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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A048994 Triangle of Stirling numbers of first kind, s(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 2, -3, 1, 0, -6, 11, -6, 1, 0, 24, -50, 35, -10, 1, 0, -120, 274, -225, 85, -15, 1, 0, 720, -1764, 1624, -735, 175, -21, 1, 0, -5040, 13068, -13132, 6769, -1960, 322, -28, 1, 0, 40320, -109584, 118124, -67284, 22449, -4536, 546, -36, 1, 0, -362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45, 1
Offset: 0

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Author

Keywords

Comments

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.
Mirror image of the triangle A054654. - Philippe Deléham, Dec 30 2006
Also the triangle gives coefficients T(n,k) of x^k in the expansion of C(x,n) = (a(k)*x^k)/n!. - Mokhtar Mohamed, Dec 04 2012
From Wolfdieter Lang, Nov 14 2018: (Start)
This is the Sheffer triangle of Jabotinsky type (1, log(1 + x)). See the e.g.f. of the triangle below.
This is the inverse Sheffer triangle of the Stirling2 Sheffer triangle A008275.
The a-sequence of this Sheffer triangle (see a W. Lang link in A006232)
is from the e.g.f. A(x) = x/(exp(x) -1) a(n) = Bernoulli(n) = A027641(n)/A027642(n), for n >= 0. The z-sequence vanishes.
The Boas-Buck sequence for the recurrences of columns has o.g.f. B(x) = Sum_{n>=0} b(n)*x^n = 1/((1 + x)*log(1 + x)) - 1/x. b(n) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), b = {-1/2, 5/12, -3/8, 251/720, -95/288, 19087/60480,...}. For the Boas-Buck recurrence of Riordan and Sheffer triangles see the Aug 10 2017 remark in A046521, adapted to the Sheffer case, also for two references. See the recurrence and example below. (End)
Let G(n,m,k) be the number of simple labeled graphs on [n] with m edges and k components. Then T(n,k) = Sum (-1)^m*G(n,m,k). See the Read link below. Equivalently, T(n,k) = Sum mu(0,p) where the sum is over all set partitions p of [n] containing k blocks and mu is the Moebius function in the incidence algebra associated to the set partition lattice on [n]. - Geoffrey Critzer, May 11 2024

Examples

			Triangle begins:
  n\k 0     1       2       3      4      5      6    7    8   9 ...
  0   1
  1   0     1
  2   0    -1       1
  3   0     2      -3       1
  4   0    -6      11      -6      1
  5   0    24     -50      35    -10      1
  6   0  -120     274    -225     85    -15      1
  7   0   720   -1764    1624   -735    175    -21    1
  8   0 -5040   13068  -13132   6769  -1960    322  -28    1
  9   0 40320 -109584  118124 -67284  22449  -4536  546  -36   1
  ... - _Wolfdieter Lang_, Aug 22 2012
------------------------------------------------------------------
From _Wolfdieter Lang_, Nov 14 2018: (Start)
Recurrence: s(5,2)= s(4, 1) - 4*s(4, 2) = -6 - 4*11 = -50.
Recurrence from the a- and z-sequences: s(6, 3) = 2*(1*1*(-50) + 3*(-1/2)*35 + 6*(1/6)*(-10) + 10*0*1) = -225.
Boas-Buck recurrence for column k = 3, with b = {-1/2, 5/12, -3/8, ...}:
s(6, 3) = 6!*((-3/8)*1/3! + (5/12)*(-6)/4! + (-1/2)*35/5!) = -225. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974; Chapter V, also p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 93.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 245.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

Crossrefs

See especially A008275 which is the main entry for this triangle. A132393 is an unsigned version, and A008276 is another version.
A000142(n) = Sum_{k=0..n} |s(n, k)| for n >= 0.
Row sums give A019590(n+1).

Programs

  • Haskell
    a048994 n k = a048994_tabl !! n !! k
    a048994_row n = a048994_tabl !! n
    a048994_tabl = map fst $ iterate (\(row, i) ->
    (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 0)
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    A048994:= proc(n,k) combinat[stirling1](n,k) end: # R. J. Mathar, Feb 23 2009
    seq(print(seq(coeff(expand(k!*binomial(x,k)),x,i),i=0..k)),k=0..9); # Peter Luschny, Jul 13 2009
    A048994_row := proc(n) local k; seq(coeff(expand(pochhammer(x-n+1,n)), x,k), k=0..n) end: # Peter Luschny, Dec 30 2010
  • Mathematica
    Table[StirlingS1[n, m], {n, 0, 9}, {m, 0, n}] (* Peter Luschny, Dec 30 2010 *)
  • Maxima
    create_list(stirling1(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    a(n,k) = if(k<0 || k>n,0, if(n==0,1,(n-1)*a(n-1,k)+a(n-1,k-1)))
    
  • PARI
    trg(nn)=for (n=0, nn-1, for (k=0, n, print1(stirling(n,k,1), ", ");); print();); \\ Michel Marcus, Jan 19 2015
    

Formula

s(n, k) = A008275(n,k) for n >= 1, k = 1..n; column k = 0 is {1, repeat(0)}.
s(n, k) = s(n-1, k-1) - (n-1)*s(n-1, k), n, k >= 1; s(n, 0) = s(0, k) = 0; s(0, 0) = 1.
The unsigned numbers a(n, k)=|s(n, k)| satisfy a(n, k)=a(n-1, k-1)+(n-1)*a(n-1, k), n, k >= 1; a(n, 0) = a(0, k) = 0; a(0, 0) = 1.
Triangle (signed) = [0, -1, -1, -2, -2, -3, -3, -4, -4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; Triangle(unsigned) = [0, 1, 1, 2, 2, 3, 3, 4, 4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} (-m)^(n-k)*s(n, k) = A000142(n), A001147(n), A007559(n), A007696(n), ... for m = 1, 2, 3, 4, ... .- Philippe Deléham, Oct 29 2005
A008275*A007318 as infinite lower triangular matrices. - Gerald McGarvey, Aug 20 2009
T(n,k) = n!*[x^k]([t^n]exp(x*log(1+t))). - Peter Luschny, Dec 30 2010, updated Jun 07 2020
From Wolfdieter Lang, Nov 14 2018: (Start)
Recurrence from the Sheffer a-sequence (see a comment above): s(n, k) = (n/k)*Sum_{j=0..n-k} binomial(k-1+j, j)*Bernoulli(j)*s(n-1, k-1+j), for n >= 1 and k >= 1, with s(n, 0) = 0 if n >= 1, and s(0,0) = 1.
Boas-Buck type recurrence for column k: s(n, k) = (n!*k/(n - k))*Sum_{j=k..n-1} b(n-1-j)*s(j, k)/j!, for n >= 1 and k = 0..n-1, with input s(n, n) = 1. For sequence b see the Boas-Buck comment above. (End)
T(n,k) = Sum_{j=k..n} (-1)^(n-j)*A271705(n,j)*A216294(j,k). - Mélika Tebni, Feb 23 2023

Extensions

Offset corrected by R. J. Mathar, Feb 23 2009
Formula corrected by Philippe Deléham, Sep 10 2009

A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856
Offset: 1

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Author

Keywords

Comments

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 111.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 148 (the numbers |C^{2n-1}|).
  • J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 282.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1, p. 20.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • 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).
  • E. van Fossen Conrad, Some continued fraction expansions of elliptic functions, PhD thesis, The Ohio State University, 2002, p. 28.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 267-268.

Crossrefs

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009

Programs

  • Maple
    series(tan(x),x,40);
    with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
    A000182_list := proc(n) local T,k,j; T[1] := 1;
    for k from 2 to n do T[k] := (k-1)*T[k-1] od;
       for k from 2 to n do
           for j from k to n do
               T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
    seq(T[j], j=1..n)  end:
    A000182_list(15);  # Peter Luschny, Apr 02 2012
  • Mathematica
    Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
    v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
    Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
    t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
    a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
    a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
    Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
    a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
    a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)
  • Maxima
    a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */
    
  • Maxima
    a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
    makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */
    
  • PARI
    {a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};
    
  • PARI
    {a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */
    
  • PARI
    {a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */
    
  • PARI
    {a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */
    
  • PARI
    /* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
    {a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}
    
  • PARI
    /* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
    {a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [0, a(1), a(2), ..., a(n)] for n > 0.
    def A000182_list(n):
        T = [0 for i in range(1, n+2)]
        T[1] = 1
        for k in range(2, n+1):
            T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1):
                T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        return T
    print(A000182_list(100)) # Peter Luschny, Aug 07 2011
    
  • Python
    from sympy import bernoulli
    def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023
    
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A000182_list(len) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..2*len-1) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            if e > 0 : R.append(A[i//2])
        return R
    A000182_list(15) # Peter Luschny, Mar 31 2012

Formula

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024

A000367 Numerators of Bernoulli numbers B_2n.

Original entry on oeis.org

1, 1, -1, 1, -1, 5, -691, 7, -3617, 43867, -174611, 854513, -236364091, 8553103, -23749461029, 8615841276005, -7709321041217, 2577687858367, -26315271553053477373, 2929993913841559, -261082718496449122051
Offset: 0

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Keywords

Examples

			B_{2n} = [ 1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, ... ].
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 230.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.6.
  • F. Lemmermeyer, Reciprocity Laws From Euler to Eisenstein, Springer-Verlag, 2000, p. 330.
  • H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.
  • 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

B_n gives A027641/A027642. See A027641 for full list of references, links, formulas, etc.
See A002445 for denominators.

Programs

  • Maple
    A000367 := n -> numer(bernoulli(2*n)):
    # Illustrating an algorithmic approach:
    S := proc(n,k) option remember; if k=0 then `if`(n=0,1,0) else S(n, k-1) + S(n-1, n-k) fi end: Bernoulli2n := n -> `if`(n = 0,1,(-1)^n * S(2*n-1,2*n-1)*n/(2^(2*n-1)*(1-4^n))); A000367 := n -> numer(Bernoulli2n(n)); seq(A000367(n),n=0..20); # Peter Luschny, Jul 08 2012
  • Mathematica
    Numerator[ BernoulliB[ 2*Range[0, 20]]] (* Jean-François Alcover, Oct 16 2012 *)
    Table[Numerator[(-1)^(n+1) 2 Gamma[2 n + 1] Zeta[2 n]/(2 Pi)^(2 n)], {n, 0, 20}] (* Artur Jasinski, Dec 29 2020 *)
  • Maxima
    B(n):=if n=0 then 1 else 2*n*sum((2*n+k-2)!*sum(((-1)^(j+1)*stirling1(2*n+j,j))/ ((k-j)!*(2*n+j)!),j,1,k),k,0,2*n);
    makelist(num(B(n)),n,0,10); /* Vladimir Kruchinin, Mar 15 2013, fixed by Vaclav Kotesovec, Oct 22 2014 */
  • PARI
    a(n)=numerator(bernfrac(2*n))
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [a(0), a(1), ..., a(n)] for n > 0.
    from fractions import Fraction
    def A000367_list(n):  # Bernoulli numerators
        T = [0 for i in range(1, n+2)]
        T[0] = 1; T[1] = 1
        for k in range(2, n+1):
            T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1):
                T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        a = 0; b = 6; s = 1
        for k in range(1, n+1):
            T[k] = s*Fraction(T[k]*k, b).numerator
            h = b; b = 20*b - 64*a; a = h; s = -s
        return T
    print(A000367_list(100)) # Peter Luschny, Aug 09 2011
    

Formula

E.g.f: x/(exp(x) - 1); take numerators of even powers.
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k>=1} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/(2*Pi)^(2*n).
If n >= 3 is prime, then 12*|a((n+1)/2)| == (-1)^((n-1)/2)*A002445((n+1)/2) (mod n). - Vladimir Shevelev, Sep 04 2010
a(n) = numerator(-i*(2*n)!/(Pi*(1-2*n))*Integral_{t=0..1} log(1-1/t)^(1-2*n) dt). - Gerry Martens, May 17 2011, corrected by Vaclav Kotesovec, Oct 22 2014
a(n) = numerator((-1)^(n+1)*(2*Pi)^(-2*n)*(2*n)!*Li_{2*n}(1)) for n > 0. - Peter Luschny, Jun 29 2012
E.g.f.: G(0) where G(k) = 2*k + 1 - x*(2*k+1)/(x + (2*k+2)/(1 + x/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 13 2013
a(n) = numerator(2*n*Sum_{k=0..2*n} (2*n+k-2)! * Sum_{j=1..k} ((-1)^(j+1) * Stirling1(2*n+j,j)) / ((k-j)!*(2*n+j)!)), n > 0. - Vladimir Kruchinin, Mar 15 2013
E.g.f.: E(0) where E(k) = 2*k+1 - x/(2 + x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 16 2013
E.g.f.: E(0) - x, where E(k) = x + k + 1 - x*(k+1)/E(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 14 2013
a(n) = numerator((-1)^(n+1)*2*Gamma(2*n + 1)*zeta(2*n)/(2*Pi)^(2*n)). - Artur Jasinski, Dec 29 2020
a(n) = numerator(-2*n*zeta(1 - 2*n)) for n > 0. - Artur Jasinski, Jan 01 2021

A002445 Denominators of Bernoulli numbers B_{2n}.

Original entry on oeis.org

1, 6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, 1806, 690, 282, 46410, 66, 1590, 798, 870, 354, 56786730, 6, 510, 64722, 30, 4686, 140100870, 6, 30, 3318, 230010, 498, 3404310, 6, 61410, 272118, 1410, 6, 4501770, 6, 33330, 4326, 1590, 642, 209191710, 1518, 1671270, 42
Offset: 0

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Comments

From the von Staudt-Clausen theorem, denominator(B_2n) = product of primes p such that (p-1)|2n.
Row products of A138239. - Mats Granvik, Mar 08 2008
Equals row products of even rows in triangle A143343. In triangle A080092, row products = denominators of B1, B2, B4, B6, ... . - Gary W. Adamson, Aug 09 2008
Julius Worpitzky's 1883 algorithm for generating Bernoulli numbers is shown in A028246. - Gary W. Adamson, Aug 09 2008
There is a relation between the Euler numbers E_n and the Bernoulli numbers B_{2*n}, for n>0, namely, B_{2*n} = A000367(n)/a(n) = ((-1)^n/(2*(1-2^{2*n}))) * Sum_{k = 0..n-1} (-1)^k*2^{2*k}*C(2*n,2*k)*A000364(n-k)*A000367(k)/a(k). (See Bucur, et al.) - L. Edson Jeffery, Sep 17 2012
a(n) is the product of all primes of the form (k + n)/(k - n). - Thomas Ordowski, Jul 24 2025

Examples

			B_{2n} = [ 1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, ... ].
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 136.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • 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).
  • See A000367 for further references and links (there are a lot).

Crossrefs

Cf. A090801 (distinct numbers appearing as denominators of Bernoulli numbers)
B_n gives A027641/A027642. See A027641 for full list of references, links, formulas, etc.
Cf. A160014 for a generalization.

Programs

  • Magma
    [Denominator(Bernoulli(2*n)): n in [0..60]]; // Vincenzo Librandi, Nov 16 2014
    
  • Maple
    A002445 := n -> mul(i,i=select(isprime,map(i->i+1,numtheory[divisors] (2*n)))): seq(A002445(n),n=0..40); # Peter Luschny, Aug 09 2011
    # Alternative
    N:= 1000: # to get a(0) to a(N)
    A:= Vector(N,2):
    for p in select(isprime,[seq(2*i+1,i=1..N)]) do
      r:= (p-1)/2;
      for n from r to N by r do
        A[n]:= A[n]*p
      od
    od:
    1, seq(A[n],n=1..N); # Robert Israel, Nov 16 2014
  • Mathematica
    Take[Denominator[BernoulliB[Range[0,100]]],{1,-1,2}] (* Harvey P. Dale, Oct 17 2011 *)
  • PARI
    a(n)=prod(p=2,2*n+1,if(isprime(p),if((2*n)%(p-1),1,p),1)) \\ Benoit Cloitre
    
  • PARI
    A002445(n,P=1)=forprime(p=2,1+n*=2,n%(p-1)||P*=p);P \\ M. F. Hasler, Jan 05 2016
    
  • PARI
    a(n) = denominator(bernfrac(2*n)); \\ Michel Marcus, Jul 16 2021
    
  • Sage
    def A002445(n):
        if n == 0:
            return 1
        M = (i + 1 for i in divisors(2 * n))
        return prod(s for s in M if is_prime(s))
    [A002445(n) for n in (0..57)] # Peter Luschny, Feb 20 2016

Formula

E.g.f: x/(exp(x) - 1); take denominators of even powers.
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k=1..inf} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/ (2*Pi)^(2*n).
If n>=3 is prime,then a((n+1)/2)==(-1)^((n-1)/2)*12*|A000367((n+1)/2)|(mod n). - Vladimir Shevelev, Sep 04 2010
a(n) = denominator(-I*(2*n)!/(Pi*(1-2*n))*integral(log(1-1/t)^(1-2*n) dt, t=0..1)). - Gerry Martens, May 17 2011
a(n) = 2*denominator((2*n)!*Li_{2*n}(1)) for n > 0. - Peter Luschny, Jun 28 2012
a(n) = gcd(2!S(2n+1,2),...,(2n+1)!S(2n+1,2n+1)). Here S(n,k) is the Stirling number of the second kind. See the paper of Komatsu et al. - Istvan Mezo, May 12 2016
a(n) = 2*A001897(n) = A027642(2*n) = 3*A277087(n) for n>0. - Jonathan Sondow, Dec 14 2016

A164555 Numerators of the "original" Bernoulli numbers; also the numerators of the Bernoulli polynomials at x=1.

Original entry on oeis.org

1, 1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051
Offset: 0

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Author

Paul Curtz, Aug 15 2009

Keywords

Comments

Apart from a sign flip in a(1), the same as A027641.
a(n) is also the numerator of the n-th term of the binomial transform of the sequence of Bernoulli numbers, i.e., of the sequence of fractions A027641(n)/A027642(n).
a(n)/A027642(n) with e.g.f. x/(1-exp(-x)) is the a-sequence for the Sheffer matrix A094645, see the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jun 20 2011
a(n)/A027642(n) also give the row sums of the rational triangle of the coefficients of the Bernoulli polynomials A053382/A053383 (falling powers) or A196838/A196839 (rising powers). - Wolfdieter Lang, Oct 25 2011
Given M = the beheaded Pascal's triangle, A074909; with B_n as a vector V, with numerators shown: (1, 1, 1, ...). Then M*V = [1, 2, 3, 4, 5, ...]. If the sign in a(1) is negative in V, then M*V = [1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 09 2012
One might interpret the term "'original' Bernoulli numbers" as numbers given by the e.g.f. x/(1-exp(-x)). - Peter Luschny, Jun 17 2012
Let B(n) = a(n)/A027642(n) then B(n) = Integral_{x=0..1} F_n(x) where F_n(x) are the signed Fubini polynomials F_n(x) = Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k (see illustration). - Peter Luschny, Jan 09 2017

Examples

			From _Peter Luschny_, Aug 13 2017: (Start)
Integral_{x=0..1} 1 = 1,
Integral_{x=0..1} x = 1/2,
Integral_{x=0..1} 2*x^2 - x = 1/6,
Integral_{x=0..1} 6*x^3 - 6*x^2 + x = 0,
Integral_{x=0..1} 24*x^4 - 36*x^3 + 14*x^2 - x = -1/30,
Integral_{x=0..1} 120*x^5 - 240*x^4 + 150*x^3 - 30*x^2 + x = 0,
...
Integral_{x=0..1} Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k = Bernoulli(n). (End)
		

References

  • Jacob Bernoulli, Ars Conjectandi, Basel: Thurneysen Brothers, 1713. See page 97.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 106-108.

Crossrefs

Programs

  • Haskell
    a164555 n = a164555_list !! n
    a164555_list = 1 : map (numerator . sum) (zipWith (zipWith (%))
       (zipWith (map . (*)) (tail a000142_list) a242179_tabf) a106831_tabf)
    -- Reinhard Zumkeller, Jul 04 2014
    
  • Maple
    A164555 := proc(n) if n <= 2 then 1; else numer(bernoulli(n)) ; fi; end: # R. J. Mathar, Aug 26 2009
    seq(numer(n!*coeff(series(t/(1-exp(-t)),t,n+2),t,n)),n=0..40); # Peter Luschny, Jun 17 2012
  • Mathematica
    CoefficientList[ Series[ x/(1 - Exp[-x]), {x, 0, 40}], x]*Range[0, 40]! // Numerator (* Jean-François Alcover, Mar 04 2013 *)
    Table[Numerator[BernoulliB[n,1]], {n, 0, 40}] (* Vaclav Kotesovec, Jan 03 2021 *)
  • Sage
    a = lambda n: bernoulli_polynomial(1,n).numerator()
    [a(n) for n in (0..40)] # Peter Luschny, Jan 09 2017

Formula

a(n) = numerator(B(n)) with B(n) = Sum_{k=0..n} (-1)^(n-k) * C(n+1, k+1) * S(n+k, k) / C(n+k, k) and S the Stirling set numbers. - Peter Luschny, Jun 25 2016
a(n) = numerator(n*EulerPolynomial(n-1, 1)/(2*(2^n-1))) for n>=1. - Peter Luschny, Sep 01 2017
From Artur Jasinski, Jan 01 2021: (Start)
a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n) for n != 1.
a(n) = numerator(-n*zeta(1-n)) for n >= 1. In the case n = 0 take the limit. (End)

Extensions

Edited and extended by R. J. Mathar, Sep 03 2009
Name extended by Peter Luschny, Jan 09 2017

A001550 a(n) = 1^n + 2^n + 3^n.

Original entry on oeis.org

3, 6, 14, 36, 98, 276, 794, 2316, 6818, 20196, 60074, 179196, 535538, 1602516, 4799354, 14381676, 43112258, 129271236, 387682634, 1162785756, 3487832978, 10462450356, 31385253914, 94151567436, 282446313698, 847322163876
Offset: 0

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Comments

a(n)*(-1)^n, n>=0, gives the z-sequence for the Sheffer triangle A049458 ((signed) 3-restricted Stirling1 numbers), which is the inverse triangle of A143495 with offset [0,0] (3-restricted Stirling2 numbers). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The a-sequence for each (signed) r-restricted Stirling1 Sheffer triangle is A027641/A027642 (Bernoulli numbers). - Wolfdieter Lang, Oct 10 2011

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • 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

Programs

  • Haskell
    a001550 n = sum $ map (^ n) [1..3]  -- Reinhard Zumkeller, Mar 01 2012
    
  • Magma
    [1^n + 2^n + 3^n : n in [0..30]]; // Wesley Ivan Hurt, Jun 25 2020
    
  • Maple
    A001550:=-(3-12*z+11*z^2)/(z-1)/(3*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    Table[1^n + 2^n + 3^n, {n, 0, 30}]
    CoefficientList[Series[(3-12x+11x^2)/(1-6x+11x^2-6x^3),{x,0,30}],x] (* or *) LinearRecurrence[{6,-11,6},{3,6,14},31] (* Harvey P. Dale, Apr 30 2011 *)
    Total[Range[3]^#]&/@Range[0,30] (* Harvey P. Dale, Sep 23 2019 *)
  • PARI
    a(n)=1+2^n+3^n \\ Charles R Greathouse IV, Jun 10 2011
    
  • Python
    def A001550(n): return 3**n+(1<Chai Wah Wu, Nov 01 2024

Formula

From Michael Somos: (Start)
G.f.: (3 -12*x +11*x^2)/(1 -6*x +11*x^2 -6*x^3).
a(n) = 5*a(n-1) - 6*a(n-2) + 2. (End)
E.g.f.: exp(x) + exp(2*x) + exp(3*x). - Mohammad K. Azarian, Dec 26 2008
a(0)=3, a(1)=6, a(2)=14, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3). - Harvey P. Dale, Apr 30 2011
a(n) = A007689(n) + 1. - Reinhard Zumkeller, Mar 01 2012
From Kai Wang, May 18 2020: (Start)
a(n) = 3*A000392(n+3) - 12*A000392(n+2) + 11*A000392(n+1).
A000392(n) = (3*a(n+1) - 12*a(n) + 10*a(n-1))/2. (End)

Extensions

Additional terms from Michael Somos
Attribute "conjectured" removed from Simon Plouffe's g.f. by R. J. Mathar, Mar 11 2009

A131689 Triangle of numbers T(n,k) = k!*Stirling2(n,k) = A000142(k)*A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 0, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880
Offset: 0

Views

Author

Philippe Deléham, Sep 14 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,6,...] where DELTA is the operator defined in A084938; another version of A019538.
See also A019538: version with n > 0 and k > 0. - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 21 2014: (Start)
T(n,k) gives the number of (k-1)-dimensional faces in the interior of the first barycentric subdivision of the standard (n-1)-dimensional simplex. For example, the barycentric subdivision of the 1-simplex is o--o--o, with 1 interior vertex and 2 interior edges, giving T(2,1) = 1 and T(2,2) = 2.
This triangle is used when calculating the face vectors of the barycentric subdivision of a simplicial complex. Let S be an n-dimensional simplicial complex and write f_k for the number of k-dimensional faces of S, with the usual convention that f_(-1) = 1, so that F := (f_(-1), f_0, f_1,...,f_n) is the f-vector of S. If M(n) denotes the square matrix formed from the first n+1 rows and n+1 columns of the present triangle, then the vector F*M(n) is the f-vector of the first barycentric subdivision of the simplicial complex S (Brenti and Welker, Lemma 2.1). For example, the rows of Pascal's triangle A007318 (but with row and column indexing starting at -1) are the f-vectors for the standard n-simplexes. It follows that A007318*A131689, which equals A028246, is the array of f-vectors of the first barycentric subdivision of standard n-simplexes. (End)
This triangle T(n, k) appears in the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} (x^k/(1 - x)^(k+2))*T(n, k). See also the Eulerian triangle A008292 with a Mar 31 2017 comment for a rewritten form. For the e.g.f. see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
T(n,k) = the number of alignments of length k of n strings each of length 1. See Slowinski. An example is given below. Cf. A122193 (alignments of strings of length 2) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018
The row polynomials R(n,x) are the Fubini polynomials. - Emanuele Munarini, Dec 05 2020
From Gus Wiseman, Feb 18 2022: (Start)
Also the number of patterns of length n with k distinct parts (or with maximum part k), where we define a pattern to be a finite sequence covering an initial interval of positive integers. For example, row n = 3 counts the following patterns:
(1,1,1) (1,2,2) (1,2,3)
(2,1,2) (1,3,2)
(2,2,1) (2,1,3)
(1,1,2) (2,3,1)
(1,2,1) (3,1,2)
(2,1,1) (3,2,1)
(End)
Regard A048994 as a lower-triangular matrix and divide each term A048994(n,k) by n!, then this is the matrix inverse. Because Sum_{k=0..n} (A048994(n,k) * x^n / n!) = A007318(x,n), Sum_{k=0..n} (A131689(n,k) * A007318(x,k)) = x^n. - Natalia L. Skirrow, Mar 23 2023
T(n,k) is the number of ordered partitions of [n] into k blocks. - Alois P. Heinz, Feb 21 2025

Examples

			The triangle T(n,k) begins:
  n\k 0 1    2     3      4       5        6        7        8        9      10 ...
  0:  1
  1:  0 1
  2:  0 1    2
  3:  0 1    6     6
  4:  0 1   14    36     24
  5:  0 1   30   150    240     120
  6:  0 1   62   540   1560    1800      720
  7:  0 1  126  1806   8400   16800    15120     5040
  8:  0 1  254  5796  40824  126000   191520   141120    40320
  9:  0 1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 0 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... reformatted and extended. - _Wolfdieter Lang_, Mar 31 2017
From _Peter Bala_, Feb 04 2018: (Start)
T(4,2) = 14 alignments of length 2 of 4 strings of length 1. Examples include
  (i) A -    (ii) A -    (iii) A -
      B -         B -          - B
      C -         - C          - C
      - D         - D          - D
There are C(4,1) = 4 alignments of type (i) with a single gap character - in column 1, C(4,2) = 6 alignments of type (ii) with two gap characters in column 1 and C(4,3) = 4 alignments of type (iii) with three gap characters in column 1, giving a total of 4 + 6 + 4 = 14 alignments. (End)
		

Crossrefs

Case m=1 of the polynomials defined in A278073.
Cf. A000142 (diagonal), A000670 (row sums), A000012 (alternating row sums), A210029 (central terms).
Cf. A008292, A028246 (o.g.f. and e.g.f. of sums of powers).
A version for partitions is A116608, or by maximum A008284.
A version for compositions is A235998, or by maximum A048004.
Classes of patterns:
- A000142 = strict
- A005649 = anti-run, complement A069321
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A226316 = (1,2,3)-avoiding, weakly A052709, complement A335515
- A296975 = aperiodic
- A345194 = alternating, up/down A350354, complement A350252
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths

Programs

  • Julia
    function T(n, k)
        if k < 0 || k > n return 0 end
        if n == 0 && k == 0 return 1 end
        k*(T(n-1, k-1) + T(n-1, k))
    end
    for n in 0:7
        println([T(n, k) for k in 0:n])
    end
    # Peter Luschny, Mar 26 2020
    
  • Maple
    A131689 := (n,k) -> Stirling2(n,k)*k!: # Peter Luschny, Sep 17 2011
    # Alternatively:
    A131689_row := proc(n) 1/(1-t*(exp(x)-1)); expand(series(%,x,n+1)); n!*coeff(%,x,n); PolynomialTools:-CoefficientList(%,t) end:
    for n from 0 to 9 do A131689_row(n) od; # Peter Luschny, Jan 23 2017
  • Mathematica
    t[n_, k_] := k!*StirlingS2[n, k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
    T[n_, k_] := If[n <= 0 || k <= 0, Boole[n == 0 && k == 0], Sum[(-1)^(i + k) Binomial[k, i] i^(n + k), {i, 0, k}]]; (* Michael Somos, Jul 08 2018 *)
  • PARI
    {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))};
    /* Michael Somos, Jul 08 2018 */
    
  • SageMath
    @cached_function
    def F(n): # Fubini polynomial
        R. = PolynomialRing(ZZ)
        if n == 0: return R(1)
        return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
    for n in (0..9): print(F(n).list()) # Peter Luschny, May 21 2021

Formula

T(n,k) = k*(T(n-1,k-1) + T(n-1,k)) with T(0,0)=1. Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A000629(n), A033999(n), A000007(n), A000670(n), A004123(n+1), A032033(n), A094417(n), A094418(n), A094419(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. [corrected by Philippe Deléham, Feb 11 2013]
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A000670(n), A122704(n) for x=-1, 0, 1, 2 respectively. - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} (-1)^k*T(n,k)/(k+1) = Bernoulli numbers A027641(n)/A027642(n). - Peter Luschny, Sep 17 2011
G.f.: F(x,t) = 1 + x*t + (x+x^2)*t^2/2! + (x+6*x^2+6*x^3)*t^3/3! + ... = Sum_{n>=0} R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x. - Philippe Deléham, Feb 11 2013
T(n,k) = [t^k] (n! [x^n] (1/(1-t*(exp(x)-1)))). - Peter Luschny, Jan 23 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See also Bala, Example E8. - Peter Bala, Jan 08 2018

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11
Offset: 0

Views

Author

Wouter Meeussen, Oct 01 2002

Keywords

Comments

This sequence counts the "almost triangular" partitions of n. A partition is triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An "almost triangular" partition is a triangular partition with at most 1 added to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4. Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of "Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A function f is acyclic if for every subset B of the domain the image of B under f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)}, f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. - Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) = 1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2, 1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2, 1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688. Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) = RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . - Paul Curtz, Jun 21 2010
A054143 is the fission of the polynomial sequence (p(n,x)) given by p(n,x) = x^n + x^(n-1) + ... + x + 1 by the polynomial sequence ((x+1)^n). See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Reversal of A135278. - Philippe Deléham, Feb 11 2012
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From A238363, the operator equation d/d(:xD:)f(xD)={exp[d/d(xD)]-1}f(xD) = f(xD+1)-f(xD) follows. Choosing f(x) = x^n and using :xD:^n/n! = binomial(xD,n) and (xD)^n = Bell(n,:xD:), the Bell polynomials of A008277, it follows that the lower triangular matrix [padded A074909]
A) = [St2]*[dP]*[St1] = A048993*A132440*[padded A008275]
B) = [St2]*[dP]*[St2]^(-1)
C) = [St1]^(-1)*[dP]*[St1],
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 whereas [padded A074909]=A007318-I with I=identity matrix. - Tom Copeland, Apr 25 2014
T(n,k) generated by m-gon expansions in the case of odd m with "vertex to side" version or even m with "vertex to vertes" version. Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as a triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to side" version or (ii) m is even and expansion is "vertex to vertex" version. m*Sum_{i=1..k} T(n,k) gives the total label value at the n-th iteration. See also A247976. Vertex to vertex: A061777, A247618, A247619, A247620. Vertex to side: A101946, A247903, A247904, A247905. - Kival Ngaokrajang Sep 28 2014
From Tom Copeland, Nov 12 2014: (Start)
With P(n,x) = [(x+1)^(n+1)-x^(n+1)], the row polynomials of this entry, Up(n,x) = P(n,x)/(n+1) form an Appell sequence of polynomials that are the umbral compositional inverses of the Bernoulli polynomials B(n,x), i.e., B[n,Up(.,x)] = x^n = Up[n,B(.,x)] under umbral substitution, e.g., B(.,x)^n = B(n,x).
The e.g.f. for the Bernoulli polynomials is [t/(e^t - 1)] e^(x*t), and for Up(n,x) it's exp[Up(.,x)t] = [(e^t - 1)/t] e^(x*t).
Another g.f. is G(t,x) = log[(1-x*t)/(1-(1+x)*t)] = log[1 + t /(1 + -(1+x)t)] = t/(1-t*Up(.,x)) = Up(0,x)*t + Up(1,x)*t^2 + Up(2,x)*t^3 + ... = t + (1+2x)/2 t^2 + (1+3x+3x^2)/3 t^3 + (1+4x+6x^2+4x^3)/4 t^4 + ... = -log(1-t*P(.,x)), expressed umbrally.
The inverse, Ginv(t,x), in t of the g.f. may be found in A008292 from Copeland's list of formulas (Sep 2014) with a=(1+x) and b=x. This relates these two sets of polynomials to algebraic geometry, e.g., elliptic curves, trigonometric expansions, Chebyshev polynomials, and the combinatorics of permutahedra and their duals.
Ginv(t,x) = [e^((1+x)t) - e^(xt)] / [(1+x) * e^((1+x)t) - x * e^(xt)] = [e^(t/2) - e^(-t/2)] / [(1+x)e^(t/2) - x*e^(-t/2)] = (e^t - 1) / [1 + (1+x) (e^t - 1)] = t - (1 + 2 x) t^2/2! + (1 + 6 x + 6 x^2) t^3/3! - (1 + 14 x + 36 x^2 + 24 x^3) t^4/4! + ... = -exp[-Perm(.,x)t], where Perm(n,x) are the reverse face polynomials, or reverse f-vectors, for the permutahedra, i.e., the face polynomials for the duals of the permutahedra. Cf. A090582, A019538, A049019, A133314, A135278.
With L(t,x) = t/(1+t*x) with inverse L(t,-x) in t, and Cinv(t) = e^t - 1 with inverse C(t) = log(1 + t). Then Ginv(t,x) = L[Cinv(t),(1+x)] and G(t,x) = C[L[t,-(1+x)]]. Note L is the special linear fractional (Mobius) transformation.
Connections among the combinatorics of the permutahedra, simplices (cf. A135278), and the associahedra can be made through the Lagrange inversion formula (LIF) of A133437 applied to G(t,x) (cf. A111785 and the Schroeder paths A126216 also), and similarly for the LIF A134685 applied to Ginv(t,x) involving the simplicial Whitehouse complex, phylogenetic trees, and other structures. (See also the LIFs A145271 and A133932). (End)
R = x - exp[-[B(n+1)/(n+1)]D] = x - exp[zeta(-n)D] is the raising operator for this normalized sequence UP(n,x) = P(n,x) / (n+1), that is, R UP(n,x) = UP(n+1,x), where D = d/dx, zeta(-n) is the value of the Riemann zeta function evaluated at -n, and B(n) is the n-th Bernoulli number, or constant B(n,0) of the Bernoulli polynomials. The raising operator for the Bernoulli polynomials is then x + exp[-[B(n+1)/(n+1)]D]. [Note added Nov 25 2014: exp[zeta(-n)D] is abbreviation of exp(a.D) with (a.)^n = a_n = zeta(-n)]. - Tom Copeland, Nov 17 2014
The diagonals T(n, n-m), for n >= m, give the m-th iterated partial sum of the positive integers; that is A000027(n+1), A000217(n), A000292(n-1), A000332(n+1), A000389(n+1), A000579(n+1), A000580(n+1), A000581(n+1), A000582(n+1), ... . - Wolfdieter Lang, May 21 2015
The transpose gives the numerical coefficients of the Maurer-Cartan form matrix for the general linear group GL(n,1) (cf. Olver, but note that the formula at the bottom of p. 6 has an error--the 12 should be a 15). - Tom Copeland, Nov 05 2015
The left invariant Maurer-Cartan form polynomial on p. 7 of the Olver paper for the group GL^n(1) is essentially a binomial convolution of the row polynomials of this entry with those of A133314, or equivalently the row polynomials generated by the product of the e.g.f. of this entry with that of A133314, with some reindexing. - Tom Copeland, Jul 03 2018
From Tom Copeland, Jul 10 2018: (Start)
The first column of the inverse matrix is the sequence of Bernoulli numbers, which follows from the umbral definition of the Bernoulli polynomials (B.(0) + x)^n = B_n(x) evaluated at x = 1 and the relation B_n(0) = B_n(1) for n > 1 and -B_1(0) = 1/2 = B_1(1), so the Bernoulli numbers can be calculated using Cramer's rule acting on this entry's matrix and, therefore, from the ratios of volumes of parallelepipeds determined by the columns of this entry's square submatrices. - Tom Copeland, Jul 10 2018
Umbrally composing the row polynomials with B_n(x), the Bernoulli polynomials, gives (B.(x)+1)^(n+1) - (B.(x))^(n+1) = d[x^(n+1)]/dx = (n+1)*x^n, so multiplying this entry as a lower triangular matrix (LTM) by the LTM of the coefficients of the Bernoulli polynomials gives the diagonal matrix of the natural numbers. Then the inverse matrix of this entry has the elements B_(n,k)/(k+1), where B_(n,k) is the coefficient of x^k for B_n(x), and the e.g.f. (1/x) (e^(xt)-1)/(e^t-1). (End)

Examples

			T(4,2) = 0+0+1+3+6 = 10 = binomial(5, 2).
Triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7   8   9 10 11
0:  1
1:  1  2
2:  1  3  3
3:  1  4  6   4
4:  1  5 10  10   5
5:  1  6 15  20  15   6
6:  1  7 21  35  35  21   7
7:  1  8 28  56  70  56  28   8
8:  1  9 36  84 126 126  84  36  9
9:  1 10 45 120 210 252 210 120 45   10
10: 1 11 55 165 330 462 462 330 165  55 11
11: 1 12 66 220 495 792 924 792 495 220 66 12
... Reformatted. - _Wolfdieter Lang_, Nov 04 2014
.
Can be seen as the square array A(n, k) = binomial(n + k + 1, n) read by descending antidiagonals. A(n, k) is the number of monotone nondecreasing functions f: {1,2,..,k} -> {1,2,..,n}. - _Peter Luschny_, Aug 25 2019
[0]  1,  1,   1,   1,    1,    1,     1,     1,     1, ... A000012
[1]  2,  3,   4,   5,    6,    7,     8,     9,    10, ... A000027
[2]  3,  6,  10,  15,   21,   28,    36,    45,    55, ... A000217
[3]  4, 10,  20,  35,   56,   84,   120,   165,   220, ... A000292
[4]  5, 15,  35,  70,  126,  210,   330,   495,   715, ... A000332
[5]  6, 21,  56, 126,  252,  462,   792,  1287,  2002, ... A000389
[6]  7, 28,  84, 210,  462,  924,  1716,  3003,  5005, ... A000579
[7]  8, 36, 120, 330,  792, 1716,  3432,  6435, 11440, ... A000580
[8]  9, 45, 165, 495, 1287, 3003,  6435, 12870, 24310, ... A000581
[9] 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, ... A000582
		

Crossrefs

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n+1,k)))); # Muniru A Asiru, Jul 10 2018
    
  • Haskell
    a074909 n k = a074909_tabl !! n !! k
    a074909_row n = a074909_tabl !! n
    a074909_tabl = iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Feb 25 2012
    
  • Magma
    /* As triangle */ [[Binomial(n+1,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2018
    
  • Maple
    A074909 := proc(n,k)
        if k > n or k < 0 then
            0;
        else
            binomial(n+1,k) ;
        end if;
    end proc: # Zerinvary Lajos, Nov 09 2006
  • Mathematica
    Flatten[Join[{1}, Table[Sum[Binomial[k, m], {k, 0, n}], {n, 0, 12}, {m, 0, n}] ]] (* or *) Flatten[Join[{1}, Table[Binomial[n, m], {n, 12}, {m, n}]]]
  • PARI
    print1(1);for(n=1,10,for(k=1,n,print1(", "binomial(n,k)))) \\ Charles R Greathouse IV, Mar 26 2013
    
  • Python
    from math import comb, isqrt
    def A074909(n): return comb(r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),n-comb(r,2)) # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)*e^t - x) e^(x*t). The row polynomials p_n(x) satisfy dp_n(x)/dx = (n+1)*p_(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. - Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)*y^(n-k), where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the n-th row polynomial of this entry. dUp(n,x)/dx = n * Up(n-1,x) and dP(n,x)/dx = (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-t*x)*(1-(1+t)x)) = x + (1+2t)*x^2 + (1+3t+3t^2)*x^3 + ... has the inverse GFinv(x,t) = (1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f., i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...) associated with the moments defined by GFinv, and, in fact, these free cumulants generate these moments through the noncrossing partitions of A134264. The associated e.g.f. and relations to Grassmannians are described in A248727, whose polynomials are the basis for an Appell sequence of polynomials that are umbral compositional inverses of the Appell sequence formed from this entry's polynomials (distinct from the one described in the comments above, without the normalizing reciprocal). - Tom Copeland, Jan 07 2015
T(n, k) = (1/k!) * Sum_{i=0..k} Stirling1(k,i)*(n+1)^i, for 0<=k<=n. - Ridouane Oudra, Oct 23 2022

Extensions

I added an initial 1 at the suggestion of Paul Barry, which makes the triangle a little nicer but may mean that some of the formulas will now need adjusting. - N. J. A. Sloane, Feb 11 2003
Formula section edited, checked and corrected by Wolfdieter Lang, Nov 04 2014

A153641 Nonzero coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent, and Bernoulli numbers (triangle read by rows).

Original entry on oeis.org

1, 1, 1, -1, 1, -3, 1, -6, 5, 1, -10, 25, 1, -15, 75, -61, 1, -21, 175, -427, 1, -28, 350, -1708, 1385, 1, -36, 630, -5124, 12465, 1, -45, 1050, -12810, 62325, -50521, 1, -55, 1650, -28182, 228525, -555731, 1, -66, 2475, -56364, 685575, -3334386, 2702765, 1
Offset: 0

Views

Author

Peter Luschny, Dec 29 2008

Keywords

Comments

In the following the expression [n odd] is 1 if n is odd, 0 otherwise.
(+) W_n(0) = E_n are the Euler (or secant) numbers A122045.
(+) W_n(1) = T_n are the signed tangent numbers, see A009006.
(+) W_{n-1}(1) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli number A027641/A027642.
(+) W_n(-1) 2^{-n}(n+1) = G_n the Genocchi number A036968.
(+) W_n(1/2) 2^{n} are the signed generalized Euler (Springer) number, see A001586.
(+) | W_n([n odd]) | the number of alternating permutations A000111.
(+) | W_n([n odd]) / n! | for 0<=n the Euler zeta number A099612/A099617 (see Wikipedia on Bernoulli number). - Peter Luschny, Dec 29 2008
The diagonals in the full triangle (with zero coefficients) of the polynomials have the general form E(k)*binomial(n+k,k) (k>=0 fixed, n=0,1,...) where E(n) are the Euler numbers in the enumeration A122045. For k=2 we find the triangular numbers A000217 and for k=4 A154286. - Peter Luschny, Jan 06 2009
From Peter Bala, Jun 10 2009: (Start)
The Swiss-Knife polynomials W_n(x) may be expressed in terms of the Bernoulli polynomials B(n,x) as
... W_n(x) = 4^(n+1)/(2*n+2)*[B(n+1,(x+3)/4) - B(n+1,(x+1)/4)].
The Swiss-Knife polynomials are, apart from a multiplying factor, examples of generalized Bernoulli polynomials.
Let X be the Dirichlet character modulus 4 defined by X(4*n+1) = 1, X(4*n+3) = -1 and X(2*n) = 0. The generalized Bernoulli polynomials B(X;n,x), n = 1,2,..., associated with the character X are defined by means of the generating function
... t*exp(x*t)*(exp(t)-exp(3*t))/(exp(4*t)-1) = sum {n = 1..inf} B(X;n,x)*t^n/n!.
The first few values are B(X;1,x) = -1/2, B(X;2,x) = -x, B(X,3,x) = -3/2*(x^2-1) and B(X;4,x) = -2*(x^3-3*x).
In general, W_n(x) = -2/(n+1)*B(X;n+1,x).
For the theory of generalized Bernoulli polynomials associated to a periodic arithmetical function see [Cohen, Section 9.4].
The generalized Bernoulli polynomials may be used to evaluate twisted sums of k-th powers. For the present case the result is
sum{n = 0..4*N-1} X(n)*n^k = 1^k - 3^k + 5^k - 7^k + ... - (4*N-1)^k
= [B(X;k+1,4*N) - B(X;k+1,0)]/(k+1) = [W_k(0) - W_k(4*N)]/2.
For the proof apply [Cohen, Corollary 9.4.17 with m = 4 and x = 0].
The generalized Bernoulli polynomials and the Swiss-Knife polynomials are also related to infinite sums of powers through their Fourier series - see the formula section below. For a table of the coefficients of generalized Bernoulli polynomials attached to a Dirichlet character modulus 8 see A151751.
(End)
The Swiss-Knife polynomials provide a general formula for alternating sums of powers similar to the formula which are provided by the Bernoulli polynomials for non-alternating sums of powers (see the Luschny link). Sequences covered by this formula include A001057, A062393, A062392, A011934, A144129, A077221, A137501, A046092. - Peter Luschny, Jul 12 2009
The greatest common divisor of the nonzero coefficients of the decapitated Swiss-Knife polynomials is exp(Lambda(n)), where Lambda(n) is the von Mangoldt function for odd primes, symbolically:
gcd(coeffs(SKP_{n}(x) - x^n)) = A155457(n) (n>1). - Peter Luschny, Dec 16 2009
Another version is at A119879. - Philippe Deléham, Oct 26 2013

Examples

			1
x
x^2  -1
x^3  -3x
x^4  -6x^2   +5
x^5 -10x^3  +25x
x^6 -15x^4  +75x^2  -61
x^7 -21x^5 +175x^3 -427x
		

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag. [From Peter Bala, Jun 10 2009]

Crossrefs

W_n(k), k=0,1,...
W_0: 1, 1, 1, 1, 1, 1, ........ A000012
W_1: 0, 1, 2, 3, 4, 5, ........ A001477
W_2: -1, 0, 3, 8, 15, 24, ........ A067998
W_3: 0, -2, 2, 18, 52, 110, ........ A121670
W_4: 5, 0, -3, 32, 165, 480, ........
W_n(k), n=0,1,...
k=0: 1, 0, -1, 0, 5, 0, -61, ... A122045
k=1: 1, 1, 0, -2, 0, 16, 0, ... A155585
k=2: 1, 2, 3, 2, -3, 2, 63, ... A119880
k=3: 1, 3, 8, 18, 32, 48, 128, ... A119881
k=4: 1, 4, 15, 52, 165, 484, ........ [Peter Luschny, Jul 07 2009]

Programs

  • Maple
    w := proc(n,x) local v,k,pow,chen; pow := (a,b) -> if a = 0 and b = 0 then 1 else a^b fi; chen := proc(m) if irem(m+1,4) = 0 then RETURN(0) fi; 1/((-1)^iquo(m+1,4) *2^iquo(m,2)) end; add(add((-1)^v*binomial(k,v)*pow(v+x+1,n)*chen(k),v=0..k), k=0..n) end:
    # Coefficients with zeros:
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)*sech(t),t,16),t,i),x,i-n),n=0..i)), i=0..8);
    # Recursion
    W := proc(n,z) option remember; local k,p;
    if n = 0 then 1 else p := irem(n+1,2);
    z^n - p + add(`if`(irem(k,2)=1,0,
    W(k,0)*binomial(n,k)*(power(z,n-k)-p)),k=2..n-1) fi end:
    # Peter Luschny, edited and additions Jul 07 2009, May 13 2010, Oct 24 2011
  • Mathematica
    max = 9; rows = (Reverse[ CoefficientList[ #, x]] & ) /@ CoefficientList[ Series[ Exp[x*t]*Sech[t], {t, 0, max}], t]*Range[0, max]!; par[coefs_] := (p = Partition[ coefs, 2][[All, 1]]; If[ EvenQ[ Length[ coefs]], p, Append[ p, Last[ coefs]]]); Flatten[ par /@ rows] (* Jean-François Alcover, Oct 03 2011, after g.f. *)
    sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; Table[CoefficientList[sk[n, x], x] // Reverse // Select[#, # =!= 0 &] &, {n, 0, 13}] // Flatten (* Jean-François Alcover, May 21 2013 *)
    Flatten@Table[Binomial[n, 2k] EulerE[2k], {n, 0, 12}, {k, 0, n/2}](* Oliver Seipel, Jan 14 2025 *)
  • Sage
    def A046978(k):
        if k % 4 == 0:
            return 0
        return (-1)**(k // 4)
    def A153641_poly(n, x):
        return expand(add(2**(-(k // 2))*A046978(k+1)*add((-1)**v*binomial(k,v)*(v+x+1)**n for v in (0..k)) for k in (0..n)))
    for n in (0..7): print(A153641_poly(n, x))  # Peter Luschny, Oct 24 2011

Formula

W_n(x) = Sum_{k=0..n}{v=0..k} (-1)^v binomial(k,v)*c_k*(x+v+1)^n where c_k = frac((-1)^(floor(k/4))/2^(floor(k/2))) [4 not div k] (Iverson notation).
From Peter Bala, Jun 10 2009: (Start)
E.g.f.: 2*exp(x*t)*(exp(t)-exp(3*t))/(1-exp(4*t))= 1 + x*t + (x^2-1)*t^2/2! + (x^3-3*x)*t^3/3! + ....
W_n(x) = 1/(2*n+2)*Sum_{k=0..n+1} 1/(k+1)*Sum_{i=0..k} (-1)^i*binomial(k,i)*((x+4*i+3)^(n+1) - (x+4*i+1)^(n+1)).
Fourier series expansion for the generalized Bernoulli polynomials:
B(X;2*n,x) = (-1)^n*(2/Pi)^(2*n)*(2*n)! * {sin(Pi*x/2)/1^(2*n) - sin(3*Pi*x/2)/3^(2*n) + sin(5*Pi*x/2)/5^(2*n) - ...}, valid for 0 <= x <= 1 when n >= 1.
B(X;2*n+1,x) = (-1)^(n+1)*(2/Pi)^(2*n+1)*(2*n+1)! * {cos(Pi*x/2)/1^(2*n+1) - cos(3*Pi*x/2)/3^(2*n+1) + cos(5*Pi*x/2)/5^(2*n+1) - ...}, valid for 0 <= x <= 1 when n >= 1 and for 0 <= x < 1 when n = 0.
(End)
E.g.f.: exp(x*t) * sech(t). - Peter Luschny, Jul 07 2009
O.g.f. as a J-fraction: z/(1-x*z+z^2/(1-x*z+4*z^2/(1-x*z+9*z^2/(1-x*z+...)))) = z + x*z^2 + (x^2-1)*z^3 + (x^3-3*x)*z^4 + .... - Peter Bala, Mar 11 2012
Conjectural o.g.f.: Sum_{n >= 0} (1/2^((n-1)/2))*cos((n+1)*Pi/4)*( Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - (k + x)*t) ) = 1 + x*t + (x^2 - 1)*t^2 + (x^3 - 3*x)*t^3 + ... (checked up to O(t^13)), which leads to W_n(x) = Sum_{k = 0..n} 1/2^((k - 1)/2)*cos((k + 1)*Pi/4)*( Sum_{j = 0..k} (-1)^j*binomial(k, j)*(j + x)^n ). - Peter Bala, Oct 03 2016

A007850 Giuga numbers: composite numbers n such that p divides n/p - 1 for every prime divisor p of n.

Original entry on oeis.org

30, 858, 1722, 66198, 2214408306, 24423128562, 432749205173838, 14737133470010574, 550843391309130318, 244197000982499715087866346, 554079914617070801288578559178, 1910667181420507984555759916338506
Offset: 1

Views

Author

D. Borwein, J. M. Borwein, P. B. Borwein and R. Girgensohn

Keywords

Comments

There are no other Giuga numbers with 8 or fewer prime factors. I did an exhaustive search using a PARI script which implemented Borwein and Girgensohn's method for finding n factor solutions given n - 2 factors. - Fred Schneider, Jul 04 2006
One further Giuga number is known with 10 prime factors, namely:
420001794970774706203871150967065663240419575375163060922876441614\
2557211582098432545190323474818 =
2 * 3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519 * 8491659218261819498490029296021 * 58254480569119734123541298976556403 but this may not be the next term. (See the Butske et al. paper.)
Conjecture: Giuga numbers are the solution of the differential equation n' = n + 1, where n' is the arithmetic derivative of n. - Paolo P. Lava, Nov 16 2009
n is a Giuga number if and only if n' = a*n + 1 for some integer a > 0 (see our preprint in arXiv:1103.2298). - José María Grau Ribas, Mar 19 2011
A composite number n is a Giuga number if and only if Sum_{i = 1..n-1} i^phi(n) == -1 (mod n), where phi(n) = A000010(n). - Jonathan Sondow, Jan 03 2014
A composite number n is a Giuga number if and only if Sum_{prime p|n} 1/p = 1/n + an integer. (In fact, all known Giuga numbers n satisfy Sum_{prime p|n} 1/p = 1/n + 1.) - Jonathan Sondow, Jan 08 2014
The prime factors of a(n) are listed as n-th row of A236434. - M. F. Hasler, Jul 13 2015
Conjecture: let k = a(n) and k be the product of x(n) distinct prime factors where x(n) <= x(n+1). Then, for any even n, n/2 + 2 <= x(n) <= n/2 + 3 and, for any odd n, (n+1)/2 + 2 <= x(n) <= (n+1)/2 + 3. For any n > 1, there are y "old" distinct prime factors o(1)...o(y) such that o(1) = 2, o(2) = 3, and z "new" distinct prime factors n(1)...n(z) such that none of them - unlike the "old" ones - can be a divisor of a(q) while q < n; n(1) > o(y), y = x(n) - z >= 2, 2 <= z <= b where b is either 4, or 1/2*n. - Sergey Pavlov, Feb 24 2017
Conjecture: a composite n is a Giuga number if and only if Sum_{k=1..n-1} k^lambda(n) == -1 (mod n), where lambda(n) = A002322(n). - Thomas Ordowski and Giovanni Resta, Jul 25 2018
A composite number n is a Giuga number if and only if A326690(n) = 1. - Jonathan Sondow, Jul 19 2019
A composite n is a Giuga number if and only if n * A027641(phi(n)) == - A027642(phi(n)) (mod n^2). Note: Euler's phi function A000010 can be replaced by the Carmichael lambda function A002322. - Thomas Ordowski, Jun 07 2020
By von Staudt and Clausen theorem, a composite n is a Giuga number if and only if n * A027759(phi(n)) == A027760(phi(n)) (mod n^2). Note: Euler's phi function can be replaced by the Carmichael lambda function. - Thomas Ordowski, Aug 01 2020

Examples

			From _M. F. Hasler_, Jul 13 2015: (Start)
The prime divisors of 30 are {2, 3, 5}, and 2 divides 30/2-1 = 14, 3 divides 30/3-1 = 9, and 5 divides 30/5-1 = 5.
The prime divisors of 858 are {2, 3, 11, 13} and 858/2-1 = 428 is even, 858/3-1 = 285 is divisible by 3, 858/11-1 = 77 is a multiple of 11, and 858/13-1 = 65 = 13*5.
(End)
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 30, pp 11, Ellipses, Paris 2008.

Crossrefs

Programs

  • Mathematica
    fQ[n_] := AllTrue[First /@ FactorInteger@ n, Divisible[n/# - 1, #] &]; Select[Range@ 100000, CompositeQ@ # && fQ@ # &] (* Michael De Vlieger, Oct 05 2015 *)
  • PARI
    is(n)=if(isprime(n), return(0)); my(f=factor(n)[,1]); for(i=1,#f, if((n/f[i])%f[i]!=1, return(0))); n>1 \\ Charles R Greathouse IV, Apr 28 2015
    
  • Python
    from itertools import count, islice
    from sympy import isprime, primefactors
    def A007850_gen(startvalue=2): # generator of terms >= startvalue
        return filter(lambda x: not isprime(x) and all((x//p-1) % p == 0 for p in primefactors(x)), count(max(startvalue,2)))
    A007850_list = list(islice(A007850_gen(),4)) # Chai Wah Wu, Feb 19 2022

Formula

Sum_{i = 1..a(n)-1} i^phi(a(n)) == -1 (mod a(n)). - Jonathan Sondow, Jan 03 2014

Extensions

a(12) from Fred Schneider, Jul 04 2006
Further references from Fred Schneider, Aug 19 2006
Definition corrected by Jonathan Sondow, Sep 16 2012
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