,
1, 4, 3, 5, 2, 5, 3, 4,
1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5,
1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6,
To find a(5) we consider the entries 1, 5, 4, 7, 3, 8, 5, 7, 2 in row 5. Ignoring the initial 1 and the final 2, the maximal multiplicity is 2 (for example, 5 appears twice), so a(5) = 2.
From _Don Reble_, Nov 04 2017: (Start)
The initial values of a(n) for n >= 3 together with the terms that have the highest multiplicity are:
3    1 [3]
4    1 [3 4 5]
5    2 [5 7]
6    2 [5 7 9 11]
7    4 [11]
8    5 [13 17]
9    6 [19 23 31 41]
10    8 [23 37 43]
11   12 [71]
12   16 [71]
13   22 [127]
14   29 [109]
15   36 [199 251]
16   48 [263]
17   67 [433]
18   84 [701]
19  118 [839]
20  151 [1193]
21  203 [1801]
22  270 [2693]
23  362 [4229]
24  472 [4349]
25  636 [7759]
26  846 [11287]
27 1142 [14627]
28 1526 [20929]
29 2024 [37243]
30 2736 [43133]
31 3666 [67231]
32 4918 [90227]
33 6550 [127819]
34 8776 [181031]
35 11796 [251071]
36 15824 [394549]
(End)

Cf. A002487, A049456, A135510, A293160.

A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end:
ans:=[];
for n from 3 to 18 do
b1:=2^(n-1); b2:=2^n-1; b3:=2^(n-2)-1; mx:=0;
ar:=Array(0..b1-1,0);
for k from 1 to b3 do
kk:=b1+k;
v:=A002487(kk);
ar[v]:=ar[v]+1;
od:
   for k from 0 to b1-1 do if ar[k]>mx then mx:=ar[k]; fi; od:
ans:=[op(ans),mx];
od:
ans;

from itertools import chain, product
from collections import Counter
from functools import reduce
def A293957(n): return 0 if n <= 2 else max(Counter(m for m in (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)) if m != 1 and m != 2).values()) # Chai Wah Wu, Jun 20 2022

a(19)-a(36) from Don Reble, Nov 04 2017

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

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A178031 Consider the Farey tree A049455/A049456; a(n) = row at which the denominator n first appears (assumes first row is labeled row 1).

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A293957 When A002487 is written as a triangle the n-th row has length 2^(n-1); a(n) is the maximal multiplicity of any entry in that row, considering the entries strictly between the initial 1 and the central 2.

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Maple

Python

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