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 1-10 of 75 results. Next

A358654 a(n) = A025480(A353654(n+1) - 1).

Original entry on oeis.org

0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 9, 10, 23, 12, 27, 29, 30, 63, 8, 19, 21, 22, 47, 25, 26, 55, 28, 59, 61, 62, 127, 17, 18, 39, 20, 43, 45, 46, 95, 24, 51, 53, 54, 111, 57, 58, 119, 60, 123, 125, 126, 255, 16, 35, 37, 38, 79, 41, 42, 87, 44, 91, 93
Offset: 0

Views

Author

Mikhail Kurkov, Nov 25 2022

Keywords

Comments

Permutation of the nonnegative integers.
Conjecture: A247648(n) with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n).

Crossrefs

Formula

Conjecture: a(n) = A348366(A343152(n)) for n > 0 with a(0) = 1.

A110962 Fractalization of A025480, zero-based version of Kimberling's paraphrases sequence.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 0, 3, 0, 0, 0, 4, 2, 2, 1, 5, 1, 1, 0, 6, 3, 3, 0, 7, 0, 0, 0, 8, 4, 4, 2, 9, 2, 2, 1, 10, 5, 5, 1, 11, 1, 1, 0, 12, 6, 6, 3, 13, 3, 3, 0, 14, 7, 7, 0, 15, 0, 0, 0, 16, 8, 8, 4, 17, 4, 4, 2, 18, 9, 9, 2, 19, 2, 2, 1, 20, 10, 10, 5, 21, 5, 5, 1, 22, 11, 11, 1, 23, 1, 1, 0, 24, 12, 12
Offset: 0

Views

Author

Alexandre Wajnberg, Sep 26 2005

Keywords

Comments

Self-descriptive sequence: the terms at odd indices are the sequence itself, while the terms at even indices (the skeleton of this sequence) are the terms of A025480, which is a zero-based sequence of Kimberling's paraphrases sequence, A003602.

Crossrefs

One less than A110963 (note also the different starting offsets).

Programs

Formula

For even n, a(n) = A025480(n/2), for odd n, a(n) = a((n-1)/2). - Antti Karttunen, Apr 18 2022
a(2n+1) = a(4n+3) = a(n).
a(2n) = a(4n+1) = a(4n+2) = A025480(n/2).
a(4n) = a(8n+1) = a(8n+2) = n.
a(n) = A110963(1+n) - 1.

Extensions

Entry edited and more terms added by Antti Karttunen, Apr 18 2022

A364950 Lexicographically earliest infinite sequence such that a(i) = a(j) => A025480(i) = A025480(j) and A348717(i) = A348717(j) for all i, j >= 1.

Original entry on oeis.org

1, 2, 3, 4, 2, 5, 3, 6, 4, 7, 2, 8, 9, 10, 11, 12, 13, 14, 15, 16, 7, 17, 2, 18, 19, 20, 21, 22, 23, 24, 3, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 2, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 23, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 13, 62, 63, 64, 14, 65, 66, 67, 15, 68, 69, 70, 34, 71
Offset: 1

Views

Author

Antti Karttunen, Aug 17 2023

Keywords

Comments

Restricted growth sequence transform of the ordered pair [A025480(n), A348717(n)], or equally, of the ordered pair [A003602(1+n), A246277(n)].
For all i, j:
a(i) = a(j) => A364949(i) = A364949(j),
a(i) = a(j) => A364951(i) = A364951(j).

Crossrefs

Programs

  • PARI
    up_to = 65537;
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    A025480(n) = (n>>valuation(n*2+2, 2));
    A348717(n) = if(1==n, 1, my(f = factor(n), k = primepi(f[1, 1])-1); for (i=1, #f~, f[i, 1] = prime(primepi(f[i, 1])-k)); factorback(f));
    Aux364950(n) = [A025480(n), A348717(n)];
    v364950 = rgs_transform(vector(up_to, n, Aux364950(n)));
    A364950(n) = v364950[n];

A364951 Lexicographically earliest infinite sequence such that a(i) = a(j) => A025480(i) = A025480(j) and A046523(i) = A046523(j) for all i, j >= 1.

Original entry on oeis.org

1, 2, 3, 4, 2, 5, 3, 6, 4, 7, 2, 8, 9, 10, 11, 12, 13, 14, 15, 16, 7, 17, 2, 18, 19, 20, 21, 22, 23, 24, 3, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 2, 40, 41, 42, 43, 44, 45, 46, 5, 47, 48, 49, 23, 50, 51, 52, 53, 54, 55, 56, 57, 58, 27, 59, 13, 60, 61, 62, 14, 63, 31, 64, 15, 65, 66, 67, 34, 68
Offset: 1

Views

Author

Antti Karttunen, Aug 17 2023

Keywords

Comments

Restricted growth sequence transform of the ordered pair [A025480(n), A046523(n)], or equally, of the ordered pair [A003602(1+n), A101296(n)].
For all i, j: A364950(i) = A364950(j) => a(i) = a(j).

Crossrefs

Programs

  • PARI
    up_to = 65537;
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    A025480(n) = (n>>valuation(n*2+2, 2));
    A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); }; \\ From A046523
    Aux364951(n) = [A025480(n), A046523(n)];
    v364951 = rgs_transform(vector(up_to, n, Aux364951(n)));
    A364951(n) = v364951[n];

A108685 Hidden fractal sequence: increasing sequence all of whose successive digits are the digits of the fractal sequence A025480 (which is built upon the natural counting numbers).

Original entry on oeis.org

0, 10, 21, 30, 42, 51, 63, 70, 84, 92, 105, 111, 126, 133, 147, 150, 168, 174, 189, 192, 2010, 2152, 2112, 3124, 12256, 26132, 73281, 429730, 1531032, 1633834, 1735436, 1837938, 1939240, 2041104, 2214354, 4224511
Offset: 0

Views

Author

Alexandre Wajnberg, Jun 18 2005

Keywords

Crossrefs

A366601 a(0) = 0. For a(n-1) a novel term a(n) = a(A025480(a(n-1))), and if a(n-1) has occurred k (>1) times, a(n) = k-1 (the number of repetitions of a(n-1) in all prior terms).

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 3, 0, 4, 1, 1, 2, 1, 3, 1, 4, 1, 5, 0, 5, 1, 6, 0, 6, 1, 7, 0, 7, 1, 8, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 1, 9, 1, 10, 0, 8, 2, 9, 1, 11, 0, 9, 2, 10, 1, 12, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 2, 11, 1, 13, 0, 10, 3, 11, 2, 12, 1, 14, 0
Offset: 0

Views

Author

Keywords

Comments

An experimental sequence in which fractal sequence A025480 is used to select the response to a novel term, whereas a repeat term is followed by the number of repeats of that term. Every number appears infinitely many times, suggesting the possibility of there being multiple proper subsequences identical to the original, although no specific example (expressible by formula) has been found. The scatterplot shows fractal like features and the records subsequence is A001477. Shelly's formula (using A001511) for A025480 has been used in the computation of terms. a(n) <= n (equality when n = 0).

Examples

			a(0) = 0 is a novel term, therefore a(1) = a(A025480(0)) = a(0) = 0.
Since 0 has now been repeated once, a(2) = 1, another novel term, so a(3) = a(A025480(a(2))) = a(A025480(1)) = a(0) = 0.
Now 0 has been repeated 2 times, so a(4) = 2.
The data can be shown as an irregular table in which each row begins with a record term:
0,0;
1,0;
2,0;
3,0;
4,1,1,2,1,3,1,4,1;
5,0,5,1;
6,0,6,1;
7,0,7,1;
8,2,2,3,2,4,2,5,2,6,2,7,2,8,1;
		

Crossrefs

Programs

  • Mathematica
    nn = 120; f[x_] := Floor[x/2^IntegerExponent[2 (x + 1), 2]]; a[0] = 0; c[_] := 0; Do[a[n] = If[c[#] == 0, c[#]++; a[f[#]], c[#]++; c[#] - 1] &[a[n - 1]], {n, nn}]; Array[a, nn, 0]

A379818 a(2n+1) = a(n) for n >= 0, a(2n) = a(n) + a(n - 2^f(n)) + a(2n - 2^f(n)) + a(A025480(n-1)) for n > 0 with a(0) = 1 where f(n) = A007814(n).

Original entry on oeis.org

1, 1, 4, 1, 10, 4, 10, 1, 22, 10, 28, 4, 49, 10, 22, 1, 46, 22, 64, 10, 118, 28, 64, 4, 190, 49, 118, 10, 190, 22, 46, 1, 94, 46, 136, 22, 256, 64, 148, 10, 424, 118, 292, 28, 478, 64, 136, 4, 661, 190, 478, 49, 796, 118, 256, 10, 1177, 190, 424, 22, 661, 46
Offset: 0

Views

Author

Mikhail Kurkov, Jan 03 2025

Keywords

Crossrefs

Programs

  • PARI
    upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1] + v1[i\(4*A)+1])); v1

Formula

Conjecture: a(2^m*(2k+1)) = Sum_{j=0..m} (binomial(m+2, j+1) - binomial(m, j))*a(2^j*k) for m >= 0, k >= 0 with a(0) = 1.

A108715 First differences of A025480.

Original entry on oeis.org

0, 1, -1, 2, -1, 2, -3, 4, -2, 3, -4, 5, -3, 4, -7, 8, -4, 5, -7, 8, -5, 6, -10, 11, -6, 7, -10, 11, -7, 8, -15, 16, -8, 9, -13, 14, -9, 10, -17, 18, -10, 11, -16, 17, -11, 12, -22, 23, -12, 13, -19, 20, -13, 14, -24, 25, -14, 15, -22, 23, -15, 16, -31, 32, -16, 17, -25, 26, -17, 18, -31, 32, -18, 19, -28, 29, -19, 20, -37, 38, -20
Offset: 0

Views

Author

Alexandre Wajnberg, Jun 20 2005

Keywords

Comments

Pattern: cycles of four successive terms. [this needs to be clarified, Joerg Arndt, Apr 16 2011]

Crossrefs

Formula

From Paul Curtz, Apr 15 2011: (Start)
a(A000225(n)) = A131577(n).
2 * a(n) = A000265(n+1) - A000265(n).
a(2*n) + a(1+2*n) = 1.
a(4*n) = -n, a(1+4*n) = n.
a(2+8*n) = 1 + 3*n, a(3+8*n) = 2+3*n.
a(2+2*n) = a(n) - a(1+2*n). (End)

A275875 Subadditive triangle read by rows associated with the Grundy function A025480.

Original entry on oeis.org

2, 1, 3, 1, 2, 3, 1, 1, 2, 4, 1, 1, 2, 2, 4, 1, 1, 1, 2, 3, 4, 1, 1, 1, 2, 2, 3, 4, 1, 1, 1, 1, 2, 2, 3, 5, 1, 1, 1, 1, 2, 2, 2, 3, 5, 1, 1, 1, 1, 1, 2, 2, 3, 3, 5
Offset: 0

Views

Author

N. J. A. Sloane, Aug 14 2016

Keywords

Comments

See Levine 2004/2006 for definition. His Theorem 3.6 shows that there is a one-to-one correspondence between fractal sequences and subadditive triangles.

Examples

			Triangle begins:
2,
1,3,
1,2,3,
1,1,2,4,
1,1,2,2,4,
1,1,1,2,3,4,
1,1,1,2,2,3,4,
1,1,1,1,2,2,3,5,
1,1,1,1,2,2,2,3,5,
1,1,1,1,1,2,2,3,3,5,
...
		

References

  • L. Levine, Fractal sequences and restricted Nim, Ars Combin. 80 (2006), 113-127.

Crossrefs

Cf. A025480.

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

Original entry on oeis.org

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0
Offset: 1

Views

Author

John Tromp, Dec 11 1996

Keywords

Comments

This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre, Mar 06 2003
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ...), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - Philippe Deléham, Mar 15 2004
Fixed point of the morphism 0->010, 1->2, 2->3, ..., n->(n+1), .... - Joerg Arndt, Apr 29 2014
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r = k mod 2^n and r != 1. - T. D. Noe, Jul 12 2007
The following relation holds: 2^A007814(n)*(2*A025480(n-1)+1) = A001477(n) = n. (See functions hd, tl and cons in [Paul Tarau 2009].)
a(n) is the number of 0's at the end of n when n is written in base 2.
a(n+1) is the number of 1's at the end of n when n is written in base 2. - M. F. Hasler, Aug 25 2012
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 0). That is, A003188(n) XOR A003188(n+1) == 2^A007814(n). - Russ Cox, Dec 04 2010
The sequence is squarefree (in the sense of not containing any subsequence of the form XX) [Allouche and Shallit]. Of course it contains individual terms that are squares (such as 4). - Comment expanded by N. J. A. Sloane, Jan 28 2019
a(n) is the number of zero coefficients in the n-th Stern polynomial, A125184. - T. D. Noe, Mar 01 2011
Lemma: For n < m with r = a(n) = a(m) there exists n < k < m with a(k) > r. Proof: We have n=b2^r and m=c2^r with b < c both odd; choose an even i between them; now a(i2^r) > r and n < i2^r < m. QED. Corollary: Every finite run of consecutive integers has a unique maximum 2-adic valuation. - Jason Kimberley, Sep 09 2011
a(n-2) is the 2-adic valuation of A000166(n) for n >= 2. - Joerg Arndt, Sep 06 2014
a(n) = number of 1's in the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} p_j-th prime (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24)=3; indeed, the partition having Heinz number 24 = 2*2*2*3 is [1,1,1,2]. - Emeric Deutsch, Jun 04 2015
a(n+1) is the difference between the two largest parts in the integer partition having viabin number n (0 is assumed to be a part). Example: a(20) = 2. Indeed, we have 19 = 10011_2, leading to the Ferrers board of the partition [3,1,1]. For the definition of viabin number see the comment in A290253. - Emeric Deutsch, Aug 24 2017
Apart from being squarefree, as noted above, the sequence has the property that every consecutive subsequence contains at least one number an odd number of times. - Jon Richfield, Dec 20 2018
a(n+1) is the 2-adic valuation of Sum_{e=0..n} u^e = (1 + u + u^2 + ... + u^n), for any u of the form 4k+1 (A016813). - Antti Karttunen, Aug 15 2020
{a(n)} represents the "first black hat" strategy for the game of countably infinitely many hats, with a probability of success of 1/3; cf. the Numberphile link below. - Frederic Ruget, Jun 14 2021
a(n) is the least nonnegative integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022

Examples

			2^3 divides 24, so a(24)=3.
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
  0;
  1,0;
  2,0,1,0;
  3,0,1,0,2,0,1,0;
  4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 27.
  • K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

Cf. A011371 (partial sums), A094267 (first differences), A001511 (bisection), A346070 (mod 4).
Bisection of A050605 and |A088705|. Pairwise sums are A050603 and A136480. Difference of A285406 and A281264.
This is Guy Steele's sequence GS(1, 4) (see A135416). Cf. A053398(1,n). Column/row 1 of table A050602.
Cf. A007949 (3-adic), A235127 (4-adic), A112765 (5-adic), A122841 (6-adic), A214411 (7-adic), A244413 (8-adic), A122840 (10-adic).
Cf. A086463 (Dgf at s=2).

Programs

  • Haskell
    a007814 n = if m == 0 then 1 + a007814 n' else 0
                where (n', m) = divMod n 2
    -- Reinhard Zumkeller, Jul 05 2013, May 14 2011, Apr 08 2011
    
  • Haskell
    a007814 n | odd n = 0 | otherwise = 1 + a007814 (n `div` 2)
    --  Walt Rorie-Baety, Mar 22 2013
    
  • Magma
    [Valuation(n, 2): n in [1..120]]; // Bruno Berselli, Aug 05 2013
    
  • Maple
    ord := proc(n) local i,j; if n=0 then return 0; fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end proc: seq(ord(n), n=1..111);
    A007814 := n -> padic[ordp](n,2): seq(A007814(n), n=1..111); # Peter Luschny, Nov 26 2010
  • Mathematica
    Table[IntegerExponent[n, 2], {n, 64}] (* Eric W. Weisstein *)
    IntegerExponent[Range[64], 2] (* Eric W. Weisstein, Feb 01 2024 *)
    p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
    DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (* Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004 *)
    Nest[Join[ #, ReplacePart[ #, Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (* N. J. Gunther, May 23 2009 *)
    Nest[ Flatten[# /. a_Integer -> {0, a + 1}] &, {0}, 7] (* Robert G. Wilson v, Jan 17 2011 *)
  • PARI
    A007814(n)=valuation(n,2);
    
  • Python
    import math
    def a(n): return int(math.log(n - (n & n - 1), 2)) # Indranil Ghosh, Apr 18 2017
    
  • Python
    def A007814(n): return (~n & n-1).bit_length() # Chai Wah Wu, Jul 01 2022
    
  • R
    sapply(1:100,function(x) sum(gmp::factorize(x)==2)) # Christian N. K. Anderson, Jun 20 2013
    
  • Scheme
    (define (A007814 n) (let loop ((n n) (e 0)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A001511(n) - 1.
a(2*n) = A050603(2*n) = A001511(n).
a(n) = A091090(n-1) + A036987(n-1) - 1.
a(n) = 0 if n is odd, otherwise 1 + a(n/2). - Reinhard Zumkeller, Aug 11 2001
Sum_{k=1..n} a(k) = n - A000120(n). - Benoit Cloitre, Oct 19 2002
G.f.: A(x) = Sum_{k>=1} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Apr 10 2002
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Franklin T. Adams-Watters, Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a(k) = c(0) + c(0)*c(1) + c(0)*c(1)*c(2) + ... + c(0)*c(1)...c(n-1); a(k+1) = b(0) + b(0)*b(1) + b(0)*b(1)*b(2) + ... + b(0)*b(1)...b(n-1). - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
Sum_{k=1..n} (-1)^A000120(n-k)*a(k) = (-1)^(A000120(n)-1)*(A000120(n) - A000035(n)). - Vladimir Shevelev, Mar 17 2009
a(A001147(n) + A057077(n-1)) = a(2*n). - Vladimir Shevelev, Mar 21 2009
For n>=1, a(A004760(n+1)) = a(n). - Vladimir Shevelev, Apr 15 2009
2^(a(n)) = A006519(n). - Philippe Deléham, Apr 22 2009
a(n) = A063787(n) - A000120(n). - Gary W. Adamson, Jun 04 2009
a(C(n,k)) = A000120(k) + A000120(n-k) - A000120(n). - Vladimir Shevelev, Jul 19 2009
a(n!) = n - A000120(n). - Vladimir Shevelev, Jul 20 2009
v_{2}(n) = Sum_{r>=1} (r / 2^(r+1)) Sum_{k=0..2^(r+1)-1} e^(2(k*Pi*i(n+2^r))/(2^(r+1))). - A. Neves, Sep 28 2010, corrected Oct 04 2010
a(n) mod 2 = A096268(n-1). - Robert G. Wilson v, Jan 18 2012
a(A005408(n)) = 1; a(A016825(n)) = 3; A017113(a(n)) = 5; A051062(a(n)) = 7; a(n) = (A037227(n)-1)/2. - Reinhard Zumkeller, Jun 30 2012
a((2*n-1)*2^p) = p, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 04 2013
a(n) = A067255(n,1). - Reinhard Zumkeller, Jun 11 2013
a(n) = log_2(n - (n AND n-1)). - Gary Detlefs, Jun 13 2014
a(n) = 1 + A000120(n-1) - A000120(n), where A000120 is the Hamming weight function. - Stanislav Sykora, Jul 14 2014
A053398(n,k) = a(A003986(n-1,k-1)+1); a(n) = A053398(n,1) = A053398(n,n) = A053398(2*n-1,n) = Min_{k=1..n} A053398(n,k). - Reinhard Zumkeller, Aug 04 2014
a((2*x-1)*2^n) = a((2*y-1)*2^n) for positive n, x and y. - Juri-Stepan Gerasimov, Aug 04 2016
a(n) = A285406(n) - A281264(n). - Ralf Steiner, Apr 18 2017
a(n) = A000005(n)/(A000005(2*n) - A000005(n)) - 1. - conjectured by Velin Yanev, Jun 30 2017, proved by Nicholas Stearns, Sep 11 2017
Equivalently to above formula, a(n) = A183063(n) / A001227(n), i.e., a(n) is the number of even divisors of n divided by number of odd divisors of n. - Franklin T. Adams-Watters, Oct 31 2018
a(n)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Feb 16 2019
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Jul 11 2020
a(n) = 2*Sum_{j=1..floor(log_2(n))} frac(binomial(n, 2^j)*2^(j-1)/n). - Dario T. de Castro, Jul 08 2022
a(n) = A070939(n) - A070939(A030101(n)). - Andrew T. Porter, Dec 16 2022
a(n) = floor((gcd(n, 2^n)^(n+1) mod (2^(n+1)-1)^2)/(2^(n+1)-1)) (see Lemma 3.4 from Mazzanti's 2002 article). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 - A088705(n). - Chai Wah Wu, Sep 18 2024

Extensions

Formula index adapted to the offset of A025480 by R. J. Mathar, Jul 20 2010
Edited by Ralf Stephan, Feb 08 2014
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