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 142 results. Next

A143589 Kolakoski fan based on A000034 with initial row 1.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Aug 25 2008

Keywords

Comments

Conjecture (following Benoit Cloitre's conjecture at A111090): if L(n) is the number (assumed finite) of terms in row n of K, then L(n)*(2/3)^n approaches a constant. (L= A143590.)

Examples

			s=(1,2,1,2,1,2,1,2,...) and w=1, so the first 7 rows are
1
2
1 1
2 1
1 1 2
2 1 2 2
1 1 2 1 1 2 2
		

Crossrefs

Formula

Introduced here is an array K called the "Kolakoski fan based on a sequence s with initial row w": suppose that s=(s(1),s(2),...) is a sequence of 1's and 2's and that w=(w(1),w(2),...) is a finite or infinite sequence of 1's and 2's. Assume that s(1)=w(1) and that if w(1)=1 then s contains at least one 2. Row 1 of the array K is w. Subsequent rows are defined inductively: the first term of row n is s(n) and the remaining terms are defined by Kolakoski substitution; viz., each number in row n-1 tells the string-length (1 or 2) of the next string in row n, each term being either 1 or 2.

A191662 a(n) = n! / A000034(n-1).

Original entry on oeis.org

1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000
Offset: 1

Views

Author

Paul Curtz, Jun 10 2011

Keywords

Comments

The a(n) are the denominators in the formulas of the k-dimensional square pyramidal numbers:
A005408 = (2*n+1)/1 = 1, 3, 5, 7, 9, ... (k=1)
A000290 = (n^2)/1 = 1, 4, 9, 16, 25, ... (k=2)
A000330 = n*(n+1)*(2*n+1)/6 = 1, 5, 14, 30, 55, ... (k=3)
A002415 = (n^2)*(n^2-1)/12 = 1, 6, 20, 50, 105, ... (k=4)
A005585 = n*(n+1)*(n+2)*(n+3)*(2*n+3)/120 = 1, 7, 27, 77, 182, ... (k=5)
A040977 = (n^2)*(n^2-1)*(n^2-4)/360 = 1, 8, 35, 112, 294, ... (k=6)
A050486 (k=7), A053347 (k=8), A054333 (k=9), A054334 (k=10), A057788 (k=11).
The first superdiagonal of this array appears in A029651. - Paul Curtz, Jul 04 2011
The general formula for the k-dimensional square pyramidal numbers is (2*n+k)*binomial(n+k-1,k-1)/k, k >= 1, n >= 0, see A097207. - Johannes W. Meijer, Jun 22 2011

Crossrefs

Programs

Formula

a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
a(n+1) = A064680(n+1) * a(n).
From Amiram Eldar, Jul 06 2022: (Start)
Sum_{n>=1} 1/a(n) = sinh(1) + 2*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = sinh(1) - 2*cosh(1) + 2. (End)
D-finite with recurrence: a(n) - (n-1)*n*a(n-2) = 0 for n >= 3 with a(1)=a(2)=1. - Georg Fischer, Nov 25 2022
a(n) = A052612(n)/2 for n >= 1. - Alois P. Heinz, Sep 05 2023

Extensions

More terms from Harvey P. Dale, Mar 14 2014

A113497 Ascending descending base exponent transform of sequence A000034(n) = 1 + n mod 2.

Original entry on oeis.org

1, 3, 6, 6, 11, 9, 16, 12, 21, 15, 26, 18, 31, 21, 36, 24, 41, 27, 46, 30, 51, 33, 56, 36, 61, 39, 66, 42, 71, 45, 76, 48, 81, 51, 86, 54, 91, 57, 96, 60, 101, 63, 106, 66, 111, 69, 116, 72, 121, 75, 126, 78, 131, 81, 136, 84, 141, 87, 146, 90, 151, 93, 156, 96, 161, 99, 166, 102, 171
Offset: 1

Views

Author

Jonathan Vos Post, Jan 10 2006

Keywords

Comments

A000034 = 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ... = continued fraction for (sqrt(3)+1)/2 (cf. A040001) = base 3 digital root of n+1. In general, the ascending descending base exponent transform of any simple periodic sequence can be written as a periodic set of interleaved sequences.

Examples

			a(1) = 1^1 = 1.
a(2) = 1^2 + 2^1 = 3.
a(3) = 1^1 + 2^2 + 1^1 = 6.
a(4) = 1^2 + 2^1 + 1^2 + 2^1 = 6.
a(5) = 1^1 + 2^2 + 1^1 + 2^2 + 1^1 = 11.
a(6) = 1^2 + 2^1 + 1^2 + 2^1 + 1^2 + 2^1 = 9.
		

Crossrefs

Programs

  • Mathematica
    Table[(-3 + 3*(-1)^n + 8*n - 2*(-1)^n*n)/4, {n,1,50}] (* G. C. Greubel, Mar 12 2017 *)
  • PARI
    x='x +O('x^50); Vec(x*(1+3*x+4*x^2)/((1-x)^2*(1+x)^2)) \\ G. C. Greubel, Mar 12 2017

Formula

a(n) = Sum_{i=1..n} A000034(i)^A000034(n-i+1).
a(2*n) = 3*n; a(2*n+1) = 5*n+1.
From Colin Barker, Jun 16 2012: (Start)
a(n) = (-3+3*(-1)^n+8*n-2*(-1)^n*n)/4.
a(n) = 2*a(n-2)-a(n-4).
G.f.: x*(1+3*x+4*x^2)/((1-x)^2*(1+x)^2). (End)
E.g.f.: (1/2)*(3*(x-1)*sinh(x) + 5*x*cosh(x)). - G. C. Greubel, Mar 12 2017

Extensions

Definition improved by M. F. Hasler, Jan 13 2012

A143110 Triangle read by rows, A051731 * A000034 * 0^(n-k), 1<=k<=n.

Original entry on oeis.org

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

Views

Author

Gary W. Adamson & Mats Granvik, Jul 25 2008

Keywords

Comments

Row sums = A069735: (1, 3, 2, 5, 2, 6, 2, 7,...).
T(n,k) = 2 if k is even and divides n, 1 if k is odd and divides n; 0 otherwise.

Examples

			First few rows of the triangle are:
1;
1, 2;
1, 0, 1;
1, 2, 0, 2;
1, 0, 0, 0, 1;
1, 2, 1, 0, 0, 2;
1, 0, 0, 0, 0, 0, 1;
...
		

Crossrefs

Formula

Triangle read by rows, A051731 * A000034 * 0^(n-k), 1<=k<=n; where A051731 = the inverse Mobius transform and A000034 = (1, 2, 1, 2, 1, 2,...).

A203150 (n-1)-st elementary symmetric function of the first n terms of (1,2,1,2,1,2,1,2,1,2,...)=A000034.

Original entry on oeis.org

1, 3, 5, 12, 16, 36, 44, 96, 112, 240, 272, 576, 640, 1344, 1472, 3072, 3328, 6912, 7424, 15360, 16384, 33792, 35840, 73728, 77824, 159744, 167936, 344064, 360448, 737280, 770048, 1572864, 1638400, 3342336, 3473408, 7077888, 7340032
Offset: 1

Views

Author

Clark Kimberling, Dec 29 2011

Keywords

Examples

			Let esf abbreviate "elementary symmetric function".  Then
0th esf of {1}:  1
1st esf of {1,2}:  1+2=3
2nd esf of {1,2,1} is 1*2+1*1+2*1=5
		

Crossrefs

Cf. A000034, A167667 (bisection?), A053220 (bisection?)

Programs

  • Mathematica
    f[k_] := 1 + Mod[k + 1, 2];
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 33}]  (* A203150 *)

Formula

Empirical G.f.: x*(1+3*x+x^2)/(1-4*x^2+4*x^4). - Colin Barker, Jan 03 2012
Conjecture: a(n) = (6*r*n+(1+3*(1-r)*n)*(1-(-1)^n))*r^(n-1)/8, where r=sqrt(2). - Bruno Berselli, Jan 03 2011

A144021 Eigentriangle by rows, T(n,k) = A000034(n-k+1)*A105476(k-1).

Original entry on oeis.org

1, 2, 1, 1, 2, 3, 2, 1, 6, 6, 1, 2, 3, 12, 15, 2, 1, 6, 6, 30, 33, 1, 2, 3, 12, 15, 66, 78, 2, 1, 6, 6, 30, 33, 156, 177, 1, 2, 3, 12, 15, 66, 78, 354, 411, 2, 1, 6, 6, 30, 33, 156, 177, 822, 942
Offset: 1

Views

Author

Gary W. Adamson, Sep 07 2008

Keywords

Comments

Row sums = A105476: (1, 3, 6, 15, 33, 78,...).
Left column = A000034: (1, 2, 1, 2, 1, 2,...).
Right border = A105476 shifted: (1, 1, 3, 6, 15, 33, 78,...).

Examples

			First few rows of the triangle =
1;
2, 1;
1, 2, 3
2, 1, 6, 6
1, 2, 3, 12, 15
2, 1, 6, 6, 30, 33;
1, 2, 3, 12, 15, 66, 78
2, 1, 6, 6, 30, 33, 156, 177;
1, 2, 3, 12, 15, 66, 78, 354, 411;
...
Row 4 = (2, 1, 6, 6) = termwise product of (2, 1, 2, 1) and (1, 1, 3, 6) = (2*1, 1*1, 2*3, 1*6).
		

Crossrefs

Formula

Eigentriangle by rows, A000034(n-k+1)*A105476(k-1); where A105476(k-1) = A105476 shifted = (1, 1, 3, 6, 15, 33, 78, 177,...).

A169590 Triangle T(n,k) with : column n = A000034 if n even and column n = A000007 if n odd.

Original entry on oeis.org

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

Views

Author

Philippe Deléham, Dec 02 2009

Keywords

Comments

Row sums : A145051.

Examples

			Triangle begins: 1 ; 2,1 ; 1,0,1 ; 2,0,2,1 ; 1,0,1,0,1 ; 2,0,2,0,2,1 ; ...
		

Formula

Sum_{k, 0<=k<=n} T(n,k)= A145051(n+1).

A001477 The nonnegative integers.

Original entry on oeis.org

0, 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, 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, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 0

Views

Author

Keywords

Comments

Although this is a list, and lists normally have offset 1, it seems better to make an exception in this case. - N. J. A. Sloane, Mar 13 2010
The subsequence 0,1,2,3,4 gives the known values of n such that 2^(2^n)+1 is a prime (see A019434, the Fermat primes). - N. J. A. Sloane, Jun 16 2010
Also: The identity map, defined on the set of nonnegative integers. The restriction to the positive integers yields the sequence A000027. - M. F. Hasler, Nov 20 2013
The number of partitions of 2n into exactly 2 parts. - Colin Barker, Mar 22 2015
The number of orbits of Aut(Z^7) as function of the infinity norm n of the representative lattice point of the orbit, when the cardinality of the orbit is equal to 8960 or 168.- Philippe A.J.G. Chevalier, Dec 29 2015
Partial sums give A000217. - Omar E. Pol, Jul 26 2018
First differences are A000012 (the "all 1's" sequence). - M. F. Hasler, May 30 2020
See A061579 for the transposed infinite square matrix, or triangle with rows reversed. - M. F. Hasler, Nov 09 2021
This is the unique sequence (a(n)) that satisfies the inequality a(n+1) > a(a(n)) for all n in N. This simple and surprising result comes from the 6th problem proposed by Bulgaria during the second day of the 19th IMO (1977) in Belgrade (see link and reference). - Bernard Schott, Jan 25 2023

Examples

			Triangular view:
   0
   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  32  33  34  35
  36  37  38  39  40  41  42  43  44
  45  46  47  48  49  50  51  52  53  54
		

References

  • Maurice Protat, Des Olympiades à l'Agrégation, suite vérifiant f(n+1) > f(f(n)), Problème 7, pp. 31-32, Ellipses, Paris 1997.

Crossrefs

Cf. A000027 (n>=1).
Cf. A000012 (first differences).
Partial sums of A057427. - Jeremy Gardiner, Sep 08 2002
Cf. A038608 (alternating signs), A001787 (binomial transform).
Cf. A055112.
Cf. Boustrophedon transforms: A231179, A000737.
Cf. A245422.
Number of orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008586, A008585, A005843, A000217.
When written as an array, the rows/columns are A000217, A000124, A152948, A152950, A145018, A167499, A166136, A167487... and A000096, A034856, A055998, A046691, A052905, A055999... (with appropriate offsets); cf. analogous lists for A000027 in A185787.
Cf. A000290.
Cf. A061579 (transposed matrix / reversed triangle).

Programs

Formula

a(n) = n.
a(0) = 0, a(n) = a(n-1) + 1.
G.f.: x/(1-x)^2.
Multiplicative with a(p^e) = p^e. - David W. Wilson, Aug 01 2001
When seen as array: T(k, n) = n + (k+n)*(k+n+1)/2. Main diagonal is 2*n*(n+1) (A046092), antidiagonal sums are n*(n+1)*(n+2)/2 (A027480). - Ralf Stephan, Oct 17 2004
Dirichlet generating function: zeta(s-1). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: x*e^x. - Franklin T. Adams-Watters, Sep 11 2005
a(0)=0, a(1)=1, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
Alternating partial sums give A001057 = A000217 - 2*(A008794). - Eric Desbiaux, Oct 28 2008
a(n) = 2*A080425(n) + 3*A008611(n-3), n>1. - Eric Desbiaux, Nov 15 2009
a(n) = A007966(n)*A007967(n). - Reinhard Zumkeller, Jun 18 2011
a(n) = Sum_{k>=0} A030308(n,k)*2^k. - Philippe Deléham, Oct 20 2011
a(n) = 2*A028242(n-1) + (-1)^n*A000034(n-1). - R. J. Mathar, Jul 20 2012
a(n+1) = det(C(i+1,j), 1 <= i, j <= n), where C(n,k) are binomial coefficients. - Mircea Merca, Apr 06 2013
a(n-1) = floor(n/e^(1/n)) for n > 0. - Richard R. Forberg, Jun 22 2013
a(n) = A000027(n) for all n>0.
a(n) = floor(cot(1/(n+1))). - Clark Kimberling, Oct 08 2014
a(0)=0, a(n>0) = 2*z(-1)^[( |z|/z + 3 )/2] + ( |z|/z - 1 )/2 for z = A130472(n>0); a 1 to 1 correspondence between integers and naturals. - Adriano Caroli, Mar 29 2015
G.f. as triangle: x*(1 + (x^2 - 5*x + 2)*y + x*(2*x - 1)*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Jul 22 2025

A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Also called fusc(n) [Dijkstra].
a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf].
If the terms are written as an array:
column 0 1 2 3 4 5 6 7 8 9 ...
row 0: 0
row 1: 1
row 2: 1,2
row 3: 1,3,2,3
row 4: 1,4,3,5,2,5,3,4
row 5: 1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5
row 6: 1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,...
...
then (ignoring row 0) the sum of the k-th row is 3^(k-1), each column is an arithmetic progression and the steps are nothing but the original sequence. - Takashi Tokita (butaneko(AT)fa2.so-net.ne.jp), Mar 08 2003
From N. J. A. Sloane, Oct 15 2017: (Start)
The above observation can be made more precise. Let A(n,k), n >= 0, 0 <= k <= 2^(n-1)-1 for k > 0, denote the entry in row n and column k of the left-justified array above.
The equations for columns 0,1,2,3,4,... are successively (ignoring row 0):
1 (n >= 1),
n (n >= 2),
n-1 (n >= 3),
2n-3 (n >= 3),
n-2 (n >= 4),
3n-7 (n >= 4),
...
and in general column k > 0 is given by
A(n,k) = a(k)*n - A156140(k) for n >= ceiling(log_2(k+1))+1, and 0 otherwise.
(End)
a(n) is the number of odd Stirling numbers S_2(n+1, 2r+1) [Carlitz].
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 - x). The successor function f(x) = 1/(floor(x) + 1 - frac(x)) can also be used.
a(n+1) = number of alternating bit sets in n [Finch].
If f(x) = 1/(1 + floor(x) - frac(x)) then f(a(n-1)/a(n)) = a(n)/a(n+1) for n >= 1. If T(x) = -1/x and f(x) = y, then f(T(y)) = T(x) for x > 0. - Michael Somos, Sep 03 2006
a(n+1) is the number of ways of writing n as a sum of powers of 2, each power being used at most twice (the number of hyperbinary representations of n) [Carlitz; Lind].
a(n+1) is the number of partitions of the n-th integer expressible as the sum of distinct even-subscripted Fibonacci numbers (= A054204(n)), into sums of distinct Fibonacci numbers [Bicknell-Johnson, theorem 2.1].
a(n+1) is the number of odd binomial(n-k, k), 0 <= 2*k <= n. [Carlitz], corrected by Alessandro De Luca, Jun 11 2014
a(2^k) = 1. a(3*2^k) = a(2^(k+1) + 2^k) = 2. Sequences of terms between a(2^k) = 1 and a(2^(k+1)) = 1 are palindromes of length 2^k-1 with a(2^k + 2^(k-1)) = 2 in the middle. a(2^(k-1) + 1) = a(2^k - 1) = k+1 for k > 1. - Alexander Adamchuk, Oct 10 2006
The coefficients of the inverse of the g.f. of this sequence form A073469 and are related to binary partitions A000123. - Philippe Flajolet, Sep 06 2008
It appears that the terms of this sequence are the number of odd entries in the diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Aug 06 2009
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column shifted down twice:
1;
1, 0;
1, 1, 0;
0, 1, 0, 0;
0, 1, 1, 0, 0;
0, 0, 1, 0, 0, 0;
0, 0, 1, 1, 0, 0, 0;
...
Then this sequence A002487 (without initial 0) is the first column of lim_{n->oo} M^n. (Cf. A026741.) - Gary W. Adamson, Dec 11 2009 [Edited by M. F. Hasler, Feb 12 2017]
Member of the infinite family of sequences of the form a(n) = a(2*n); a(2*n+1) = r*a(n) + a(n+1), r = 1 for A002487 = row 1 in the array of A178239. - Gary W. Adamson, May 23 2010
Equals row 1 in an infinite array shown in A178568, sequences of the form
a(2*n) = r*a(n), a(2*n+1) = a(n) + a(n+1); r = 1. - Gary W. Adamson, May 29 2010
Row sums of A125184, the Stern polynomials. Equivalently, B(n,1), the n-th Stern polynomial evaluated at x = 1. - T. D. Noe, Feb 28 2011
The Kn1y and Kn2y triangle sums, see A180662 for their definitions, of A047999 lead to the sequence given above, e.g., Kn11(n) = A002487(n+1) - A000004(n), Kn12(n) = A002487(n+3) - A000012(n), Kn13(n) = A002487(n+5) - A000034(n+1) and Kn14(n) = A002487(n+7) - A157810(n+1). For the general case of the knight triangle sums see the Stern-Sierpiński triangle A191372. This triangle not only leads to Stern's diatomic series but also to snippets of this sequence and, quite surprisingly, their reverse. - Johannes W. Meijer, Jun 05 2011
Maximum of terms between a(2^k) = 1 and a(2^(k+1)) = 1 is the Fibonacci number F(k+2). - Leonid Bedratyuk, Jul 04 2012
Probably the number of different entries per antidiagonal of A223541. That would mean there are exactly a(n+1) numbers that can be expressed as a nim-product 2^x*2^y with x + y = n. - Tilman Piesk, Mar 27 2013
Let f(m,n) be the frequency of the integer n in the interval [a(2^(m-1)), a(2^m-1)]. Let phi(n) be Euler's totient function (A000010). Conjecture: for all integers m,n n<=m f(m,n) = phi(n). - Yosu Yurramendi, Sep 08 2014
Back in May 1995, it was proved that A000360 is the modulo 3 mapping, (+1,-1,+0)/2, of this sequence A002487 (without initial 0). - M. Jeremie Lafitte (Levitas), Apr 24 2017
Define a sequence chf(n) of Christoffel words over an alphabet {-,+}: chf(1) = '-'; chf(2*n+0) = negate(chf(n)); chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))). Then the length of the chf(n) word is fusc(n) = a(n); the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n); the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below. - I. V. Serov, Jun 01 2017
The sequence can be extended so that a(n) = a(-n), a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1) for all n in Z. - Michael Somos, Jun 25 2019
Named after the German mathematician Moritz Abraham Stern (1807-1894), and sometimes also after the French clockmaker and amateur mathematician Achille Brocot (1817-1878). - Amiram Eldar, Jun 06 2021
It appears that a(n) is equal to the multiplicative inverse of A007305(n+1) mod A007306(n+1). For example, a(12) is 2, the multiplicative inverse of A007305(13) mod A007306(13), where A007305(13) is 4 and A007306(13) is 7. - Gary W. Adamson, Dec 18 2023

Examples

			Stern's diatomic array begins:
  1,1,
  1,2,1,
  1,3,2,3,1,
  1,4,3,5,2,5,3,4,1,
  1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,
  1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,...
  ...
a(91) = 19, because 91_10 = 1011011_2; b_6=b_4=b_3=b_1=b_0=1, b_5=b_2=0;  L=5; m_1=0, m_2=1, m_3=3, m_4=4, m_5=6; c_1=2, c_2=3, c_3=2, c_4=3; f(1)=1, f(2)=2, f(3)=5, f(4)=8, f(5)=19. - _Yosu Yurramendi_, Jul 13 2016
From _I. V. Serov_, Jun 01 2017: (Start)
a(n) is the length of the Christoffel word chf(n):
n  chf(n) A070939(n)   a(n)
1   '-'       1          1
2   '+'       2          1
3   '+-'      2          2
4   '-'       3          1
5   '--+'     3          3
6   '-+'      3          2
... (End)
G.f. = x + x^2 + 2*x^3 + x^4 + 3*x^5 + 2*x^6 + 3*x^7 + x^8 + ... - _Michael Somos_, Jun 25 2019
		

References

  • M. Aigner and G. M. Ziegler, Proofs from THE BOOK, 3rd ed., Berlin, Heidelberg, New York: Springer-Verlag, 2004, p. 97.
  • Elwyn R. Berlekamp, John H. Conway and Richard K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 114.
  • Krishna Dasaratha, Laure Flapan, Chansoo Lee, Cornelia Mihaila, Nicholas Neumann-Chun, Sarah Peluse and Matthew Stroegeny, A family of multi-dimensional continued fraction Stern sequences, Abtracts Amer. Math. Soc., Vol. 33 (#1, 2012), #1077-05-2543.
  • Edsger W. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232 (sequence is called fusc).
  • F. G. M. Eisenstein, Eine neue Gattung zahlentheoretischer Funktionen, welche von zwei Elementen abhaengen und durch gewisse lineare Funktional-Gleichungen definirt werden, Verhandlungen der Koenigl. Preuss. Akademie der Wiss. Berlin (1850), pp. 36-42, Feb 18, 1850. Werke, II, pp. 705-711.
  • Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.16.3; pp. 148-149.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 117.
  • Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Record values are in A212289.
If the 1's are replaced by pairs of 1's we obtain A049456.
Inverse: A020946.
Cf. a(A001045(n)) = A000045(n). a(A062092(n)) = A000032(n+1).
Cf. A064881-A064886 (Stern-Brocot subtrees).
A column of A072170.
Cf. A049455 for the 0,1 version of Stern's diatomic array.
Cf. A000119, A262097 for analogous sequences in other bases and A277189, A277315, A277328 for related sequences with similar graphs.
Cf. A086592 and references therein to other sequences related to Kepler's tree of fractions.

Programs

  • Haskell
    a002487 n = a002487_list !! n
    a002487_list = 0 : 1 : stern [1] where
       stern fuscs = fuscs' ++ stern fuscs' where
         fuscs' = interleave fuscs $ zipWith (+) fuscs $ (tail fuscs) ++ [1]
       interleave []     ys = ys
       interleave (x:xs) ys = x : interleave ys xs
    -- Reinhard Zumkeller, Aug 23 2011
    
  • Julia
    using Nemo
    function A002487List(len)
        a, A = QQ(0), [0,1]
        for n in 1:len
            a = next_calkin_wilf(a)
            push!(A, denominator(a))
        end
    A end
    A002487List(91) |> println # Peter Luschny, Mar 13 2018
    
  • Magma
    [&+[(Binomial(k, n-k-1) mod 2): k in [0..n]]: n in [0..100]]; // Vincenzo Librandi, Jun 18 2019
    
  • Maple
    A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end: seq(A002487(n),n=0..91);
    A002487 := proc(m) local a,b,n; a := 1; b := 0; n := m; while n>0 do if type(n,odd) then b := a+b else a := a+b end if; n := floor(n/2); end do; b; end proc: seq(A002487(n),n=0..91); # Program adapted from E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. - Igor Urbiha (urbiha(AT)math.hr), Oct 28 2002. Since A007306(n) = a(2*n+1), this program can be adapted for A007306 by replacing b := 0 by b := 1.
    A002487 := proc(n::integer) local k; option remember; if n = 0 then 0 elif n=1 then 1 else add(K(k,n-1-k)*procname(n - k), k = 1 .. n) end if end proc:
    K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n and n-k <= 2 then KC:=1; else KC:= 0; end if; end proc: seq(A002487(n),n=0..91); # Thomas Wieder, Jan 13 2008
    # next Maple program:
    a:= proc(n) option remember; `if`(n<2, n,
          (q-> a(q)+(n-2*q)*a(n-q))(iquo(n, 2)))
        end:
    seq(a(n), n=0..100);  # Alois P. Heinz, Feb 11 2021
    fusc := proc(n) local a, b, c; a := 1; b := 0;
        for c in convert(n, base, 2) do
            if c = 0 then a := a + b else b := a + b fi od;
        b end:
    seq(fusc(n), n = 0..91); # Peter Luschny, Nov 09 2022
    Stern := proc(n, u) local k, j, b;
        b := j -> nops({seq(Bits:-Xor(k, j-k), k = 0..j)}):
        ifelse(n=0, 1-u, seq(b(j), j = 2^(n-1)-1..2^n-1-u)) end:
    seq(print([n], Stern(n, 1)), n = 0..5); # As shown in the comments.
    seq(print([n], Stern(n, 0)), n = 0..5); # As shown in the examples. # Peter Luschny, Sep 29 2024
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[ EvenQ[n], a[n/2], a[(n-1)/2] + a[(n+1)/2]]; Table[ a[n], {n, 0, 100}] (* end of program *)
    Onemore[l_] := Transpose[{l, l + RotateLeft[l]}] // Flatten;
    NestList[Onemore, {1}, 5] // Flatten  (*gives [a(1), ...]*) (* Takashi Tokita, Mar 09 2003 *)
    ToBi[l_] := Table[2^(n - 1), {n, Length[l]}].Reverse[l]; Map[Length,
    Split[Sort[Map[ToBi, Table[IntegerDigits[n - 1, 3], {n, 500}]]]]]  (*give [a(1), ...]*) (* Takashi Tokita, Mar 10 2003 *)
    A002487[m_] := Module[{a = 1, b = 0, n = m}, While[n > 0, If[OddQ[n], b = a+b, a = a+b]; n = Floor[n/2]]; b]; Table[A002487[n], {n, 0, 100}] (* Jean-François Alcover, Sep 06 2013, translated from 2nd Maple program *)
    a[0] = 0; a[1] = 1;
    Flatten[Table[{a[2*n] = a[n], a[2*n + 1] = a[n] + a[n + 1]}, {n, 0, 50}]] (* Horst H. Manninger, Jun 09 2021 *)
    nmax = 100; CoefficientList[Series[x*Product[(1 + x^(2^k) + x^(2^(k+1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2022 *)
  • PARI
    {a(n) = n=abs(n); if( n<2, n>0, a(n\2) + if( n%2, a(n\2 + 1)))};
    
  • PARI
    fusc(n)=local(a=1,b=0);while(n>0,if(bitand(n,1),b+=a,a+=b);n>>=1);b \\ Charles R Greathouse IV, Oct 05 2008
    
  • PARI
    A002487(n,a=1,b=0)=for(i=0,logint(n,2),if(bittest(n,i),b+=a,a+=b));b \\ M. F. Hasler, Feb 12 2017, updated Feb 14 2019
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def a(n): return n if n<2 else a(n//2) if n%2==0 else a((n - 1)//2) + a((n + 1)//2) # Indranil Ghosh, Jun 08 2017; corrected by Reza K Ghazi, Dec 27 2021
    
  • Python
    def a(n):
        a, b = 1, 0
        while n > 0:
            if n & 1:
                b += a
            else:
                a += b
            n >>= 1
        return b
    # Reza K Ghazi, Dec 29 2021
    
  • Python
    def A002487(n): return sum(int(not (n-k-1) & ~k) for k in range(n)) # Chai Wah Wu, Jun 19 2022
    
  • Python
    # (fast way for big vectors)
    from math import log, ceil
    import numpy
    how_many_terms = 2**20  # (Powers of 2 recommended but other integers are also possible.)
    A002487, A002487[1]  = numpy.zeros(2**(ce:=ceil(log(how_many_terms,2))), dtype=object), 1
    for exponent in range(1,ce):
        L, L2 = 2**exponent, 2**(exponent+1)
        A002487[L2 - 1] = exponent + 1
        A002487[L:L2][::2] = A002487[L >> 1: L]
        A002487[L + 1:L2 - 2][::2] = A002487[L:L2 - 3][::2]  +  A002487[L + 2:L2 - 1][::2]
    print(list(A002487[0:100])) # Karl-Heinz Hofmann, Jul 22 2025
  • R
    N <- 50 # arbitrary
    a <- 1
    for (n in 1:N)
    {
      a[2*n    ] = a[n]
      a[2*n + 1] = a[n] + a[n+1]
      a
    }
    a
    # Yosu Yurramendi, Oct 04 2014
    
  • R
    # Given n, compute a(n) by taking into account the binary representation of n
    a <- function(n){
      b <- as.numeric(intToBits(n))
      l <- sum(b)
      m <- which(b == 1)-1
      d <- 1
      if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
      f <- c(0,1)
      if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
      return(f[l+1])
    } # Yosu Yurramendi, Dec 13 2016
    
  • R
    # computes the sequence as a vector A, rather than function a() as above.
    A <- c(1,1)
    maxlevel <- 5 # by choice
    for(m in 1:maxlevel) {
      A[2^(m+1)] <- 1
      for(k in 1:(2^m-1)) {
        r <- m - floor(log2(k)) - 1
        A[2^r*(2*k+1)] <- A[2^r*(2*k)] + A[2^r*(2*k+2)]
    }}
    A # Yosu Yurramendi, May 08 2018
    
  • Sage
    def A002487(n):
        M = [1, 0]
        for b in n.bits():
            M[b] = M[0] + M[1]
        return M[1]
    print([A002487(n) for n in (0..91)])
    # For a dual see A174980. Peter Luschny, Nov 28 2017
    
  • Scheme
    ;; An implementation of memoization-macro definec can be found for example in: http://oeis.org/wiki/Memoization
    (definec (A002487 n) (cond ((<= n 1) n) ((even? n) (A002487 (/ n 2))) (else (+ (A002487 (/ (- n 1) 2)) (A002487 (/ (+ n 1) 2))))))
    ;; Antti Karttunen, Nov 05 2016
    

Formula

a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). - David S. Newman, Mar 04 2001
Let e(n) = A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002
Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? - Benoit Cloitre, Jan 18 2004. Coons and Tyler show the limit is A246765 = 0.9588... - Kevin Ryde, Jan 09 2021
a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). - Paul Barry, Sep 13 2004
a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). - Paul Barry, Mar 26 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. - Michael Somos, May 02 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. - Michael Somos, May 02 2005
G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). - Mike Stay, Nov 06 2006
A079978(n) = (1 + e^(i*Pi*A002487(n)))/2, i=sqrt(-1). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at A002083 = Narayana-Zidek-Capell numbers.) - Thomas Wieder, Jan 13 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) = A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008
a(n) = A126606(n + 1) / 2. - Reikku Kulon, Oct 05 2008
Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. - Mats Granvik and Gary W. Adamson, Oct 02 2009; corrected by Mats Granvik, Oct 10 2009
a(2^(p+2)*n+2^(p+1)-1) - a(2^(p+1)*n+2^p-1) = A007306(n+1), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 07 2013
a(2*n-1) = A007306(n), n > 0. - Yosu Yurramendi, Jun 23 2014
a(n*2^m) = a(n), m>0, n > 0. - Yosu Yurramendi, Jul 03 2014
a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) = A007306(j), j > 0, k >= 0 (see Adamchuk's comment). - Yosu Yurramendi, Mar 05 2016
a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. - Yosu Yurramendi, Jul 13 2016
From Yosu Yurramendi, Jul 13 2016: (Start)
Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(n) = A001222(A260443(n)) = A000120(A277020(n)). Also a(n) = A000120(A101624(n-1)) for n >= 1. - Antti Karttunen, Nov 05 2016
(a(n-1) + a(n+1))/a(n) = A037227(n) for n >= 1. - Peter Bala, Feb 07 2017
a(0) = 0; a(3n) = 2*A000360(3n-1); a(3n+1) = 2*A000360(3n) - 1; a(3n+2) = 2*A000360(3n+1) + 1. - M. Jeremie Lafitte (Levitas), Apr 24 2017
From I. V. Serov, Jun 14 2017: (Start)
a(n) = A287896(n-1) - 1*A288002(n-1) for n > 1;
a(n) = A007306(n-1) - 2*A288002(n-1) for n > 1. (End)
From Yosu Yurramendi, Feb 14 2018: (Start)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m, a(2^m) = 1, a(0) = 0. (End)
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR (j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). - Peter Luschny, Sep 29 2024
a(n) = A000120(A168081(n)). - Karl-Heinz Hofmann, Jun 16 2025

Extensions

Additional references and comments from Len Smiley, Joshua Zucker, Rick L. Shepherd and Herbert S. Wilf
Typo in definition corrected by Reinhard Zumkeller, Aug 23 2011
Incorrect formula deleted and text edited by Johannes W. Meijer, Feb 07 2013

A204892 Least k such that n divides s(k)-s(j) for some j in [1,k), where s(k)=prime(k).

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 20 2012

Keywords

Comments

Suppose that (s(i)) is a strictly increasing sequence in the set N of positive integers. For i in N, let r(h) be the residue of s(i+h)-s(i) mod n, for h=1,2,...,n+1. There are at most n distinct residues r(h), so that there must exist numbers h and h' such that r(h)=r(h'), where 0<=h
Corollary: for each n, there are infinitely many pairs (j,k) such that n divides s(k)-s(j), and this result holds if s is assumed unbounded, rather than strictly increasing.
Guide to related sequences:
...
s(n)=prime(n), primes
... k(n), j(n): A204892, A204893
... s(k(n)),s(j(n)): A204894, A204895
... s(k(n))-s(j(n)): A204896, A204897
s(n)=prime(n+1), odd primes
... k(n), j(n): A204900, A204901
... s(k(n)),s(j(n)): A204902, A204903
... s(k(n))-s(j(n)): A109043(?), A000034(?)
s(n)=prime(n+2), primes >=5
... k(n), j(n): A204908, A204909
... s(k(n)),s(j(n)): A204910, A204911
... s(k(n))-s(j(n)): A109043(?), A000034(?)
s(n)=prime(n)*prime(n+1) product of consecutive primes
... k(n), j(n): A205146, A205147
... s(k(n)),s(j(n)): A205148, A205149
... s(k(n))-s(j(n)): A205150, A205151
s(n)=(prime(n+1)+prime(n+2))/2: averages of odd primes
... k(n), j(n): A205153, A205154
... s(k(n)),s(j(n)): A205372, A205373
... s(k(n))-s(j(n)): A205374, A205375
s(n)=2^(n-1), powers of 2
... k(n), j(n): A204979, A001511(?)
... s(k(n)),s(j(n)): A204981, A006519(?)
... s(k(n))-s(j(n)): A204983(?), A204984
s(n)=2^n, powers of 2
... k(n), j(n): A204987, A204988
... s(k(n)),s(j(n)): A204989, A140670(?)
... s(k(n))-s(j(n)): A204991, A204992
s(n)=C(n+1,2), triangular numbers
... k(n), j(n): A205002, A205003
... s(k(n)),s(j(n)): A205004, A205005
... s(k(n))-s(j(n)): A205006, A205007
s(n)=n^2, squares
... k(n), j(n): A204905, A204995
... s(k(n)),s(j(n)): A204996, A204997
... s(k(n))-s(j(n)): A204998, A204999
s(n)=(2n-1)^2, odd squares
... k(n), j(n): A205378, A205379
... s(k(n)),s(j(n)): A205380, A205381
... s(k(n))-s(j(n)): A205382, A205383
s(n)=n(3n-1), pentagonal numbers
... k(n), j(n): A205138, A205139
... s(k(n)),s(j(n)): A205140, A205141
... s(k(n))-s(j(n)): A205142, A205143
s(n)=n(2n-1), hexagonal numbers
... k(n), j(n): A205130, A205131
... s(k(n)),s(j(n)): A205132, A205133
... s(k(n))-s(j(n)): A205134, A205135
s(n)=C(2n-2,n-1), central binomial coefficients
... k(n), j(n): A205010, A205011
... s(k(n)),s(j(n)): A205012, A205013
... s(k(n))-s(j(n)): A205014, A205015
s(n)=(1/2)C(2n,n), (1/2)*(central binomial coefficients)
... k(n), j(n): A205386, A205387
... s(k(n)),s(j(n)): A205388, A205389
... s(k(n))-s(j(n)): A205390, A205391
s(n)=n(n+1), oblong numbers
... k(n), j(n): A205018, A205028
... s(k(n)),s(j(n)): A205029, A205030
... s(k(n))-s(j(n)): A205031, A205032
s(n)=n!, factorials
... k(n), j(n): A204932, A204933
... s(k(n)),s(j(n)): A204934, A204935
... s(k(n))-s(j(n)): A204936, A204937
s(n)=n!!, double factorials
... k(n), j(n): A204982, A205100
... s(k(n)),s(j(n)): A205101, A205102
... s(k(n))-s(j(n)): A205103, A205104
s(n)=3^n-2^n
... k(n), j(n): A205000, A205107
... s(k(n)),s(j(n)): A205108, A205109
... s(k(n))-s(j(n)): A205110, A205111
s(n)=Fibonacci(n+1)
... k(n), j(n): A204924, A204925
... s(k(n)),s(j(n)): A204926, A204927
... s(k(n))-s(j(n)): A204928, A204929
s(n)=Fibonacci(2n-1)
... k(n), j(n): A205442, A205443
... s(k(n)),s(j(n)): A205444, A205445
... s(k(n))-s(j(n)): A205446, A205447
s(n)=Fibonacci(2n)
... k(n), j(n): A205450, A205451
... s(k(n)),s(j(n)): A205452, A205453
... s(k(n))-s(j(n)): A205454, A205455
s(n)=Lucas(n)
... k(n), j(n): A205114, A205115
... s(k(n)),s(j(n)): A205116, A205117
... s(k(n))-s(j(n)): A205118, A205119
s(n)=n*(2^(n-1))
... k(n), j(n): A205122, A205123
... s(k(n)),s(j(n)): A205124, A205125
... s(k(n))-s(j(n)): A205126, A205127
s(n)=ceiling[n^2/2]
... k(n), j(n): A205394, A205395
... s(k(n)),s(j(n)): A205396, A205397
... s(k(n))-s(j(n)): A205398, A205399
s(n)=floor[(n+1)^2/2]
... k(n), j(n): A205402, A205403
... s(k(n)),s(j(n)): A205404, A205405
... s(k(n))-s(j(n)): A205406, A205407

Examples

			Let s(k)=prime(k).  As in A204890, the ordering of differences s(k)-s(j), follows from the arrangement shown here:
k...........1..2..3..4..5...6...7...8...9
s(k)........2..3..5..7..11..13..17..19..23
...
s(k)-s(1)......1..3..5..9..11..15..17..21..27
s(k)-s(2).........2..4..8..10..14..16..20..26
s(k)-s(3)............2..6..8...12..14..18..24
s(k)-s(4)...............4..6...10..12..16..22
...
least (k,j) such that 1 divides s(k)-s(j) for some j is (2,1), so a(1)=2.
least (k,j) such that 2 divides s(k)-s(j): (3,2), so a(2)=3.
least (k,j) such that 3 divides s(k)-s(j): (3,1), so a(3)=3.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := s[n] = Prime[n]; z1 = 400; z2 = 50;
    Table[s[n], {n, 1, 30}]          (* A000040 *)
    u[m_] := u[m] = Flatten[Table[s[k] - s[j],
       {k, 2, z1}, {j, 1, k - 1}]][[m]]
    Table[u[m], {m, 1, z1}]          (* A204890 *)
    v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]
    w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]
    d[n_] := d[n] = First[Delete[w[n],
       Position[w[n], 0]]]
    Table[d[n], {n, 1, z2}]          (* A204891 *)
    k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]
    m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]
    j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2
    Table[k[n], {n, 1, z2}]          (* A204892 *)
    Table[j[n], {n, 1, z2}]          (* A204893 *)
    Table[s[k[n]], {n, 1, z2}]       (* A204894 *)
    Table[s[j[n]], {n, 1, z2}]       (* A204895 *)
    Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204896 *)
    Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204897 *)
    (* Program 2: generates A204892 and A204893 rapidly *)
    s = Array[Prime[#] &, 120];
    lk = Table[NestWhile[# + 1 &, 1, Min[Table[Mod[s[[#]] - s[[j]], z], {j, 1, # - 1}]] =!= 0 &], {z, 1, Length[s]}]
    Table[NestWhile[# + 1 &, 1, Mod[s[[lk[[j]]]] - s[[#]], j] =!= 0 &], {j, 1, Length[lk]}]
    (* Peter J. C. Moses, Jan 27 2012 *)
  • PARI
    a(n)=forprime(p=n+2,,forstep(k=p%n,p-1,n,if(isprime(k), return(primepi(p))))) \\ Charles R Greathouse IV, Mar 20 2013
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