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A166133 After initial 1,2,4, a(n+1) is the smallest divisor of a(n)^2-1 that has not yet appeared in the sequence.

1, 2, 4, 3, 8, 7, 6, 5, 12, 11, 10, 9, 16, 15, 14, 13, 21, 20, 19, 18, 17, 24, 23, 22, 69, 28, 27, 26, 25, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 201, 80, 79

Offset: 1

Franklin T. Adams-Watters, Oct 07 2009

The initial 1,2,4 provides the smallest example with this rule that is not simply the integers in order, nor (apparently) ends with all divisors of a(n)^2-1 already present.
Apparently the sequence is infinite and includes every positive integer.
Apr 05 2015: John Mason has computed the first ten million terms. See link to zipped file. - N. J. A. Sloane, Apr 06 2015
The sequence contains many runs of incrementing and decrementing values. In the 1200 steps following the 4, there are 136 increments, 706 decrements, and 358 larger steps. What is the limiting distribution for these steps? [Click the "listen" button to appreciate these runs. - N. J. A. Sloane, Apr 03 2015]
After 3, 198, 270, 570, 522, 600, 822, and 882, we have a(n+1) = a(n)^2-1. Does this happen infinitely often? Cf. A256406, A256407.
A256543 gives numbers m such that a(m+1) = a(m)-1 or a(m+1) = a(m)+1. - Reinhard Zumkeller, Apr 01 2015
If this is a permutation, then A255833 is the inverse permutation. - M. F. Hasler, Apr 01 2015
a(A256703(n)+1) = a(A256703(n))^2 - 1. - Reinhard Zumkeller, Apr 08 2015
For n > 3: a(n) = A027750(a(n-1)^2-1, A256751(n)). - Reinhard Zumkeller, Apr 09 2015

			After a(24) = 22, the divisors of 22^2-1 = 483 are 1, 3, 7, 21, 23, 69, 161, and 483; 1, 3, 7, 21, and 23 have already occurred, so a(25) = 69.

Franklin T. Adams-Watters and N. J. A. Sloane, Table of n, a(n) for n = 1..20000 (first 1203 terms from Franklin T. Adams-Watters)
Hans Havermann, Log plot of 450000+ terms [Produced by Mathematica's ListLogPlot command]
Hans Havermann, Over-sized point-joined (over 250000 terms) graph of the sequence [Heavily clipped, which explains the strange appearance. - _N. J. A. Sloane_, Apr 01 2015]
John Mason, Table of n, a(n) for n = 1..711888 [10 megabytes]
John Mason, Table of n, a(n) for n = 1..2000000 [32 megabytes]
John Mason, Ten million terms (zipped file)
John Mason, Java program to generate this sequence, used to generate 10M terms, and some other associated sequences; it requires splitting into single classes for use.
N. J. A. Sloane and others, "Blog" about A166133

Cf. A166134, A000027, A122280, A005563, A256406, A256407, A027750, A005563, A256557, A256559.
For records see A256403, A256404.
Smallest missing numbers: A256405, A256408, A256409.
Cf. A256541 (first differences), A256543.
Inverse (conjectured): A255833.
Cf. A256564 (smallest prime factors), A244080 (largest prime factors), A256578 (largest proper divisors), A256542 (number of divisors).
Upper envelope: the sequence of pairs (A256422(n),A256423(n)).
Cf. A256703.
Cf. A256751.

import Data.List (delete); import Data.List.Ordered (isect)
a166133 n = a166133_list !! (n-1)
a166133_list = 1 : 2 : 4 : f (3:[5..]) 4 where
   f zs x = y : f (delete y zs) y where
            y = head $ isect (a027750_row' (x ^ 2 - 1)) zs
-- Reinhard Zumkeller, Apr 01 2015

s = {1, 2, 4}; e = 4; Do[d = Divisors[e^2 - 1]; i = 1;
While[MemberQ[s, d[[i]]], i++]; e = d[[i]]; AppendTo[s, e], {19997}]; s (* Hans Havermann, Apr 03 2015 *)

al(n,m=4,u=6)={local(ds,db);
u=bitor(u,1<

A089911 a(n) = Fibonacci(n) mod 12.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1

Offset: 0

Casey Mongoven, Nov 14 2003

From Reinhard Zumkeller, Jul 05 2013: (Start)
Sequence has been applied by several composers to 12-tone equal temperament pitch structure. The complete Fibonacci mod 12 system (a set of 10 periodic sequences) exhausts all possible ordered dyads; that is, every possible combination of two pitches is found in these sets.
a(A008594(n)) = 0;
a(A227144(n)) = 1;
a(3*A047522(n)) = 2;
a(A017569(n)) = a(2*A016933(n)) = a(4*A016777(n)) = 3;
a(2*A017629(n)) = a(3*A017137(n)) = a(6*A004767(n)) = 4;
a(A227146(n)) = 5;
a(nonexistent) = 6;
a(2*A017581(n)) = 7;
a(2*A017557(n)) = a(4*A016813(n)) = 8;
a(A017617(n)) = a(2*A016957(n)) = a(4*A016789(n)) = 9;
a(3*A047621(n)) = 10;
a(2*A017653(n)) = 11. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1199
Ron Knott, Fibonacci Numbers and the Golden Section
M. Renault, The Fibonacci Sequence Modulo M
A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quarterly, Vol.6, No.2 (1968), 139-141.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Index entries for linear recurrences with constant coefficients, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1).

Cf. A000045, A001175, A003893, A010881.

a089911 n = a089911_list !! n
a089911_list = 0 : 1 : zipWith (\u v -> (u + v) `mod` 12)
                       (tail a089911_list) a089911_list
-- Reinhard Zumkeller, Jul 01 2013

[Fibonacci(n) mod 12: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

with(combinat,fibonacci); A089911 := proc(n) fibonacci(n) mod 12; end;

Table[Mod[Fibonacci[n], 12], {n, 0, 100}] (* Vincenzo Librandi, Feb 04 2014 *)

a(n)=fibonacci(n)%12 \\ Charles R Greathouse IV, Feb 03 2014

Has period of 24, restricted period 12 and multiplier 5.

a(n) = (a(n-1) + a(n-2)) mod 12, a(0) = 0, a(1) = 1.

More terms from Ray Chandler, Nov 15 2003

A108618 A quaternion-generated sequence calculated using the rules given in the comment box with initial seed x = .5'i + .5'j + .5'k + .5e; version: "tes".

Original entry on oeis.org

1, 2, -1, -2, -3, -6, -6, 1, 4, 3, 0, -5, -10, -8, 3, 8, 5, -2, -9, -12, -6, 7, 16, 10, -9, -18, -11, 4, 15, 14, -2, -16, -20, -3, 14, 17, 6, -12, -24, -11, 10, 21, 14, -8, -22, -20, 3, 20, 17, -2, -21, -24, -6, 19, 28, 10, -21, -36, -18, 19, 40, 22, -21, -42, -23, 16, 39, 26, -14, -40, -32, 9, 38, 29, -8, -39, -36, 2, 36, 38, -1, -38

Offset: 0

Creighton Dement, Jun 12 2005

Set y = x = .5'i + .5'j + .5'k + .5e Define a(0) = 1 (this is twice the coefficient of the unit e in x), then "loop" steps 1-5, below. a(n) is given by twice the coefficient of e (the unit) in y from step 4 inside the n-th loop. Step 1 (Loop 1): Calculate x*y Result: x*y = .5'i + .5'j + .5'k - .5e Step 2 (Loop 1): Add the fractional parts of the real coefficient basis vectors of x*y (i.e. 'i, 'j, 'k, e) Result: .5 + .5 + .5 - .5 = 1 = s Step 3 (Loop 1): Calculate x + x*y + se Result .5'i + .5'j + .5'k + .5e + (.5'i + .5'j + .5'k - .5e) + se = 'i + 'j + 'k + e. Step 4 (Loop 1): Set y equal to the result from Step 3. Result: y = 'i + 'j + 'k + e; thus a(1) = 2*1 = 2 Step 5 (Loop 1): Return to Step 1 Step 1 (Loop 2): Result: x*y = 'i + 'j + 'k - e Step 2 (Loop 2): Result: s = 0 Step 3 (Loop 2): 1.5'i + 1.5'j + 1.5'k -.5e Step 4 (Loop 2): y = 1.5'i + 1.5'j + 1.5'k -.5e; thus a(2) = 2*(-.5) = -1 **Loop 1** + 'i + 'j + 'k + e **Loop 2** + 1.5'i + 1.5'j + 1.5'k - .5e **Loop 3** + 'i + 'j + 'k - e **Loop 4** + .5'i + .5'j + .5'k - 1.5e **Loop 5** - 3e **Loop 6** - 'i - 'j - 'k - 3e **Loop 7** - 1.5'i - 1.5'j - 1.5'k + .5e **Loop 8** + 2e **Loop 9** + 1.5'i + 1.5'j + 1.5'k + 1.5e **Loop 10** + 2'i + 2'j + 2'k **Loop 11** + 1.5'i + 1.5'j + 1.5'k - 2.5e **Loop 12** - 5e
Notice the horizontal line segments in the graph of (a(n)) against the natural numbers. These may be referred to as "Gerald's diamonds" (after Gerald McGarvey, who pointed them out shortly after this sequence was submitted). It could be an interesting task to find the approximate area of these diamonds and compare to the approximate area of the other diamonds.
From Benoit Jubin, Aug 12 2009: (Start)
Define the function f on the integers to be the odd function such that for n>=0, f(2n)=0 and f(2n+1)=1. Define the sequences a and b by
a(0)=b(0)=0,
a(n+1) = 1 + (a(n)-3b(n))/2 + f((a(n)-3b(n))/2) + 3 f((a(n)+b(n))/2),
b(n+1) = 1 + (a(n)+b(n))/2.
Then (with an offset shifted by 1), a=A108618 and b=A108619. (End)

Creighton Dement, Table of n, a(n) for n = 0..10000
Creighton Dement, Plot of A108618 against A108619 (patch on)
Creighton Dement, Plot of A108618 against A108619 (patch off)
Creighton Dement, Floretions 2009, DRAFT. See section 4.1 Algorithms.
Rémy Sigrist, Colored scatterplot of a(n) for n = 0..10000 (where the color is function of n mod 6)

Cf. A108619, A108620, A108621, A272693.

a[0] = b[0] = 0;
f[n_] := Sign[n]*Mod[n, 2];
a[n_] := a[n] = (1/2)*(a[n-1] - 3*b[n-1]) + 3*f[(1/2)*(a[n-1] + b[n-1])] + f[(1/2)*(a[n-1] - 3*b[n-1])] + 1;
b[n_] := b[n] = (1/2)*(a[n-1] + b[n-1]) + 1;
A108618 = Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Feb 25 2015, after Benoit Jubin *)

A261422 Number of ordered triples (u,v,w) of palindromes such that u+v+w=n.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 63, 72, 79, 84, 87, 88, 87, 84, 79, 72, 66, 55, 51, 45, 40, 36, 33, 31, 30, 30, 30, 33, 27, 34, 33, 33, 33, 33, 33, 33, 33, 33, 36, 27, 39, 36, 36, 36, 36, 36, 36, 36, 36, 39, 27, 45, 39, 39, 39, 39, 39, 39, 39, 39, 42, 27, 52, 42, 42, 42, 42, 42, 42, 42, 42, 45

Offset: 0

N. J. A. Sloane, Aug 27 2015

It is known that a(n)>0 for all n.

			4 can be written as a sum of three palindromes in 15 ways: 4+0+0 (3 ways), 3+1+0 (6 ways), 2+2+0 (3 ways), and 2+1+1 (3 ways), so a(4)=15.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000 (See also the extended file with 200000 terms below)
William D. Banks, Every natural number is the sum of forty-nine palindromes, arXiv:1508.04721 [math.NT], 2015. [Establishes the much weaker result that the coefficients of P(x)^49 are positive (see Formula section below).]
Javier Cilleruelo and Florian Luca, Every positive integer is a sum of three palindromes, arXiv: 1602.06208 [math.NT], 2016-2017.
Erich Friedman, Problem of the Month (June 1999)
N. J. A. Sloane, Table of n, a(n) for n=0..200000
N. J. A. Sloane, Maple program to produce 200000 terms

Cf. A002113. Differs from A261132, which assumes 0 <= u <= v <= w.

For records see A262544, A262545.

(* This program is not suitable to compute a large number of terms. *)
compositions[n_, k_] := Flatten[Permutations[PadLeft[#, k]]& /@ IntegerPartitions[n, k], 1];
a[n_] := Select[compositions[n, 3], AllTrue[#, PalindromeQ]&] // Length;
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 05 2018 *)

G.f. = P(x)^3, where P(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^11 + x^22 + x^33 + x^44 + x^55 + x^66 + x^77 + x^88 + x^99 + x^101 + x^111 + ... = Sum_{palindromes p} x^p.

A255582 a(n)=n when n <= 3, otherwise a(n) is the smallest positive number not yet in the sequence such that gcd(a(n), a(n-1)) <= gcd(a(n), a(n-2)) > 1.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 12, 5, 14, 15, 7, 18, 21, 16, 27, 20, 33, 24, 11, 26, 22, 13, 28, 39, 32, 30, 34, 25, 17, 35, 51, 40, 42, 38, 36, 19, 44, 57, 46, 45, 23, 48, 69, 50, 54, 52, 58, 56, 29, 49, 87, 63, 60, 77, 62, 55, 31, 65, 93, 70, 66, 64, 74, 68, 37

Offset: 1

Bob Selcoe, Feb 26 2015

This is a permutation of the natural numbers: the proof for A098550 applies with essentially no changes. [Confirmed by N. J. A. Sloane, Feb 27 2015]
For n > 3, all primes first appear in order as composites with one smaller prime (proof similar to that in A098550).
For any given set S of primes, the subsequence consisting of numbers whose prime factors are exactly the primes in S appears in increasing order. For example, if S = {2,3}, 6 appears first, followed by 12, 18, 24, 36, 48, 54, 72, etc.
Appears to be very similar to A064413. Compare the respective inverses A255479 and A064664; see also A255482. Speaking very loosely, the ratio a(n)/n seems to be about 1/2, 1, or 3/2, just as for A064413, although this is a long way from being proved for either sequence. David Applegate points out that this is (presumably) because primes p >= 13 always occur as part of a subsequence 2p X p Y 3p, and subsequences 2p X p Y 5p, 2p X p Y 7p, etc. that produced the extra curves in the graph of A098550 just do not happen. - N. J. A. Sloane, Feb 27 2015, Mar 05 2015.
First differs from A254077 at a(29). - Omar E. Pol, May 21 2015

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..10000
David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669, 2015.

Cf. A098550, A254077, A064413, A064664, A255480, A255481, A255482, A256974.
A255479 is the inverse permutation.
A256424 is a smoothed version.
A256529 gives the partial sums.

import Data.List (delete)
a255582 n = a255582_list !! (n-1)
a255582_list = 1 : 2 : 3 : f 2 3 [4..] where
   f u v ws = y : f v y (delete y ws) where
              y = head [z | z <- ws, let d = gcd u z, d > 1, gcd v z <= d]
-- Reinhard Zumkeller, Mar 10 2015

a[n_] := a[n] = If[n<5, n, For[k=5, True, k++, If[FreeQ[Array[a, n-1], k], If[GCD[k, a[n-2]]>1 && GCD[k, a[n-1]] <= GCD[k, a[n-2]], Return[k]]]]];
Array[a, 100] (* Jean-François Alcover, Jul 31 2018 *)

a(41)-a(67) from Hiroaki Yamanouchi, Feb 27 2015

A333216 Lengths of maximal subsequences without adjacent equal terms in the sequence of prime gaps.

Original entry on oeis.org

2, 13, 21, 3, 7, 8, 1, 18, 29, 5, 3, 8, 11, 31, 4, 20, 3, 7, 5, 19, 21, 32, 1, 19, 48, 19, 29, 32, 7, 38, 1, 43, 12, 33, 46, 6, 16, 8, 4, 34, 15, 1, 19, 7, 1, 23, 28, 30, 22, 8, 1, 7, 1, 52, 14, 56, 10, 26, 2, 30, 65, 5, 71, 12, 44, 39, 37, 6, 19, 47, 11, 10

Offset: 1

Gus Wiseman, Mar 15 2020

Prime gaps are differences between adjacent prime numbers.

Essentially the same as A145024. - R. J. Mathar, Mar 16 2020

			The prime gaps split into the following subsequences without adjacent equal terms: (1,2), (2,4,2,4,2,4,6,2,6,4,2,4,6), (6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,10,2,6), (6,4,6), (6,2,10,2,4,2,12), (12,4,2,4,6,2,10,6), ...

First differences of A064113.
The version for the Kolakoski sequence is A306323.
The weakly decreasing version is A333212.
The weakly increasing version is A333215.
The strictly decreasing version is A333252.
The strictly increasing version is A333253.
The equal version is A333254.
Cf. A000040, A001223, A084758, A106356, A124762, A124767, A333214.

Length/@Split[Differences[Array[Prime,100]],UnsameQ]//Most

Ones correspond to balanced prime quartets (A054800), so the sum of terms up to but not including the n-th one is A000720(A054800(n - 1)) = A090832(n).

A325225 Lesser of the number of prime factors of n counted with multiplicity and the maximum prime index of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 3, 2, 2, 1, 2, 2, 2, 2, 3, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 3, 1, 3, 1, 3, 3, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 4, 2, 2, 1, 3, 1, 2, 3, 1, 2, 3, 1, 3, 2, 3, 1, 2, 1, 2, 3, 3, 2, 3, 1, 3, 2, 2, 1, 4, 2, 2, 2, 4, 1, 3, 2, 3, 2, 2, 2, 2, 1, 3, 3, 3, 1, 3, 1, 4, 3

Offset: 1

Gus Wiseman, Apr 12 2019

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			88 has 4 prime indices {1,1,1,5}, the maximum of which is 5, so a(88) = min(4,5) = 4.

Antti Karttunen, Table of n, a(n) for n = 1..65537
FindStat, St000533: The maximal number of non-attacking rooks on a Ferrers shape

Positions of 1's are A174090. Positions of 2's are A325229.

Cf. A001222, A056239, A061395, A065770, A093641, A112798, A252464, A257541, A257990, A263297, A325169, A325224, A325227, A325232.

Table[Min[PrimeOmega[n],PrimePi[FactorInteger[n][[-1,1]]]],{n,100}]

A061395(n) = if(1==n, 0, primepi(vecmax(factor(n)[, 1])));
A325225(n) = min(bigomega(n), A061395(n)); \\ Antti Karttunen, Apr 14 2019

a(n) = min(A001222(n), A061395(n)).

More terms from Antti Karttunen, Apr 14 2019

A093873 Numerators in Kepler's tree of harmonic fractions.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 4, 3, 4, 2, 5, 3, 5, 1, 4, 3, 4, 2, 5, 3, 5, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 6, 5, 6, 4, 9, 5, 9, 3, 10, 7, 10, 4, 11, 7, 11, 2, 9, 7, 9, 5, 12, 7, 12, 3, 11, 8, 11, 5, 13, 8, 13, 1, 6

Offset: 1

N. J. A. Sloane and Reinhard Zumkeller, May 24 2004

Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j).

			The first few fractions are:
1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ...
1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8

R. Zumkeller, Table of n, a(n) for n = 1..10000
Johannes Kepler, Harmonices Mundi, Liber III, see p. 27.
Index entries for fraction trees
Index entries for sequences related to music

The denominators are in A093875. Usually one only considers the left-hand half of the tree, which gives the fractions A020651/A086592. See A086592 for more information, references to Kepler, etc.

See A294442 for another version of Kepler's tree of fractions.

{-# LANGUAGE ViewPatterns #-}
import Data.Ratio((%), numerator, denominator)
rat :: Rational -> (Integer,Integer)
rat r = (numerator r, denominator r)
data Harmony = Harmony Harmony Rational Harmony
rows :: Harmony -> [[Rational]]
rows (Harmony hL r hR) = [r] : zipWith (++) (rows hL) (rows hR)
kepler :: Rational -> Harmony
kepler r = Harmony (kepler (i%(i+j))) r (kepler (j%(i+j)))
           where (rat -> (i,j)) = r
-- Full tree of Kepler's harmonic fractions:
k = rows $ kepler 1 :: [[Rational]] -- as list of lists
h = concat k :: [Rational] -- flattened
a093873 n = numerator $ h !! (n - 1)
a093875 n = denominator $ h !! (n - 1)
a011782 n = numerator $ (map sum k) !! n -- denominator == 1
-- length (k !! n) == 2^n
-- numerator $ (map last k) !! n == fibonacci (n + 1)
-- denominator $ (map last k) !! n == fibonacci (n + 2)
-- numerator $ (map maximum k) !! n == n
-- denominator $ (map maximum k) !! n == n + 1
-- eop.
-- Reinhard Zumkeller, Oct 17 2010

M:= 8: # to get a(1) .. a(2^M-1)
gen[1]:= [1];
for n from 2 to M do
  gen[n]:= map(t -> (numer(t)/(numer(t)+denom(t)),
      denom(t)/(numer(t)+denom(t))), gen[n-1]);
od:
seq(op(map(numer,gen[i])),i=1..M): # Robert Israel, Jan 11 2016

num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093873 = Table[num[n], {n, 1, 97}] (* Jean-François Alcover, Dec 16 2011 *)

a(n) = a([n/2])*(1 - n mod 2) + A093875([n/2])*(n mod 2).
a(A029744(n-1)) = 1; a(A070875(n-1)) = 2; a(A123760(n-1)) = 3. - Reinhard Zumkeller, Oct 13 2006
A011782(k) = SUM(a(n)/A093875(n): 2^k<=n<2^(k+1)), k>=0. [Reinhard Zumkeller, Oct 17 2010]
a(1) =  1. For all n>0  a(2n) =  a(n), a(2n+1) =  A093875(n). - Yosu Yurramendi, Jan 09 2016
a(4n+3) = a(4n+1), a(4n+2) = a(4n+1) - a(4n), a(4n+1) = A071585(n). - Yosu Yurramendi, Jan 11 2016
G.f. G(x) satisfies G(x) = x + (1+x) G(x^2) + Sum_{k>=2} x (1+x^(2^(k-1))) G(x^(2^k)). - Robert Israel, Jan 11 2016
a(2^(m+1)+k) = a(2^(m+1)+2^m+k) = A020651(2^m+k), m>=0, 0<=k<2^m. - Yosu Yurramendi, May 18 2016
a(k) = A020651(2^(m+1)+k) - A020651(2^m+k), m>0, 0Yosu Yurramendi, Jun 01 2016
a(2^(m+1)+k) - a(2^m+k) = a(k) , m >=0, 0 <= k < 2^m. For k=0 a(0)=0 is needed. - Yosu Yurramendi, Jul 22 2016
a(2^(m+2)-1-k) = a(2^(m+1)-1-k) + a(2^m-1-k), m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Jul 22 2016
a(2^m-1-(2^r -1)) = A000045(m-r), m >= 1, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
a(2^m+2^r) = m-r, , m >= 1, 0 <= r <= m-1 ; a(2^m+2^r+2^(r-1)) = m-(r-1), m >= 2, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
A093875(2n) - a(2n) = A093875(n), n > 0; A093875(2n+1) - a(2n+1) = a(n), n > 0. - Yosu Yurramendi, Jul 23 2016

A248034 a(n+1) gives the number of occurrences of the last digit of a(n) so far, up to and including a(n), with a(0)=0.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 4, 10, 5, 5, 6, 5, 7, 5, 8, 5, 9, 5, 10, 6, 6, 7, 6, 8, 6, 9, 6, 10, 7, 7, 8, 7, 9, 7, 10, 8, 8, 9, 8, 10, 9, 9, 10

Offset: 0

Eric Angelini and M. F. Hasler, Oct 11 2014

In other words, the number to the right of a comma gives the number of occurrences of the digit immediately to the left of the comma, counting from the beginning up to that digit or comma.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Eric Angelini, Digit-counters updating themselves, SeqFan list, Oct 11 2014.
Zak Seidov, Graph of 10^6 terms

Cf. A249068 (analogous sequence in base 8).
Cf. A249009 (analogous sequence which uses the first, not the last digit).
Cf. A249069, A249070, A249144, A249148, A364788.

a:= proc(n) option remember; `if`(n=0, 0,
      coeff(b(n-1), x, irem(a(n-1), 10)))
    end:
b:= proc(n) option remember; `if`(n=0, 1, b(n-1)+
      add(x^i, i=convert(a(n), base, 10)))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Oct 18 2014

nn = 120; a[0] = j = 0; c[] := 0; Do[Map[c[#]++ &, IntegerDigits[j]]; a[n] = j = c[Mod[j, 10]], {n, nn}]; Array[a, nn, 0] (* _Michael De Vlieger, Aug 07 2023 *)

c=vector(10);print1(a=0);for(n=1,99,apply(d->c[d+1]++,if(a,digits(a)));print1(","a=c[1+a%10]))
(MIT/GNU Scheme)
;; An implementation of memoization-macro definec can be found for example from: http://oeis.org/wiki/Memoization
(definec (A248034 n) (if (zero? n) n (vector-ref (A248034aux_digit_counts (- n 1)) (modulo (A248034 (- n 1)) 10))))
(definec (A248034aux_digit_counts n) (cond ((zero? n) (vector 1 0 0 0 0 0 0 0 0 0)) (else (let loop ((digcounts-for-n (vector-copy (A248034aux_digit_counts (- n 1)))) (n (A248034 n))) (cond ((zero? n) digcounts-for-n) (else (vector-set! digcounts-for-n (modulo n 10) (+ 1 (vector-ref digcounts-for-n (modulo n 10)))) (loop digcounts-for-n (floor->exact (/ n 10)))))))))
;; Antti Karttunen, Oct 22 2014

from itertools import islice
def A248034_gen(): # generator of terms
    c, clist = 0, [1]+[0]*9
    while True:
        yield c
        c = clist[c%10]
        for d in str(c):
            clist[int(d)] += 1
A248034_list = list(islice(A248034_gen(),30)) # Chai Wah Wu, Dec 13 2022

A268289 a(0)=0; thereafter a(n) = a(n-1) - A037861(n).

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 4, 7, 5, 5, 5, 7, 7, 9, 11, 15, 12, 11, 10, 11, 10, 11, 12, 15, 14, 15, 16, 19, 20, 23, 26, 31, 27, 25, 23, 23, 21, 21, 21, 23, 21, 21, 21, 23, 23, 25, 27, 31, 29, 29, 29, 31, 31, 33, 35, 39, 39, 41, 43, 47, 49

Offset: 0

Mark Moore, Jan 30 2016

The graph of this sequence is related to the Takagi (blancmange) curve: see Lagarias (2012), Section 9, especially Theorem 9.1. [Corrected by Laura Monroe, Oct 21 2020]
Theorem: a(n) is the cardinality of the set { 1<= m <= n, ((n-m) mod 2^floor(log_2(m)+1)) < 2^floor(log_2(m)) }. See links.
From Laura Monroe, Jun 11 2020: (Start)
Consider a full balanced binary tree with n unlabeled leaves such that for each internal node, the number of leaf descendants of the two children differs by at most 1. Call a tree with this even distribution of leaves "pairwise".
Apply labels to the internal nodes, where an internal node is labeled S if its two children have the same number of leaf descendants, and D if its two children have a different number of leaf descendants, and call this an SD-tree. (For a pairwise tree, this is equivalent to saying that a node is an S-node iff it has an even number of leaf descendants.)
a(n) is then the number of S-nodes on a pairwise SD-tree with n+1 leaves.
This is proved in Props. 17 and 18 of the Monroe et al. article in the links.
One example of such a tree is the summation tree generated by a pairwise summation on n+1 summands (see example below). Another example is the tree representing a neutral single-elimination tournament on n+1 teams, as in A096351.
(End)
From Laura Monroe, Oct 23 2020: (Start)
Subtracting a(n) from n gives a sequence of dilations of increasing length on the dyadic rational points of the Takagi function. The number of points in each dilation is 2^k and the scale of each dilation in both the x and y directions is 2^k, where k = floor(log_2(n+1)).
2^(a(n)) is the number of tree automorphisms on the pairwise (i.e., divide-and-conquer) tree with n+1 leaves.
(End)

			From _Laura Monroe_, Jun 11 2020: (Start)
For n=2, the pairwise summation on 2+1=3 summands takes the form ((a+b)+c). The corresponding summation tree and SD-tree look like:
       +            D
      / \          / \
     +   c        S   c
    / \          / \
   a   b        a   b
and exactly 1 internal node has an even number of leaf descendants, hence is an S-node.
For n=3, the pairwise summation on 3+1=4 summands takes the form ((a+b)+(c+d)). The corresponding summation tree and SD-tree look like:
       +            S
      / \          / \
     +   +        S   S
    /|   |\      /|   |\
   a b   c d    a b   c d
and exactly 3 internal nodes have an even number of leaf descendants, hence are S-nodes.
(End)

Laura Monroe, Table of n, a(n) for n = 0..16383 (first 10000 terms from N. J. A. Sloane)
Thomas Baruchel, Flattening Karatsuba's recursion tree into a single summation, arXiv:1902.08982 [math.NT], 2019. See (5) conjecture of theorem.
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019. See 2.4 proof of theorem.
Thomas Baruchel, Flattening Karatsuba's Recursion Tree into a Single Summation, SN Computer Science (2020) Vol. 1, Article No. 48.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Jeffrey C. Lagarias, The Takagi function and its properties, arXiv:1112.4205 [math.CA], 2011-2012.
Jeffrey C. Lagarias, The Takagi function and its properties, In Functions in number theory and their probabilistic aspects, 153--189, RIMS Kôkyûroku Bessatsu, B34, Res. Inst. Math. Sci. (RIMS), Kyoto, 2012. MR3014845.
Laura Monroe, A Few Identities of the Takagi Function on Dyadic Rationals, arXiv:2111.05996 [math.CO], 2021.
Laura Monroe, Takagi Function Identities on Dyadic Rationals, J. Int. Seq (2024) Vol. 27, Art. 24.2.7.
Laura Monroe and Vanessa Job, Computationally Inequivalent Summations and Their Parenthetic Forms, arXiv:2005.05387 [cs.DM], 2020. See Prop. 26, the alternate proof.

Partial sums of A145037.

Cf. A000120, A000788, A023416, A037861, A096351, A181132, A269735, A233271, A296062.

int a(int n)   {
    int m=n+1;
    int result=0;
    int i=0;
    while (n) {
        int ith_bit_set = m&(1<>= 1;
    }
   return result;
}
/* Laura Monroe, Jun 17 2020 */

function A268289List(len)
    A = zeros(Int, len)
    for n in 1:len-1
        a, b, c = n, n & 1, 1
        while (a >>= 1) != 0
            b += a & 1
            c += 1
        end
        A[n+1] = A[n] + <<(b, 1) - c
    end
    A
end; println(A268289List(61)) # Peter Luschny, Jun 22 2020

a000120 := proc(n) add(i, i=convert(n, base, 2)) end:
a023416 := proc(n) if n = 0 then 1; else add(1-e, e=convert(n, base, 2)) ; end if; end proc:
a268289:=proc(n) option remember; global a000120, a023416;
if n=0 then 0 else a268289(n-1)+a000120(n)-a023416(n); fi; end;
[seq(a268289(n),n=0..132)];
# N. J. A. Sloane, Mar 07 2016
# second Maple program:
a:= proc(n) option remember; `if`(n<0, 0,
      a(n-1)+add(2*i-1, i=Bits[Split](n)))
    end:
seq(a(n), n=0..60);  # Alois P. Heinz, Jan 18 2022

Join[{0}, Table[DigitCount[n, 2, 1] - DigitCount[n, 2, 0], {n, 1, 100}] // Accumulate] (* Jean-François Alcover, Oct 24 2016 *)

a(n) = if (n==0, 0, if (n%2, 2*a((n-1)/2)+1, a(n/2) + a(n/2-1))); \\ Michel Marcus, Jun 16 2020

a(n) = my(v=binary(n+1),s=-1); for(i=1,#v, v[i]=if(v[i],s++,s--;1)); fromdigits(v,2); \\ Kevin Ryde, Jun 16 2020

def A268289(n): return (sum(i.bit_count() for i in range(1,n+1))<<1)-1-(n+1)*(m:=(n+1).bit_length())+(1<Chai Wah Wu, Mar 01 2023

def A268289(n): return sum((n+1)%m if (n+1)&(m:=1<Chai Wah Wu, Nov 11 2024

From N. J. A. Sloane, Mar 11 2016: (Start)
a(0)=0; for n > 0, a(n) = a(n-1) + A000120(n) - A023416(n) = A000788(n) - A181132(n).
a(0)=0; thereafter a(2*n) = a(n) + a(n-1), a(2*n+1) = 2*a(n) + 1.
G.f.: (1/(1-x)^2) * Sum_{k >= 0} x^(2^k)*(1-x^(2^k))/(1+x^(2^k)).
a(2^k-1) = 2^k-1, a(3*2^k-1) = 2^(k+1)-1, a(5*2^k-1) = 3*2^k-1, etc.
(End)
From Laura Monroe, Jun 11 2020: (Start)
a(n-1) = Sum_{i=0..floor(log_2(n))} (((floor(n/(2^i))+1) mod 2)*(2^i)+(-1)^((floor(n/(2^i))+1) mod 2)*(n mod (2^i))), for n>=1.
This is an explicit formula for this sequence, and is O(log(n)). This formula is proven in Prop. 18, in the Monroe et al. reference in the links. (End)
From Laura Monroe, Oct 23 2020: (Start)
a(n) = n - A296062(n).
a(n+1) = (n+1) - (2^k)*tau(x/(2^k)), where tau is the Takagi function and n+1 = (2^k)+x with x < 2^k. (End)

Simplified definition following a suggestion from Michel Marcus. Corrected start, added more terms. - N. J. A. Sloane, Mar 07 2016

cp's OEIS Frontend

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A089911 a(n) = Fibonacci(n) mod 12.

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A108618 A quaternion-generated sequence calculated using the rules given in the comment box with initial seed x = .5'i + .5'j + .5'k + .5e; version: "tes".

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A261422 Number of ordered triples (u,v,w) of palindromes such that u+v+w=n.

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A255582 a(n)=n when n <= 3, otherwise a(n) is the smallest positive number not yet in the sequence such that gcd(a(n), a(n-1)) <= gcd(a(n), a(n-2)) > 1.

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A333216 Lengths of maximal subsequences without adjacent equal terms in the sequence of prime gaps.

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A325225 Lesser of the number of prime factors of n counted with multiplicity and the maximum prime index of n.

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