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 20 results. Next

A227740 Integers from 0 to A037834(n) followed by integers from 0 to A037834(n+1) and so on.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jul 25 2013

Keywords

Comments

Equivalently, integers from 0 to A005811(n)-1 followed by integers from 0 to A005811(n+1)-1 and so on.

Crossrefs

Programs

  • Mathematica
    Table[Range[0, #] &@ Total@ Flatten@ Map[Abs@ Differences@ # &,
    Partition[IntegerDigits[n, 2], 2, 1]], {n, 34}] // Flatten (* Michael De Vlieger, May 09 2017 *)
  • Scheme
    (define (A227740 n) (- n (+ 1 (A173318 (- (A227737 n) 1)))))

Formula

a(n) = n - (1 + A173318(A227737(n)-1)).

A005811 Number of runs in binary expansion of n (n>0); number of 1's in Gray code for n.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 5, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 5, 4, 3, 4, 5, 6, 5, 6, 7, 6, 5, 4, 5, 6, 5, 4, 5
Offset: 0

Views

Author

Keywords

Comments

Starting with a(1) = 0 mirror all initial 2^k segments and increase by one.
a(n) gives the net rotation (measured in right angles) after taking n steps along a dragon curve. - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
This sequence generates A082410: (0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) and A014577; identical to the latter except starting 1, 1, 0, ...; by writing a "1" if a(n+1) > a(n); if not, write "0". E.g., A014577(2) = 0, since a(3) < a(2), or 1 < 2. - Gary W. Adamson, Sep 20 2003
Starting with 1 = partial sums of A034947: (1, 1, -1, 1, 1, -1, -1, 1, 1, 1, ...). - Gary W. Adamson, Jul 23 2008
The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.
Can be used as a binomial transform operator: Let a(n) = the n-th term in any S(n); then extract 2^k strings, adding the terms. This results in the binomial transform of S(n). Say S(n) = 1, 3, 5, ...; then we obtain the strings: (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), ...; = the binomial transform of (1, 3, 5, ...) = (1, 4, 12, 32, 80, ...). Example: the 8-bit string has a sum of 32 with a distribution of (1, 3, 3, 1) or one 1, three 3's, three 5's, and one 7; as expected. - Gary W. Adamson, Jun 21 2012
Considers all positive odd numbers as nodes of a graph. Two nodes are connected if and only if the sum of the two corresponding odd numbers is a power of 2. Then a(n) is the distance between 2n + 1 and 1. - Jianing Song, Apr 20 2019

Examples

			Considered as a triangle with 2^k terms per row, the first few rows are:
  1
  2, 1
  2, 3, 2, 1
  2, 3, 4, 3, 2, 3, 2, 1
  ...
The n-th row becomes right half of next row; left half is mirrored terms of n-th row increased by one. - _Gary W. Adamson_, Jun 20 2012
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A037834 (-1), A088748 (+1), A246960 (mod 4), A034947 (first differences), A000975 (indices of record highs), A173318 (partial sums).
Partial sums of A112347. Recursion depth of A035327.

Programs

  • Haskell
    import Data.List (group)
    a005811 0 = 0
    a005811 n = length $ group $ a030308_row n
    a005811_list = 0 : f [1] where
       f (x:xs) = x : f (xs ++ [x + x `mod` 2, x + 1 - x `mod` 2])
    -- Reinhard Zumkeller, Feb 16 2013, Mar 07 2011
    
  • Maple
    A005811 := proc(n)
        local i, b, ans;
        if n = 0 then
            return 0 ;
        end if;
        ans := 1;
        b := convert(n, base, 2);
        for i from nops(b)-1 to 1 by -1 do
            if b[ i+1 ]<>b[ i ] then
                ans := ans+1
            fi
        od;
        return ans ;
    end proc:
    seq(A005811(i), i=1..50) ;
    # second Maple program:
    a:= n-> add(i, i=Bits[Split](Bits[Xor](n, iquo(n, 2)))):
    seq(a(n), n=0..100);  # Alois P. Heinz, Feb 01 2023
  • Mathematica
    Table[ Length[ Length/@Split[ IntegerDigits[ n, 2 ] ] ], {n, 1, 255} ]
    a[n_] := DigitCount[BitXor[n, Floor[n/2]], 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 11 2024 *)
  • PARI
    a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2))
    
  • PARI
    a(n)=if(n<1,0,a(n\2)+(a(n\2)+n)%2) \\ Benoit Cloitre, Jan 20 2014
    
  • PARI
    a(n) = hammingweight(bitxor(n, n>>1));  \\ Gheorghe Coserea, Sep 03 2015
    
  • Python
    def a(n): return bin(n^(n>>1))[2:].count("1") # Indranil Ghosh, Apr 29 2017

Formula

a(2^k + i) = a(2^k - i + 1) + 1 for k >= 0 and 0 < i <= 2^k. - Reinhard Zumkeller, Aug 14 2001
a(2n+1) = 2a(n) - a(2n) + 1, a(4n) = a(2n), a(4n+2) = 1 + a(2n+1).
a(j+1) = a(j) + (-1)^A014707(j). - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^2^k/(1+x^2^(k+1)). - Ralf Stephan, May 02 2003
Delete the 0, make subsets of 2^n terms; and reverse the terms in each subset to generate A088696. - Gary W. Adamson, Oct 19 2003
a(0) = 0, a(2n) = a(n) + [n odd], a(2n+1) = a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(n) = Sum_{k=1..n} (-1)^((k/2^A007814(k)-1)/2) = Sum_{k=1..n} (-1)^A025480(k-1). - Ralf Stephan, Oct 29 2003
a(n) = A069010(n) + A033264(n). - Ralf Stephan, Oct 29 2003
a(0) = 0 then a(n) = a(floor(n/2)) + (a(floor(n/2)) + n) mod 2. - Benoit Cloitre, Jan 20 2014
a(n) = A037834(n) + 1.
a(n) = A000120(A003188(n)). - Amiram Eldar, Jul 11 2024

Extensions

Additional description from Wouter Meeussen

A297330 Total variation of base-10 digits of n; see Comments.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 8, 7, 6, 5, 4, 3, 2
Offset: 1

Views

Author

Clark Kimberling, Jan 17 2018

Keywords

Comments

Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). Guide to related sequences and partitions of the natural numbers:
***
Base b {DV(n,b)} {UV(n,b)} {TV(n,b)}
For each b, let u = {n : UV(n,b) < DV(n,b)}, e = {n : UV(n,b) = DV(n,b)}, and d = {n : UV(n,b) > DV(n,b)}. The sets u,e,d partition the natural numbers. A guide to the matching sequences for u, e, d follows:
***
Base b Sequence u Sequence e Sequence d
2 A005843 A005408 (none)
Not a duplicate of A151950: e.g., a(100)=1 but A151950(100)=11. - Robert Israel, Feb 06 2018

Examples

			13684632 has DV = 8-4 + 6-3 + 3-2 = 8 and has UV = 3-1 + 6-3 + 8-6 + 6-4 = 9, so that a(13684632) = DV + UV = 17.
		

Crossrefs

Programs

  • Maple
    f:= proc(n) local L,i; L:= convert(n,base,10);
    add(abs(L[i+1]-L[i]),i=1..nops(L)-1) end proc:
    map(f, [$1..100]); # Robert Israel, Feb 04 2018
    # alternative
    A297330 := proc(n)
        A037860(n)+A037851(n) ;
    end proc: # R. J. Mathar, Sep 27 2021
  • Mathematica
    b = 10; z = 120; t = Table[Total@Flatten@Map[Abs@Differences@# &, Partition[ IntegerDigits[n, b], 2, 1]], {n, z}] (* after Michael De Vlieger, e.g. A037834 *)
  • Python
    def A297330(n):
        s = str(n)
        return sum(abs(int(s[i])-int(s[i+1])) for i in range(len(s)-1)) # Chai Wah Wu, May 31 2022

A278219 Filter-sequence related to base-2 run-length encoding: a(n) = A046523(A243353(n)).

Original entry on oeis.org

1, 2, 4, 2, 4, 8, 6, 2, 4, 12, 16, 8, 6, 12, 6, 2, 4, 12, 36, 12, 16, 32, 24, 8, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 16, 48, 64, 32, 24, 72, 24, 8, 6, 30, 60, 30, 24, 48, 60, 12, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 36, 180, 144, 72, 60, 180, 60, 12, 16, 48, 144, 48, 64, 128, 96, 32, 24, 120, 216, 72, 24, 72
Offset: 0

Views

Author

Antti Karttunen, Nov 16 2016

Keywords

Crossrefs

Other base-2 related filter sequences: A278217, A278222.
Sequences that (seem to) partition N into same or coarser equivalence classes are at least these: A005811, A136004, A033264, A037800, A069010, A087116, A090079 and many others like A105500, A106826, A166242, A246960, A277561, A037834, A225081 although these have not been fully checked yet.

Programs

  • Mathematica
    f[n_, i_, x_] := Which[n == 0, x, EvenQ@ n, f[n/2, i + 1, x], True, f[(n - 1)/2, i, x Prime@ i]]; g[n_] := If[n == 1, 1, Times @@ MapIndexed[ Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]]];
    Table[g@ f[BitXor[n, Floor[n/2]], 1, 1], {n, 0, 93}] (* Michael De Vlieger, May 09 2017 *)
  • Python
    from sympy import prime, factorint
    import math
    def A(n): return n - 2**int(math.floor(math.log(n, 2)))
    def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
    def a005940(n): return b(n - 1)
    def P(n):
        f = factorint(n)
        return sorted([f[i] for i in f])
    def a046523(n):
        x=1
        while True:
            if P(n) == P(x): return x
            else: x+=1
    def a003188(n): return n^int(n/2)
    def a243353(n): return a005940(1 + a003188(n))
    def a(n): return a046523(a243353(n)) # Indranil Ghosh, May 07 2017
  • Scheme
    (define (A278219 n) (A046523 (A243353 n)))
    

Formula

a(n) = A046523(A243353(n)).
a(n) = A278222(A003188(n)).
a(n) = A278220(1+A075157(n)).

A082410 a(1)=0. Thereafter, the sequence is constructed using the rule: for any k >= 0, if a(1), a(2), ..., a(2^k+1) are known, the next 2^k terms are given as follows: a(2^k+1+i) = 1 - a(2^k+1-i) for 1 <= i <= 2^k.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Apr 24 2003

Keywords

Comments

a(n) is A014577 shifted right twice (the definition here is similar to one of the constructions for A034947). - N. J. A. Sloane, Jul 27 2012
Complement of characteristic function of A060833.
From Tanya Khovanova, Apr 21 2020: (Start)
Suppose you have a deck of cards face down with 2^n cards such that the color pattern corresponds to this sequence: 0 for one color, 1 for the other. Then you proceed in the following manner: transfer to top card to the bottom of the deck, deal the next card, then repeat. The dealt cards will have alternating colors.
Even terms of this sequence alternate: 1, 0, 1, 0 and so on.
Removing even-indexed terms doesn't change the sequence. (End)

Examples

			First 3 terms are 0,1,1; therefore, a(4) = a(3+1) = 1 - a(3-1) = 1 - a(2) = 0, a(5) = a(3+2) = 1 - a(3-2) = 1 - a(1) = 1 and the sequence begins 0, 1, 1, 0, 1, ...
		

Crossrefs

The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

Programs

  • Python
    def A082410(n):
        if n == 1:
            return 0
        s = bin(n-1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

For n >= 2, Sum_{k=1..n} a(k) = (n + A037834(n-1))/2.
a(1) = 0, a(4*n+2) = 1, a(4*n+4) = 0, a(2*n+1) = a(n+1) for n >= 0. - A.H.M. Smeets, Jul 27 2018

A173318 Partial sums of A005811.

Original entry on oeis.org

0, 1, 3, 4, 6, 9, 11, 12, 14, 17, 21, 24, 26, 29, 31, 32, 34, 37, 41, 44, 48, 53, 57, 60, 62, 65, 69, 72, 74, 77, 79, 80, 82, 85, 89, 92, 96, 101, 105, 108, 112, 117, 123, 128, 132, 137, 141, 144, 146, 149, 153, 156, 160, 165, 169, 172, 174, 177, 181, 184, 186, 189
Offset: 0

Views

Author

Jonathan Vos Post, Feb 16 2010

Keywords

Comments

Partial sums of number of runs in binary expansion of n (n>0). Partial sums of number of 1's in Gray code for n. The subsequence of squares in this partial sum begins 0, 1, 4, 9, 144, 169, 256, 289, 324 (since we also have 32 and 128 I wonder about why so many powers). The subsequence of primes in this partial sum begins: 3, 11, 17, 29, 31, 37, 41, 53, 79, 89, 101, 137, 149, 181, 191, 197, 229, 271.
Note: A227744 now gives the squares, which occur at positions given by A227743. - Antti Karttunen, Jul 27 2013

Examples

			1 has 1 run in its binary representation "1".
2 has 2 runs in its binary representation "10".
3 has 1 run in its binary representation "11".
4 has 2 runs in its binary representation "100".
5 has 3 runs in its binary representation "101".
Thus a(1) = 1, a(2) = 1+2 = 3, a(3) = 1+2+1 = 4, a(4) = 1+2+1+2 = 6, a(5) = 1+2+1+2+3 = 9.
		

Crossrefs

Cf. also A227737, A227741, A227742.
Cf. A227744 (squares occurring), A227743 (indices of squares).

Programs

  • Mathematica
    Accumulate[Join[{0},Table[Length[Split[IntegerDigits[n,2]]],{n,110}]]] (* Harvey P. Dale, Jul 29 2013 *)
  • PARI
    a(n) = my(v=binary(n+1),d=0,e=4); for(i=1,#v, if(v[i], v[i]=#v-i+d;d+=e;e=0, e=4)); fromdigits(v,2)>>1; \\ Kevin Ryde, Aug 27 2021

Formula

a(n) = sum(i=0..n) A005811(i) = sum(i=0..n) (A037834(i)+1) = sum(i=0..n) (A069010(i) + A033264(i)).
a(A000225(n)) = A001787(n) = A000788(A000225(n)). - Antti Karttunen, Jul 27 2013 & Aug 09 2013
a(2n) = 2*a(n) + n - 2*(ceiling(A005811(n)/2) - (n mod 2)), a(2n+1) = 2*a(n) + n + 1. - Ralf Stephan, Aug 11 2013

A297233 Total variation of base-11 digits of n; see Comments.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2
Offset: 1

Views

Author

Clark Kimberling, Jan 17 2018

Keywords

Comments

Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See A297330 for a guide to related sequences and partitions of the natural numbers:

Examples

			2^20 in base 11:  6, 5, 6, 8, 10, 1; here, DV = 12 and UV = 5, so that a(2^20) = 17.
		

Crossrefs

Programs

  • Mathematica
    b = 11; z = 120; t = Table[Total@Flatten@Map[Abs@Differences@# &,      Partition[IntegerDigits[n, b], 2, 1]], {n, z}] (* cf. Michael De Vlieger, e.g. A037834 *)

A297239 Total variation of base-13 digits of n; see Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 17 2018

Keywords

Comments

Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See A297330 for a guide to related sequences and partitions of the natural numbers:

Examples

			2^20 in base 13:  2, 10, 9, 3, 7, 9; here, DV = 12 and UV = 9, so that a(2^20) = 21.
		

Crossrefs

Programs

  • Mathematica
    b = 13; z = 120; t = Table[Total@Flatten@Map[Abs@Differences@# &,      Partition[IntegerDigits[n, b], 2, 1]], {n, z}] (* cf. Michael De Vlieger, e.g. A037834 *)

A297242 Total variation of base-14 digits of n; see Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 17 2018

Keywords

Comments

Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See A297330 for a guide to related sequences and partitions of the natural numbers:

Examples

			2^20 in base 14:  1, 13, 4, 1, 12, 4; here, DV = 20 and UV = 23, so that a(2^20) = 43.
		

Crossrefs

Programs

  • Mathematica
    b = 14; z = 120; t = Table[Total@Flatten@Map[Abs@Differences@# &,      Partition[IntegerDigits[n, b], 2, 1]], {n, z}] (* cf. Michael De Vlieger, e.g. A037834 *)

A297245 Total variation of base-15 digits of n; see Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 17 2018

Keywords

Comments

Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See A297330 for a guide to related sequences and partitions of the natural numbers:

Examples

			2^20 in base 15:  1, 5, 10, 10, 5, 1; here, DV = 9 and UV = 9, so that a(2^20) = 18.
		

Crossrefs

Programs

  • Mathematica
    b = 15; z = 120; t = Table[Total@Flatten@Map[Abs@Differences@# &,      Partition[IntegerDigits[n, b], 2, 1]], {n, z}] (* cf. Michael De Vlieger, e.g. A037834 *)
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