Displaying 1-10 of 16 results found.
0, 2, 0, 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
COMMENTS
Offset 0 is chosen instead of -1. (The offset 0 corresponds to A176327(n), -1 to 0 followed by A176327(n).) Denominators: b(n) = 1 followed by A141459(n). Difference table of a(n)/b(n):
0, 2, 0, 1/3, 0, -1/15, 0, ...
2, -2, 1/3, -1/3, -1/15, 1/15, ...
-4, 7/3, -2/3, 4/15, 2/15, ...
19/3, -3, 14/15, -2/15, ...
-28/3, 59/15, -16/15, ...
199/15, -5, ...
-274/15, ...
etc.
Without the first column, the antidiagonal sums are (-1)^n * A254667(n+1).
FORMULA
a(2n) = 0. a(2n+1) = A172086(2n), from the main pure Bernoulli twin numbers.
Numerator of Bernoulli number B_n.
+10
245
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
COMMENTS
a(n)/ A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/ A130190(n). Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008 Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012 The sequence of the Bernoulli numbers B_n = a(n)/ A027642(n) is the inverse binomial transform of the sequence { A164555(n)/ A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016 Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023
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.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
Harold 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.
Harold M. Edwards, Riemann's Zeta Function, Academic Press, NY, 1974; see p. 11.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.1.
Herman H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.6.
L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 137.
Hans Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
E.g.f: x/(exp(x) - 1); take numerators.
Recurrence: B^n = (1+B)^n, n >= 2 (interpreting B^j as B_j).
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).
Sum_{i=1..n-1} i^k = ((n+B)^(k+1)-B^(k+1))/(k+1) (interpreting B^j as B_j).
B_{n-1} = - Sum_{r=1..n} (-1)^r binomial(n, r) r^(-1) Sum_{k=1..r} k^(n-1). More concisely, B_n = 1 - (1-C)^(n+1), where C^r is replaced by the arithmetic mean of the first r n-th powers of natural numbers in the expansion of the right-hand side. [Bergmann]
Sum_{i>=1} 1/i^(2k) = zeta(2k) = (2*Pi)^(2k)*|B_{2k}|/(2*(2k)!).
B_{2n} = (-1)^(m-1)/2^(2m+1) * Integral{-inf..inf, [d^(m-1)/dx^(m-1) sech(x)^2 ]^2 dx} (see Grosset/Veselov).
Let B(s,z) = -2^(1-s)(i/Pi)^s s! PolyLog(s,exp(-2*i*Pi/z)). Then B(2n,1) = B_{2n} for n >= 1. Similarly the numbers B(2n+1,1), which might be called Co-Bernoulli numbers, can be considered, and it is remarkable that Leonhard Euler in 1755 already calculated B(3,1) and B(5,1) (Opera Omnia, Ser. 1, Vol. 10, p. 351). (Cf. the Luschny reference for a discussion.) - Peter Luschny, May 02 2009 The B_n sequence is the left column of the inverse of triangle A074909, the "beheaded" Pascal's triangle. - Gary W. Adamson, Mar 05 2012 E.g.f. E(x)= 2 - x/(tan(x) + sec(x) - 1)= Sum_{n>=0} a(n)*x^n/n!, a(n)=|B(n)|, where B(n) is Bernoulli number B_n.
E(x)= 2 + x - B(0), where B(k)= 4*k+1 + x/(2 + x/(4*k+3 - x/(2 - x/B(k+1)))); (continued fraction, 4-step). (End)
E.g.f.: x/(exp(x)-1)= U(0); U(k)= 2*k+1 - x(2*k+1)/(x + (2*k+2)/(1 + x/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 05 2012 E.g.f.: 2*(x-1)/(x*Q(0)-2) where Q(k) = 1 + 2*x*(k+1)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)^2/(x*(2*k+3) + 4*(k+1)*(k+2)/Q(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 26 2013 a(n) = numerator(B(n)), B(n) = (-1)^n*Sum_{k=0..n} Stirling1(n,k) * Stirling2(n+k,n) / binomial(n+k,k). - Vladimir Kruchinin, Mar 16 2013 E.g.f.: x/(exp(x)-1) = E(0) where E(k) = 2*k+1 - x/(2 + x/E(k+1)); (continued fraction). - Sergei N. Gladkovskii, Mar 16 2013 G.f. for Bernoulli(n) = a(n)/ A027642(n): psi_1(1/x)/x - x, where psi_n(z) is the polygamma function, psi_n(z) = (d/dz)^(n+1) log(Gamma(z)). - Vladimir Reshetnikov, Apr 24 2013 E.g.f.: 2*E(0) - 2*x, where E(k)= x + (k+1)/(1 + 1/(1 - x/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 10 2013 B_n = Sum_{m=0..n} (-1)^m * A131689(n, m)/(m + 1), n >= 0. See one of the Maple programs. - Wolfdieter Lang, May 05 2017 a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n), for n=0 and n>1.
a(n) = numerator(-n*zeta(1-n)), for n=0 and n>1. (End)
EXAMPLE
B_n sequence begins 1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, -691/2730, 0, 7/6, 0, -3617/510, ...
MAPLE
B := n -> add((-1)^m*m!*Stirling2(n, m)/(m+1), m=0..n);
B := n -> bernoulli(n);
MATHEMATICA
Numerator[ Range[0, 40]! CoefficientList[ Series[x/(E^x - 1), {x, 0, 40}], x]]
Numerator[CoefficientList[Series[PolyGamma[1, 1/x]/x - x, {x, 0, 40}, Assumptions -> x > 0], x]] (* Vladimir Reshetnikov, Apr 24 2013 *)
PROG
(PARI) a(n)=numerator(bernfrac(n))
(Maxima) B(n):=(-1)^((n))*sum((stirling1(n, k)*stirling2(n+k, n))/binomial(n+k, k), k, 0, n);
(SageMath)
[bernoulli(n).numerator() for n in range(41)] # Peter Luschny, Feb 19 2016 (SageMath) # Alternatively:
f, R, C = 1, [1], [1]+[0]*(len-1)
for n in (1..len-1):
f *= n
for k in range(n, 0, -1):
C[k] = C[k-1] / (k+1)
C[0] = -sum(C[k] for k in (1..n))
R.append((C[0]*f).numerator())
return R
(Python)
from sympy import bernoulli
from fractions import Fraction
[bernoulli(i).as_numer_denom()[0] for i in range(51)] # Indranil Ghosh, Mar 18 2017 (Python)
from sympy import bernoulli
def A027641(n): return bernoulli(n).p
CROSSREFS
This is the main entry for the Bernoulli numbers and has all the references, links and formulas. Sequences A027642 (the denominators of B_n) and A000367/ A002445 = B_{2n} are also important!
Pascal's triangle, with the first two columns removed.
+10
20
1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715
COMMENTS
A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle. A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)). For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11. This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
1;
1, 3;
1, 6, 4;
1, 10, 10, 5;
1, 15, 20, 15, 6;
...
(End)
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014
FORMULA
a(n,k) = binomial(n,k), for 2 <= k <= n.
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014 A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!. B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014 O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015
EXAMPLE
The triangle a(n, k) begins:
n\k 2 3 4 5 6 7 8 9 10 11 12 13
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 1
12: 66 220 495 792 924 792 495 220 66 12 1
13: 78 286 715 1287 1716 1716 1287 715 286 78 13 1
MATHEMATICA
t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)
PROG
(PARI) for(n=2, 10, for(k=2, n, print1(binomial(n, k), ", "))) \\ G. C. Greubel, May 15 2018 (Magma) /* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018
Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.
+10
12
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
COMMENTS
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. 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) ). 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).
LINKS
A. F. Horadam, Oresme Numbers, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.
EXAMPLE
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,
...
MATHEMATICA
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 *)
Numerators of the rational sequence with e.g.f. (x/2)*(exp(-x) + 1)/(exp(x) - 1).
+10
11
1, -1, 7, -3, 59, -5, 127, -7, 119, -9, 335, -11, 15689, -13, 49, -15, 463, -17, 51049, -19, -171311, -21, 856031, -23, -236331331, -25, 8553181, -27, -23749448849, -29, 8615841490835, -31, -7709321033057, -33, 2577687858469
COMMENTS
Numerator of the n-term of the inverse binomial transform of the modified Bernoulli sequence A176327(k)/ A027642(k). The sequence of modified Bernoulli numbers A176327(k)/ A027642(k) is defined to be the same as the Bernoulli sequence, except the term at index k=1 which is zero. Its inverse binomial transform is 1, -1, 7/6, -3/2, 59/30, -5/2, 127/42, -7/2, 119/30, -9/2, 335/66, -11/2, ...; the numerators define this sequence here.
FORMULA
Conjecture: a(2*n+1) = -2*n-1.
a(n) = numerator((-1)^n*(bernoulli(n, 1) + bernoulli(n, 2))/2. - Peter Luschny, Jun 17 2012 (-1)^n*a(n) are the numerators of the polynomials generated by cosh(x*z)*z/(1-exp(-z)) evaluated x=1 (see the example section). The denominators of these values are A141056. - Peter Luschny, Aug 18 2018
EXAMPLE
The first few of the polynomials mentioned in the formula section are: 1, 1/2, 1/6 + x^2, (3/2)*x^2, -1/30 + x^2 + x^4, (5/2)*x^4, 1/42 - (1/2)*x^2 +(5/2)*x^4 + x^6, (7/2)*x^6, -1/30 + (2/3)*x^2 - (7/3)*x^4 + (14/3)*x^6 + x^8, (9/2)*x^8, ... The values of these polynomials at x=1 start 1, 1/2, 7/6, 3/2, 59/30, 5/2, 127/42, 7/2, ... - Peter Luschny, Aug 18 2018
MAPLE
read("transforms") ; evb := [1, 0, seq(bernoulli(n), n=2..50)] ; BINOMIALi(evb) ; apply(numer, %) ; # R. J. Mathar, Dec 01 2010 seq(numer((-1)^n*(bernoulli(n, 1)+bernoulli(n, 2))/2), n=0..34); # Peter Luschny, Jun 17 2012 gf := cosh(x*z)*z/(1-exp(-z)): ser := series(gf, z, 35):
seq((-1)^n*numer(subs(x=1, n!*coeff(ser, z, n))), n=0..34); # Peter Luschny, Aug 19 2018
MATHEMATICA
terms = 35; egf = (x/2)*((Exp[-x] + 1)/(Exp[x] - 1)) + O[x]^(terms);
PROG
(PARI) my(x = 'x + O('x^50)); apply(x->numerator(x), Vec(serlaplace((x/2)*(exp(-x) + 1)/(exp(x) - 1)))) \\ Michel Marcus, Aug 19 2018
CROSSREFS
Cf. A176591 (denominators), A141056 (denominators for the unsigned variant).
EXTENSIONS
Apparently incorrect claims concerning the inverse binomial transform of the B_n removed by R. J. Mathar, Dec 01 2010
Denominators of the rational sequence with e.g.f. (x/2)*(1+exp(-x))/(1-exp(-x)).
+10
8
1, 1, 6, 1, 30, 1, 42, 1, 30, 1, 66, 1, 2730, 1, 6, 1, 510, 1, 798, 1, 330, 1, 138, 1, 2730, 1, 6, 1, 870, 1, 14322, 1, 510, 1, 6, 1, 1919190, 1, 6, 1, 13530, 1, 1806, 1, 690, 1, 282, 1, 46410, 1, 66, 1, 1590, 1, 798, 1, 870, 1, 354, 1, 56786730, 1, 6, 1, 510
COMMENTS
Denominator of the Bernoulli number B_n, except a(1)=1. A minor variant of the Bernoulli denominators A027642. The sequence of fractions A164555(n)/ A027642(n) = 1/1, 1/2, 1/6, 0/1, -1/30, ... and the sequence of fractions A027641(n)/ A027642(n) = B_n = 1/1, -1/2, 1/6, 0/1, -1/30, ... differ only (by a sign) at n=1. The arithmetic mean of both sequences is 1/1, 0/1, 1/6, 0/1, -1/30, ..., equal to the aerated sequence A000367(n)/ A002445(n). The definition here provides the denominators of this sequence of arithmetic means.
MAPLE
seq(denom((bernoulli(i, 0)+bernoulli(i, 1))/2), i=0..64); # Peter Luschny, Jun 17 2012
MATHEMATICA
Join[{1, 1}, Rest[Denominator[BernoulliB[Range[80]]]]] (* Harvey P. Dale, Jun 18 2012 *)
a(2*n) = 1 + 6*n, a(2*n+1) = A165367(n).
+10
5
1, 1, 7, 5, 13, 4, 19, 11, 25, 7, 31, 17, 37, 10, 43, 23, 49, 13, 55, 29, 61, 16, 67, 35, 73, 19, 79, 41, 85, 22, 91, 47, 97, 25, 103, 53, 109, 28, 115, 59, 121, 31, 127, 65, 133, 34, 139, 71, 145, 37, 151, 77, 157, 40, 163, 83, 169, 43, 175, 89, 181, 46, 187, 95, 193
COMMENTS
Motivation: Start an array from a left column of fractions 0, 1/6, 0, -1/30, 0, ... = A176327(.)/ A176592(.), which is zero followed by the Bernoulli numbers from B_2 onwards. Construct more columns of the array by iteration of the Akiyama-Tanigawa algorithm working backwards through the rows of the table. In our case, the array starts with column indices k>=0:
0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, ...
1/6, 1/6, 3/20, 2/15, 5/42, 3/28, 7/72, 4/45, 9/110, ...
0, 1/30, 1/20, 2/35, 5/84, 5/84, 7/120, 28/495, ...
-1/30, -1/30, -3/140, -1/105, 0, 1/140, 49/3960, ...
0, -1/42, -1/28, -4/105, -1/28, -29/924, ...
1/42, 1/42, 1/140, -1/105, -5/231, -9/308, -343/10296, ...
The fractions of the top row are - A060819(n)/ A145979(n). The current sequence contains essentially the difference between numerator and denominator of each fraction, a(2)=6+1, a(3)=4+1, a(4)=10+3, ... The sum of numerator and denominator is essentially A060819. Also, numerators of (3*n + 1)/4. - Altug Alkan, Apr 17 2018
FORMULA
a(n) = 2*a(n-4) - a(n-8).
G.f.: (1 + x + 7*x^2 + 5*x^3 + 11*x^4 + 2*x^5 + 5*x^6 + x^7) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2). (End)
a(n) = (2*(3*n + 1)*(11 + 5*(-1)^n) + (6*n + 5 + 3*(-1)^n)*(1 - (-1)^n)*(-1)^((2*n + 3 + (-1)^n)/4))/32. - Luce ETIENNE, Jan 28 2015
Numerators of rationals with e.g.f. D(4,x), a Debye function.
+10
5
1, -2, 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
COMMENTS
The denominators are given in A227574. For general remarks on the e.g.f.s D(n,x), the Debye function with index n = 1, 2, 3, ... see the W. Lang link under A120080. D(4,x) := (4/x^4)*int(t^4/(exp(x) - 1), t=0..x) is the e.g.f. of the rationals r(4,n) = 4*B(n)/(n+4), n >= 0, with the Bernoulli numbers B(n) = A027641(n)/ A027642(n). See the Abramowitz-Stegun reference for the integral appearing in
D(4,x) and a series expansion valid for |x| < 2*Pi.
Differs from these sequences for n = 1328, 2660, 2828, 2880... - Andrey Zabolotskiy, Dec 08 2023
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 998, equ. 27.1.1 for n=4, with a factor (x^4)/4 extracted.
FORMULA
a(n) = numerator(4*B(n)/(n+4)), n >= 0, with the Bernoulli numbers B(n).
EXAMPLE
The rationals r(4,n), n=0..15 are: 1, -2/5, 1/9, 0, -1/60, 0, 1/105, 0, -1/90, 0, 5/231, 0, -691/10920, 0, 7/27, 0.
MATHEMATICA
A227573[n_]:=Numerator[4BernoulliB[n]/(n+4)];
PROG
(Sage)
print([(bernoulli(n)*4/(n+4)).numerator() for n in range(30)]) # Andrey Zabolotskiy, Dec 08 2023
Array read by antidiagonals: denominators of the core of the classical Bernoulli numbers.
+10
5
15, 15, 15, 105, 105, 105, 21, 105, 105, 21, 105, 105, 105, 105, 105, 15, 105, 105, 105, 105, 15, 165, 165, 1155, 231, 1155, 165, 165, 33, 165, 165, 231, 231, 165, 165, 33, 15015, 15015, 15015, 15015, 15015, 15015, 15015, 15015, 15015
COMMENTS
We consider the autosequence A164555(n)/ A027642(n) (see A190339(n)) and its difference table without the first two rows and the first two columns: 2/15, 1/15, -1/105, -1/21, -1/105, 1/15, 7/165, -5/33,...
-1/15, -8/105, -4/105, 4/105, 8/105, -4/165, -32/165,...
-1/105, 4/105, 8/105, 4/105, -116/1155, -28/165,...
1/21, 4/105, -4/105, -32/231, -16/231,...
-1/105, -8/105, -116/1155, 16/231,...
-1/15, -4/165, 28/165,...
7/165, 32/165,...
5/33,... etc.
This is an autosequence of the second kind.
The antidiagonals are palindromes in absolute values.
a(n) are the denominators. Multiples of 3.
Sum of odd antidiagonals: 2/15, -2/21, 2/15, -10/33, 1382/1365,... = -2* A000367(n+2)/ A001897(n+2). The sum of the even antidiagonals is A000004.
EXAMPLE
As a triangle:
15,
15, 15,
105, 105, 105,
21, 105, 105, 21,
105, 105, 105, 105, 105,
etc.
MATHEMATICA
max = 12; tb = Table[BernoulliB[n], {n, 0, max}]; td = Table[Differences[tb, n][[3 ;; -1]], {n, 2, max - 1}]; Table[td[[n - k + 1, k]] // Denominator, {n, 1, max - 3}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 11 2014 *)
Numerators of rationals with e.g.f. D(3,x), a Debye function.
+10
4
1, -3, 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
COMMENTS
The denominators are given in A227571. For general remarks on the e.g.f.s D(n,x), the Debye function with index n = 1, 2, 3, ... see the W. Lang link under A120080. D(3,x) := (3/x^3)*int(t^3/(exp(x) - 1), t=0..x) is the e.g.f. of the rationals r(3,n) = 3*B(n)/(n+3), n >= 0, with the Bernoulli numbers B(n) = A027641(n)/ A027642(n). See the Abramowitz-Stegun link for the integral appearing in
D(3,x) and a series expansion valid for |x| < 2*Pi.
Differs from these sequences at n = 1292, 2624, 2770, 2778.... - Andrey Zabolotskiy, Dec 08 2023
REFERENCES
L. D. Landau, E. M. Lifschitz: Lehrbuch der Theoretischen Physik, Band V: Statistische Physik, Akademie Verlag, Leipzig, p. 195, equ. (63.5), and footnote 1 on p. 197.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 998, equ. 27.1.1 for n=3, with a factor (x^3)/3 extracted.
FORMULA
a(n) = numerator(3*B(n)/(n+3)), n >= 0, with the Bernoulli numbers B(n).
EXAMPLE
The rationals r(3,n), n=0..15 are: 1, -3/8, 1/10, 0, -1/70, 0, 1/126, 0, -1/110, 0, 5/286, 0, -691/13650, 0, 7/34, 0.
MATHEMATICA
A227570[n_]:=Numerator[3BernoulliB[n]/(n+3)];
PROG
(Sage)
print([(bernoulli(n)*3/(n+3)).numerator() for n in range(30)]) # Andrey Zabolotskiy, Dec 08 2023
Search completed in 0.017 seconds
| 