A174341 -id:A174341 - cp's OEIS frontend

A309132 a(n) is the denominator of F(n) = A027641(n-1)/n + A027642(n-1)/n^2.

1, 1, 1, 16, 1, 36, 1, 64, 27, 100, 1, 144, 1, 196, 75, 256, 1, 324, 1, 400, 49, 484, 1, 576, 125, 676, 243, 784, 1, 900, 1, 1024, 363, 1156, 1225, 1296, 1, 1444, 169, 1600, 1, 1764, 1, 1936, 135, 2116, 1, 2304, 343, 2500, 867, 2704, 1, 2916, 3025, 3136, 361, 3364, 1, 3600, 1, 3844, 1323, 4096, 845, 4356, 1

Offset: 1

Thomas Ordowski, Jul 14 2019

It seems that the numerator of F(n) is the numerator of (B(n-1) + 1/n), where B(k) is the k-th Bernoulli number; if so, for n > 2, the numerator of F(n) is A174341(n-1). How to prove it?
Conjecture: for n > 1, a(n) = 1 if and only if n is prime.
Is this conjecture equivalent to the Agoh-Giuga conjecture?
Theorem 1. If p is prime, then a(p) = 1. Proof. a(2) = 1, so let p be an odd prime. By the von Staudt-Clausen theorem, if k is even, then B(k) = A(k) - Sum_{prime q, q-1 | k} 1/q, where A(k) is an integer and the sum is over all primes q such that q-1 divides k. Thus B(k) = N(k)/D(k) with D(k) = Product_{prime q, q-1 | k} q. Now let k = p-1. Then N(p-1)/D(p-1) = B(p-1) = A(p-1) - 1/p - Sum_{prime q < p, q-1 | p-1} 1/q (*). Add 1/p to both sides of (*) and multiply by p*D(p-1) to get p*N(p-1) + D(p-1) = p*D(p-1)*(A(p-1) - Sum_{prime q < p, q-1 | p-1} 1/q) (**). Now p | D(p-1), so p^2 | p*D(p-1) in (**). The denominators on the right side of (**) are all of the form q < p. Therefore, p^2 divides both sides of (**). Hence F(p) = N(p-1)/p + D(p-1)/p^2 is an integer, so a(p) = 1. - Jonathan Sondow, Jul 14 2019
Conjecture: composite numbers n such that a(n) is squarefree are only the Carmichael numbers A002997. Cf. A309235. - Thomas Ordowski, Jul 15 2019
Conjecture checked up to n = 101101. - Amiram Eldar, Jul 16 2019
Theorem 2. If n is a prime or a Carmichael number, then a(n) = A326690(n) = denominator of (Sum_{prime p | n} 1/p - 1/n). The proof is a generalization of that of Theorem 1. (Note that Theorem 2 implies Theorem 1, since if n is prime, then (Sum_{prime p | n} 1/p - 1/n) = 1/n - 1/n = 0/1, so a(p) = A326690(n) = 1.) For n a prime or a Carmichael number, an application of Theorem 2 is computing a(n) without calculating Bernoulli(n-1) which may be huge; see A309268 and A326690. - Jonathan Sondow, Jul 19 2019
The values of F(n) when n is prime are A327033. - Jonathan Sondow, Aug 16 2019

			F(n) = 2/1, 0/1, 1/1, 1/16, 1/1, 1/36, 1/1, 1/64, 7/27, 1/100, 1/1, 1/144, -37/1, 1/196, 37/75, 1/256, -211/1, 1/324, 2311/1, 1/400, -407389/49, ...

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem
Wikipedia, Agoh-Giuga conjecture
Wikipedia, Bernoulli number: Related sequences

Cf. A000040, A000146, A002997, A027641, A027642, A110936, A166062, A174341, A174342, A309235, A326690, A327033.

[Denominator(Numerator(Bernoulli(n-1))/n + Denominator(Bernoulli(n-1))/n^2): n in [1..70]]; // Vincenzo Librandi, Jul 14 2019

Table[Denominator[Numerator[BernoulliB[n - 1]] / n + Denominator[ BernoulliB[ n - 1]] / n^2], {n, 70}] (* Vincenzo Librandi, Jul 14 2019 *)

a(n) = denominator(numerator(bernfrac(n-1))/n + denominator(bernfrac(n-1))/n^2); \\ Michel Marcus, Jul 14 2019

a(p) = 1 for prime p.
a(2k) = (2k)^2 for k > 1.
Conjecture: for k > 0, a(2k+1) = (2k+1)^2 iff 2k+1 is in A121707.
Denominator(F(p)/p) = 1 for the primes p = 2 and p = 1277 but for no other prime p < 1.5 * 10^4. Does denominator(F(p)/p) = 1 for any prime p > 1.5 * 10^4? - Jonathan Sondow, Jul 14 2019
Similarly, Sum_{k=1..p-1} k^(p-1) == -1 (mod p^2) for the prime p = 1277. - Thomas Ordowski, Jul 15 2019
a(n) = denominator(Sum_{prime p | n} 1/p - 1/n) if n is a prime or a Carmichael number. - Jonathan Sondow, Jul 19 2019

A181722 Numerator of (1/n - Bernoulli number A164555(n)/A027642(n)).

Original entry on oeis.org

0, 0, 1, 1, 7, 1, 5, 1, 13, 1, 1, 1, 901, 1, -11, 1, 3647, 1, -43825, 1, 1222387, 1, -854507, 1, 1181821001, 1, -76977925, 1, 23749461059, 1, -8615841275543, 1, 28267510484519, 1

Offset: 1

Paul Curtz, Nov 17 2010

An autosequence is a sequence whose inverse binomial transform is the sequence signed. In integers, the oldest example is Fibonacci A000045. In fractions, A164555/A027642 is the son of 1/n via the Akiyama-Tanigawa algorithm; grandson is (A174110/A174111) = 1/2, 2/3, 1/2, 2/15, ...; see A164020. See A174341/A174342. All are from the same family.

			Fractions are 0, 0, 1/6, 1/4, 7/30, 1/6, 5/42, 1/8, 13/90, 1/10, 1/66, 1/12, 901/2730, ...

G. C. Greubel, Table of n, a(n) for n = 1..625
OEIS Wiki, Autosequence.

Cf. A000045, A027642, A164020, A164555, A174110, A174111, A174341, A174342.

A181722:= func< n | n le 2 select 0 else Numerator(1/n - Bernoulli(n-1)) >;
[A181722(n): n in [1..40]]; // G. C. Greubel, Mar 25 2024

a[n_] := If[n <= 2, 0, Numerator[1/n - BernoulliB[n-1]]];
Table[a[n], {n, 1, 34}] (* Jean-François Alcover, Jun 07 2017 *)

def A181722(n): return 0 if n<3 else numerator(1/n - bernoulli(n-1))
[A181722(n) for n in range(1,41)] # G. C. Greubel, Mar 25 2024

A174342 Denominator of ( A164555(n)/A027642(n) + 1/(n+1) ).

Original entry on oeis.org

1, 1, 2, 4, 6, 6, 6, 8, 90, 10, 6, 12, 210, 14, 30, 16, 30, 18, 42, 20, 770, 22, 6, 24, 13650, 26, 54, 28, 30, 30, 462, 32, 5610, 34, 210, 36, 51870, 38, 26, 40, 330, 42, 42, 44, 2070, 46, 6, 48, 324870, 50, 1122, 52, 30, 54, 43890, 56, 5510, 58, 6, 60, 930930

Offset: 0

Paul Curtz, Mar 16 2010

The sequence A174341(n)/a(n) = 2, 1, 1/2, 1/4, 1/6, 1/6, 1/6, ... becomes 2, -1, 1/2, -1/4, 1/6,.. under inverse binomial transform: an autosequence, where each second term flips the sign.

OEIS Wiki, Autosequence

Cf. A174341 (numerators).

B(n)=if(n!=1, bernfrac(n), -bernfrac(n));
a(n)=denominator(B(n) + 1/(n + 1));
for(n=0, 60, print1(a(n),", ")) \\ Indranil Ghosh, Jun 19 2017

from sympy import bernoulli, Rational
def B(n):
    return bernoulli(n) if n != 1 else -bernoulli(n)
def a(n):
    return (B(n) + Rational(1, n + 1)).as_numer_denom()[1]
[a(n) for n in range(61)] # Indranil Ghosh, Jun 19 2017

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 *)

A189731 a(n) = numerator of B(0,n) where B(n,n) = 0, B(n-1,n) = 1/n, and B(m,n) = B(m-1,n+1) - B(m-1,n).

Original entry on oeis.org

0, 1, 1, 3, 2, 17, 4, 23, 25, 61, 18, 107, 40, 421, 1363, 1103, 210, 5777, 492, 7563, 24475, 19801, 2786, 103681, 33552, 135721, 146401, 355323, 39650, 1860497, 97108, 2435423, 2627065, 6376021, 20633238, 11128427, 1459960, 43701901

Offset: 0

Paul Curtz, Apr 26 2011

Square array B(m,n) begins:
    0,    1/1,   1/1,   3/2,   2/1,  17/6,  ...
   1/1,    0,    1/2,   1/2,   5/6,   7/6,  ...
  -1/1,   1/2,    0,    1/3,   1/3,   7/12, ...
   3/2,  -1/2,   1/3,    0,    1/4,   1/4,  ...
  -2/1,   5/6,  -1/3,   1/4,    0,    1/5,  ...
  17/6,  -7/6,   7/12, -1/4,   1/5,    0,   ...
The inverse binomial transform of B(0,n) gives B(n,0) and thus it is an eigensequence in the sense that it remains the same (up to a sign) under inverse binomial transform.
The bisection of B(0,n) (odd part) gives A175385/A175386, and thus a(2*n+1) = A175385(n+1).

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000204, A242926 (denominators).

Cf. A174341, A177690, A181722.

B:= proc(m, n) option remember;
      if m=n then 0
    elif n=m+1 then 1/n
    elif n>m then B(m, n-1) +B(m+1, n-1)
    else B(m-1, n+1) -B(m-1, n)
      fi
    end:
a:= n-> numer(B(0, n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 29 2011

Rest[Numerator[Abs[CoefficientList[Normal[Series[Log[1 - x^2/(1 + x)], {x, 0, 40}]], x]]]] (* Vaclav Kotesovec, Jul 07 2020 *)
Table[Numerator[(LucasL[n]-1)/n],{n,1,38}] (* Artur Jasinski, Oct 21 2022 *)

Numerator of (A000204(n) - 1)/n. - Artur Jasinski, Oct 21 2022

A327033 N(p-1)/p + D(p-1)/p^2 with p the n-th prime and B(k) = N(k)/D(k) the k-th Bernoulli number.

Original entry on oeis.org

0, 1, 1, 1, 1, -37, -211, 2311, 37153, -818946931, 277930363757, -711223555487930419, -6367871182840222481, 35351107998094669831, 12690449182849194963361, -15116334304443206742413679091, 1431925649981017658678758915153153, -19921854762028779869513196624259348280501

Offset: 1

Jonathan Sondow, Aug 15 2019

sign

a(n) is an integer, as conjectured by Thomas Ordowski and proved by the author in A309132 and A326690.

Ordowski also conjectured that the sequence is a subsequence of A174341.

			Prime(6) = 13 and B(12) = -691/2730, so a(6) = -691/13 + 2730/13^2 = -37.

Wikipedia, Bernoulli number

Cf. A174341, A309132, A326690.

a[n_] := With[{p = Prime[n]}, With[{b = BernoulliB[p - 1]}, (p  Numerator[b] + Denominator[b])/p^2]];
Table[a[n], {n, 1, 18}]

a(n) = my(p = prime(n), b = bernfrac(p-1)); numerator(b)/p + denominator(b)/p^2; \\ Michel Marcus, Aug 16 2019

cp's OEIS Frontend

A309132 a(n) is the denominator of F(n) = A027641(n-1)/n + A027642(n-1)/n^2.

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A181722 Numerator of (1/n - Bernoulli number A164555(n)/A027642(n)).

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A174342 Denominator of ( A164555(n)/A027642(n) + 1/(n+1) ).

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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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A189731 a(n) = numerator of B(0,n) where B(n,n) = 0, B(n-1,n) = 1/n, and B(m,n) = B(m-1,n+1) - B(m-1,n).

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A327033 N(p-1)/p + D(p-1)/p^2 with p the n-th prime and B(k) = N(k)/D(k) the k-th Bernoulli number.

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