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.

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A101211 Triangle read by rows: n-th row is length of run of leftmost 1's, followed by length of run of 0's, followed by length of run of 1's, etc., in the binary representation of n.

Original entry on oeis.org

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

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Author

Leroy Quet, Dec 13 2004

Keywords

Comments

Row n has A005811(n) elements. In rows 2^(k-1)..2^k-1 we have all the compositions (ordered partitions) of k. Other orderings of compositions: A066099, A108244, and A124734. - Jason Kimberley, Feb 09 2013
A043276(n) = largest term in n-th row. - Reinhard Zumkeller, Dec 16 2013
From the first comment it follows that we have a bijection between the positive integers and the set of all compositions. - Emeric Deutsch, Jul 11 2017
From Robert Israel, Jan 23 2018: (Start)
If n is even, row 2*n is row n with its last element incremented by 1, and row 2*n+1 is row n with 1 appended.
If n is odd, row 2*n+1 is row n with its last element incremented by 1, and row 2*n is row n with 1 appended. (End)

Examples

			Since 9 is 1001 in binary, the 9th row is 1,2,1.
Since 11 is 1011 in binary, the 11th row is 1,1,2.
Triangle begins:
  1;
  1,1;
  2;
  1,2;
  1,1,1;
  2,1;
  3;
  1,3;
		

Crossrefs

A070939(n) gives the sum of terms in row n, while A167489(n) gives the product of its terms. A090996 gives the first column. A227736 lists the terms of each row in reverse order.
Cf. also A227186.
Cf. A318927 (concatenation of each row), A318926 (concatenations of reversed rows).
Cf. A382255 (Heinz numbers of the rows: Product_k prime(T(n,k))).

Programs

  • Haskell
    import Data.List (group)
    a101211 n k = a101211_tabf !! (n-1) !! (k-1)
    a101211_row n = a101211_tabf !! (n-1)
    a101211_tabf = map (reverse . map length . group) $ tail a030308_tabf
    -- Reinhard Zumkeller, Dec 16 2013
    
  • Maple
    # Maple program due to W. Edwin Clark:
    Runs := proc (L) local j, r, i, k; j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: # Row n is obtained with the command c(n). - Emeric Deutsch, Jul 03 2017
    # Maple program due to W. Edwin Clark, yielding the integer ind corresponding to a given composition (the index of the composition):
    ind := proc (x) local X, j, i: X := NULL: for j to nops(x) do if type(j, odd) then X := X, seq(1, i = 1 .. x[j]) end if: if type(j, even) then X := X, seq(0, i = 1 .. x[j]) end if end do: X := [X]: add(X[i]*2^(nops(X)-i), i = 1 .. nops(X)) end proc; # Clearly, ind(c(n))= n. - Emeric Deutsch, Jan 23 2018
  • Mathematica
    Table[Length /@ Split@ IntegerDigits[n, 2], {n, 38}] // Flatten (* Michael De Vlieger, Jul 11 2017 *)
  • PARI
    apply( {A101211_row(n)=Vecrev((n=vecextract([-1..exponent(n)], bitxor(2*n, bitor(n,1))))[^1]-n[^-1])}, [1..19]) \\ replacing older code by M. F. Hasler, Mar 24 2025
  • Python
    from itertools import groupby
    def arow(n): return [len(list(g)) for k, g in groupby(bin(n)[2:])]
    def auptorow(rows):
        alst = []
        for i in range(1, rows+1): alst.extend(arow(i))
        return alst
    print(auptorow(38)) # Michael S. Branicky, Oct 02 2021
    

Formula

a(n) = A227736(A227741(n)) = A227186(A056539(A227737(n)),A227740(n)) - Antti Karttunen, Jul 27 2013

Extensions

More terms from Emeric Deutsch, Apr 12 2005

A006472 a(n) = n!*(n-1)!/2^(n-1).

Original entry on oeis.org

1, 1, 3, 18, 180, 2700, 56700, 1587600, 57153600, 2571912000, 141455160000, 9336040560000, 728211163680000, 66267215894880000, 6958057668962400000, 834966920275488000000, 113555501157466368000000, 17373991677092354304000000, 2970952576782792585984000000
Offset: 1

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Author

Keywords

Comments

Product of first (n-1) positive triangular numbers. - Amarnath Murthy, May 19 2002, corrected by Alex Ratushnyak, Dec 03 2013
Number of ways of transforming n distinguishable objects into n singletons via a sequence of n-1 refinements. Example: a(3)=3 because we have XYZ->X|YZ->X|Y|Z, XYZ->Y|XZ->X|Y|Z and XYZ->Z|XY->X|Y|Z. - Emeric Deutsch, Jan 23 2005
In other words, a(n) is the number of maximal chains in the lattice of set partitions of {1, ..., n} ordered by refinement. - Gus Wiseman, Jul 22 2018
From David Callan, Aug 27 2009: (Start)
With offset 0, a(n) = number of unordered increasing full binary trees of 2n edges with leaf set {n,n+1,...,2n}, where full binary means each nonleaf vertex has two children, increasing means the vertices are labeled 0,1,2,...,2n and each child is greater than its parent, unordered might as well mean ordered and each pair of sibling vertices is increasing left to right. For example, a(2)=3 counts the trees with edge lists {01,02,13,14}, {01,03,12,14}, {01,04,12,13}.
PROOF. Given such a tree of size n, to produce a tree of size n+1, two new leaves must be added to the leaf n. Choose any two of the leaf set {n+1,...,2n,2n+1,2n+2} for the new leaves and use the rest to replace the old leaves n+1,...,2n, maintaining relative order. Thus each tree of size n yields (n+2)-choose-2 trees of the next size up. Since the ratio a(n+1)/a(n)=(n+2)-choose-2, the result follows by induction.
Without the condition on the leaves, these trees are counted by the reduced tangent numbers A002105. (End)
a(n) = Sum(M(t)N(t)), where summation is over all rooted trees t with n vertices, M(t) is the number of ways to take apart t by sequentially removing terminal edges (see A206494) and N(t) is the number of ways to build up t from the one-vertex tree by adding successively edges to the existing vertices (the Connes-Moscovici weight; see A206496). See Remark on p. 3801 of the Hoffman reference. Example: a(3) = 3; indeed, there are two rooted trees with 3 vertices: t' = the path r-a-b and t" = V; we have M(t')=N(t')=1 and M(t") =1, N(t")=2, leading to M(t')N(t') + M(t")N(t")=3. - Emeric Deutsch, Jul 20 2012
Number of coalescence sequences or labeled histories for n lineages: the number of sequences by which n distinguishable leaves can coalesce to a single sequence. The coalescence process merges pairs of lineages into new lineages, labeling each newly formed lineage l by a subset of the n initial lineages corresponding to the union of all initial lineages that feed into lineage l. - Noah A Rosenberg, Jan 28 2019
Conjecture: For n > 1, n divides 2*a(n-1) + 4 if and only if n is prime. - Werner Schulte, Oct 04 2020
For a proof of the above conjecture see Himane. The list of primes p such that p^2 divides (2*a(p-1) + 4) (analog of A007540 - Wilson primes) begins [239, 24049, ...]. - Peter Bala, Nov 06 2024
a(n) is the number of maximal chains in the poset of set of permutations of {1, ..., n} ordered by containment. - Rajesh Kumar Mohapatra, Sep 03 2025

Examples

			From _Gus Wiseman_, Jul 22 2018: (Start)
The a(3) = 3 maximal chains in the lattice of set partitions of {1,2,3}:
  {{1},{2},{3}} < {{1},{2,3}} < {{1,2,3}}
  {{1},{2},{3}} < {{2},{1,3}} < {{1,2,3}}
  {{1},{2},{3}} < {{3},{1,2}} < {{1,2,3}} (End)
From _Rajesh Kumar Mohapatra_, Sep 03 2025: (Start)
The a(3) = 3 maximal chains in the poset of the set of permutations of {1,2,3}:
  {(1)(2)(3)} < (12)(3) < (123)}
  {(1)(2)(3)} < (1)(23) < (123)}
  {(1)(2)(3)} < (13)(2) < (132)} (End)
		

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 148.
  • László Lovász, Combinatorial Problems and Exercises, North-Holland, 1979, p. 165.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Mike Steel, Phylogeny: Discrete and Random Processes in Evolution, SIAM, 2016, p. 47.

Crossrefs

Cf. A000110, A000258, A002846, A005121, A213427, A317145, A363679 (sum of reciprocals).
For the type B and D analogs, see A001044 and A123385.

Programs

  • Magma
    [Factorial(n)*Factorial(n-1)/2^(n-1): n in [1..20]]; // Vincenzo Librandi, Aug 23 2018
    
  • Maple
    A006472 := n -> n!*(n-1)!/2^(n-1):
  • Mathematica
    FoldList[Times,1,Accumulate[Range[20]]] (* Harvey P. Dale, Jan 10 2013 *)
  • PARI
    a(n) = n*(n-1)!^2/2^(n-1) \\ Charles R Greathouse IV, May 18 2015
    
  • Python
    from math import factorial
    def A006472(n): return n*factorial(n-1)**2 >> n-1 # Chai Wah Wu, Jun 22 2022

Formula

a(n) = a(n-1)*A000217(n-1).
a(n) = A010790(n-1)/2^(n-1).
a(n) = polygorial(n, 3) = (A000142(n)/A000079(n))*A000142(n+1) = (n!/2^n)*Product_{i=0..n-1} (i+2) = (n!/2^n)*Pochhammer(2, n) = (n!^2/2^n)*(n+1) = polygorial(n, 4)/2^n*(n+1). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
a(n-1) = (-1)^(n+1)/(n^2*det(M_n)) where M_n is the matrix M_(i, j) = abs(1/i - 1/j). - Benoit Cloitre, Aug 21 2003
From Ilya Gutkovskiy, Dec 15 2016: (Start)
a(n) ~ 4*Pi*n^(2*n)/(2^n*exp(2*n)).
Sum_{n>=1} 1/a(n) = BesselI(1,2*sqrt(2))/sqrt(2) = 2.3948330992734... (End)
D-finite with recurrence 2*a(n) -n*(n-1)*a(n-1)=0. - R. J. Mathar, May 02 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = BesselJ(1,2*sqrt(2))/sqrt(2). - Amiram Eldar, Jun 25 2022
From Rajesh Kumar Mohapatra, Sep 03 2025: (Start)
a(n) = A331955(n,n)
a(n) = A331956(n,n)
a(n) = A375835(n,n)
a(n) = A375837(n,n) (End)

A005282 Mian-Chowla sequence (a B_2 sequence): a(1) = 1; for n>1, a(n) = smallest number > a(n-1) such that the pairwise sums of elements are all distinct.

Original entry on oeis.org

1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312, 1395, 1523, 1572, 1821, 1896, 2029, 2254, 2379, 2510, 2780, 2925, 3155, 3354, 3591, 3797, 3998, 4297, 4433, 4779, 4851
Offset: 1

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Author

Keywords

Comments

An alternative definition is to start with 1 and then continue with the least number such that all pairwise differences of distinct elements are all distinct. - Jens Voß, Feb 04 2003. [However, compare A003022 and A227590. - N. J. A. Sloane, Apr 08 2016]
Rachel Lewis points out [see link] that S, the sum of the reciprocals of this sequence, satisfies 2.158435 <= S <= 2.158677. Similarly, the sum of the squares of reciprocals of this sequence converges to approximately 1.33853369 and the sum of the cube of reciprocals of this sequence converges to approximately 1.14319352. - Jonathan Vos Post, Nov 21 2004; comment changed by N. J. A. Sloane, Jan 02 2020
Let S denote the reciprocal sum of a(n). Then 2.158452685 <= S <= 2.158532684. - Raffaele Salvia, Jul 19 2014
From Thomas Ordowski, Sep 19 2014: (Start)
Known estimate: n^2/2 + O(n) < a(n) < n^3/6 + O(n^2).
Conjecture: a(n) ~ n^3 / log(n)^2. (End)

Examples

			The second term is 2 because the 3 pairwise sums 1+1=2, 1+2=3, 2+2=4 are all distinct.
The third term cannot be 3 because 1+3 = 2+2. But it can be 4, since 1+4=5, 2+4=6, 4+4=8 are distinct and distinct from the earlier sums 1+1=2, 1+2=3, 2+2=4.
		

References

  • P. Erdős and R. Graham, Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique (1980).
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.20.2.
  • R. K. Guy, Unsolved Problems in Number Theory, E28.
  • A. M. Mian and S. D. Chowla, On the B_2-sequences of Sidon, Proc. Nat. Acad. Sci. India, A14 (1944), 3-4.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 2 of A347570.
Cf. A051788, A080200 (for differences between terms).
Different from A046185. Cf. A011185.
See also A003022, A227590.
A259964 has a greater sum of reciprocals.
Cf. A002858.

Programs

  • Haskell
    import Data.Set (Set, empty, insert, member)
    a005282 n = a005282_list !! (n-1)
    a005282_list = sMianChowla [] 1 empty where
       sMianChowla :: [Integer] -> Integer -> Set Integer -> [Integer]
       sMianChowla sums z s | s' == empty = sMianChowla sums (z+1) s
                            | otherwise   = z : sMianChowla (z:sums) (z+1) s
          where s' = try (z:sums) s
                try :: [Integer] -> Set Integer -> Set Integer
                try []     s                      = s
                try (x:sums) s | (z+x) `member` s = empty
                               | otherwise        = try sums $ insert (z+x) s
    -- Reinhard Zumkeller, Mar 02 2011
    
  • Maple
    a[1]:= 1: P:= {2}: A:= {1}:
    for n from 2 to 100 do
      for t from a[n-1]+1 do
        Pt:= map(`+`,A union {t},t);
        if Pt intersect P = {} then break fi
      od:
      a[n]:= t;
      A:= A union {t};
      P:= P union Pt;
    od:
    seq(a[n],n=1..100); # Robert Israel, Sep 21 2014
  • Mathematica
    t = {1}; sms = {2}; k = 1; Do[k++; While[Intersection[sms, k + t] != {}, k++]; sms = Join[sms, t + k, {2 k}]; AppendTo[t, k], {49}]; t (* T. D. Noe, Mar 02 2011 *)
  • PARI
    A005282_vec(N, A=[1], U=[0], D(A, n=#A)=vector(n-1, k, A[n]-A[n-k]))={ while(#A2 && U=U[k-1..-1]);A} \\ M. F. Hasler, Oct 09 2019
    
  • PARI
    aupto(L)= my(S=vector(L), A=[1]); for(i=2, L, for(j=1, #A, if(S[i-A[j]], next(2))); for(j=1, #A, S[i-A[j]]=1); A=concat(A, i)); A \\ Ruud H.G. van Tol, Jun 30 2025
    
  • Python
    from itertools import count, islice
    def A005282_gen(): # generator of terms
        aset1, aset2, alist = set(), set(), []
        for k in count(1):
            bset2 = {k<<1}
            if (k<<1) not in aset2:
                for d in aset1:
                    if (m:=d+k) in aset2:
                        break
                    bset2.add(m)
                else:
                    yield k
                    alist.append(k)
                    aset1.add(k)
                    aset2 |= bset2
    A005282_list = list(islice(A005282_gen(),30)) # Chai Wah Wu, Sep 05 2023

Formula

a(n) = A025582(n) + 1.
a(n) = (A034757(n)+1)/2.

Extensions

Examples added by N. J. A. Sloane, Jun 01 2008

A005823 Numbers whose ternary expansion contains no 1's.

Original entry on oeis.org

0, 2, 6, 8, 18, 20, 24, 26, 54, 56, 60, 62, 72, 74, 78, 80, 162, 164, 168, 170, 180, 182, 186, 188, 216, 218, 222, 224, 234, 236, 240, 242, 486, 488, 492, 494, 504, 506, 510, 512, 540, 542, 546, 548, 558, 560, 564, 566, 648, 650, 654, 656, 666, 668, 672, 674
Offset: 1

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Author

Keywords

Comments

The set of real numbers between 0 and 1 that contain no 1's in their ternary expansion is the well-known Cantor set with Hausdorff dimension log 2 / log 3.
Complement of A081606. - Reinhard Zumkeller, Mar 23 2003
Numbers k such that the k-th Apery number is congruent to 1 (mod 3) (cf. A005258). - Benoit Cloitre, Nov 30 2003
Numbers k such that the k-th central Delannoy number is congruent to 1 (mod 3) (cf. A001850). - Benoit Cloitre, Nov 30 2003
Numbers k such that there exists a permutation p_1, ..., p_k of 1, ..., k such that i + p_i is a power of 3 for every i. - Ray Chandler, Aug 03 2004
Subsequence of A125292. - Reinhard Zumkeller, Nov 26 2006
The first 2^n terms of the sequence could be obtained using the Cantor process for the segment [0,3^n-1]. E.g., for n=2 we have [0,{1},2,{3,4,5},6,{7},8]. The numbers outside of braces are the first 4 terms of the sequence. Therefore the terms of the sequence could be called "Cantor's numbers". - Vladimir Shevelev, Jun 13 2008
Mahler proved that positive a(n) is never a square. - Michel Marcus, Nov 12 2012
Define t: Z -> P(R) so that t(k) is the translated Cantor ternary set spanning [k, k+1], and let T be the union of t(a(n)) for all n. T = T * 3 = T / 3 is the closure of the Cantor ternary set under multiplication by 3. - Peter Munn, Oct 30 2019

References

  • K. J. Falconer, The Geometry of Fractal Sets, Cambridge, 1985; p. 14.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Twice A005836.
Cf. A088917 (characteristic function), A306556.

Programs

  • Maple
    a:= proc(n) option remember;
          `if`(n=1, 0, `if`(irem (n, 2, 'q')=0, 3*a(q)+2, 3*a(q+1)))
        end:
    seq(a(n), n=1..100); # Alois P. Heinz, Apr 19 2012
  • Mathematica
    Select[ Range[ 0, 729 ], (Count[ IntegerDigits[ #, 3 ], 1 ]==0)& ]
    Select[Range[0,700],DigitCount[#,3,1]==0&] (* Harvey P. Dale, Mar 12 2016 *)
    FromDigits[#,3]&/@Tuples[{0,2},6] (* Harvey P. Dale, Sep 08 2025 *)
  • PARI
    is(n)=while(n,if(n%3==1,return(0),n\=3));1 \\ Charles R Greathouse IV, Apr 20 2012
    
  • PARI
    a(n)=n=binary(n-1);sum(i=1,#n,2*n[i]*3^(#n-i)) \\ Charles R Greathouse IV, Apr 20 2012
    
  • PARI
    a(n)=2*fromdigits(binary(n-1),3) \\ Charles R Greathouse IV, Aug 24 2016
    
  • Python
    def A005823(n):
        return 2*int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015

Formula

a(n) = 2 * A005836(n).
a(2n) = 3*a(n)+2, a(2n+1) = 3*a(n+1), a(1) = 0.
a(n) = Sum_{k = 1..n} 1 + 3^A007814(k). - Philippe Deléham, Jul 09 2005
A125291(a(n)) = 1 for n>0. - Reinhard Zumkeller, Nov 26 2006
From Reinhard Zumkeller, Mar 02 2008: (Start)
A062756(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 1 then y else if x mod 3 = 1 then f(y+1, y+1) else f(floor(x/3), y). (End)
G.f. g(x) satisfies g(x) = 3*g(x^2)*(1+1/x) + 2*x^2/(1-x^2). - Robert Israel, Jan 04 2015
Sum_{n>=2} 1/a(n) = 1.341426555483087715426958452292349687410838545707857407585878304836140592352... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Extensions

More terms from Sascha Kurz, Mar 24 2002
Offset corrected by N. J. A. Sloane, Mar 02 2008. This may require some of the formulas to be adjusted.

A014551 Jacobsthal-Lucas numbers.

Original entry on oeis.org

2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025, 2047, 4097, 8191, 16385, 32767, 65537, 131071, 262145, 524287, 1048577, 2097151, 4194305, 8388607, 16777217, 33554431, 67108865, 134217727, 268435457, 536870911, 1073741825, 2147483647, 4294967297, 8589934591
Offset: 0

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Author

Keywords

Comments

Also gives the number of points of period n in the subshift of finite type corresponding to the square matrix A=[1,2;1,0] (this is then given by trace(A^n)). - Thomas Ward, Mar 07 2001
Sequence is identical to its signed inverse binomial transform (autosequence of the second kind). - Paul Curtz, Jul 11 2008
a(n) can be expressed in terms of values of the Fibonacci polynomials F_n(x), computed at x=1/sqrt(2). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
Pisano period lengths: 1, 1, 2, 2, 4, 2, 6, 2, 6, 4, 10, 2, 12, 6, 4, 2, 8, 6, 18, 4, ... - R. J. Mathar, Aug 10 2012
Let F(x) = Product_{n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/2) = 2.83717 78068 73232 99799 ... = 2 + 1/(1 + 1/(5 + 1/(7 + 1/(17 + ...)))). See A111317. - Peter Bala, Dec 26 2012
With different signs, 2, -1, 5, -7, 17, -31, 65, -127, 257, -511, 1025, -2047, ... is the Lucas V(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
The identity 2 = 2/2 + 2^2/(2*1) - 2^3/(2*1*5) - 2^4/(2*1*5*7) + 2^5/(2*1*5*7*17) + 2^6/(2*1*5*7*17*31) - - + + can be viewed as a generalized Engel-type expansion of the number 2 to the base 2. Compare with A062510. - Peter Bala, Nov 13 2013
For n >= 2, a(n) is the number of ways to tile a 2 X n strip, where the first two columns have an extra cell at the top, with 1 X 2 dominoes and 2 X 2 squares. Shown here is one of the a(7)=127 ways for the n=7 case:
._.
|_|_________.
| | | |_| |
||__|_|_|_|. - Greg Dresden, Sep 26 2021
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965) and the French mathematician Édouard Lucas (1842-1891). - Amiram Eldar, Oct 02 2023

References

  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 180, 255.
  • Lind and Marcus, An Introduction to Symbolic Dynamics and Coding, Cambridge University Press, 1995. (General material on subshifts of finite type)
  • Kritkhajohn Onphaeng and Prapanpong Pongsriiam. Jacobsthal and Jacobsthal-Lucas Numbers and Sums Introduced by Jacobsthal and Tverberg. Journal of Integer Sequences, Vol. 20 (2017), Article 17.3.6.
  • Abdelmoumène Zekiri, Farid Bencherif, Rachid Boumahdi, Generalization of an Identity of Apostol, J. Int. Seq., Vol. 21 (2018), Article 18.5.1.

Crossrefs

Cf. A001045 (companion "autosequence"), A019322, A066845, A111317.
Cf. A135440 (first differences), A166920 (partial sums).
Cf. A006995.

Programs

Formula

a(n+1) = 2 * a(n) - (-1)^n * 3.
From Len Smiley, Dec 07 2001: (Start)
a(n) = 2^n + (-1)^n.
G.f.: (2-x)/(1-x-2*x^2). (End)
E.g.f.: exp(x) + exp(-2*x) produces a signed version. - Paul Barry, Apr 27 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-1, 2*k)*3^(2*k)/2^(n-2). - Paul Barry, Feb 21 2003
0, 1, 5, 7 ... is 2^n - 2*0^n + (-1)^n, the 2nd inverse binomial transform of (2^n-1)^2 (A060867). - Paul Barry, Sep 05 2003
a(n) = 2*T(n, i/(2*sqrt(2))) * (-i*sqrt(2))^n with i^2=-1. - Paul Barry, Nov 17 2003
a(n) = A078008(n) + A001045(n+1). - Paul Barry, Feb 12 2004
a(n) = 2*A001045(n+1) - A001045(n). - Paul Barry, Mar 22 2004
a(0)=2, a(1)=1, a(n) = a(n-1) + 2*a(n-2) for n > 1. - Philippe Deléham, Nov 07 2006
a(2*n+1) = Product_{d|(2*n+1)} cyclotomic(d,2). a(2^k*(2*n+1)) = Product_{d|(2*n+1)} cyclotomic(2*d,2^(2^k)). - Miklos Kristof, Mar 12 2007
a(n) = 2^{(n-1)/2}F_{n-1}(1/sqrt(2)) + 2^{(n+2)/2}F_{n-2}(1/sqrt(2)). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
E.g.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k*(k+1)/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
G.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
a(n) = sqrt(9*(A001045)^2 + (-1)^n*2^(n+2)). - Vladimir Shevelev, Mar 13 2013
G.f.: 2 + G(0)*x*(1+4*x)/(2-x), where G(k) = 1 + 1/(1 - x*(9*k-1)/( x*(9*k+8) - 2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 13 2013
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 9*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
For n >= 1: a(n) = A006995(2^((n+2)/2)) when n is even, a(n) = A006995(3*2^((n-1)/2) - 1) when n is odd. - Bob Selcoe, Sep 04 2017
a(n) = J(n) + 4*J(n-1), a(0)=2, where J is A001045. - Yuchun Ji, Apr 23 2019
For n >= 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 06 2020
From Kai Wang, May 30 2020: (Start)
(2 - a(n+1)/a(n))/9 = Sum_{m>=n} (-2)^m/(a(m)*a(m+1)).
a(n) = 2*A001045(n+1) - A001045(n).
a(n)^2 = a(2*n) + 2*(-2)^n.
a(n)^2 = 9*A001045(n)^2 + 4*(-2)^n.
a(2*n) = 9*A001045(n)^2 + 2*(-2)^n.
2*A001045(m+n) = A001045(m)*a(n) + a(m)*A001045(n).
2*(-2)^n*A001045(m-n) = A001045(m)*a(n) - a(m)*A001045(n).
A001045(m+n) + (-2)^n*A001045(m-n) = A001045(m)*a(n).
A001045(m+n) - (-2)^n*A001045(m-n) = a(m)*A001045(n).
2*a(m+n) = 9*A001045(m)*A001045(n) + a(m)*a(n).
2*(-2)^n*a(m-n) = a(m)*a(n) - 9*A001045(m)*A001045(n).
a(m+n) - (-2)^n*a(m-n) = 9*A001045(m)*A001045(n).
a(m+n) + (-2)^n*a(m-n) = a(m)*a(n).
a(m+n)*a(m-n) - a(m)*a(m) = 9*(-2)^(m-n)*A001045(n)^2.
a(m+1)*a(n) - a(m)*a(n+1) = 9*(-2)^n*A001045(m-n). (End)
a(n) = F(n+1) + F(n-1) + Sum_{k=0..(n-2)} a(k)*F(n-1-k) for F(n) the Fibonacci numbers and for n > 1. - Greg Dresden, Jun 03 2020

A073004 Decimal expansion of exp(gamma).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Aug 03 2002

Keywords

Comments

See references and additional links in A094644.
The Riemann hypothesis holds if and only if the inequality sigma(n)/(n*log(log(n))) < exp(gamma) is valid for all n >= 5041, (G. Robin, 1984). - Peter Luschny, Oct 18 2020
From Peter Bala, Aug 24 2025: (Start)
By definition, gamma = lim_{n -> oo} s(n), where s(n) = Sum_{k = 1..n} 1/k - log(n). The convergence is slow. For example, s(50) = 0.5(87...) is only correct to 1 decimal digit. Let S(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*s(n+k). Elsner shows that S(n) converges to gamma much more rapidly. For example, S(50) = 0.57721566490153286060651209008(02...) gives gamma correct to 29 decimal digits.
Define E(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*exp(s(n+k)). Then it appears that E(n) converges rapidly to exp(gamma). For example, E(50) = 1.78107241799019798523650410310(43...) gives exp(gamma) correct to 29 decimal digits. Cf. A002389. (End)

Examples

			Exp(gamma) = 1.7810724179901979852365041031071795491696452143034302053...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 1.5.1 and 2.27.2, pp. 31, 187.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 166, 191, 208.

Crossrefs

Cf. A001620 (Euler-Mascheroni constant, gamma).
Cf. A001113, A002389, A067698, A080130, A091901, A094644 (continued fraction for exp(gamma)), A155969, A246499.

Programs

  • Magma
    R:=RealField(100); Exp(EulerGamma(R)); // G. C. Greubel, Aug 27 2018
  • Mathematica
    RealDigits[ E^(EulerGamma), 10, 110] [[1]]
  • PARI
    exp(Euler)
    

Formula

By Mertens theorem, equals lim_{m->infinity}(1/log(prime(m))*Product_{k=1..m} 1/(1-1/prime(k))). - Stanislav Sykora, Nov 14 2014
Equals limsup_{n->oo} sigma(n)/(n*log(log(n))) (Gronwall, 1913). - Amiram Eldar, Nov 07 2020
Equals limsup_{n->oo} (Sum_{d|n} log(d)/d)/(log(log(n)))^2 (Erdős and Zaremba, 1973). - Amiram Eldar, Mar 03 2021
Equals Product_{k>=1} (1-1/(k+1))*exp(1/k). - Amiram Eldar, Mar 20 2022
Equals lim_{n->oo} n * Product_{prime p<=n} p^(1/(1-p)). - Thomas Ordowski, Jan 30 2023
Equals Product_{k>=1} (k/sqrt(2))^((-1)^k/(k*log(2))). - Antonio Graciá Llorente, Oct 11 2024
Equals lim_{n->oo} (1/log(n))*Product_{prime p<=n} p/(p - 1) [Mertens] (see Finch at p. 31). - Stefano Spezia, Oct 27 2024

A020760 Decimal expansion of 1/sqrt(3).

Original entry on oeis.org

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

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Author

Keywords

Comments

If the sides of a triangle form an arithmetic progression in the ratio 1:1+d:1+2d then when d=1/sqrt(3) it uniquely maximizes the area of the triangle. This triangle has approximate internal angles 25.588 degs, 42.941 degs, 111.471 degs. - Frank M Jackson, Jun 15 2011
When a cylinder is completely enclosed by a sphere, it occupies a fraction f of the sphere volume. The value of f has a trivial lower bound of 0, and an upper bound which is this constant. It is achieved iff the cylinder diameter is sqrt(2) times its height, and the sphere is circumscribed to it. A similar constant can be associated with any n-dimensional geometric shape. For 3D cuboids it is A165952. - Stanislav Sykora, Mar 07 2016
The ratio between the thickness and diameter of a dynamically fair coin having an equal probability, 1/3, of landing on each of its two faces and on its side after being tossed in the air. The calculation is based on the dynamic of rigid body (Yong and Mahadevan, 2011). See A020765 for a simplified geometrical solution. - Amiram Eldar, Sep 01 2020
The coefficient of variation (relative standard deviation) of natural numbers: Limit_{n->oo} sqrt((n-1)/(3*n+3)) = 1/sqrt(3). - Michal Paulovic, Mar 21 2023

Examples

			0.577350269189625764509148780501957455647601751270126876018602326....
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 8.4.3 and 8.17, pp. 495, 531.

Crossrefs

Cf. A002194 (sqrt(3)), A010701 (1/3).

Programs

Formula

Equals 1/A002194. - Michel Marcus, Oct 12 2014
Equals cosine of the magic angle: cos(A195696). - Stanislav Sykora, Mar 07 2016
Equals square root of A010701. - Michel Marcus, Mar 07 2016
Equals 1 + Sum_{k>=0} -(4*k+1)^(-1/2) + (4*k+3)^(-1/2) + (4*k+5)^(-1/2) - (4*k+7)^(-1/2). - Gerry Martens, Nov 22 2022
Equals (1/2)*(2 - zeta(1/2,1/4) + zeta(1/2,3/4) + zeta(1/2,5/4) - zeta(1/2,7/4)). - Gerry Martens, Nov 22 2022
Has periodic continued fraction expansion [0, 1; 1, 2] (A040001). - Michal Paulovic, Mar 21 2023
Equals Product_{k>=1} (1 + (-1)^k/A047235(k)). - Amiram Eldar, Nov 22 2024
Equals tan(Pi/6) = (1/2)/A010527. - R. J. Mathar, Aug 31 2025

A058313 Numerator of the n-th alternating harmonic number, Sum_{k=1..n} (-1)^(k+1)/k.

Original entry on oeis.org

1, 1, 5, 7, 47, 37, 319, 533, 1879, 1627, 20417, 18107, 263111, 237371, 52279, 95549, 1768477, 1632341, 33464927, 155685007, 166770367, 156188887, 3825136961, 3602044091, 19081066231, 18051406831, 57128792093, 7751493599, 236266661971
Offset: 1

Views

Author

N. J. A. Sloane, Dec 09 2000

Keywords

Comments

A Wolstenholme-like theorem: for prime p > 3, if p = 6k-1, then p divides a(4k-1), otherwise if p = 6k+1, then p divides a(4k). - T. D. Noe, Apr 01 2004
For the limit n -> infinity of the partial sums of the alternating harmonic series see A002162. - Wolfdieter Lang, Sep 08 2015
a(n)/A058312(n) appears in the locker puzzle (see the links in A364317) as the probability of failures with the strategy used for n lockers and opening of up to floor(n/2) lockers. Note the alternative formula given below for a(n)/A058312(n) using only positive fractions. - Wolfdieter Lang, Aug 12 2023

Examples

			1, 1/2, 5/6, 7/12, 47/60, 37/60, 319/420, 533/840, 1879/2520, ...
For n=4: a(n)/A058312(n) = 7/12 because 1/1 - 1/2 + 1/3 - 1/4 = 7/12 = 1/4 + 1/3. - _Wolfdieter Lang_, Aug 12 2023
		

References

  • L. B. W. Jolley, Summation of Series, Dover Publications, 1961, page 14, #71.

Crossrefs

Denominators are A058312. Cf. A025530.
Apart from leading term, same as A075830.
Cf. A001008 (numerator of n-th harmonic number).
Bisections are A049281 and A082687.
Cf. A181983.

Programs

  • Haskell
    import Data.Ratio((%), numerator)
    a058313 n = a058313_list !! (n-1)
    a058313_list = map numerator $ scanl1 (+) $ map (1 %) $ tail a181983_list
    -- Reinhard Zumkeller, Mar 20 2013
  • Maple
    A058313 := n->numer(add((-1)^(k+1)/k,k=1..n));
    # Alternatively:
    a := n -> numer(harmonic(n) - harmonic((n-modp(n,2))/2)):
    seq(a(n), n=1..29); # Peter Luschny, May 03 2016
  • Mathematica
    Numerator[Table[Sum[(-1)^(k + 1)/k, {k, n}], {n, 30}]] (* Harvey P. Dale, Jul 18 2012 *)
    a[n_]:= (-1)^n (HarmonicNumber[n/2 - 1/2] - HarmonicNumber[n/2] + (-1)^n Log[4])/2; Table[a[n] // FullSimplify, {n, 29}] // Numerator (* Gerry Martens, Jul 05 2015 *)
    Rest[Numerator[CoefficientList[Series[Log[1 + x]/(1 - x), {x, 0, 33}], x]]] (* Vincenzo Librandi, Jul 06 2015 *)
    Table[Log[2] - (-1)^n LerchPhi[-1, 1, n + 1], {n, 20}] // Numerator (* Eric W. Weisstein, Aug 25 2023 *)
  • PARI
    a(n)=(-1)^n*numerator(polcoeff(log(1-x)/(x+1)+O(x^(n+1)), n))
    

Formula

G.f. for a(n)/A058312(n): log(1+x)/(1-x). - Benoit Cloitre, Jun 15 2003
a(n) = (n*a(n-1) + (-1)^(n+1)*A058312(n-1))/gcd(n*a(n-1) + (-1)^(n+1)*A058312(n-1), n*A058312(n-1)). - Robert Israel, Jul 06 2015
From Peter Luschny, May 03 2016: (Start)
Let H(n) denote the harmonic numbers, AH(n) denote the alternating harmonic numbers, Psi the polygamma function and euler(n,x) the Euler polynomials. Then:
AH(n) = H(n) - H((n - n mod 2)/2).
AH(z) = log(2)+(1/2)*cos(Pi*z)*(Psi(z/2+1/2)-Psi(z/2+1)).
AH(z) ~ log(2)+(1/2)*cos(Pi*z)*(-1/z+1/(2*z^2)-1/(4*z^4)+1/(2*z^6)-...).
AH(z) ~ log(2)-(1/2)*cos(Pi*z)*Sum_{n>=0} Euler(n,0)/z^(n+1). (End)
Sum_{k>=1} (-1)^(k+1)*AH(k)/k = Pi^2/12 + log(2)^2/2 (Boyadzhiev, 2013). - Amiram Eldar, Oct 04 2021
a(n)/A058312(n) = Sum_{j=0..ceiling(n/2) - 1} 1/(n-j), for n >= 1. (Proof by comparing the recurrences for even and odd n.) - Wolfdieter Lang, Aug 12 2023
For n >= 1, log(2) = a(n)/A058312(n) + (-1)^n*n!*Sum_{k >= 1} 1/(k*(k + 1)* ...*(k + n)*2^k). - Peter Bala, Dec 07 2023
a(n) = the (reduced) numerator of the continued fraction 1/(1 + 1^2/(1 + 2^2/(1 + 3^2/(1 + ... + (n-1)^2/(1))))). - Peter Bala, Feb 18 2024

A060594 Number of solutions to x^2 == 1 (mod n), that is, square roots of unity modulo n.

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 4, 2, 2, 8, 2, 2, 2, 4, 2, 4, 2, 4, 4, 2, 4, 4, 2, 2, 4, 8, 2, 4, 2, 4, 4, 2, 2, 8, 2, 2, 4, 4, 2, 2, 4, 8, 4, 2, 2, 8, 2, 2, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 2, 4, 4, 4, 4, 2, 8, 2, 2, 2, 8, 4, 2, 4, 8, 2, 4, 4, 4, 4, 2, 4, 8, 2, 2, 4, 4, 2, 4, 2
Offset: 1

Views

Author

Jud McCranie, Apr 11 2001

Keywords

Comments

Sum_{k=1..n} a(k) appears to be asymptotic to C*n*log(n) with C = 0.6... - Benoit Cloitre, Aug 19 2002
a(q) is the number of real characters modulo q. - Benoit Cloitre, Feb 02 2003
Also number of real Dirichlet characters modulo n and Sum_{k=1..n}a(k) is asymptotic to (6/Pi^2)*n*log(n). - Steven Finch, Feb 16 2006
Let P(n) be the product of the numbers less than and coprime to n. By theorem 59 in Nagell (which is Gauss's generalization of Wilson's theorem): for n > 2, P == (-1)^(a(n)/2) (mod n). - T. D. Noe, May 22 2009
Shadow transform of A005563. - Michel Marcus, Jun 06 2013
For n > 2, a(n) = 2 iff n is in A033948. - Max Alekseyev, Jan 07 2015
For n > 1, number of square numbers on the main diagonal of an (n-1) X (n-1) square array whose elements are the numbers from 1..n^2, listed in increasing order by rows. - Wesley Ivan Hurt, May 19 2021
a(n) is the number of linear Alexander quandles (Z/nZ, k) that are kei (equivalently, that have good involutions); see Ta, Prop. 5.6 and cf. A387317. - Luc Ta, Aug 26 2025

Examples

			The four numbers 1^2, 3^2, 5^2 and 7^2 are congruent to 1 modulo 8, so a(8) = 4.
		

References

  • Trygve Nagell, Introduction to Number Theory, AMS Chelsea, 1981, p. 100. [From T. D. Noe, May 22 2009]
  • Gérald Tenenbaum, Introduction à la théorie analytique et probabiliste des nombres, Cours spécialisé, 1995, Collection SMF, p. 260.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 196-197.

Crossrefs

Cf. A000010, A005087, A033948, A046073, A073103 (x^4 == 1 (mod n)).
Cf. A387317.

Programs

  • Maple
    A060594 := proc(n)
       option remember;
       local a,b,c;
       if type(n,even) then
         a:= padic:-ordp(n,2);
         b:= 2^a;
         c:= n/b;
         min(b/2, 4) * procname(c)
       else
         2^nops(numtheory:-factorset(n))
       fi
    end proc:
    map(A060594, [$1 .. 100]); # Robert Israel, Jan 05 2015
  • Mathematica
    a[n_] := Sum[ Boole[ Mod[k^2 , n] == 1], {k, 1, n}]; a[1] = 1; Table[a[n], {n, 1, 103}] (* Jean-François Alcover, Oct 21 2011 *)
    a[n_] := Switch[Mod[n, 8], 2|6, 2^(PrimeNu[n]-1), 1|3|4|5|7, 2^PrimeNu[n], 0, 2^(PrimeNu[n]+1)]; Array[a, 103] (* Jean-François Alcover, Apr 09 2016 *)
  • PARI
    a(n)=sum(i=1,n,if((i^2-1)%n,0,1))
    
  • PARI
    a(n)=my(o=valuation(n,2));2^(omega(n>>o)+max(min(o-1,2),0)) \\ Charles R Greathouse IV, Jun 06 2013
    
  • PARI
    a(n)=if(n<=2, 1, 2^#znstar(n)[3] ); \\ Joerg Arndt, Jan 06 2015
    
  • Python
    from sympy import primefactors
    def a007814(n): return 1 + bin(n - 1).count('1') - bin(n).count('1')
    def a(n):
        if n%2==0:
            A=a007814(n)
            b=2**A
            c=n//b
            return min(b//2, 4)*a(c)
        else: return 2**len(primefactors(n))
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 18 2017, after the Maple program
    
  • Python
    from sympy import primefactors
    def A060594(n): return (1<>(s:=(~n & n-1).bit_length()))))*(1 if n&1 else 1<Chai Wah Wu, Oct 26 2022
  • Sage
    print([len(Integers(n).square_roots_of_one()) for n in range(1,100)]) # Ralf Stephan, Mar 30 2014
    

Formula

If n == 0 (mod 8), a(n) = 2^(A005087(n) + 2); if n == 4 (mod 8), a(n) = 2^(A005087(n) + 1); otherwise a(n) = 2^(A005087(n)). - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 29 2001
a(n) = 2^omega(n)/2 if n == +/-2 (mod 8), a(n) = 2^omega(n) if n== +/-1, +/-3, 4 (mod 8), a(n) = 2*2^omega(n) if n == 0 (mod 8), where omega(n) = A001221(n). - Benoit Cloitre, Feb 02 2003
For n >= 2 A046073(n) * a(n) = A000010(n) = phi(n). This gives a formula for A046073(n) using the one in A060594(n). - Sharon Sela (sharonsela(AT)hotmail.com), Mar 09 2002
Multiplicative with a(2) = 1; a(2^2) = 2; a(2^e) = 4 for e > 2; a(q^e) = 2 for q an odd prime. - Eric M. Schmidt, Jul 09 2013
a(n) = 2^A046072(n) for n>2, in accordance with the above formulas by Ahmed Fares. - Geoffrey Critzer, Jan 05 2015
a(n) = Sum_{k=1..n} floor(sqrt(1+n*(k-1)))-floor(sqrt(n*(k-1))). - Wesley Ivan Hurt, May 19 2021
From Amiram Eldar, Dec 30 2022: (Start)
Dirichlet g.f.: (1-1/2^s+2/4^s)*zeta(s)^2/zeta(2*s).
Sum_{k=1..n} a(k) ~ (6/Pi^2)*n*(log(n) + 2*gamma - 1 - log(2)/2 - 2*zeta'(2)/zeta(2)), where gamma is Euler's constant (A001620). (End)

A002383 Primes of form k^2 + k + 1.

Original entry on oeis.org

3, 7, 13, 31, 43, 73, 157, 211, 241, 307, 421, 463, 601, 757, 1123, 1483, 1723, 2551, 2971, 3307, 3541, 3907, 4423, 4831, 5113, 5701, 6007, 6163, 6481, 8011, 8191, 9901, 10303, 11131, 12211, 12433, 13807, 14281, 17293, 19183, 20023, 20593, 21757, 22651, 23563
Offset: 1

Views

Author

Keywords

Comments

Also primes p such that 4p-3 is square. - Giovanni Teofilatto, Sep 07 2005
Also these primes are sums of 1 and some consecutive even numbers starting at 2; e.g., 31 = 1+2+4+6+8+10. - Labos Elemer, Apr 15 2003
Also primes of form n^2 - n + 1 (Prime central polygonal numbers, A002061). - Zak Seidov, Jan 26 2006
Also primes which are of the form TriangularNumber(n) + TriangularNumber(n+2): 7 = 1+6, 13 = 3+10, 31 = 10+21, 43 = 15+28, 73 = 28+45, ... - Vladimir Joseph Stephan Orlovsky, Apr 03 2009
It is not known whether there are infinitely many primes of the form n^2+n+1. See Rose reference. - Daniel Tisdale, Jun 27 2009
These numbers when >= 7 are prime repunits 111_n in a base n >= 2, so except for 3, they are all Brazilian primes belonging to A085104. (See Links "Les nombres brésiliens", Sections V.4 - V.5.) A002383 is generated by A002384 which lists the bases n of 111_n. A002383 = A053183 Union A185632. - Bernard Schott, Dec 22 2012
Conjecture: the set of these numbers, except 3, is the intersection of sets A085104 and A059055. See A225148. - Thomas Ordowski, May 02 2013
For a(n)>13, the fractional part of square root of a(n) starts with digit 5 (see A034101). - Charles Kusniec, Sep 06 2022

References

  • D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 46.
  • L. Poletti, Le serie dei numeri primi appartenente alle due forme quadratiche (A) n^2+n+1 e (B) n^2+n-1 per l'intervallo compreso entro 121 milioni, e cioè per tutti i valori di n fino a 11000, Atti della Reale Accademia Nazionale dei Lincei, Memorie della Classe di Scienze Fisiche, Matematiche e Naturali, s. 6, v. 3 (1929), pages 193-218.
  • H. E. Rose, A Course in Number Theory, Clarendon Press, 1988, p. 217.
  • 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

Cf. A002384 (corresponding k-values), A088503 ((n^2+3)/4 is prime), A110284 (4p-3 is prime), A085104 (primes (n^k-1)/(n-1)).
Cf. A237037, A237038, A237039, A237040 (from semiprimes of form n^3 + 1).
See also A034101 (frac(sqrt(n)) starts with digit 5).

Programs

  • Magma
    [ a: n in [1..100] | IsPrime(a) where a is n^2+n+1 ]; // Wesley Ivan Hurt, Jun 16 2014
    
  • Maple
    select(isprime, [j^2+j+1$j=1..200])[];  # Alois P. Heinz, Apr 20 2022
  • Mathematica
    Select[Table[n^2+n+1, {n,250}], PrimeQ] (* Harvey P. Dale, Mar 23 2012 *)
  • PARI
    list(lim)=select(n->isprime(n),vector((sqrt(4*lim-3)-1)\2,k,k^2+k+1)) \\ Charles R Greathouse IV, Jul 25 2011
    
  • Python
    from sympy import isprime
    print(list(filter(isprime, (n**2 + n + 1 for n in range(150))))) # Michael S. Branicky, Apr 20 2022

Formula

a(n) = A002384(n)^2 + A002384(n) + 1 = (A088503(n-1)^2 + 3)/4 = (A110284(n) + 3)/4. - Ray Chandler, Sep 07 2005

Extensions

Extended by Ray Chandler, Sep 07 2005
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