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 1-9 of 9 results.

A227643 a(0)=1; for n > 0, a(n) = 1 + Sum_{i=A228086(n)..A228087(n)} [A092391(i) = n]*a(i), where [] is the Iverson bracket, resulting in 1 when i + A000120(i) = n and 0 otherwise.

Original entry on oeis.org

1, 1, 2, 3, 1, 5, 1, 6, 2, 3, 7, 4, 8, 1, 13, 1, 2, 16, 1, 18, 2, 1, 21, 1, 2, 22, 3, 2, 23, 4, 1, 26, 1, 6, 2, 7, 29, 1, 37, 1, 2, 38, 3, 2, 39, 4, 1, 42, 1, 5, 3, 1, 48, 4, 1, 50, 1, 5, 2, 2, 51, 6, 3, 1, 54, 55, 7, 59, 8, 2, 68, 1, 3, 69, 4, 2, 70, 5, 1, 73, 1
Offset: 0

Views

Author

Andres M. Torres, Jul 18 2013

Keywords

Comments

Each a(n) = 1 + the count of nodes in the finite subtree defined by the edge relation parent = child + A000120(child). In other words, one more than the count of n's descendants, by which we mean the whole transitive closure of all children emanating from the parent at n. The subtree is finite because successive descendant values get smaller and approach zero.

Examples

			0 has no children distinct from itself (we only have A092391(0)=0), so we define a(0) = (0+1) = 1,
1 has no children (it is one of the terms of A010061), so a(1) = (0+1) = 1,
4 and 6 are also members of A010061, so both a(4) and a(6) = (0+1) = 1,
7 has 1,2,3,4 and 5 among its descendants (as A092391(5)=7, A092391(3)=A092391(4)=5, A092391(2)=3, A092391(1)=2), so a(7) = (5+1) = 6,
8 has 6 as a child value,        so a(8) = (1+1) = 2,
9 has 6 and 8 as descendants,    so a(9) = (2+1) = 3,
10 has {1,2,3,4,5,7}             so a(10) = (6+1) = 7.
		

Crossrefs

Cf. A010061 (gives the positions of ones), A000120, A092391, A228082, A228083, A228085, A227359, A227361, A227408.
Cf. also A213727 for a descendant counts for a similar tree defined by the edge relation parent = child - A000120(child).

Programs

  • Scheme
    ;; A deficient definition which works only up to n=128:
    (definec (A227643deficient n) (cond ((zero? n) 1) ((zero? (A228085 n)) 1) ((= 1 (A228085 n)) (+ 1 (A227643deficient (A228086 n)))) ((= 2 (A228085 n)) (+ 1 (A227643deficient (A228086 n)) (A227643deficient (A228087 n)))) (else (error "Not yet implemented for cases where n has more than two immediate children!"))))
    ;; Another definition that works for all n, but is somewhat slower:
    (definec (A227643full n) (cond ((zero? n) 1) (else (+ 1 (add (lambda (i) (if (= (A092391 i) n) (A227643full i) 0)) (A228086 n) (A228087 n))))))
    ;; Auxiliary function add 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)))))))
    ;; by Antti Karttunen, Aug 16 2013, macro definec can be found in his IntSeq-library.

Formula

From Antti Karttunen, Aug 16 2013: (Start)
a(0)=1; and for n > 0, if A228085(n)=0 then a(n)=1; if A228085(n)=1 then a(n)=1+a(A228086(n)); if A228085(n)=2 then a(n)=1+a(A228086(n))+a(A228087(n)); otherwise (when A228085(n)>2) cannot be computed with this formula, which works only up to n=128.
a(0)=1; and for n > 0, a(n) = 1+Sum_{i=A228086(n)..A228087(n)} [A092391(i) = n]*a(i). (Here [...] denotes the Iverson bracket, resulting in 1 when i+A000120(i) = n and 0 otherwise. This formula works with all n.) (End)

A092391 a(n) = n + wt(n), where wt(n) = A000120(n) = binary weight of n.

Original entry on oeis.org

0, 2, 3, 5, 5, 7, 8, 10, 9, 11, 12, 14, 14, 16, 17, 19, 17, 19, 20, 22, 22, 24, 25, 27, 26, 28, 29, 31, 31, 33, 34, 36, 33, 35, 36, 38, 38, 40, 41, 43, 42, 44, 45, 47, 47, 49, 50, 52, 50, 52, 53, 55, 55, 57, 58, 60, 59, 61, 62, 64, 64, 66, 67, 69, 65, 67, 68, 70, 70, 72, 73, 75
Offset: 0

Views

Author

Reinhard Zumkeller, May 08 2004

Keywords

Crossrefs

A010061 gives the numbers not occurring in this sequence. A228082 gives the terms of this sequence sorted into ascending order, with duplicates removed. A228085(n) gives the number of times n occurs in this sequence.

Programs

Formula

a(n) = n + A000120(n).
A010062(n+1) = a(A010062(n)).
G.f.: (1/(1 - x))*Sum_{k>=0} (2^k + 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

A228085 a(n) = number of distinct k which satisfy n = k + wt(k), where wt(k) (A000120) gives the number of 1's in binary representation of a nonnegative integer k.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 2, 1, 1, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 2, 1, 1, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0
Offset: 0

Views

Author

Antti Karttunen, Aug 09 2013

Keywords

Comments

wt(k) is also called bitcount(k).
a(n) = number of times n occurs in A092391.
The first 2 occurs at n = A230303(2) = 5 (as we have two solutions A092391(3) = A092391(4) = 5).
The first 3 occurs at n = A230303(3) = 129 (as we have three solutions A092391(123) = A092391(124) = A092391(128) = 129).
The first 4 occurs at n = A230303(4) = 4102, where we have solutions A092391(4091) = A092391(4092) = A092391(4099) = A092391(4100) = 4102.
For n>=1, a(2^n) = a(n-1) since an integer k = m is a solution to n-1 = m + wt(m) if and only if k = 2^n - 1 - m is a solution to 2^n = k + wt(k). - Max Alekseyev, Feb 23 2021

Crossrefs

A010061 gives the position of zeros, A228082 the positions of nonzeros, A228088 the positions of ones.
Cf. A000120, A010062, A092391, A228086, A228087, A228091 (positions of 2's), A227643, A230058, A230092 (positions of 3's), A230093, A227915 (positions of 4's), A070939, A230303.

Programs

  • Haskell
    a228085 n = length $ filter ((== n) . a092391) [n - a070939 n .. n]
    -- Reinhard Zumkeller, Oct 13 2013
  • Maple
    For Maple code see A230091. - N. J. A. Sloane, Oct 10 2013
    # Find all inverses of m under x -> x + wt(x) - N. J. A. Sloane, Oct 19 2013
    A000120 := proc(n) local w, m, i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
    F:=proc(m) local ans,lb,n,i;
    lb:=m-ceil(log(m+1)/log(2)); ans:=[];
    for n from max(1,lb) to m do if (n+wt(n)) = m then ans:=[op(ans),n]; fi; od:
    [seq(ans[i],i=1..nops(ans))];
    end;
  • Mathematica
    nmax = 8191; Clear[a]; a[_] = 0;
    Scan[Set[a[#[[1]]], #[[2]]]&, Tally[Table[n + DigitCount[n, 2, 1], {n, 0, nmax}]]];
    a /@ Range[0, nmax] (* Jean-François Alcover, Oct 29 2019 *)
    a[n_] := Module[{k, cnt = 0}, For[k = n - Floor[Log[2, n]] - 1, k < n, k++, If[n == k + DigitCount[k, 2, 1], cnt++]]; cnt];
    a /@ Range[0, 100] (* Jean-François Alcover, Nov 28 2020 *)

A228087 a(n) = largest k which satisfies n = k + bitcount(k), or 0 if no such k exists. Here bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

Original entry on oeis.org

0, 0, 1, 2, 0, 4, 0, 5, 6, 8, 7, 9, 10, 0, 12, 0, 13, 16, 0, 17, 18, 0, 20, 0, 21, 22, 24, 23, 25, 26, 0, 28, 0, 32, 30, 33, 34, 0, 36, 0, 37, 38, 40, 39, 41, 42, 0, 44, 0, 45, 48, 0, 49, 50, 0, 52, 0, 53, 54, 56, 55, 57, 58, 0, 60, 64, 61, 65, 66, 63, 68, 0
Offset: 0

Views

Author

Antti Karttunen, Aug 09 2013

Keywords

Comments

A083058(n)+1 gives a lower bound for nonzero terms, n-1 an upper bound.

Crossrefs

Cf. A228086, A228085. A010061 gives the positions of zeros after a(0). The union of A010061 and A228088 gives the positions where a(n) = A228086(n).
Cf. also A213724, A227643.

Programs

  • Scheme
    (define (A228087 n) (let loop ((k n)) (cond ((<= k (A083058 n)) 0) ((= n (A092391 k)) k) (else (loop (- k 1))))))

A083058 Number of eigenvalues equal to 1 of n X n matrix A(i,j)=1 if j=1 or i divides j.

Original entry on oeis.org

1, 0, 1, 1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 10, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66
Offset: 1

Views

Author

Michael Somos, Apr 18 2003

Keywords

Comments

All numbers occur at least once, but terms > 1 of A000295 appear twice. - Robert G. Wilson v, Apr 19 2006
It appears that a(n) = Sum_{k=0..n-1} (1 + (-1)^A000108(k))/2 (n > 1). - Paul Barry, Mar 31 2008
Barry's observation above is true because A000108 obtains odd values only at points (2^j)-1 (A000225) and here the repeated values (A000295) occur precisely at positions given by A000225 and A000079. - Antti Karttunen, Aug 17 2013
a(n)+1 gives a lower bound for nonzero terms of A228086 and A228087. - Antti Karttunen, Aug 17 2013

Crossrefs

Programs

  • Maple
    A083058 := proc(n)
        if n = 1 then
            1;
        else
            n-floor(log[2](n))-1 ;
        end if;
    end proc:
    seq(A083058(n),n=1..40) ; # R. J. Mathar, Jul 23 2017
  • Mathematica
    a[1] = 1; a[n_] := n - Floor[Log[2, n]] - 1;
    Array[a, 100] (* Jean-François Alcover, Feb 27 2019 *)
  • PARI
    a(n)=if(n<2,n>0,n-floor(log(n)/log(2))-1)
    
  • PARI
    a(n)= if(n<1, 0, valuation( subst( charpoly( matrix(n, n, i, j, (j==1) || (0==j%i))), x, x+1), x))
    
  • Python
    def a(n): return n - n.bit_length() + (n == 1)  # Matthew Andres Moreno, Jan 04 2024
  • Scheme
    (define (A083058 n) (if (< n 2) n (- n (A070939 n)))) ;; Antti Karttunen, Aug 17 2013
    

Formula

a(n) = n - A070939(n), n > 1.
a(1)=1, else a(n)=b(n) with b(0)=0, b(2n)=b(n)+n-1, b(2n+1)=b(n)+n. - Ralf Stephan, Oct 11 2003
Except for a(1), a(n) = n - 1 - floor(log(2,n)). - Robert G. Wilson v, Apr 19 2006
It seems that a(n) = A182220(n+1)-1 for all n > 1. - Antti Karttunen, Aug 17 2013

A228083 Table of binary Self-numbers and their descendants; square array T(r,c), with row r>=1, column c>=1, read by antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 5, 6, 5, 7, 8, 13, 7, 10, 9, 16, 15, 10, 12, 11, 17, 19, 18, 12, 14, 14, 19, 22, 20, 21, 14, 17, 17, 22, 25, 22, 24, 23, 17, 19, 19, 25, 28, 25, 26, 27, 30, 19, 22, 22, 28, 31, 28, 29, 31, 34, 32, 22, 25, 25, 31, 36, 31, 33, 36, 36, 33, 37
Offset: 1

Views

Author

Antti Karttunen, Aug 09 2013

Keywords

Examples

			The top-left corner of the square array:
   1,  2,  3,  5,  7, 10, 12, 14, ...
   4,  5,  7, 10, 12, 14, 17, 19, ...
   6,  8,  9, 11, 14, 17, 19, 22, ...
  13, 16, 17, 19, 22, 25, 28, 31, ...
  15, 19, 22, 25, 28, 31, 36, 38, ...
  18, 20, 22, 25, 28, 31, 36, 38, ...
  21, 24, 26, 29, 33, 35, 38, 41, ...
  23, 27, 31, 36, 38, 41, 44, 47, ...
  ...
The non-initial terms on each row are obtained by adding to the preceding term the number of 1-bits in its binary representation (A000120).
		

Crossrefs

First column: A010061. First row: A010062. Transpose: A228084. See A151942 for decimal analog.

Programs

  • Mathematica
    nmax0 = 100;
    nmax := Length[col[1]];
    col[1] = Table[n + DigitCount[n, 2, 1], {n, 0, nmax0}] // Complement[Range[Last[#]], #]&;
    col[k_] := col[k] = col[k - 1] + DigitCount[col[k-1], 2, 1];
    T[n_, k_] := col[k][[n]];
    Table[T[n-k+1, k], {n, 1, nmax}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 28 2020 *)

Formula

T(r,1) are those numbers not of form n + sum of binary digits of n (binary Self numbers) = A010061(r);
T(r,c) = T(r,c-1) + sum of binary digits of T(r,c-1) = A092391(T(r,c-1)).

A228091 Numbers n for which there exists such a natural number k < n that k + bitcount(k) = n + bitcount(n), where bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

Original entry on oeis.org

4, 12, 16, 17, 20, 28, 32, 34, 36, 44, 48, 49, 52, 60, 65, 68, 76, 80, 81, 84, 92, 96, 98, 100, 108, 112, 113, 116, 124, 128, 129, 130, 131, 132, 140, 144, 145, 148, 156, 160, 162, 164, 172, 176, 177, 180, 188, 193, 196, 204, 208, 209, 212, 220, 224, 226, 228
Offset: 1

Views

Author

Antti Karttunen, Aug 09 2013

Keywords

Comments

In other words, all such terms A228236(n) which satisfy A228236(n) > A228086(A092391(A228236(n))), which means that the sequence contains all natural numbers n such that A228085(A092391(n)) > 1 and n > A228086(A092391(n)).
Note: 124 is the first term that occurs both here and in A228237.

Examples

			For cases 0 + A000120(0) = 0, 1 + A000120(1) = 2, 2 + A000120(2) = 3, 3 + A000120(3) = 5 there are no smaller solutions yielding the same result.
However, for 4 + A000120(4) = 5, we already saw the case 3+A000120(3) giving the same result, thus 4 is the first term of this sequence.
Next time this occurs for 12, as 12 + A000120(12) = 14 = 11 + A000120(11), and 11 < 12.
		

Crossrefs

Subset of A228236. Cf. also A228237. Complement of this sequence gives the nonzero terms of A228086 in ascending order.

A228089 Integers k for which a sum k + bitcount(k) cannot be obtained as a sum k2 + bitcount(k2) for any other k2<>k.

Original entry on oeis.org

0, 1, 2, 5, 6, 8, 7, 9, 10, 13, 18, 21, 22, 24, 23, 25, 26, 30, 33, 37, 38, 40, 39, 41, 42, 45, 50, 53, 54, 56, 55, 57, 58, 64, 61, 66, 63, 69, 70, 72, 71, 73, 74, 77, 82, 85, 86, 88, 87, 89, 90, 94, 97, 101, 102, 104, 103, 105, 106, 109, 114, 117, 118, 120, 119
Offset: 1

Views

Author

Antti Karttunen, Aug 17 2013

Keywords

Comments

The values of k's are sorted here according to the magnitude of the sum k + bitcount(k), where bitcount(k) (= A000120) gives the number of 1's in binary representation of nonnegative integer k; a(n) = A228086(A228088(n)).

Examples

			6 is in this sequence because the sum 6+A000120(6)=8 cannot be obtained with any other value of k than k=6.
8 is in this sequence because the sum 8+A000120(8)=9 cannot be obtained with any other value of k than k=8.
7 is in this sequence because the sum 7+A000120(7)=10 cannot be obtained with any other value of k than k=7.
In this sequence 8 becomes before 7 because 8+A000120(8) < 7+A000120(7).
		

Crossrefs

A228090 gives the same terms sorted into ascending order.

Programs

Formula

a(n) = A228086(A228088(n)).
A092391(a(n)) = A228088(n).

A335599 Sequence is limit_{k->oo} s_k, where s_k = s_{k-1}, s_{k-1}[k-1] + 2^(k-1), ..., s_{k-1}[end] + 2^(k-1) starting with s_0 = s_0[0..1] = 0,0.

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 3, 5, 5, 6, 7, 7, 9, 10, 11, 11, 13, 13, 14, 15, 15, 18, 19, 19, 21, 21, 22, 23, 23, 25, 26, 27, 27, 29, 29, 30, 31, 31, 35, 35, 37, 37, 38, 39, 39, 41, 42, 43, 43, 45, 45, 46, 47, 47, 50, 51, 51, 53, 53, 54, 55, 55, 57, 58, 59, 59, 61, 61
Offset: 0

Views

Author

Richard Aime Blavy, Jun 15 2020

Keywords

Comments

In binary 0, 0, 1, 1, 10, 11, 11, 101, 101, 110, 111, 111, 1001, 1010, 1011, 1101, 1110, 1111, 1111, 10010, 10011, 10011, 10101, ...
a(n) = m is the smallest solution to m + bitcount(m) = n or n-1. So a(n) = smaller nonzero of A228086(n) and A228086(n-1) (for n>=2). - Kevin Ryde, Jul 05 2020

Crossrefs

Programs

  • Maple
    s:= proc(n) option remember; `if`(n=0, [0, 0][], (l->
         [l[], map(x-> x+2^(n-1), l[n..-1])[]][])([s(n-1)]))
        end:
    s(7);  # gives 136 = A005126(7) terms;  # Alois P. Heinz, Jul 04 2020
  • Mathematica
    a[n_] := If[n == 0, 0,
    Module[{m = n, k = Floor@Log2[n]}, m -= k + 1; While[k >= 0,
         If[BitGet[m, k] == 0, m++;
         If[BitGet[m, k] == 1, Return[m-1]]]; k--]; m]];
    Table[a[n], {n, 0, 67}] (* Jean-François Alcover, May 30 2022, after Kevin Ryde *)
  • PARI
    a(n) = { if(n, my(k=logint(n,2)); n-=k+1;
      while(k>=0, if(!bittest(n,k), n++; if(bittest(n,k), return(n-1))); k--));
      n; }  \\ Kevin Ryde, Jul 05 2020

Formula

a(n) + bitcount(a(n)) + A334820(n) = n for n>=0.
Showing 1-9 of 9 results.