A293519 -id:A293519 - cp's OEIS frontend

A293517 a(n) = A293518(n) - A293519(n); how many more surviving even nodes than surviving (but not bifurcating) odd nodes there are at generation n in the binary tree of persistently squarefree numbers.

0, 1, 1, 1, 1, 2, 1, 3, 4, 5, 9, 8, 10, 20, 17, 36, 32, 50, 66, 83, 118, 171, 219, 291, 410, 511, 730, 952, 1325, 1665, 2389, 3100, 4147, 5631, 7591, 10093, 13756, 18390, 24540, 33288, 44391, 60052, 80291, 108096, 145226, 194764, 262091, 352096, 473452, 635336, 854332, 1147668

Offset: 0

Antti Karttunen, Oct 16 2017

nonn

As long as there are at least as many surviving even than surviving (but not bifurcating) odd nodes at each generation in the tree of persistently squarefree numbers (see illustration in A293230), this sequence also stays nonnegative, and being also the first differences of A293441 guarantees its monotonicity. If A293441 is monotonic, then A293230 is also, which in turn implies also that A293430 has infinite number of terms and that there will be nonzero terms arbitrary far in A293233.

The surviving children of even vertices are all of the form 4k+1, while the surviving children (those without an odd sibling) of odd vertices are all of the form 4k+2.

Cf. A293230, A293233, A293430, A293518, A293519.
First differences of A293441.
Cf. also A293428.

(define (A293517 n) (- (A293518 n) (A293519 n)))

a(n) = A293518(n) - A293519(n).

a(n) = A293441(1+n) - A293441(n).

A293230 a(n) is the number of integers k in range [2^n, (2^(n+1))-1] such that all terms in finite sequence [k, floor(k/2), floor(k/4), floor(k/8), ..., 1] are squarefree.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 12, 15, 19, 26, 35, 49, 66, 84, 114, 151, 204, 272, 354, 470, 619, 820, 1109, 1499, 2009, 2710, 3631, 4872, 6554, 8831, 11821, 15875, 21364, 28611, 38389, 51611, 69295, 93144, 125290, 168220, 226048, 303727, 408170, 548513, 736900, 990222, 1330212, 1787067, 2401254, 3226802, 4335590, 5825258

Offset: 0

Antti Karttunen and Michael De Vlieger, Oct 10 2017

nonn

Question: Is this sequence monotonic? If monotonic, then it certainly cannot settle to zero, which implies that A293430 is infinite and that there are nonzero terms arbitrary far in A293233.
If there are no zero terms, then in a simple binary tree illustrated below (where the left hand child is obtained as 2*parent, and the right hand child is 1 + 2*parent) there are arbitrary long trajectories starting from 1 that consist squarefree numbers (A005117) only. All numbers k that are in such trajectories are marked as  (terms of A293430). a(n) = the number of marked numbers at level n, where level 0 is the root 1, level 1 has nodes 2 and 3, level 2 nodes 5, 6, 7, etc.
                                    <1>
                                     |
                 .................../ \...................
                <2>                                     <3>
       4......../ \.......<5>                 <6>......./ \.......<7>
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
   /       \           /       \           /       \           /       \
  8         9        <10>     <11>        12      <13>       <14>     <15>
16 17     18 19    20  <21> <22> <23>   24  25  <26>  27   28 <29> <30> <31>
etc.
---

			In range [2^0 .. (2^1)-1] = [1], all terms (namely 1) are in A293430, thus a(0) = 1.
In range [2^1 .. (2^2)-1] = [2 .. 3] all terms are in A293430, thus a(1) = 2.
In range [2^2 .. (2^3)-1] = [4 .. 7] the terms 5, 6, 7 are in A293430 (because they themselves are squarefree and when applying x -> floor(x/2) to them, give either 2 or 3, numbers that are also included in A293430), thus a(2) = 3.

Cf. A005117, A293233, A293430, A293441, A293518, A293519, A293520, A293521, A293522.

Cf. A293440 (the first differences).

Table[Count[Range[2^n, (2^(n + 1)) - 1], ?(AllTrue[Table[Floor[#/2^e], {e, 0, n}], SquareFreeQ] &)], {n, 0, 20}] (* _Michael De Vlieger, Oct 10 2017 *)

\\ A naive algorithm that computes A293233, A293430 and A293230 at the same time:
allocatemem(2^30);
up_to_level = 23;
up_to = (2^(1+up_to_level))-1;
v293233 = vector(up_to);
v293233[1] = 1;
write("b293430.txt", 1, " ", 1);
countsA293230 = 1; kA293430 = 2; for(n=2,up_to,if(!bitand(n,n-1), print1(countsA293230,", "); countsA293230 = 0); v293233[n] = moebius(n)* v293233[n\2];if(v293233[n],write("b293430.txt", kA293430, " ", n); kA293430++; countsA293230++)); print1(countsA293230);

\\ Much faster algorithm:
allocatemem(2^30);
next_living_bud_or_zero(n) = if(issquarefree(n),n,0);
nextA293230generation(tops) = { my(new_tops = vecsort(vector(2*#tops,i,next_living_bud_or_zero((2*tops[(i+1)\2])+(i%2))),,8)); if(0==new_tops[1], vector(#new_tops-1,i,new_tops[1+i]), new_tops); }
tops_of_tree = [1];
write("b293230.txt", 0, " ", 1);
print1(1, ", ");
for(n=1,64,tops_of_tree = nextA293230generation(tops_of_tree); write("b293230.txt", n, " ", k = length(tops_of_tree)); print1(k, ", "));

a(n) = Sum_{k=2^n..2^(1+n)-1} abs(A293233(k)).
For n >= 1, a(n) = A293441(n) + A293441(n-1).
a(n) = A293520(n) + A293521(n) + A293522(n). [sum of number of withering, surviving and bifurcating nodes at each level.]
a(n) = A293520(n) + (A293518(n) + A293519(n)) + A293522(n).
It seems that lim_{n ->oo} A293441(n+1)/a(n) ~= 0.770... (if it exists) and similarly lim_{n ->oo} a(n+1)/a(n) ~= 1.34...

A293441 a(n) is the number of odd numbers k in range [2^n, (2^(n+1))-1] such that all terms in finite sequence [k, floor(k/2), floor(k/4), floor(k/8), ..., 1] are squarefree.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 7, 8, 11, 15, 20, 29, 37, 47, 67, 84, 120, 152, 202, 268, 351, 469, 640, 859, 1150, 1560, 2071, 2801, 3753, 5078, 6743, 9132, 12232, 16379, 22010, 29601, 39694, 53450, 71840, 96380, 129668, 174059, 234111, 314402, 422498, 567724, 762488, 1024579, 1376675, 1850127, 2485463, 3339795

Offset: 0

Antti Karttunen, Oct 10 2017

nonn

For n > 0, a(n-1) is the number of even numbers in the same range satisfying the same condition. This follows because the alive even nodes in any generation (or level) of the binary tree illustrated in A293230 are all offspring of the odd nodes of the previous generation. (Even nodes cannot have even offspring simply because no number divisible by 4 can be squarefree). On the other hand, each odd node has an alive even child, because if an odd number k is squarefree, then 2k is squarefree as well.

The necessary and sufficient condition for this sequence to stay monotonic is that A293517(n) = A293518(n) - A293519(n) >= 0 (because A293517(n) = a(1+n) - a(n) also), in other words, that for every generation #{even nodes that survive} >= #{odd nodes that just survive, i.e., do not bifurcate}. If this sequence is monotonic then surely A293230 is also.

Cf. A293230, A293520, A293521, A293522.

Cf. A293517 (the first differences), A293518, A293519.

Table[Count[Range[2^n + 1, (2^(n + 1)) - 1, 2], ?(AllTrue[ Table[Floor[#/2^e], {e, 0, n}], SquareFreeQ] &)], {n, 0, 20}] (* _Michael De Vlieger, Oct 10 2017 *)

\\ A naive algorithm:
up_to_level = 28;
up_to = (2^(1+up_to_level));
is_persistently_squarefree(n,base) = { while(n>1, if(!issquarefree(n),return(0)); n \= base); (1); };
is_oddA293430(n) = ((n%2)&&is_persistently_squarefree(n,2));
countsA293441 = 1; k=1; for(n=2,up_to,if(!bitand(n,n-1), write("b293441.txt", k, " ", countsA293441); print1(countsA293441,", "); countsA293441 = 0; k++); if(is_oddA293430(n),countsA293441++));

\\ Faster way, compute A293441, A293518 and A293519 at the same time:
allocatemem(2^30);
next_living_bud_or_zero(n) = if(issquarefree(n),n,0);
nextA293230generation(tops) = { my(new_tops = vecsort(vector(2*#tops,i,next_living_bud_or_zero((2*tops[(i+1)\2])+((i+1)%2))),,8)); if(0==new_tops[1], vector(#new_tops-1,i,new_tops[1+i]), new_tops); }
writeA293441etc_counts(n,tops) = { my(os=0, es=0, k=0); for(i=1,#tops, if((tops[i]%2), k++; if(!issquarefree(1+(2*tops[i])), os++), if(issquarefree(1+(2*tops[i])), es++));); write("b293441.txt", n, " ", k); write("b293518.txt", n, " ", es); write("b293519.txt", n, " ", os); print1(k, ", ");}
tops_of_tree = [1];
write("b293441.txt", 0, " ", 1);
write("b293518.txt", 0, " ", 0);
write("b293519.txt", 0, " ", 0);
print1(1, ", ");
for(n=1,51,tops_of_tree = nextA293230generation(tops_of_tree); writeA293441etc_counts(n,tops_of_tree););

(define (A293441 n) (add (lambda (k) (* (A000035 k) (abs (A293233 k)))) (A000079 n) (+ -1 (A000079 (+ 1 n)))))

a(n) = Sum_{k=2^n..2^(1+n)-1} A000035(k)*abs(A293233(k)).
For n >= 1, A293230(n) = a(n) + a(n-1).
For n >= 1, if a(n) > a(n-1) then A293230(n) > A293230(n-1) and thus also A293522(n) > A293520(n). [If this sequence is monotonic, then so is A293230.]
For n >= 1, if a(n) > a(n-1) then a(n) > A293520(n). [Because only even nodes may die.]
A293522(n) <= a(n) <= A293521(n) + A293522(n). [Because no even node can bifurcate but all odd nodes either survive or bifurcate.]

A293521 Number of surviving (but not bifurcating) nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

Original entry on oeis.org

0, 1, 1, 3, 3, 2, 5, 9, 8, 11, 15, 24, 30, 42, 51, 76, 94, 126, 158, 217, 298, 403, 539, 731, 970, 1305, 1748, 2322, 3179, 4225, 5715, 7596, 10259, 13731, 18357, 24771, 33184, 44448, 59968, 80764, 107973, 145638, 195237, 262446, 352904, 474964, 637081, 856232, 1149966, 1543986, 2076534, 2789516

Offset: 0

Antti Karttunen, Oct 12 2017

nonn

			a(2) = 1 because in the binary tree illustrated in A293230, there is only one node at the level (namely, the node 6) that spawns just one offspring.

Cf. A293230, A293441, A293518, A293519, A293520, A293522.

PARI
```
\\ See program at A293520.
```

a(n) = Sum_{k=(2^n)..(2^(1+n)-1)} abs(A293233(k)) * [1 == A008966(2*k)+A008966(1+2*k)].

a(n) = A293518(n) + A293519(n). [even survivors + odd survivors.]

A293518 Number of surviving even nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 6, 6, 8, 12, 16, 20, 31, 34, 56, 63, 88, 112, 150, 208, 287, 379, 511, 690, 908, 1239, 1637, 2252, 2945, 4052, 5348, 7203, 9681, 12974, 17432, 23470, 31419, 42254, 57026, 76182, 102845, 137764, 185271, 249065, 334864, 449586, 604164, 811709, 1089661, 1465433, 1968592

Offset: 0

Antti Karttunen, Oct 16 2017

nonn

			a(3) = 2 because in the binary tree illustrated in A293230, there are two even nodes at the level 3 (namely, the nodes 10 and 14) that spawn just one offspring each.

Cf. A008966, A293230, A293233, A293517, A293519, A293521.

\\ Compute the sequences A293441, A293518 and A293519 at the same time:
allocatemem(2^30);
next_living_bud_or_zero(n) = if(issquarefree(n),n,0);
nextA293230generation(tops) = { my(new_tops = vecsort(vector(2*#tops,i,next_living_bud_or_zero((2*tops[(i+1)\2])+((i+1)%2))),,8)); if(0==new_tops[1], vector(#new_tops-1,i,new_tops[1+i]), new_tops); }
writeA293441etc_counts(n,tops) = { my(os=0, es=0, k=0); for(i=1,#tops, if((tops[i]%2), k++; if(!issquarefree(1+(2*tops[i])), os++), if(issquarefree(1+(2*tops[i])), es++));); write("b293441.txt", n, " ", k); write("b293518.txt", n, " ", es); write("b293519.txt", n, " ", os); print1(k, ", ");}
tops_of_tree = [1];
write("b293441.txt", 0, " ", 1);
write("b293518.txt", 0, " ", 0);
write("b293519.txt", 0, " ", 0);
print1(1, ", ");
for(n=1,51,tops_of_tree = nextA293230generation(tops_of_tree); writeA293441etc_counts(n,tops_of_tree););

(define (A293518 n) (add (lambda (k) (* (if (and (= 0 (A008966 (+ k k))) (= 1 (A008966 (+ 1 k k)))) 1 0) (abs (A293233 k)))) (A000079 n) (+ -1 (A000079 (+ 1 n)))))
;; Implements sum_{i=lowlim..uplim} intfun(i)
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

a(n) = Sum_{k=(2^n)..(2^(1+n)-1)} abs(A293233(k)) * [0==A008966(2*k)] * [1==A008966(1+2*k)].
a(n) + A293519(n) = A293521(n).
a(n) - A293519(n) = A293517(n).

cp's OEIS Frontend

A293517 a(n) = A293518(n) - A293519(n); how many more surviving even nodes than surviving (but not bifurcating) odd nodes there are at generation n in the binary tree of persistently squarefree numbers.

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A293230 a(n) is the number of integers k in range [2^n, (2^(n+1))-1] such that all terms in finite sequence [k, floor(k/2), floor(k/4), floor(k/8), ..., 1] are squarefree.

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A293441 a(n) is the number of odd numbers k in range [2^n, (2^(n+1))-1] such that all terms in finite sequence [k, floor(k/2), floor(k/4), floor(k/8), ..., 1] are squarefree.

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A293521 Number of surviving (but not bifurcating) nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

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A293518 Number of surviving even nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

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