A177427 -id:A177427 - cp's OEIS frontend

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387

Offset: 1

N. J. A. Sloane

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
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).

Kenny Lau, Table of n, a(n) for n = 1..2295 (first 200 terms provided by T. D. Noe)
David H. Bailey, Jonathan M. Borwein, and Roland Girgensohn, Experimental evaluation of Euler sums, Exper. Math. 3(1) (1994), 17-30; they evaluate the constants Sum_{k>=1} H_k^m/(k+1)^n.
Hongwei Chen, Evaluations of Some Variant Euler Sums, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.3.
Hongwei Chen, Interesting Series Associated with Central Binomial Coefficients, Catalan Numbers and Harmonic Numbers, J. Int. Seq. 19 (2016), #16.1.5.
R. M. Dickau, Harmonic numbers and the book-stacking problem.
Leo Goldmakher, A short(er) proof of the divergence of the harmonic series.
Antal Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013), 54-82.
Fredrik Johansson, How (not) to compute harmonic numbers. Feb 21 2009.
Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), #00.2.9.
Romeo Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.
Jerry Metzger and Thomas Richards, A Prisoner Problem Variation, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.7.
Hisanori Mishima, Factorizations of Wolstenholme numbers, n=1..100, n=101..200, n=201..300.
Mike Paterson et al., Maximum Overhang.
Maxie D. Schmidt, Generalized j-Factorial Functions, Polynomials, and Applications, J. Int. Seq. 13 (2010), 10.6.7, Section 4.3.
Peter Shiu, The denominators of harmonic numbers, arXiv:1607.02863 [math.NT], 2016.
N. J. A. Sloane, Illustration of initial terms.
Jonathan Sondow and Eric W. Weisstein, MathWorld: Harmonic Number.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4.
Eric Weisstein's World of Mathematics, Book Stacking Problem, Wolstenholme's Theorem, Harmonic Mean, Digamma Function.
Wikipedia, Harmonic number.

Cf. A002805 (denominators), A007406, A007408, A007410, A075135, A001220, A125854, A121999, A014566, A056903, A067657, A177427, A177690.
Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018

import Data.Ratio ((%), numerator)
a001008 = numerator . sum . map (1 %) . enumFromTo 1
a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
-- Reinhard Zumkeller, Jul 03 2012

[Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016

A001008 := proc(n)
    add(1/k,k=1..n) ;
    numer(%) ;
end proc:
seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016

Table[Numerator[HarmonicNumber[n]], {n, 30}]
(* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)

A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009

H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019

from sympy import Integer
[sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017

def harmonic(a, b):  # See the F. Johansson link.
    if b - a == 1:
        return 1, a
    m = (a+b)//2
    p, q = harmonic(a,m)
    r, s = harmonic(m,b)
    return p*s+q*r, q*s
def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
[A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n  = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Edited by Max Alekseyev, Oct 21 2011

Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 4, 4, 8, 8, 1, 4, 8, 4, 16, 2, 2, 1, 8, 32, 32, 1, 2, 4, 4, 16, 32, 64, 8, 8, 16, 16, 64, 64, 128, 128, 1, 8, 16, 8, 32, 64, 128, 32, 256, 2, 2, 8, 16, 64, 64, 128, 64, 512, 512, 1, 2, 4, 8, 32, 64, 128, 16, 128, 512, 1024

Offset: 0

Paul Curtz, Jan 18 2013

1/2^n and successive rows are
1,       1/2,   1/4,   1/8,  1/16,  1/32,   1/64, 1/128, 1/256,...
1/2,     1/2,   3/8,   1/4,  5/32,  3/32,  7/128,  1/32,...       = A000265/A075101, the Oresme numbers n/2^n. Paul Curtz, Jan 18 2013 and May 11 2016
0,       1/4,   3/8,   3/8,  5/16, 15/64, 21/128,...              = (0 before A069834)/new,
-1/4,   -1/4,     0,   1/4, 25/64, 27/64,...
0,      -1/2,  -3/4, -9/16, -5/32,...
1/2,     1/2, -9/16, -13/8,...
0,      17/8, 51/16,...
-17/8, -17/8,...
0
The first column is A198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,...  in A060096.
Numerators b(n): 1, 1, 1, 0, 1, 1, -1, 1, 3, 1, ... .
Coll(n+1) - 2*Coll(n) = -1/2, -5/8, -1/2, -11/32, -7/32, -17/128, -5/64, -23/512, ... = -A075677/new, from Collatz problem.
There are three different Bernoulli numbers:
The first Bernoulli numbers are  1, -1/2, 1/6, 0,... = A027641(n)/A027642(n).
The second Bernoulli numbers are 1,  1/2, 1/6, 0,... = A164555(n)/A027642(n). These are the binomial transform of the first one.
The third Bernoulli numbers are  1,   0,  1/6, 0,... = A176327(n)/A027642(n), the half sum. Via A177427(n) and A191567(n), they yield the Balmer series A061037/A061038.
There are three different fractional Euler numbers:
1) The first are  1, -1/2, 0, 1/4, 0, -1/2,... in A060096(n).
Also Akiyama-Tanigawa algorithm for ( 1, 3/2, 7/4, 15/8, 31/16, 63/32,... = A000225(n+1)/A000079(n) ).
2) The second are 1, 1/2, 0, -1/4, 0,  1/2,... , mentioned by Wolfdieter Lang in A198631(n).
3) The third are  0, 1/2, 0, -1/4, 0,  1/2,... , half difference of 2) and 1).
Also Akiyama-Tanigawa algorithm for ( 0, -1/2, -3/4, -7/8, -15/16, -31/32,... =  A000225(n)/A000079(n) ). See A097110(n).

			Triangle begins:
  1,
  2, 2,
  1, 2,  4,
  4, 4,  8,  8,
  1, 4,  8,  4, 16,
  2, 2,  1,  8, 32, 32,
  1, 2,  4,  4, 16, 32,  64,
  8, 8, 16, 16, 64, 64, 128, 128,
  ...

G. C. Greubel, Table of n, a(n) for n = 0..5049
A. F. Horadam, Oresme Numbers, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.

Cf. Second Bernoulli numbers A164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1), A272263.

max = 10; t[0, k_] := 1/2^k; t[n_, k_] := t[n, k] = (k + 1)*(t[n - 1, k] - t[n - 1, k + 1]); denoms = Table[t[n, k] // Denominator, {n, 0, max}, {k, 0, max - n}]; Table[denoms[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 05 2013 *)

A176743 a(n) = gcd(A000217(n+1), A002378(n+2)).

Original entry on oeis.org

1, 3, 2, 10, 3, 7, 4, 18, 5, 11, 6, 26, 7, 15, 8, 34, 9, 19, 10, 42, 11, 23, 12, 50, 13, 27, 14, 58, 15, 31, 16, 66, 17, 35, 18, 74, 19, 39, 20, 82, 21, 43, 22, 90, 23, 47, 24, 98, 25, 51, 26

Offset: 0

Paul Curtz, Apr 25 2010

These greatest common divisors of (n+1)*(n+2)/2 and (n+2)*(n+3) appear in the second row of the table discussed in A177427: 0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, -2/5, ... These fractions can be written as A000217(n+1)/A002378(n+2), n >= 0, and the current sequence shows the common factors that reduces the fractions to the standard format with coprime numerator and denominator.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A000217 (triangular), A002378 (oblong), A176662.

A176743 := proc(n)
    if type(n,'even') then
        n/2+1 ;
    else
        A123167((n-1)/2+2) ;
    end if;
end proc: # R. J. Mathar, Jul 25 2013

Table[GCD[Plus@@Range[n + 1], (n + 2)(n + 3)], {n, 0, 49}] (* Alonso del Arte, Jan 16 2011 *)

a(n)=(n+2)*gcd((n+1)/2, n+3) \\ Charles R Greathouse IV, Sep 02 2015

a(2n) = n+1, a(2n+1) = A123167(n+2).
a(n) = 2*a(n-4) - a(n-8).
G.f.: (1 + 3*x + 2*x^2 + 10*x^3 + x^4 + x^5 - 2*x^7) / ( (x-1)^2*(1+x)^2*(x^2+1)^2 ). - R. J. Mathar, Jan 16 2011
a(n) = (8*n + 16 - 4*(n+2)*(-1)^n + (2*n + 5 + (-1)^n)*((1-(-1)^n)*(-1)^((2*n + 3 + (-1)^n)/4)))/8. - Luce ETIENNE, Feb 03 2015

A177690 Denominators of the Inverse Akiyama-Tanigawa transform of the aerated even-indexed Bernoulli numbers 1, 0, 1/6, 0, -1/30, 0, 1/42, ...

Original entry on oeis.org

1, 1, 12, 6, 120, 120, 280, 140, 5040, 5040, 55440, 55440, 720720, 720720, 720720, 360360, 24504480, 24504480, 155195040, 155195040, 31039008, 10346336, 237965728, 713897184, 17847429600, 17847429600, 160626866400, 22946695200

Offset: 0

Paul Curtz, May 11 2010

See A177427 (numerators) for the description of the Akiyama-Tanigawa array of this sequence of fractions, T(0,k) = 1, 1, 13/12, 7/6, 149/120, 157/120, ...
If we add a zero in front and construct an array A(n,k) with successive differences A(n,k) = A(n-1,k+1)-A(n-1,k), the array A(.,.) becomes
0, 1, 1, 13/12, 7/6, 149/120, 157/120, 383/280, 199/140, 7409/5040, ...
1, 0, 1/12, 1/12, 3/40, 1/15, 5/84, 3/56, 7/144, 2/45, 9/220, ...
-1, 1/12, 0, -1/120, -1/120, -1/140, -1/168, -5/1008, -1/240, -7/1980, ...
13/12, -1/12, -1/120, 0, 1/840, 1/840, 1/1008, 1/1260, 1/1584, ...
-7/6, 3/40, 1/120, 1/840, 0, -1/5040, -1/5040, -1/6160, -1/7920, ...
149/120, -1/15, -1/140, -1/840, -1/5040, 0, 1/27720, 1/27720, ...
-157/120, 5/84, 1/168, 1/1008, 1/5040, 1/27720, 0, -1/144144, -1/144144, ...
On the diagonal, A(n,n)=0. The left column A(n,0) = (-1)^(n+1)*A(0,k) is a signed variant of the top row, which means the sequence is some eigensequence under the inverse binomial transform (see A174341 for other examples). This eigen-feature would remain if the same number of top rows and left columns were removed from A(.,.).

Cf. A177427 (numerators).

read("transforms3") ; [seq(bernoulli(2*n),n=0..20)] ; AERATE(%,1) ; AKIYAMATANIGAWAi(%) ; apply(denom,%) ; # R. J. Mathar, Jan 16 2011

t[n_, 0] := BernoulliB[n]; t[1, 0]=0; t[n_, k_] := t[n, k] = (t[n, k-1] + (k-1)*t[n, k-1] - t[n+1, k-1])/k; Table[t[0, k], {k, 0, 27}] // Denominator (* Jean-François Alcover, Aug 09 2012 *)

A191567 Four interlaced 2nd order polynomials: a(4k) = k(1+2k); a(1+2k) = 2(1+2k)(3+2k); a(2+4k) = 4(1+k)(1+2k).

Original entry on oeis.org

0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286, 21, 390, 112, 510, 36, 646, 180, 798, 55, 966, 264, 1150, 78, 1350, 364, 1566, 105, 1798, 480, 2046, 136, 2310, 612, 2590, 171, 2886, 760, 3198, 210, 3526, 924, 3870, 253, 4230, 1104, 4606, 300, 4998, 1300, 5406, 351

Offset: 0

Paul Curtz, Jun 12 2011

a(n) = T(0,n) and differences T(n,k) = T(n-1,k+1) - T(n-1,k) define the array
0,   6,  4,  30,    3,  70,   24,  126,   10,  198,   60,  286,   21,  390,  ..
6,  -2, 26, -27,   67, -46,  102, -116,  188, -138,  226, -265,  369, -278, ..
-8, 28 -53,  94, -113, 148, -218,  304, -326,  364, -491,  634, -647,  676, ...
T(3,n) mod 9 is the sequence 1, 1, 1, 4, 4, 4, 7, 7, 7, 4, 4, 4 (and periodically repeated with period 12).
A064680(2+n) divides a(n), where b(n) = a(n)/A064680(2+n) = 0, 1, 2, 3, 1, 5, 6, 7, 2,... for n>=0, obeys b(4*k) = k and has recurrence b(n) = 2*b(n-4) - b(n-8).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A014105, A000466, A000384, A177427.

a:=[0,6,4,30,3,70,24,126,10,198,60,286];; for n in [13..60] do a[n]:= 3*a[n-4]-3*a[n-8]+a[n-12]; od; a; # G. C. Greubel, Feb 26 2019

I:=[0,6,4,30,3,70,24,126,10,198,60,286]; [n le 12 select I[n] else 3*Self(n-4)-3*Self(n-8)+Self(n-12): n in [1..60]]; // Vincenzo Librandi, Apr 23 2017

Table[Which[OddQ@ n, 2 (1 + 2 #) (3 + 2 #) &[(n - 1)/2], Mod[n, 4] == 0, # (1 + 2 #) &[n/4], True, 4 (1 + #) (1 + 2 #) &[(n - 2)/4]], {n, 0, 60}] (* or *)
CoefficientList[Series[x(6 +4x +30x^2 +3x^3 +52x^4 +12x^5 +36x^6 +x^7 +6x^8 -2x^10)/((1-x)^3*(1+x)^3*(1+x^2)^3), {x, 0, 60}], x] (* Michael De Vlieger, Apr 22 2017 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1}, {0,6,4,30,3,70,24,126,10,198,60, 286}, 80] (* Vincenzo Librandi, Apr 23 2017 *)

m=60; v=concat([0,6,4,30,3,70,24,126,10,198,60,286], vector(m-12)); for(n=13, m, v[n]=3*v[n-4]-3*v[n-8]+v[n-12]); v \\ G. C. Greubel, Feb 26 2019

(x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10)/((1-x)^3 *(1+x)^3*(1+x^2)^3 )).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) = A061037(n+2) + A181318(n). - Paul Curtz, Jul 19 2011
a(n) = A060819(n) * A145979(n). - Paul Curtz, Sep 06 2011
G.f.: x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10) /( (1-x)^3 *(1+x)^3 *(1+x^2)^3 ). - R. J. Mathar, Jun 17 2011
Let BEB(n) = a(n)/A061038(n+2) = A060819(n)/A145979(n). Then (BEB(n))^2 = A181318(n)/A061038(n+2) = BEB(n) - A061037(n+2)/A061038(n+2). - Paul Curtz, Jul 19 2011, index corrected by R. J. Mathar, Sep 09 2011
From Luce ETIENNE, Apr 18 2017: (Start)
a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*((-1)^((2*n + 1 - (-1)^n)/4) + (-1)^((2*n - 1 + (-1)^n)/4)))/32.
a(n) = n*(n+2)*(37-27*cos(n*Pi) - 6*cos(n*Pi/2))/32.
a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*(i^n + (-i)^n))/32, where i=sqrt(-1). (End)

A260196 1, -3, followed by -1's.

Original entry on oeis.org

1, -3, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1

Offset: 0

Paul Curtz, Jul 19 2015

1/(n+1) is the inverse Akiyama-Tanigawa transform of A164555(n)/A027642(n).
For more on the Akiyama-Tanigawa transform, see Links (correction: page 7 read here A164555 instead of A027641) and A177427.
Here:
   1,  -3, -1, -1, -1, -1, ...
   4,  -4,  0,  0,  0,  0, ...
   8,  -8,  0,  0,  0,  0, ...
  16, -16,  0,  0,  0,  0, ...
  etc.
Other process, using signed A130534(n), different of A008275(n):
1,                            1/1,                                      1,
1, 4,                      (  1,     -1)/1,                            -3,
1, 4, 8,                   (  2,     -3,   1)/2,                       -1,
1, 4, 8, 16,          *    (  6,    -11,   6,   -1)/6,             =   -1,
1, 4, 8, 16, 32,           ( 24,    -50,  35,  -10,  1)/24,            -1,
1, 4, 8, 16, 32, 64,       (120,   -274, 225,  -85, 15, -1)/120,       -1,
etc.                       etc.                                        etc.
Via the modified Stirling numbers of the first kind, the second triangle, Iw(n), is the inverse of Worpitzky transform A163626(n).
a(n) is the third sequence of a family beginning with
1,  1,  1,  1,  1,  1,  1,  1, ...              = A000012(n)
1,  0,  0,  0,  0,  0,  0,  0,  0, ...          = A000007(n)
1, -3, -1, -1, -1, -1, -1, -1, -1, -1, ... .
A000012(n) is the inverse Akiyama-Tanigawa transform of A000007(n), with or without its second term.
A000007(n) is the inverse Akiyama-Tanigawa transform of A000012(n), with or without its second term.
a(n) is the inverse Akiyama-Tanigawa transform of 2^n omitting the second term i.e. 2.

Colin Barker, Table of n, a(n) for n = 0..1000
Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, Journal of Integer Sequences, 3(2000), article 00.2.9
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A000007, A000012, A008275, A027642, A130534, A151821, A163626, A164555, A177427.

first(m)=vector(m,i,i--;if(i>1,-1,if(i==0,1,if(i==1,-3)))) \\ Anders Hellström, Aug 28 2015

Vec(-(2*x^2-4*x+1)/(x-1) + O(x^100)) \\ Colin Barker, Sep 11 2015

Inverse Akiyama-Tanigawa transform of A151821(n).
From Colin Barker, Sep 11 2015: (Start)
a(n) = -1 for n>1.
a(n) = a(n-1) for n>2.
G.f.: -(2*x^2-4*x+1) / (x-1).
(End)

cp's OEIS Frontend

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

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A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

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A176743 a(n) = gcd(A000217(n+1), A002378(n+2)).

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A177690 Denominators of the Inverse Akiyama-Tanigawa transform of the aerated even-indexed Bernoulli numbers 1, 0, 1/6, 0, -1/30, 0, 1/42, ...

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A191567 Four interlaced 2nd order polynomials: a(4*k) = k*(1+2*k); a(1+2*k) = 2*(1+2*k)*(3+2*k); a(2+4*k) = 4*(1+k)*(1+2*k).

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A260196 1, -3, followed by -1's.

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A191567 Four interlaced 2nd order polynomials: a(4k) = k(1+2k); a(1+2k) = 2(1+2k)(3+2k); a(2+4k) = 4(1+k)(1+2k).