A165943 -id:A165943 - cp's OEIS frontend

A166062 a(n) = denominator(Bernoulli(prime(n) - 1)).

2, 6, 30, 42, 66, 2730, 510, 798, 138, 870, 14322, 1919190, 13530, 1806, 282, 1590, 354, 56786730, 64722, 4686, 140100870, 3318, 498, 61410, 4501770, 33330, 4326, 642, 209191710, 1671270, 4357878, 8646, 4110, 274386, 4470, 2162622, 1794590070, 130074

Offset: 1

Paul Curtz, Oct 05 2009

nonn

Divisibility through terms of A008578 is a consequence of the Staudt-Clausen theorem.
(Vaguely similar divisibility properties are considered in A165248 and A165943.)
The first 250 entries are all different. Is this true in general?
Would sorting the entries yield the full A090801?
a(n) > 1 is the largest number k such that x*y^p == y*x^p (mod k) for all integers x and y, where p = prime(n). Example: x*y^19 == y*x^19 (mod 798). - Michel Lagneau, Apr 19 2012
Comment from Herbert Kociemba, May 29 2020: (Start)
For each n there is exactly one member of the sequence whose factorization has prime(n) as its largest prime factor, namely a(n). From this we conclude:
1. All elements of the sequence are different.
2. Not all denominators of Bernoulli numbers appear in this sequence. For example the denominator of B(20), 330=2*3*5*11 never appears because the unique sequence element with largest prime divisor 11=prime(5) is a(5)=2*3*11. (End)

Peter Luschny, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem.

Cf. A071772, A110936.

seq(denom(bernoulli(ithprime(n)-1)), n=1..38); # Peter Luschny, Jul 14 2019

Table[Denominator[BernoulliB[n - 1]], {n, Prime[Range[38]]}] (* Harvey P. Dale, Apr 22 2012 *)
Table[GCD @@ Table[(n^k - n), {n, 2, 13}], {k, Prime[Range[100]]}] (* Increase n to 80 and k to 1000 for first thousand terms. - Herbert Kociemba, May 05 2020 *)
a[i_] := Times @@ Select[Prime[Range[i]], Mod[Prime[i] - 1, # - 1] == 0&]; Table[a[i], {i, 1, 100}](* Herbert Kociemba, May 06 2020 *)

a(n)=denominator(bernfrac(prime(n)-1)) \\ Charles R Greathouse IV, Apr 30 2012

a(n) = A027642(A008578(n) - 1).

Edited by Peter Luschny, Jul 14 2019

A235355 0 followed by the sum of (1),(2), (3,4),(5,6), (7,8,9),(10,11,12) from the natural numbers.

Original entry on oeis.org

0, 1, 2, 7, 11, 24, 33, 58, 74, 115, 140, 201, 237, 322, 371, 484, 548, 693, 774, 955, 1055, 1276, 1397, 1662, 1806, 2119, 2288, 2653, 2849, 3270, 3495, 3976, 4232, 4777, 5066, 5679, 6003, 6688, 7049, 7810, 8210, 9051, 9492, 10417, 10901, 11914, 12443, 13548

Offset: 0

Paul Curtz, Jan 07 2014

Difference table for 0 followed by a(n):
  0,  0,  1,   2,   7,  11,  24,  33,...
  0,  1,  1,   5,   4,  13,   9,  25,... =A147685(n)
  1,  0,  4,  -1,   9,  -4,  16,  -9,... =interleave A000290(n+1),-A000290(n)
  -1, 4, -5,  10, -13,  20, -25,  34,...
  5, -9, 15, -23,  33, -45,  59, -75,... =(-1)^n*A027688(n+2).
a(-n) = -a(n-1).
From the second row, signature (0,3,0,-3,0,1).
Consider a(n+2k+1)+a(2k-n):
  1,     2,   6,   9,  17,  22,  34,...
  9,    12,  24,  33,  57,  72, 108,...
  35,   40,  60,  75, 115, 140, 200,...
  91,   98, 126, 147, 203, 238, 322,...
  189, 198, 234, 261, 333, 378, 486,... .
The first column is A005898(n).
The rows are successively divisible by 2*k+1. Hence
  1,   2,  6,  9, 17, 22, 34,...
  3,   4,  8, 11, 19, 24, 36,...
  7,   8, 12, 15, 23, 28, 40,...
  13, 14, 18, 21, 29, 34, 46,...
  21, 22, 26, 29, 37, 42, 54,...
The first column is A002061(n+1).
The main diagonal is A212965(n).
The first difference of every row is A022998(n+1).
Compare to the (2k+1)-sections of A061037 in A165943.

			a(1)=1, a(2)=2, a(3)=3+4=7, a(4)=5+6=11, a(5)=7+8+9=24, a(6)=10+11+12=33.

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

Cf. A075356, A234305.

LinearRecurrence[{1,3,-3,-3,3,1,-1},{0,1,2,7,11,24,33},50] (* Harvey P. Dale, Nov 22 2014 *)

Vec(x*(x^2+1)*(x^2+x+1)/((x-1)^4*(x+1)^3) + O(x^100)) \\ Colin Barker, Jan 20 2014

a(n) = 4*a(n-2) -6*a(n-4) +4*a(n-6) -a(n-8), n>7.
a(2n) = 0 followed by A085786(n). a(2n+1) =  A081436(n).
a(2n) + a(2n+1) = A005898(n).
a(2n-1) + a(2n) = A061317(n).
a(n) = (-1)*((-1+(-1)^n-2*n)*(2+n+n^2))/16. a(n) = a(n-1)+3*a(n-2)-3*a(n-3)-3*a(n-4)+3*a(n-5)+a(n-6)-a(n-7). G.f.: x*(x^2+1)*(x^2+x+1) / ((x-1)^4*(x+1)^3). - Colin Barker, Jan 20 2014

More terms from Colin Barker, Jan 20 2014

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325

Offset: 0

Paul Curtz, Mar 12 2012

a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).
a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.
(n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).
The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.
A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:
0  1  1   3   3   15...
0  3  6  45  21   60...
0 15 15  60  55  325...
0 14 28 231 105  315...
0 45 45 189 171 1035...
The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:
0 1 1  3  3  15  6  14   k=0
0 1 2 15  7  20 15  77   k=1
0 3 3 12 11  65 24  63   k=2
0 2 4 33 15  45 33 175   k=3
0 5 5 21 19 115 42 112   k=4
0 3 6 51 23  70 51 273   k=5
The entries T'(k,c) are divisible by A060819(c).
Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.
Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is  A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.
The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

G. C. Greubel, 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. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

A185138 := proc(n)
        if n mod 4 = 0 then
                return n/4*(n-1) ;
        elif n mod 2 = 1 then
                return (n-1)*(n+1)/8 ;
        else
                return (n-1)*n/2 ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012

Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))
a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* Harvey P. Dale, Nov 23 2015 *)

x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ G. C. Greubel, Jun 23 2017

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).
a(n) = 3*A208950(n)/A109613(n).
a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).
G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - R. J. Mathar, Mar 22 2012
a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - Luce ETIENNE, May 13 2016
Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - Amiram Eldar, Aug 12 2022

A174850 Quintisection A061037(5*n-2).

Original entry on oeis.org

0, 5, 15, 165, 20, 525, 195, 1085, 90, 1845, 575, 2805, 210, 3965, 1155, 5325, 380, 6885, 1935, 8645, 600, 10605, 2915, 12765, 870, 15125, 4095, 17685, 1190, 20445, 5475, 23405, 1560, 26565, 7055, 29925, 1980, 33485, 8835, 37245, 2450

Offset: 0

Paul Curtz, Dec 01 2010

All entries are multiples of 5. Like A061037(n+2), the (2k+1)-sections A061037((2*k+1)*n-2) are multiples of 2k+1; see A165248, A165943.

The sequence contains 4 interlaced second-order polynomials: a(4n) = 5n*(5n-1), a(4n+1) = 5*(4n+1)*(20n+1), a(4n+2)= 5*(2n+1)*(10n+3), a(4n+3)= 5*(4n+3)*(20n+11). - R. J. Mathar, Feb 10 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

[ Numerator(1/4-1/(5*n-2)^2): n in [0..40] ]; // Bruno Berselli, Feb 10 2011

f[n_] := n/GCD[n, 4]; Table[ f[n] f[n + 4], {n, -4, 200, 5}] (* Robert G. Wilson v, Feb 03 2011 *)

vector(50, n, n--; numerator( 1/4 - 1/(5*n-2)^2 )) \\ G. C. Greubel, Sep 19 2018

a(n) = numerator( 1/4 - 1/(5*n-2)^2 ).
From R. J. Mathar, Feb 10 2011: (Start)
a(n) = +3*a(n-4) -3*a(n-8) +a(n-12).
G.f. ( -5*x*(1+3*x+33*x^2+4*x^3+102*x^4+30*x^5+118*x^6+6*x^7+57*x^8 +7*x^9+9*x^10) )/( (x-1)^3*(1+x)^3*(x^2+1)^3 ). (End)
a(n) = 5*n*(5*n-4)*(37-27*(-1)^n-3*(-i)^n-3*i^n)/64, where i=sqrt(-1).  - Bruno Berselli, Feb 10 2011

cp's OEIS Frontend

A166062 a(n) = denominator(Bernoulli(prime(n) - 1)).

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A235355 0 followed by the sum of (1),(2), (3,4),(5,6), (7,8,9),(10,11,12) from the natural numbers.

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A185138 a(4*n) = n*(4*n-1); a(2*n+1) = n*(n+1)/2; a(4*n+2) = (2*n+1)*(4*n+1).

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A174850 Quintisection A061037(5*n-2).

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A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).