A181442 -id:A181442 - cp's OEIS frontend

A180483 Expansion of (3+3x-25x^2-3x^3+2x^4)/((1-x)(1-10x^2+x^4)).

3, 6, 11, 38, 87, 354, 839, 3482, 8283, 34446, 81971, 340958, 811407, 3375114, 8032079, 33410162, 79509363, 330726486, 787061531, 3273854678, 7791105927, 32407820274, 77123997719, 320804348042, 763448871243, 3175635660126, 7557364714691, 31435552253198

Offset: 0

Paul Weisenhorn, Jan 20 2011

Previous name was: Solutions a(n) to (a(n)-2)*(a(n)-3) = 6*b(n)*(b(n)-1).
The associated b(n) are in A181442.
Consider an urn with r red and b blue balls. Draw 4 balls without replacement. The probability of picking 4 red balls is r/(r+b) *(r-1)/(r+b-1) *(r-2)/(r+b-2) * (r-3)/(r+b-3). The probability of picking 2 red and 2 blue balls is binomial(2,2) * r*(r-1)*b*(b-1)/ ((r+b)*(r+b-1)..*(r+b-3)). For equal probability we need (r-2)*(r-3)=6*b*(b-1). The current sequence shows the r, the number of red balls which allow such scenario of equal probability.
The quadratic equation is diagonalized with a(n) = (A(n) + 5)/2 and b(n) = (B(n) + 1)/2, equivalent to the Pell equation A(n)^2 - 6*B(n)^2 = -5 with the 2 fundamental solutions (1; 1); (7; 3) and the solution (5; 2) for the unit form.

			For n=3: a(3) = 38; b(3) = 15; binomial(38,4) = 73815 and  binomial(38, 2)*binomial(15, 2) = 73815.
The 2-tuples begin (3, 1); (6, 2); (11, 4); (38, 15).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1)

Cf. A004189, A181442.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3+3*x-25*x^2-3*x^3+2*x^4)/((1-x)*(1-10*x^2+x^4)) )); // G. C. Greubel, Apr 28 2022

n:=0: for s from 1 to 100 do r:=(sqrt(24*s^2-24*s+1)+5)/2: if (r=trunc(r)) then a(n):=r: b(n):=s: n:=n+1: end if: end do:

LinearRecurrence[{1,10,-10,-1,1},{3,6,11,38,87},30] (* Harvey P. Dale, Apr 28 2018 *)

def b(n): return (1/2)*(1+(-1)^n)*chebyshev_U(n//2, 5)
def A180483(n): return (1/2)*(5 +b(n) +7*b(n-1) +7*b(n-2) +b(n-3))
[A180483(n) for n in (0..40)] # G. C. Greubel, Apr 28 2022

G.f.: ( 3+3*x-25*x^2-3*x^3+2*x^4 )/( (1-x)*(1-10*x^2+x^4) ). - R. J. Mathar, Feb 05 2011
Let r=sqrt(6), s=5+2*r, and t=5-2*r, then a(2*n) = (10+(1+r)*s^n+(1-r)*t^n)/4 and a(2*n+1) = (10+(7+3*r)*s^n+(7-3*r)*t^n)/4.
a(n) = 11*a(n-2) - 11*a(n-4) + a(n-6).
a(n) = +a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5). - R. J. Mathar, Feb 05 2011
a(n) = (1/2)*(5 +b(n) +7*b(n-1) +7*b(n-2) +b(n-3)), where b(n) = (1/2)*(1+(-1)^n)*ChebyshevU(n/2, 5). - G. C. Greubel, Apr 28 2022

New name using the g.f. by R. J. Mathar from Joerg Arndt, Apr 27 2022

A118064 Decimal expansion of the sum of the reciprocals of the palindromic primes A002385 (Honaker's constant).

Original entry on oeis.org

1, 3, 2, 3, 9, 8, 2, 1, 4, 6, 8, 0, 6

Offset: 1

Martin Renner, May 11 2006

From Robert G. Wilson v, Nov 01 2010: (Start)
n \ sum to 10^n
1.267099567099567099567099567099567099567099567099567099567099567099567
1.320723244590290964212793334437872849720871258315369002493912638038324
1.323748402250648554164425746280035962754669829327727800040192015109270
1.323964105671202458016249150576217276147952428601889817773483085610332
1.323980718065525060936354534562000413901564393192688451911141729415146
1.323982026479475203850120990923294207966175748395470136325039323549015
1.323982136437462724794656629740867909978221153827990721566573347887836
1.323982145891606234777299440047139038371441916546100653011463101470839
1.323982146724859090645464845257681674740147563533254654075059843860490
1.323982146799188851138232927173756400348958236915409881890097448921521
1.323982146805857558347279363344557427339916178257233985191868031567947 (End)

			1.323982146806...

Carlos Rivera, Problems & Puzzles: Puzzle 056 - Honaker's Constant.
Eric Weisstein, Palindromic Prime.

Cf. A002385, A160910, A181442, A050251, A118031, A194097.

(* first obtain nextPalindrome from A007632 *) s = 1/11; c = 1; pp = 1; Do[ While[pp < 10^n, If[PrimeQ@ pp, c++; s = N[s + 1/pp, 64]]; pp = NextPalindrome@ pp]; If[ OddQ@ n, pp = 10^(n + 1); Print[{s, n, c}]], {n, 17}] (* Robert G. Wilson v, May 31 2009 *)
generate[n_] := Block[{id = IntegerDigits@n, insert = {{0}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}}}, FromDigits@ Join[id, #, Reverse@ id] & /@ insert]; sm = N[Plus @@ (1/{2, 3, 5, 7, 11}), 64]; k = 1; Do [While[k < 10^n, sm = N[sm + Plus @@ (1/Select[ generate@k, PrimeQ]), 128]; k++ ]; Print[{2 n + 1, sm}], {n, 9}] (* Robert G. Wilson v, Nov 01 2010 *)

Equals Sum_{p} 1/p, where p ranges over the palindromic primes.

Corrected by Eric W. Weisstein, May 14 2006
More terms from Robert G. Wilson v, Nov 01 2010
Entry revised by N. J. A. Sloane, May 05 2013

A179934 Expansion of x(4+5x-13x^2-x^3+x^4) / ( (1-x)(1-10*x^2+x^4) ).

Original entry on oeis.org

4, 9, 36, 85, 352, 837, 3480, 8281, 34444, 81969, 340956, 811405, 3375112, 8032077, 33410160, 79509361, 330726484, 787061529, 3273854676, 7791105925, 32407820272, 77123997717, 320804348040, 763448871241, 3175635660124

Offset: 1

Paul Weisenhorn, Aug 02 2010

nonn

Previous name was: a(n) red balls and b(n) blue balls in an urn; draw 2 balls without replacement; Probability(2 red balls) = 6*Probability(2 blue balls); b(n) = A181442(n).

The last digit has the period (4,9,6,5,2,7,0,1).

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Cf. A004189, A181442.

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( x*(4+5*x-13*x^2-x^3+x^4)/((1-x)*(1-10*x^2+x^4)) )); // G. C. Greubel, Apr 27 2022

r:= sqrt(6);
for n from 0 to 20 do
    a(2*n+1):= round((2 +(7+3*r)*(5+2*r)^n)/4);
    a(2*n+2):= round((2 +(17+7*r)*(5+2*r)^n)/4);
end do;
seq(a(n), n = 1..40);

LinearRecurrence[{1,10,-10,-1,1},{4,9,36,85,352},30] (* Harvey P. Dale, Dec 23 2012 *)

def b(n): return ((1+(-1)^n)/2)*chebyshev_U(n//2, 5)
def A179934(n): return (b(n) +7*b(n-1) +7*b(n-2) +b(n-3) -2*bool(n==0) +1)/2
[A179934(n) for n in (1..50)] # G. C. Greubel, Apr 27 2022

a(n) = (1 + sqrt(1 + 24*b(n)*(b(n) - 1)))/2 where b(n) = A181442(n); this is equivalent to the Pell equation A(n)^2 - 6*B(n)^2 = -5 with the two fundamental solutions (7;3) and (17;7) and the solution (5;2) for the unit form; a(n) = (A(n) + 1)/2; b(n) = (B(n) + 1)/2. [corrected by Jason Yuen, Feb 09 2025]
a(n+4) = 10*a(n+2) - a(n) - 4.
a(n+6) = 11*(a(n+4) - a(n+2)) + a(n).
a(2*n+1) = (2 + (7 + 3*r)*(5 + 2*r)^n + (7 - 3*r)*(5 - 2*r)^n)/4, r = sqrt(6).
a(2*n+2) = (2 + (17 + 7*r)*(5 + 2*r)^n + (17 - 7*r)*(5 - 2*r)^n)/4, r = sqrt(6).
From R. J. Mathar, Aug 03 2010: (Start)
a(n) = +a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5).
G.f.: x*(4+5*x-13*x^2-x^3+x^4) / ( (1-x)*(1-10*x^2+x^4) ). (End)
a(n) = (b(n) +7*b(n-1) +7*b(n-2) +b(n-3) -2*bool(n==0) +1)/2, where b(n) = ((1 + (-1)^n)/2)*ChebyshevU(n/2, 5). - G. C. Greubel, Apr 27 2022

Edited by G. C. Greubel, Apr 27 2022

New name using g.f. by R. J. Mathar from Joerg Arndt, Apr 27 2022

A216073 The list of the a(n)-values in the common solutions to k+1=b^2 and 6*k+1=a^2.

Original entry on oeis.org

1, 7, 17, 71, 169, 703, 1673, 6959, 16561, 68887, 163937, 681911, 1622809, 6750223, 16064153, 66820319, 159018721, 661452967, 1574123057, 6547709351, 15582211849, 64815640543, 154247995433, 641608696079, 1526897742481, 6351271320247, 15114729429377, 62871104506391

Offset: 1

Paul Weisenhorn, Sep 01 2012

The equations are equivalent to the Pell equation a^2 - 6*b^2 = -5 with the 2 fundamental solutions (1;1) and (7;3) and the solution (5;2) for the unit form.
The associated b(n) are in A080806.
A181442(n) = (A080806(n) + 1)/2.
A180483(n) = (a(n) + 5)/2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Cf. A080806, A180483, A181442.

a(1)=1: a(2)=7: a(3)=17: a(4)=71:
for n from 5 to 20 do
  a(n)=10*a(n-2)-a(n-4):
  printf("%9d%20d\n",n,a(n)):
end do:

LinearRecurrence[{0,10,0,-1}, {1,7,17,71}, 50] (* G. C. Greubel, Feb 22 2017 *)

a(n) = if(n<1, 0, if(n<5, [1,7,17, 71][n], 10*a(n-2)-a(n-4) ) );
/* Joerg Arndt, Sep 03 2012 */

def b(n): return (1/2)*(1+(-1)^n)*chebyshev_U(n//2, 5)
def A216073(n): return b(n) +7*b(n-1) +7*b(n-2) +b(n-3)
[A216073(n) for n in (0..50)] # G. C. Greubel, Apr 28 2022

a(n) = 10*a(n-2) - a(n-4).
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+7*x+7*x^2+x^3)/(1-10*x^2+x^4).
a(2*n+1) = ((1+r)*(5+2*r)^n + (1-r)*(5-2*r)^n)/2  where r=sqrt(6) and 0<=n.
a(2*n+2) = ((7+3*r)*(5+2*r)^n + (7-3*r)*(5-2*r)^n)/2  where r=sqrt(6) and 0<=n.
a(n) = -((5-2*r)^(1/4)*((2*r+5)^((-1)^n/4+n/2)*(-1)^n - r*(2*r+5)^((-1)^n/4+n/2)) + (2*r+5)^(1/4)*((5-2*r)^((-1)^n/4+n/2)*(-1)^n + (5-2*r)^((-1)^n/4+n/2)*r))/(2*(5-2*r)^(1/4)*(2*r+5)^(1/4)) with r=sqrt(6) and 1<=n. - Alexander R. Povolotsky, Sep 01 2012
a(n) = b(n) +7*b(n-1) +7*b(n-2) +b(n-3), where b(n) = (1/2)*(1 +(-1)^n)* ChebyshevU(n/2, 5). - G. C. Greubel, Apr 28 2022

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A180483 Expansion of (3+3*x-25*x^2-3*x^3+2*x^4)/((1-x)*(1-10*x^2+x^4)).

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A118064 Decimal expansion of the sum of the reciprocals of the palindromic primes A002385 (Honaker's constant).

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A179934 Expansion of x*(4+5*x-13*x^2-x^3+x^4) / ( (1-x)*(1-10*x^2+x^4) ).

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A216073 The list of the a(n)-values in the common solutions to k+1=b^2 and 6*k+1=a^2.

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A180483 Expansion of (3+3x-25x^2-3x^3+2x^4)/((1-x)(1-10x^2+x^4)).

A179934 Expansion of x(4+5x-13x^2-x^3+x^4) / ( (1-x)(1-10*x^2+x^4) ).