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 1-7 of 7 results.

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

Views

Author

Keywords

Comments

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.
---

Examples

			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.
		

Crossrefs

Cf. A293440 (the first differences).

Programs

  • Mathematica
    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 *)
  • PARI
    \\ 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);
    
  • PARI
    \\ 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, ", "));

Formula

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

Views

Author

Antti Karttunen, Oct 10 2017

Keywords

Comments

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.

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    \\ 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++));
    
  • PARI
    \\ 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););
    
  • Scheme
    (define (A293441 n) (add (lambda (k) (* (A000035 k) (abs (A293233 k)))) (A000079 n) (+ -1 (A000079 (+ 1 n)))))

Formula

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.]

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

Views

Author

Antti Karttunen, Oct 16 2017

Keywords

Examples

			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.
		

Crossrefs

Programs

  • PARI
    \\ 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););
    
  • Scheme
    (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)))))))

Formula

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).

A293519 Number of surviving (but not bifurcating) odd nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

Original entry on oeis.org

0, 0, 0, 1, 1, 0, 2, 3, 2, 3, 3, 8, 10, 11, 17, 20, 31, 38, 46, 67, 90, 116, 160, 220, 280, 397, 509, 685, 927, 1280, 1663, 2248, 3056, 4050, 5383, 7339, 9714, 13029, 17714, 23738, 31791, 42793, 57473, 77175, 103839, 140100, 187495, 252068, 338257, 454325, 611101, 820924
Offset: 0

Views

Author

Antti Karttunen, Oct 16 2017

Keywords

Examples

			a(3) = 1 because in the binary tree illustrated in A293230, there is only one odd node at the level 3 (namely, the node 13) that spawns just one offspring.
		

Crossrefs

Programs

  • PARI
    \\ A naive algorithm (see A293518 for a better program):
    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); };
    { countsA293441 = 1; countsA293519 = 0; k=1; n=3; while(n <= 1+up_to,if(!bitand(n-1,n-2), write("b293441.txt", k, " ", countsA293441); write("b293519.txt", k, " ", countsA293519); print1(countsA293519,", "); countsA293441 = 0; countsA293519 = 0; k++); if(is_persistently_squarefree(n,2),countsA293441++; if(!issquarefree(1+(2*n)),countsA293519++)); n += 2); }
    
  • Scheme
    (define (A293519 n) (add (lambda (k) (* (if (and (= 1 (A008966 (+ k k))) (= 0 (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)))))))

Formula

a(n) = Sum_{k=(2^n)..(2^(1+n)-1)} abs(A293233(k)) * [1==A008966(2*k)] * [0==A008966(1+2*k)].
A293518(n) + a(n) = A293521(n).
A293518(n) - a(n) = A293517(n).

A293520 Number of dying nodes (withering branches) at generation n in the binary tree of persistently squarefree numbers (A293230).

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 2, 1, 2, 3, 3, 4, 9, 6, 13, 11, 21, 32, 40, 52, 60, 64, 90, 129, 169, 242, 321, 434, 549, 808, 1026, 1395, 1929, 2551, 3405, 4578, 6131, 8275, 11196, 14814, 20198, 26823, 36295, 48840, 65337, 87634, 118138, 158324, 212870, 287014
Offset: 0

Views

Author

Antti Karttunen, Oct 12 2017

Keywords

Comments

Provided that A293441 is strictly growing, then certainly a(n) < A293441(n), because only even nodes may die and A293441(n-1) gives the number of even nodes at level n.

Examples

			a(4) = 1 because in the binary tree illustrated in A293230, it is the only node 22 at the level 4 that does not generate any new buds as both 2*22 = 44 and 1+(2*22) = 45 are nonsquarefree numbers.
		

Crossrefs

Programs

  • PARI
    \\ Compute sequences A293230, A293520, A293521, A293522 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); }
    write_counts(n,tops) = { my(w=0, s=0, b=0, k); for(i=1,#tops,if((tops[i]%2), if(issquarefree(1+(2*tops[i])), b++, s++), if(issquarefree(1+(2*tops[i])), s++, w++)); ); write("b293520.txt", n, " ", w); write("b293521.txt", n, " ", s); write("b293522.txt", n, " ", b); write("b293230.txt", n, " ", k=length(tops)); print1(k, ", ");}
    tops_of_tree = [1];
    write("b293230.txt", 0, " ", 1);
    write("b293520.txt", 0, " ", 0);
    write("b293521.txt", 0, " ", 0);
    write("b293522.txt", 0, " ", 1);
    print1(1, ", ");
    for(n=1,52,tops_of_tree = nextA293230generation(tops_of_tree); write_counts(n,tops_of_tree););

Formula

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

A293522 Number of bifurcating nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

Original entry on oeis.org

1, 1, 2, 2, 3, 5, 5, 5, 9, 12, 17, 21, 27, 36, 50, 64, 89, 114, 156, 201, 261, 353, 480, 639, 870, 1163, 1562, 2116, 2826, 3798, 5080, 6884, 9176, 12329, 16627, 22262, 29980, 40421, 54126, 72642, 97877, 131266, 176638, 237227, 318659, 427624, 574993, 772511, 1038418, 1395802
Offset: 0

Views

Author

Antti Karttunen, Oct 12 2017

Keywords

Comments

Bifurcating node is one that branches to two alive nodes in the next generation (level) of the tree.

Examples

			a(2) = 2 because in the binary tree illustrated in A293230, there are two nodes at the level 2 (namely 5 and 7) that spawn two offspring each.
		

Crossrefs

Programs

Formula

a(n) = Sum_{k=(2^n)..(2^(1+n)-1)} abs(A293233(k)) * A008966(2*k) * A008966(1+2*k).
a(n) <= A293441(n). [Because no even node can bifurcate.]

A293440 First differences of A293230: how many more alive nodes there are in generation n+1 than in generation n in the binary tree of persistently squarefree numbers.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 3, 4, 7, 9, 14, 17, 18, 30, 37, 53, 68, 82, 116, 149, 201, 289, 390, 510, 701, 921, 1241, 1682, 2277, 2990, 4054, 5489, 7247, 9778, 13222, 17684, 23849, 32146, 42930, 57828, 77679, 104443, 140343, 188387, 253322, 339990, 456855, 614187, 825548, 1108788, 1489668
Offset: 0

Views

Author

Antti Karttunen, Oct 12 2017

Keywords

Crossrefs

Programs

Formula

a(n) = A293230(1+n)-A293230(n).
a(n) = A293522(n) - A293520(n). [Equal to the number of bifurcating nodes minus the number of withering nodes.]
Showing 1-7 of 7 results.