A165248 -id:A165248 - cp's OEIS frontend

A165943 a(n) = A061037(7*n+2).

0, 77, 63, 525, 56, 1365, 483, 2597, 210, 4221, 1295, 6237, 462, 8645, 2499, 11445, 812, 14637, 4095, 18221, 1260, 22197, 6083, 26565, 1806, 31325, 8463, 36477, 2450, 42021, 11235, 47957, 3192, 54285, 14399, 61005, 4032, 68117, 17955, 75621, 4970

Offset: 0

Paul Curtz, Oct 01 2009

The (2k+1)-sections A061037((2*k+1)*n+2) are multiples of 2k+1:
0,...21,...15,..117,...12,..285,...99,..525,...42,..837,..255, k=1, A142590
0,...45,...35,..285,...30,..725,..255,.1365,..110,.2205,..675, k=2, A165248
0,...77,...63,..525,...56,.1365,..483,.2597,..210,.4221,.1295, k=3, here
0,..117,...99,..837,...90,.2205,..783,.4221,..342,.6885,.2115, k=4,
0,..165,..143,.1221,..132,.3245,.1155,.6237,..506,10197,.3135, k=5
0,..221,..195,.1677,..182,.4485,.1599,.8645,..702,14157,.4355, k=6
After division by 2k+1 these define a table T'(k,c) :
0,....7,....5,...39,....4,...95,...33,..175,...14,..279,...85, k=1, A142883
0,....9,....7,...57,....6,..145,...51,..273,...22,..441,..135, k=2
0,...11,....9,...75,....8,..195,...69,..371,...30,..603,..185, k=3
0,...13,...11,...93,...10,..245,...87,..469,...38,..765,..235, k=4
0,...15,...13,..111,...12,..295,..105,..567,...46,..927,..285, k=5
0,...17,...15,..129,...14,..345,..123,..665,...54,.1089,..335, k=6
Differences downwards each second column in this second table are 2 = 7-5 = 9-7..; 18 = 57-39 = 75-57..; 50 = 145-95 = 195-145... = A077591(n+1) = 2*A016754.
The difference T'(k+1,c)-T'(k,c) is 0, 2, 2, 18, 2, 50, 18, 98, 8 ... = 2*A181318(c) =A061037(c-2)+A061037(c+2). - Paul Curtz, Mar 12 2012
Let b(n)= a(n) mod 11. The sequence b(n) has the property b(n+44) = b(n) with the first 43 values being {0, 0 , 8, 1, 1, 10, 1, 1, 8, 8, 0, 0, 10, 5, 5, 9, 7, 1, 5, 6, 10, 7, 0, 2, 1, 1, 8, 1, 5, 8, 2, 0, 8, 10, 6, 5, 10, 7, 9, 5, 0, 10, 0}. - G. C. Greubel, Apr 18 2016

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018

seq(numer(1/4 - 1/(7*n+2)^2), n=0..50); # Robert Israel, Apr 20 2016

Table[Numerator[1/4 - 1/(7 n + 2)^2], {n, 0, 40}] (* Michael De Vlieger, Apr 19 2016 *)
CoefficientList[Series[7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)

a(n) = numerator(1/4 - 1/(7*n+2)^2); \\  Altug Alkan, Apr 18 2016

x='x+O('x^50); concat([0], Vec(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>12. - Conjectured by R. J. Mathar, Mar 02 2010, proved by Robert Israel, Apr 20 2016
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -7*n*(7*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

Partially edited and extended by R. J. Mathar, Mar 02 2010

Removed division by 7 in definition and formula - R. J. Mathar, Mar 23 2010

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

Original entry on oeis.org

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

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

A165943 a(n) = A061037(7*n+2).

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

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A174850 Quintisection A061037(5*n-2).

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