A025547 -id:A025547 - cp's OEIS frontend

A075135 Numerator of the generalized harmonic number H(n,3,1) described below.

1, 5, 39, 209, 2857, 11883, 233057, 2632787, 13468239, 13739939, 433545709, 7488194853, 281072414761, 284780929571, 12393920563953, 288249495707519, 2038704876507433, 2058454144222533, 2077126179153173, 60750140156034617

Offset: 1

T. D. Noe, Sep 04 2002

For integers a and b, H(n,a,b) is the sum of the fractions 1/(a i + b), i = 0,1,..,n-1. This database already contains six instances of generalized harmonic numbers. Partial sums of the harmonic series 1+1/2+1/3+1/4+... are given by the sequence of harmonic numbers H(n,1,1) = A001008(n) / A002805(n).
The Jeep problem gives rise to the series H(n,2,1) = A025550(n) / A025547(n). Recent additions to the database are 3 * H(n,3,1) = A074596(n) / A051536(n), 3 * H(n,3,2) = A074597(n) / A051540(n), 4 * H(n,4,1) = A074598(n) / A051539(n) and 4 * H(n,4,3) = A074637(n) / A074638(n) . The numerator of H(n,4,1) is A075136. The fractions H(n,5,1), H(n,5,2), H(n,5,3) and H(n,5,4) are in A075137-A075144.
The sequence H(n,a,b) is of interest only when a and b are relatively prime. The sequence can also be computed as H(n,a,b) = (PolyGamma[n+1+b/a] - PolyGamma[1+b/a])/a. The sequence H(n,a,b) diverges for all a and b.
According to Hardy and Wright, if p is an odd prime, then p divides the numerator of the harmonic number H(p-1,1,1). This result can be extended to generalized harmonic numbers: for odd integer n, let q = (n-2)a + 2b. If q is prime, then q divides the numerator of H(n-1,a,b). For this sequence (a=3, b=1) we conclude that 11 divides a(4), 17 divides a(6), 29 divides a(10) and 47 divides a(16).
Graham, Knuth and Patashnik define another type of generalized harmonic number as the sum of fractions 1/i^k, i=1,...,n. For k=2, the sequence of fractions is A007406(n) / A007407(n).

			a(3)=39 because 1 + 1/4 + 1/7 = 39/28.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 263, 269, 272, 297, 302, 356.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 4th ed., Oxford Univ. Press, 1971, page 88.

Eric Weisstein's World of Mathematics, Harmonic Series
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Jeep Problem

Cf. A001008, A002805, A025550, A025547, A051536, A051539, A074637, A074638, A075136-A075144.
Cf. A051540, A074596, A074597, A074598.
Cf. A007406, A007407.

a=3; b=1; maxN=20; s=0; Numerator[Table[s+=1/(a n + b), {n, 0, maxN-1}]]
Accumulate[1/Range[1,60,3]]//Numerator (* Harvey P. Dale, Dec 30 2019 *)

A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

Original entry on oeis.org

0, 1, 2, 13, 76, 789, 7734, 110937, 1528920, 28018665, 497895210, 11110528485, 241792844580, 6361055257725, 163842638377950, 4964894559637425, 147721447995130800, 5066706567801827025, 171002070002301095250, 6548719685561840296125, 247199273204273879989500

Offset: 0

Clark Kimberling

(2*n + 1)!!/a(n+1), n>=0, is the n-th approximant for William Brouncker's continued fraction for 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))) See the C. Brezinski and J.-P. Delahaye references given under A142969 and A142970, respectively. The double factorials (2*n + 1)!! = A001147(n+1) enter. - Wolfdieter Lang, Oct 06 2008

			a(3) = (2*3 - 1)!! * Sum_{k=0..2} (-1)^k/(2k + 1) = 5!! * (1/(2*0 + 1) - 1/(2*1 + 1) + 1/(2*2 + 1)) = 5*3*1*(1/1 - 1/3 + 1/5) = 15 - 5 + 3 = 13. Notice that the first factor always cancels the common denominator of the sum. - _Michael B. Porter_, Jul 22 2016

A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Peter Luschny, Is the Gamma function misdefined? Hadamard versus Euler - Who found the better Gamma function?
Wikipedia, William Brouncker, 2nd Viscount Brouncker

Cf. A004041, A000407.
From Johannes W. Meijer, Nov 12 2009: (Start)
Cf. A007509 and A025547.
Equals first column of A167584.
Equals row sums of A167591.
Equals first right hand column of A167594.
(End)
Cf. A167576 and A135457.

[0] cat [ n le 2 select (n) else 2*Self(n-1)+(2*n-3)^2*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 17 2015

a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 else 2*a(n-1)+(2*n-3)^2* a(n-2) fi end: seq(a(n), n=0..20); # Peter Luschny, Nov 16 2016 after N. J. A. Sloane

f[k_] := (2 k - 1) (-1)^(k + 1)
t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 22}]    (* A024199 signed *)
(* Clark Kimberling, Dec 30 2011 *)
RecurrenceTable[{a[n+1] == 2*a[n] + (2*n-1)^2*a[n-1],a[0] == 0, a[1] == 1},a,{n,0,20}] (* Vaclav Kotesovec, Mar 18 2014 *)
CoefficientList[Series[Pi/4/Sqrt[1-2*x] - 1/2*Log[2*x+Sqrt[4*x^2-1]]/Sqrt[2*x-1], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Mar 18 2014 *)

a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)) where c=(-1)^(n+1) and s(k) = 2k-1 for k = 1, 2, 3, ... (was previous definition). - Clark Kimberling
D-finite with recurrence a(0) = 0, a(1) = 1, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1). - N. J. A. Sloane, Jul 19 2002
a(n) + A024200(n) = A001147(n) = (2n-1)!!. - Max Alekseyev, Sep 23 2007
a(n)/A024200(n) -> Pi/(4-Pi) as n -> oo. - Max Alekseyev, Sep 23 2007
From Wolfdieter Lang, Oct 06 2008: (Start)
E.g.f. for a(n+1), n>=0: (sqrt(1-2*x)+arcsin(2*x)*sqrt(1+2*x)/2)/((1-4*x^2)^(1/2)*(1-2*x)). From the recurrence, solving (1-4*x^2)y''(x)-2*(8*x+1)*y'(x)-9*y=0 with inputs y(0)=1, y'(0)=2.
a(n+1) = A003148(n) + A143165(n), n>=0 (from the two terms of the e.g.f.). (End)
From Johannes W. Meijer, Nov 12 2009: (Start)
a(n) = (-1)^(n-1)*(2*n-3)!! + (2*n-1)*a(n-1) with a(0) = 0.
a(n) = (2*n-1)!!*sum((-1)^(k)/(2*k+1), k=0..n-1)
(End)
E.g.f.: Pi/4/sqrt(1-2*x) - 1/2*log(2*x+sqrt(4*x^2-1))/sqrt(2*x-1). - Vaclav Kotesovec, Mar 18 2014
a(n) ~ Pi * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Mar 18 2014
a(n) = (2*H(n+1/2)-Gamma(n+1/2))*2^(n-2)*sqrt(Pi) with H(x) the Hadamard factorial (see the link section). - Cyril Damamme, Jul 19 2015
a(n) = A135457(n) - (-1)^n A001147(n-1). - Cyril Damamme, Jul 19 2015
a(n) = (Pi + (-1)^n*(Psi(n/2 + 1/4) - Psi(n/2 + 3/4)))*Gamma(n+1/2)*2^(n-2)/sqrt(Pi). - Robert Israel, Jul 20 2015
a(n) = A167576(n) - A135457(n). - Cyril Damamme, Jul 22 2015
a(n)/A001147(n) -> Pi/4 as n -> oo. - Daniel Suteu, Jul 21 2016
From Peter Bala, Nov 15 2016: (Start)
Conjecture: a(n) = 1/2*Sum_{k = 0..2*n-1} (-1)^(n-k+1)*k!*(2*n - 2*k - 3)!!, where the double factorial of an odd integer (positive or negative) may be defined in terms of the gamma function as (2*N - 1)!! = 2^((N+1)/2)*Gamma(N/2 + 1)/sqrt(Pi).
E.g.f. 1/2*arcsin(2*x)/sqrt(1 - 2*x) = x + 2*x^2/2! + 13*x^3/3! + 76*x^4/4! + .... (End)

Edited by N. J. A. Sloane, Jul 19 2002

New name from Cyril Damamme, Jul 19 2015

A007509 Numerator of Sum_{k=0..n} (-1)^k/(2*k+1).

Original entry on oeis.org

1, 2, 13, 76, 263, 2578, 36979, 33976, 622637, 11064338, 11757173, 255865444, 1346255081, 3852854518, 116752370597, 3473755390832, 3610501179557, 3481569435902, 133330680156299, 129049485078524, 5457995496252709, 227848175409504262, 234389556075339277

Offset: 0

N. J. A. Sloane

Denominators of convergents to 4/Pi. [For Brouncker's continued fraction, with numerators A025547(n+1), for n >= 0. - Wolfdieter Lang, Aug 26 2019]

See A352395 (the denominators for the present sequence) for the formula of this alternating sum, and the Abramowitz-Stegun link. - Wolfdieter Lang, Apr 06 2022

			1/1, 2/3, 13/15, 76/105, 263/315, 2578/3465, 36979/45045, 33976/45045, 622637/765765, ...

P. Beckmann, A History of Pi. Golem Press, Boulder, CO, 2nd ed., 1971, p. 131.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pi.
Eric Weisstein's World of Mathematics, Pi - Continued Fraction.
R. G. Wilson, V, Notes with attachment.

Denominators are given in A352395.
From Johannes W. Meijer, Nov 12 2009: (Start)
Cf. A157142 and A166107.
Appears in A167576, A167577, A167578, A024199, A167588 and A167589. (End)
Cf. A142969 for the numerators of Brouncker's continued fraction of 4/Pi - 1.

[Numerator(&+[(-1)^k/(2*k+1):k in [0..n]]): n in [0..23]]; // Marius A. Burtea, Aug 26 2019

A007509 := n->numer(add((-1)^k/(2*k+1),k=0..n));

Table[Numerator[FunctionExpand[(Pi + (-1)^n(HarmonicNumber[n/2 + 1/4] - HarmonicNumber[n/2 - 1/4]))/4]], {n, 0, 20}] (* Vladimir Reshetnikov, Jan 18 2011 *)
Numerator[Table[Sum[(-1)^k/(2k+1),{k,0,n}],{n,0,30}]] (* Harvey P. Dale, Oct 22 2011 *)
Table[(-1)^k/(2k+1),{k,0,30}]//Accumulate//Numerator (* Harvey P. Dale, May 03 2019 *)

a(n) = numerator((Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2), with the digamma function Psi(z). See the formula in A352395. - Wolfdieter Lang, Apr 06 2022

a(n) = numerator(Pi/4 + (-1)^n * (Psi((n + 5/2)/2) - Psi((n + 3/2)/2))/4). - Vaclav Kotesovec, May 16 2022

Crossref. corrected (A025547 replaced with A352395) by Wolfdieter Lang, Apr 06 2022

A025550 a(n) = ( 1/1 + 1/3 + 1/5 + ... + 1/(2n-1) )LCM(1, 3, 5, ..., 2*n-1).

Original entry on oeis.org

1, 4, 23, 176, 563, 6508, 88069, 91072, 1593269, 31037876, 31730711, 744355888, 3788707301, 11552032628, 340028535787, 10686452707072, 10823198495797, 10952130239452, 409741429887649, 414022624965424, 17141894231615609, 743947082888833412, 750488463554668427

Offset: 1

Clark Kimberling

Or, numerator of 1/1 + 1/3 + ... + 1/(2n-1) up to a(38).
Following similar remark by T. D. Noe in A025547, this coincides with f(n) = numerator of 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n-1) iff n <= 38. But a(39) = 18048708369314455836683437302413, f(39) = 1640791669937677803334857936583. Note that f(n) = numerator(digamma(n+1/2)/2 + log(2) + euler_gamma/2). - Paul Barry, Aug 19 2005 [See A350669(n-1).]
2*(1 + 1/3 + ... + 1/(2*n-1))/Pi = 2*a(n)/(A025547(n)*Pi) is the equivalent resistance between the points (0,0) and (n,n) on a 2-dimension infinite square grid of unit resistors. - Jianing Song, Apr 28 2025

Georg Fischer, Table of n, a(n) for n = 1..200 (first 39 terms from Jean-François Alcover)
MathPages, The Algebra of an Infinite Grid of Resistors
Physics Stack Exchange, On this infinite grid of resistors, what's the equivalent resistance?
Eric Weisstein's World of Mathematics, Jeep Problem

Cf. A025547, A075135, A002428, A350669.

a025550 n = numerator $ sum  $ map (1 %) $ take n [1, 3 ..]
-- Reinhard Zumkeller, Jan 22 2012

[&+[1/d: d in i]*Lcm(i) where i is [1..2*n-1 by 2]: n in [1..21]]; // Bruno Berselli, Apr 16 2015

a:= n-> (f-> add(1/p, p=f)*ilcm(f[]))([2*i-1$i=1..n]):
seq(a(n), n=1..40);  # Alois P. Heinz, Apr 16 2015

Table[(Total[1/Range[1,2n-1,2]])LCM@@Range[1,2n-1,2],{n,30}] (* Harvey P. Dale, Sep 09 2020 *)

a(n)=my(v=vector(n,i,2*i-1));sum(i=1,#v,1/v[i])*lcm(v) \\ Charles R Greathouse IV, Feb 28 2013

1 + 1/3 + ... + 1/(2*n-1) = a(n)/A025547(n) = A350669(n-1)/A350670(n-1). - Jianing Song, Apr 28 2025

Value of a(39) corrected by Jean-François Alcover, Apr 16 2015

A056053 a(n) = smallest odd number 2m+1 such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.

Original entry on oeis.org

1, 3, 15, 113, 837, 6183, 45691, 337607, 2494595, 18432707, 136200301, 1006391657, 7436284415, 54947122715, 406007372211, 3000011249847, 22167251422541, 163795064320249, 1210290918990281, 8942907496445513, 66079645178783351, 488266205223462461, 3607826381608149807

Offset: 0

Robert G. Wilson v, Jul 25 2000 and Jan 11 2004

nonn

a(2) = 15 and a(3) = 113 are related to the Borwein integrals. Concretely, a(2) = 15 is the smallest odd m such that the integral Integral_{x=-oo..oo} Product_{1<=k<=m, k odd} (sin(k*x)/(k*x)) dx is slightly less than Pi, and a(3) = 113 is the smallest odd m such that the integral Integral_{x=-oo..oo} cos(x) * Product_{1<=k<=m, k odd} (sin(k*x)/(k*x)) dx is slightly less than Pi/2. See the Wikipedia link and the 3Blue1Brown video link below. - Jianing Song, Dec 10 2022

Calvin C. Clawson, "Mathematical Mysteries, The Beauty and Magic of Numbers," Plenum Press, NY and London, 1996, page 64.

Robert G. Wilson v, Table of n, a(n) for n = 0..500
Grant Sanderson, Researchers thought this was a bug (Borwein integrals), 3Blue1Brown video (2022).
Wikipedia, Borwein integral

Cf. A002387, A056054, A091463, A091464, A091465.

Cf. also A092318, A092317, A092315, A281355, A074599, A025547.

s = 0; k = 1; Do[ While[s = N[s + 1/k, 24]; s <= n, k += 2]; Print[k]; k += 2, {n, 1, 11}]

a(n) ~ floor((1/2)*A002387(2n)).
The next term is approximately the previous term * e^2.
a(n) = A092315(n)*2 + 1 = floor(exp(n*2-Euler)/4+1/8)*2+1 for all n (conjectured). - M. F. Hasler, Jan 24 2017
a(n) ~ exp(2*n - A350763) = (1/2)*exp(2*n - gamma), gamma = A001620. - A.H.M. Smeets, Apr 15 2022

Corrected by N. J. A. Sloane, Feb 16 2004
More terms from Robert G. Wilson v, Apr 17 2004
a(17) corrected - see correction in A092315. - Gerhard Kirchner, Jul 25 2020
a(0) prepended by Robert G. Wilson v, Oct 23 2024

A092315 a(n) is the smallest m such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.

Original entry on oeis.org

1, 7, 56, 418, 3091, 22845, 168803, 1247297, 9216353, 68100150, 503195828, 3718142207, 27473561357, 203003686105, 1500005624923, 11083625711270, 81897532160124, 605145459495140, 4471453748222756, 33039822589391675, 244133102611731230, 1803913190804074903

Offset: 1

N. J. A. Sloane, Feb 16 2004

nonn

From Gerhard Kirchner, May 21 2020: (Start)
The terms a(n), evaluated by the formula, should pass the test OH(a(n))=n and OH(a(n)-1)=n-1, where OH(m) is the odd harmonic series, see above.
Another formula, see link Asymptotic formulas, formula 1, is OH(m) = (log(4*m)+gamma)/2+1/(2*m)-11/(48*m^2)+1/(8*m^3)-127*t/(1920*m^4), 0The Maxima code includes both tests and creates a b-file in the current directory. For n<=1000, the case "Precision too low" does not occur. (End)
a(2) = 7 and a(3) = 56 are related to the Borwein integrals. Concretely, a(2) = 7 is the smallest m such that the integral Integral_{x=-oo..oo} Product_{k=0..m} (sin((2*k+1)*x)/((2*k+1)*x)) dx is slightly less than Pi, and a(3) = 56 is the smallest m such that the integral Integral_{x=-oo..oo} cos(x) * Product_{k=0..m} (sin((2*k+1)*x)/((2*k+1)*x)) dx is slightly less than Pi/2. See the Wikipedia link and the 3Blue1Brown video link below. - Jianing Song, Dec 10 2022

Gerhard Kirchner, Table of n, a(n) for n = 1..1000
Gerhard Kirchner, Asymptotic formulas
Grant Sanderson, Researchers thought this was a bug (Borwein integrals), 3Blue1Brown video (2022).
Wikipedia, Borwein integral

Except for first term, same as A092318. Equals (A056053-1)/2.

Cf. A074599, A025547, A281355.

A092315[n_] := Floor[Exp[2*n - EulerGamma]/4]; Table[A092315[n], {n, 1, 22}] (* Robert P. P. McKone, Jul 13 2021 *)

block(
fpprec:1000, gam: %gamma, nmax:1000,
fl: openw("bfile1000.txt"),
OH(k,t):=(log(4*k)+gam)/2+1/(2*k)-11/(48*k^2)+1/(8*k^3)-127*t/(1920*k^4),
printf(fl, "1 1"),   newline(fl),
for n from 2 thru nmax do
(u: bfloat(exp(2*n-gam)/4), k: floor(u),
x0: bfloat(OH(k,0)), x01: bfloat(OH(k,1)), x1: bfloat(OH(k-1,0)),
n0: floor(x0), n01: floor(x01), n1: floor(x1),  m: n,
if n0=n and n01=n and n1=n-1 then
         (h: concat(n, " ", k), printf(fl, h),  newline(fl)) else n: nmax),
if mGerhard Kirchner, Jul 23 2020 */
/* The first nmax terms are saved as a b-file */

a(n) = floor(exp(2*n-gamma)/4+1/8) for all n >= 1 (conjectured; see also comments in A002387). - M. F. Hasler, Jan 22 2017

a(n) = floor(exp(2*n-gamma)/4). - Gerhard Kirchner, Jul 23 2020

More terms from M. F. Hasler, Jan 24 2017

a(17) in the data section and 127 terms in the b-file corrected by Gerhard Kirchner, Jul 23 2020

A355565 T(j,k) are the numerators s in the representation R = s/t + (2/Pi)*u/v of the resistance between two nodes separated by the distance vector (j,k) in an infinite square lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= j, is a triangle read by rows.

Original entry on oeis.org

0, 1, 0, 2, -1, 0, 17, -4, 1, 0, 40, -49, 6, -1, 0, 401, -140, 97, -8, 1, 0, 1042, -1569, 336, -161, 10, -1, 0, 11073, -4376, 4321, -660, 241, -12, 1, 0, 29856, -48833, 13342, -9681, 1144, -337, 14, -1, 0, 325441, -136488, 160929, -33188, 18929, -1820, 449, -16, 1, 0

Offset: 0

Hugo Pfoertner, Jul 07 2022

The recurrence given by Cserti (2000), page 5, (32) is used to calculate the resistance between two arbitrarily spaced nodes in an infinite square lattice whose edges are replaced by one-ohm resistors. The lower triangle, including the diagonal, in Table I of Atkinson and Steenwijk (1999), page 487, is reproduced. The solution to the resistor grid problem shown in the xkcd Web Comic #356 "Nerd Sniping", provided in A211074, is the special case (j,k) = (2,1).

Using the terms of A280079 and A280317 as pairs of grid indices leads to strictly increasing resistances, i.e., R(A280079(m),A280317(m)) > R(A280079(i),A280317(i)) for m > i. This implies that for grid points on the same radius the resistance increases with the circumferential angle between 0 and Pi/4. The further dependence of the resistance along the circumferential angle with a fixed radius results from symmetry. - Hugo Pfoertner, Aug 31 2022

			The triangle begins:
     0;
     1,     0;
     2,    -1,   0;
    17,    -4,   1,    0;
    40,   -49,   6,   -1,  0;
   401,  -140,  97,   -8,  1,  0;
  1042, -1569, 336, -161, 10, -1, 0
.
The combined triangles used to calculate the resistances are:
  \  k      0       |        1        |       2       |      3       |
   \    s/t     u/v |    s/t    u/v   |  s/t      u/v |  s/t    u/v  |
  j \---------------|-----------------|---------------|--------------|
  0 |   0       0   |     .      .    |   .        .  |   .      .   |
  1 |   1/2     0   |    0      1     |   .        .  |   .      .   |
  2 |   2      -2   |   -1/2    2     |  0        4/3 |   .      .   |
  3 |  17/2   -12   |   -4     23/3   |  1/2      2/3 |  0     23/15 |
  4 |  40    -184/3 | - 49/2   40     |  6    -118/15 | -1/2   12/5  |
  5 | 401/2  -940/3 | -140    3323/15 | 97/2 -1118/15 | -8    499/35 |
.
continued:
  \ k     4       |      5       |
   \  s/t   u/v   | s/t    u/v   |
  j \-------------|--------------|
  0 |  .     .    |  .      .    |
  1 |  .     .    |  .      .    |
  2 |  .     .    |  .      .    |
  3 |  .     .    |  .      .    |
  4 | 0   176/105 |  .      .    |
  5 | 1/2  20/21  | 0    563/315 |
.
E.g., the resistance for a node distance vector (4,1) is R = T(4,1)/A131406(5,2) + (2/Pi)*A355566(4,1)/A355567(4,1) = -49/2 + (2/Pi)*40/1 = 80/Pi - 49/2.

See A211074 for more references and links.

Rainer Rosenthal, Table of n, a(n) for n = 0..135, rows 0..15 of triangle, flattened.
J. Cserti, Application of the lattice Green's function for calculating the resistance of infinite networks of resistors, arXiv:cond-mat/9909120 [cond-mat.mes-hall], 1999-2000.
Hugo Pfoertner, Grid points sorted by increasing R values, (2022).
Hugo Pfoertner, PARI program for inverse problem, (2022). Finds the grid point [x,y] that leads to the best approximation of a given resistance distance R (ohms) between [0,0] and [x,y].
Physics Stack Exchange, On this infinite grid of resistors, what's the equivalent resistance? Answer by user PBS, Apr 21 2018.
Rainer Rosenthal, Maple program

A131406 are the corresponding denominators t, with indices shifted by 1.
A355566 and A355567 are u and v.
Cf. A025547, A025550, A089165, A211074, A355953, A355955, A356201, A356202.
Cf. A355585, A355586, A355587, A355588 (same problem for the infinite triangular lattice).
Cf. A280079, A280317.

Maple
```
See link.
```

alphas[beta_] :=
Log[2 - Cos[beta] + Sqrt[3 + Cos[beta]*(Cos[beta] - 4)]];
Rsqu[n_, p_] :=
Simplify[(1/Pi)*
   Integrate[(1 - Exp[-Abs[n]*alphas[beta]]*Cos[p*beta])/
     Sinh[alphas[beta]], {beta, 0, Pi}]];
Table[Rsqu[n, k], {n, 0, 4}, {k, 0, n}] // TableForm (* Hugo Pfoertner, Aug 21 2022, calculates R, after Atkinson and Steenwijk *)

R(m,p,x=pi) = {if (m==0 && p==0, return(0)); if (m==1 && p==0, return(1/2)); if (m==1 && p==1, return(2/x)); if(m==p, my(mm=m-1); return(R(mm,mm)*4*mm/(2*mm+1) - R(mm-1,mm-1)*(2*mm-1)/(2*mm+1))); if (p==(m-1), my(mm=m-1); return(2*R(mm,mm) - R(mm,mm-1))); if (p==0, my(mm=m-1); return(4*R(mm,0) - R(mm-1,0) - 2*R(mm,1))); if (p0, my(mm=m-1); return(4*R(mm,p) - R(mm-1,p) - R(mm,p+1) - R(mm,p-1)))};
for(j=0,9,for(k=0,j,my(q=pi*R(j,k,pi));print1(numerator(polcoef(q,1,pi)),", "));print())

The resistance for the distance vector (j,k) is R(j,k) = T(j,k)/(1+mod(j+k,2)) +(2/Pi)*A355566(j,k)/A355567(j,k), avoiding the use of A131406.
From Rainer Rosenthal, Aug 04 2022: (Start)
R(0,0) = 0; R(1,0) = 1/2.
R(n,n) = R(n-1,n-1) + (2/Pi)/(2*n-1) for n >= 1.
R(j,k) = R(k,j) and R(-j,k) = R(j,k).
4*R(j,k) = R(j-1,k) + R(j+1,k) + R(j,k-1) + R(j,k+1) for (j,k) != (0,0).
(End)
T(j+1,0) = A089165(j)/(1 + mod(j,2)) for j >= 0. - Hugo Pfoertner, Aug 21 2022

A074599 Numerator of 2 * H(n,2,1), a generalized harmonic number. See A075135. Also 2 * A350669.

Original entry on oeis.org

2, 8, 46, 352, 1126, 13016, 176138, 182144, 3186538, 62075752, 63461422, 1488711776, 7577414602, 23104065256, 680057071574, 21372905414144, 21646396991594, 21904260478904, 819482859775298, 828045249930848

Offset: 1

Robert G. Wilson v, Aug 27 2002

2*(1 + 1/3 + ... + 1/(2*n-1))/Pi = a(n)/(A350670(n)*Pi) is the equivalent resistance between the points (0,0) and (n,n) on a 2-dimension infinite square grid of unit resistors. - Jianing Song, Apr 28 2025

MathPages, The Algebra of an Infinite Grid of Resistors
Physics Stack Exchange, On this infinite grid of resistors, what's the equivalent resistance?

Cf. A350669. The denominators are in A350670.
Cf. A025547, A025550, A075135.
Not always equal to the second left hand column of A161198 triangle divided by A025549. - Johannes W. Meijer, Jun 08 2009

Table[ Numerator[ Sum[1/i, {i, 1/2, n}]], {n, 1, 20}]

A167576 The first column of the ED3 array A167572.

Original entry on oeis.org

1, 5, 23, 167, 1473, 16413, 211479, 3192975, 54010305, 1030249845, 21566327895, 497334999735, 12405876372225, 335591130336525, 9716331072597975, 301633179343890975, 9941514351641143425, 348336799875365041125

Offset: 1

Johannes W. Meijer, Nov 10 2009

Basically a(n) measures the difference between the Euler factorial n! and the Luschny factorial L(n) at half-integer values. For the Luschny factorial see the link. The formula given in the Maple section is a variant of a formula given by Cyril Damamme in A135457. - Peter Luschny, Jul 18 2015

			G.f. = x + 5*x^2 + 23*x^3 + 167*x^4 + 1473*x^5 + 16413*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 1..400
Peter Luschny, Is the Gamma function misdefined? Hadamard versus Euler - Who found the better Gamma function?

Equals the first column of the ED3 array A167572.
Equals the first right hand column of A167583.
Other columns are A167577 and A167578.
Cf. A097801 (the 2*(-1)^n*(2*n-5)!! factor).
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor).
Cf. A024199 and A135457.

L := x -> (1+x*(Psi(1-x/2)-Psi(1/2-x/2)))/(-x)!:
a := x -> (L(x-1/2)-(x-1/2)!)*2^(x-1)*sqrt(Pi):
seq(simplify(a(n)),n=1..18); # Peter Luschny, Jul 18 2015
a := proc(n) option remember: if n=1 then 1 else (2*n-1)*a(n-1)+2*(-1)^n*doublefactorial(2*n-5) fi: end: seq(a(n),n=1..18); # Johannes W. Meijer, Jul 20 2015

a[ n_] := If[ n < 1, 0, (2 n - 3)!! ((-1)^n - I (4 n - 2) Sum[ I^k / k, {k, 1, 2 n - 1, 2}])]; (* Michael Somos, Jul 20 2015 *)
a[ n_] := If[ n < 1, 0, (2 n - 3)!! ((-1)^n + (4 n - 2) Sum[ KroneckerSymbol[ -4, k]/ k, {k, 2 n - 1}])]; (* Michael Somos, Jan 31 2019 *)

{a(n) = if( n<1, 0, prod(k=1, n-1, 2*k - 1) * ((-1)^n - (4*n - 2) * sum(k=1, n, (-1)^k / (2*k - 1))))}; /* Michael Somos, Jul 20 2015 */

a(n) = (-1)^n*(2*n-3)!!*(1 + (4*n-2)*Sum_{k=0..n-1} (-1)^(k+n)/(2*k+1)).
a(n) = (2*n-1)*a(n-1) + 2*(-1)^n*(2*n-5)!! with a(1) = 1.
a(n) = 4*a(n-1) + (4*n^2 - 16*n + 15)*a(n-2) with a(1) = 1 and a(2) = 5 [Superseeker].
0 = a(n)*a(n+1)*(-440*a(n+2) - 220*a(n+3) + 55*a(n+4)) + a(n)*a(n+2)*(536*a(n+2) - 118*a(n+3) - 4*a(n+4)) + a(n)*a(n+3)*(-4*a(n+3) + a(n+4)) + a(n+1)^2*(-220*a(n+2) - 32*a(n+3) + 8*a(n+4)) + a(n+1)*a(n+2)*(+71*a(n+2) + 4*a(n+3) - 2*a(n+4)) + a(n+2)^2*(-4*a(n+2) + a(n+3)) if n>0. - Michael Somos, Jul 19 2015
a(n) = (-1 + (n-1/2)*LerchPhi(-1,1,n+1/2) + (-n+1/2)*LerchPhi(-1,1,-n+1/2))/(1-2*n)!!. - Johannes W. Meijer, Jul 20 2015
a(n) = A024199(n) + A135457(n). - Cyril Damamme, Jul 22 2015
a(n) = ((-1)^n/(2*n - 1) + Pi/2 - (-1)^n LerchPhi(-1, 1, n + 1/2)) (2*n - 1)!!. - Michael Somos, Jan 31 2019

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cp's OEIS Frontend

A051417 Quotients of consecutive values of lcm {1, 3, 5 ..., 2n-1} or A025547(n+1)/A025547(n).

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A075135 Numerator of the generalized harmonic number H(n,3,1) described below.

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A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

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A007509 Numerator of Sum_{k=0..n} (-1)^k/(2*k+1).

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A025550 a(n) = ( 1/1 + 1/3 + 1/5 + ... + 1/(2*n-1) )*LCM(1, 3, 5, ..., 2*n-1).

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A056053 a(n) = smallest odd number 2m+1 such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.

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A025550 a(n) = ( 1/1 + 1/3 + 1/5 + ... + 1/(2n-1) )LCM(1, 3, 5, ..., 2*n-1).