A059841 -id:A059841 - cp's OEIS frontend

A081186 4th binomial transform of (1,0,1,0,1,...), A059841.

1, 4, 17, 76, 353, 1684, 8177, 40156, 198593, 986404, 4912337, 24502636, 122336033, 611148724, 3054149297, 15265963516, 76315468673, 381534296644, 1907542343057, 9537324294796, 47685459212513, 238423809278164, 1192108586037617, 5960511549128476

Offset: 0

Paul Barry, Mar 11 2003

Binomial transform of A007582.
a(n) is a companion to A005059(n): a(n) + A005059(n) = 5^n; e.g. a(4) = A005059(4) = 353 + 272 = 625 = 5^4. - Gary W. Adamson, Jun 30 2006
Number of words of length n from an alphabet of 5 letters in which a chosen letter appears an even number of times. - James Mahoney, Feb 03 2012 [See a comment in A007582, also for crossrefs. for the 1- to 11-letter word cases. - Wolfdieter Lang, Jul 17 2017]
The sequence of fractions x(n) = a(n+1)/a(n) satisfies a simple recurrence x(n+1) = 108 - (815 - 1500 / x(n-1)) / x(n) known as Muller's recurrence. It is used for the demonstration of an unexpected failure of floating-point computations. - Andrey Zabolotskiy, Sep 17 2019

			Say the alphabet is {a,b,c,d,e} and we want to know how many words of length one and two contain c an even number of times. a(1) = 4, which we can see by the four words {(a),(b),(d),(e)} and a(2) = 17, which we can see by the seventeen words {(a,a), (a,b), (a,d), (a,e), (b,a), (b,b), (b,d), (b,e), (c,c), (d,a), (d,b), (d,d), (d,e), (e,a), (e,b), (e,d), (e,e)}. - _James Mahoney_, Feb 03 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Christian Hill, Muller's Recurrence, 2017.
Index entries for linear recurrences with constant coefficients, signature (8,-15).

Cf. A005059, A007582.

List([0..25], n-> (3^n + 5^n)/2); # G. C. Greubel, Dec 26 2019

[3^n/2+5^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013

seq( (3^n + 5^n)/2, n=0..30); # G. C. Greubel, Dec 26 2019

CoefficientList[Series[(1-4x)/((1-3x)(1-5x)), {x,0,25}], x] (* Vincenzo Librandi, Aug 07 2013 *)
LinearRecurrence[{8,-15},{1,4},30] (* Harvey P. Dale, Apr 13 2019 *)

vector(31, n,(3^(n-1) + 5^(n-1))/2 ) \\ G. C. Greubel, Dec 26 2019

[(3^n + 5^n)/2 for n in (0..25)] # G. C. Greubel, Dec 26 2019

a(n) = 8*a(n-1) - 15*a(n-2) with n>1, a(0)=1, a(1)=4.
G.f.: (1-4*x)/((1-3*x)*(1-5*x)).
a(n) = (3^n + 5^n)/2.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*4^(n-2*k).
E.g.f.: exp(4*x) * cosh(x). - Paul Barry, Oct 06 2004

A060531 9th binomial transform of (1,0,1,0,1,...), A059841.

Original entry on oeis.org

1, 9, 82, 756, 7048, 66384, 631072, 6048576, 58388608, 567108864, 5536870912, 54294967296, 534359738368, 5274877906944, 52199023255552, 517592186044416, 5140737488355328, 51125899906842624, 509007199254740992, 5072057594037927936, 50576460752303423488

Offset: 0

N. J. A. Sloane, Apr 12 2001

Binomial transform of A081190.
Number of strings of length n of the decimal digits 0..9 that contain an even number of 0's.
An equivalent formulation is: a(n) is also the number of words of length n over an alphabet of ten letters with a chosen letter appearing an even number of times. See a comment in A007582, also for the cross references for the 1- to 11-letter word cases. - Wolfdieter Lang, Jul 17 2017

			For n = 1 there are 9 strings: {1 2 3 4 5 6 7 8 9};
for n = 2 there are 82: {00 11 12 13 14 15 16 17 18 19 21 ... 96 97 98 99}.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (18,-80).

Cf. A007582, A081192.

[(8^n+10^n)/2: n in [0..20]]; // Vincenzo Librandi, Jul 18 2017

A060531 := proc(n) option remember: if n = 1 then RETURN(9) fi: 8*A060531(n-1) + 10^(n-1): end: for n from 1 to 40 do printf(`%d,`, A060531(n)) od:

Table[8^n/2 + 10^n/2, {n, 0, 19}] (* or *)
LinearRecurrence[{18, -80}, {1, 9}, 19] (* or *)
CoefficientList[Series[(1 - 9 x)/((1 - 8 x) (1 - 10 x)), {x, 0, 19}], x] (* Michael De Vlieger, Jul 17 2017 *)

a(n) = { (8^n + 10^n)/2 } \\ Harry J. Smith, Jul 06 2009

G.f.: (1 - 9*x)/((1 - 8*x)*(1 -10*x)).
E.g.f.: exp(9*x)*cosh(x).
a(n) = (8^n + 10^n)/2 = 2^(n-1)*(4^n + 5^n).
a(n) = 18*a(n-1) - 80*a(n-2), a(0) = 1, a(1) = 9.
a(n) = 8*a(n-1) + 10^(n-1), a(1) = 9.

Additional comments from Paul Barry, Mar 11 2003
Typo in definition corrected by Paolo P. Lava, Sep 18 2008
Edited by and new name from Wolfdieter Lang, Jul 18 2017

A081188 6th binomial transform of (1,0,1,0,1,.....), A059841.

Original entry on oeis.org

1, 6, 37, 234, 1513, 9966, 66637, 450834, 3077713, 21153366, 146120437, 1013077434, 7042713913, 49054856766, 342163294237, 2389039544034, 16692759230113, 116696726720166, 816114147588037, 5708984335850634

Offset: 0

Paul Barry, Mar 11 2003

Binomial transform of A081187.

a(n) is also the number of words of length n over an alphabet of seven letters, of which a chosen one appears an even number of times. See a comment in A007582, also for the crossrefs. for the 1- to 11- letter word cases. - Wolfdieter Lang, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Takao Komatsu, Some recurrence relations of poly-Cauchy numbers, J. Nonlinear Sci. Appl., (2019) Vol. 12, Issue 12, 829-845.
Index entries for linear recurrences with constant coefficients, signature (12,-35).

Cf. A007582, A081186, A081189.

[5^n/2+7^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013

CoefficientList[Series[(1 - 6 x) / ((1 - 5 x) (1 - 7 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 07 2013 *)
LinearRecurrence[{12,-35},{1,6},30] (* Harvey P. Dale, Mar 24 2016 *)

a(n)=5^n/2+7^n/2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 12*a(n-1) -35*a(n-2), a(0)=1, a(1)=6.
G.f.: (1-6*x)/((1-5*x)*(1-7*x)).
E.g.f.: exp(6*x)*cosh(x).
a(n) = 5^n/2 + 7^n/2.

A081189 7th binomial transform of (1,0,1,0,1,...), A059841.

Original entry on oeis.org

1, 7, 50, 364, 2696, 20272, 154400, 1188544, 9228416, 72147712, 567104000, 4476365824, 35448129536, 281408253952, 2238205337600, 17827278536704, 142148043309056, 1134363236564992, 9057979233075200, 72362273907933184

Offset: 0

Paul Barry, Mar 11 2003

Binomial transform of A081188.
a(n) is the number of functions f:{1,2,...,n}->{1,2,3,4,5,6,7,8} with an even number of elements mapped to 1. - Geoffrey Critzer, Dec 30 2012
For the alternative formulation in terms of words of length n over an alphabet of eight letters with a chosen letter appearing an even number of times see a comment in A007582, also for the crossrefs, for the 1- to 11- letter word cases. - Wolfdieter Lang, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (14,-48).

Cf. A007582, A081190.

[6^n/2 + 8^n/2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2013

nn=20;Range[0,nn]!CoefficientList[Series[Exp[7x]Cosh[x],{x,0,nn}],x]  (* Geoffrey Critzer, Dec 30 2012 *)
LinearRecurrence[{14, -48}, {1, 7}, 20] (* Or *)
CoefficientList[Series[(1 - 7 x)/(1 - 14 x + 48 x^2), {x, 0, 19}], x] (* Robert G. Wilson v, Jan 02 2013 *)

a(n) = 14*a(n-1) - 48*a(n-2) with n > 1, a(0)=1, a(1)=7.
G.f.: (1-7*x)/((1-6*x)*(1-8*x)).
E.g.f. exp(7*x)*cosh(x).
a(n) = 6^n/2 + 8^n/2.
a(n) = 6*a(n-1) + 8^(n-1).

A081190 8th binomial transform of (1,0,1,0,1,.....), A059841.

Original entry on oeis.org

1, 8, 65, 536, 4481, 37928, 324545, 2803256, 24405761, 213887048, 1884629825, 16679193176, 148135411841, 1319377419368, 11777507763905, 105319346802296, 943126559710721, 8454906106826888, 75861524447454785, 681125306429182616

Offset: 0

Paul Barry, Mar 11 2003

Binomial transform of A081189.

a(n) is also the number of words of length n over an alphabet of nine letters, of which a chosen one appears an even number of times. See a comment in A007582, also for the crossrefs. for the 1- to 11- letter word cases. For a formulation in terms of maps see a Geoffrey Critzer comment in A081189. - Wolfdieter Lang, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (16,-63).

Cf. A007582, A060531, A081189.

[7^n/2 + 9^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013

CoefficientList[Series[(1 - 8 x) / ((1 - 7 x) (1 - 9 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 07 2013 *)
LinearRecurrence[{16,-63},{1,8},20] (* Harvey P. Dale, Apr 04 2017 *)

a(n) = 16*a(n-1) -63*a(n-2), a(0)=1, a(1)=8.
G.f.: (1-8*x)/((1-7*x)*(1-9*x)).
E.g.f. exp(8*x) * cosh(x).
a(n) = 7^n/2 + 9^n/2.

A081192 10th binomial transform of (1,0,1,0,1,......), A059841.

Original entry on oeis.org

1, 10, 101, 1030, 10601, 110050, 1151501, 12135070, 128702801, 1372684090, 14712104501, 158346365110, 1710428956601, 18532288986130, 201313313019101, 2191569650755150, 23901375026212001, 261062105099480170

Offset: 0

Paul Barry, Mar 11 2003

Binomial transform of A060531.
Average of binomial and inverse binomial transforms of 10^n.
a(n) is also the number of words of length n over an alphabet of eleven letters with a chosen letter appearing an even number of times. See a comment in A007582, also for the crossrefs. for the 1- to 10- letter word cases. - Wolfdieter Lang, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (20,-99).

Cf. A007582, A059841, A060531.

[9^n/2 + 11^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013

A081192:=n->9^n/2 + 11^n/2: seq(A081192(n), n=0..30); # Wesley Ivan Hurt, May 03 2017

CoefficientList[Series[(1-10x)/((1-9x)(1-11x)),{x,0,200}],x] (* Vincenzo Librandi, Aug 07 2013 *)

a(n) = 20*a(n-1) -99*a(n-2), a(0)=1, a(1)=10.
G.f.: (1-10*x)/((1-9*x)*(1-11*x)).
E.g.f.: exp(10*x) * cosh(x).
a(n) = 9^n/2 + 11^n/2.
a(n) = Sum_{k=0..floor(n/2)} C(n,2*k)*10^(n-2*k).

A081187 5th binomial transform of (1,0,1,0,1,...), A059841.

Original entry on oeis.org

1, 5, 26, 140, 776, 4400, 25376, 148160, 872576, 5169920, 30757376, 183495680, 1096779776, 6563901440, 39316299776, 235629363200, 1412702437376, 8471919656960, 50814338072576, 304817308958720, 1828628975845376

Offset: 0

Paul Barry, Mar 11 2003

Binomial transform of A081187.

a(n) is also the number of words of length n over an alphabet of six letters, of which a chosen one appears an even number of times. See a comment in A007582, also for the crossrefs, for the 1- to 11-letter word cases. - Wolfdieter Lang, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (10,-24).

Cf. A007582, A074612, A081189.

List([0..25], n-> (4^n + 6^n)/2); # G. C. Greubel, Dec 26 2019

[4^n/2 + 6^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013

seq( (4^n + 6^n)/2, n=0..25); # G. C. Greubel, Dec 26 2019

CoefficientList[Series[(1-5x)/((1-4x)(1-6x)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 07 2013 *)
LinearRecurrence[{10,-24}, {1,5}, 26] (* G. C. Greubel, Dec 26 2019 *)

vector(26, n, (4^(n-1) + 6^(n-1))/2) \\ G. C. Greubel, Dec 26 2019

[(4^n + 6^n)/2 for n in (0..25)] # G. C. Greubel, Dec 26 2019

a(n) = 10*a(n-1) - 24*a(n-2) with n > 1, a(0)=1, a(1)=5.
G.f.: (1-5*x)/((1-4*x)*(1-6*x)).
E.g.f.: exp(5*x)*cosh(x).
a(n) = (4^n + 6^n)/2.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*5^(n-2k).
E.g.f.: exp(5*x)*cosh(x) = (1/2)*E(0), where E(k) = 1 + 2^k/(3^k - 6*x*9^k/(6*x*3^k + (k+1)*2^k/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
a(n) = A074612(n)/2. - G. C. Greubel, Jan 13 2024

A289846 p-INVERT of (1,0,1,0,1,0,1,0,1,...) (A059841), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 2, 4, 9, 18, 39, 80, 170, 353, 744, 1553, 3262, 6824, 14313, 29970, 62823, 131596, 275782, 577777, 1210704, 2536657, 5315210, 11136700, 23334969, 48893202, 102446199, 214654136, 449764562, 942387569, 1974580920, 4137324929, 8668915558, 18163921856

Offset: 0

Clark Kimberling, Aug 14 2017

Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A289780 for a guide to related sequences.

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

Cf. A059841, A289780.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A059841 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289846 *)

G.f.: (1 + x - x^2)/(1 - x - 3 x^2 + x^3 + x^4).

a(n) = a(n-1) + 3*a(n-2) - a(n-3) - a(n-4).

A137161 Starting at Step=0 from the sequence 1, 0, 1, 0, 1, 0, etc. (A059841) at any Step=n, with n>=1, any numbers of the previous step is inserted in the sequence any (2*n-1) numbers.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Jan 25 2008, Mar 05 2008

If the sequence is regarded as the binary representation of the number 0.1101011111110...... in decimal it is 0.843596927232253....

			Step n=0: 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, etc.
Step n=1: 2*n-1=1 thus after any 1 number -> 1 number of Step=0 is inserted
1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, etc.
Step n=2: 2*n-1=3 thus after any 3 numbers -> 1 number of Step=1 is inserted
1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, etc.
Step n=3: 2*n-1=5 thus after any 5 numbers -> 1 number of Step=2 is inserted
1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, etc.
And so on.

Cf. A137162, A137163.

A169591 Triangle T(n,k) with column n = A059841 if n even and column n = A000007 if n odd.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Philippe Deléham, Dec 02 2009

Row sums : A158416.

			Triangle begins : 1 ; 0,1 ; 1,0,1 ; 0 0 0 1 ; 1,0,1,0,1 ; 0,0,0,0,0,1 ; ...

Sum_{k, 0<=k<=n} T(n,k)= A158416(n).

cp's OEIS Frontend

A081186 4th binomial transform of (1,0,1,0,1,...), A059841.

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A060531 9th binomial transform of (1,0,1,0,1,...), A059841.

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A081188 6th binomial transform of (1,0,1,0,1,.....), A059841.

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A081189 7th binomial transform of (1,0,1,0,1,...), A059841.

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A081190 8th binomial transform of (1,0,1,0,1,.....), A059841.

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A081192 10th binomial transform of (1,0,1,0,1,......), A059841.

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A081187 5th binomial transform of (1,0,1,0,1,...), A059841.

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