cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 22 results. Next

A024036 a(n) = 4^n - 1.

Original entry on oeis.org

0, 3, 15, 63, 255, 1023, 4095, 16383, 65535, 262143, 1048575, 4194303, 16777215, 67108863, 268435455, 1073741823, 4294967295, 17179869183, 68719476735, 274877906943, 1099511627775, 4398046511103, 17592186044415, 70368744177663, 281474976710655
Offset: 0

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Author

Keywords

Comments

This sequence is the normalized length per iteration of the space-filling Peano-Hilbert curve. The curve remains in a square, but its length increases without bound. The length of the curve, after n iterations in a unit square, is a(n)*2^(-n) where a(n) = 4*a(n-1)+3. This is the sequence of a(n) values. a(n)*(2^(-n)*2^(-n)) tends to 1, the area of the square where the curve is generated, as n increases. The ratio between the number of segments of the curve at the n-th iteration (A015521) and a(n) tends to 4/5 as n increases. - Giorgio Balzarotti, Mar 16 2006
Numbers whose base-4 representation is 333....3. - Zerinvary Lajos, Feb 03 2007
From Eric Desbiaux, Jun 28 2009: (Start)
It appears that for a given area, a square n^2 can be divided into n^2+1 other squares.
It's a rotation and zoom out of a Cartesian plan, which creates squares with side
= sqrt( (n^2) / (n^2+1) ) --> A010503|A010532|A010541... --> limit 1,
and diagonal sqrt(2*sqrt((n^2)/(n^2+1))) --> A010767|... --> limit A002193.
(End)
Also the total number of line segments after the n-th stage in the H tree, if 4^(n-1) H's are added at the n-th stage to the structure in which every "H" is formed by 3 line segments. A164346 (the first differences of this sequence) gives the number of line segments added at the n-th stage. - Omar E. Pol, Feb 16 2013
a(n) is the cumulative number of segment deletions in a Koch snowflake after (n+1) iterations. - Ivan N. Ianakiev, Nov 22 2013
Inverse binomial transform of A005057. - Wesley Ivan Hurt, Apr 04 2014
For n > 0, a(n) is one-third the partial sums of A002063(n-1). - J. M. Bergot, May 23 2014
Also the cyclomatic number of the n-Sierpinski tetrahedron graph. - Eric W. Weisstein, Sep 18 2017

Examples

			G.f. = 3*x + 15*x^2 + 63*x^3 + 255*x^4 + 1023*x^5 + 4095*x^6 + ...
		

References

  • Graham Everest, Alf van der Poorten, Igor Shparlinski, and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.

Crossrefs

Programs

  • Haskell
    a024036 = (subtract 1) . a000302
    a024036_list = iterate ((+ 3) . (* 4)) 0
    -- Reinhard Zumkeller, Oct 03 2012
    
  • Maple
    A024036:=n->4^n-1; seq(A024036(n), n=0..30); # Wesley Ivan Hurt, Apr 04 2014
  • Mathematica
    Array[4^# - 1 &, 50, 0] (* Vladimir Joseph Stephan Orlovsky, Nov 03 2009 *)
    (* Start from Eric W. Weisstein, Sep 19 2017 *)
    Table[4^n - 1, {n, 0, 20}]
    4^Range[0, 20] - 1
    LinearRecurrence[{5, -4}, {0, 3}, 20]
    CoefficientList[Series[3 x/(1 - 5 x + 4 x^2), {x, 0, 20}], x]
    (* End *)
  • PARI
    for(n=0, 100, print1(4^n-1, ", ")) \\ Felix Fröhlich, Jul 04 2014
  • Sage
    [gaussian_binomial(2*n,1, 2) for n in range(21)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [stirling_number2(2*n+1, 2) for n in range(21)] # Zerinvary Lajos, Nov 26 2009
    

Formula

a(n) = 3*A002450(n). - N. J. A. Sloane, Feb 19 2004
G.f.: 3*x/((-1+x)*(-1+4*x)) = 1/(-1+x) - 1/(-1+4*x). - R. J. Mathar, Nov 23 2007
E.g.f.: exp(4*x) - exp(x). - Mohammad K. Azarian, Jan 14 2009
a(n) = A000051(n)*A000225(n). - Reinhard Zumkeller, Feb 14 2009
A079978(a(n)) = 1. - Reinhard Zumkeller, Nov 22 2009
a(n) = A179857(A000225(n)), for n > 0; a(n) > A179857(m), for m < A000225(n). - Reinhard Zumkeller, Jul 31 2010
a(n) = 4*a(n-1) + 3, with a(0) = 0. - Vincenzo Librandi, Aug 01 2010
A000120(a(n)) = 2*n. - Reinhard Zumkeller, Feb 07 2011
a(n) = (3/2)*A020988(n). - Omar E. Pol, Mar 15 2012
a(n) = (Sum_{i=0..n} A002001(i)) - 1 = A178789(n+1) - 3. - Ivan N. Ianakiev, Nov 22 2013
a(n) = n*E(2*n-1,1)/B(2*n,1), for n > 0, where E(n,x) denotes the Euler polynomials and B(n,x) the Bernoulli polynomials. - Peter Luschny, Apr 04 2014
a(n) = A000302(n) - 1. - Sean A. Irvine, Jun 18 2019
Sum_{n>=1} 1/a(n) = A248721. - Amiram Eldar, Nov 13 2020
a(n) = A080674(n) - A002450(n). - Elmo R. Oliveira, Dec 02 2023

Extensions

More terms Wesley Ivan Hurt, Apr 04 2014

A100221 Decimal expansion of Product_{k>=1} (1-1/4^k).

Original entry on oeis.org

6, 8, 8, 5, 3, 7, 5, 3, 7, 1, 2, 0, 3, 3, 9, 7, 1, 5, 4, 5, 6, 5, 1, 4, 3, 5, 7, 2, 9, 3, 5, 0, 8, 1, 8, 4, 6, 7, 5, 5, 4, 9, 8, 1, 9, 3, 7, 8, 3, 3, 5, 7, 3, 5, 3, 4, 0, 1, 5, 7, 2, 3, 2, 5, 7, 7, 5, 3, 3, 1, 9, 8, 4, 5, 0, 7, 9, 8, 6, 7, 5, 1, 2, 4, 8, 0, 3, 3, 4, 6, 0, 4, 8, 1, 4, 2, 8, 8, 7, 9, 0, 5
Offset: 0

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Author

Eric W. Weisstein, Nov 09 2004

Keywords

Examples

			0.68853753712033971545651435729350818467554981937833...
		

Crossrefs

Programs

  • Mathematica
    EllipticThetaPrime[1, 0, 1/2]^(1/3)/2^(1/4)
    N[QPochhammer[1/4]] (* G. C. Greubel, Nov 30 2015 *)
    RealDigits[Fold[Times,1-1/4^Range[1000]],10,110][[1]] (* Harvey P. Dale, Sep 27 2024 *)
  • PARI
    prodinf(x=1, 1-1/4^x) \\ Altug Alkan, Dec 01 2015

Formula

Equals exp(-Sum_{k>0} sigma_1(k)/(k*4^k)) where sigma_1() is A000203(). - Hieronymus Fischer, Aug 07 2007
Equals (1/4; 1/4){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 30 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(Pi/log(2)) * exp(log(2)/12 - Pi^2/(12*log(2))) * Product_{k>=1} (1 - exp(-2*k*Pi^2/log(2))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027637(n). (End)

A022168 Triangle of Gaussian binomial coefficients [ n,k ] for q = 4.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 21, 1, 1, 85, 357, 85, 1, 1, 341, 5797, 5797, 341, 1, 1, 1365, 93093, 376805, 93093, 1365, 1, 1, 5461, 1490853, 24208613, 24208613, 1490853, 5461, 1, 1, 21845, 23859109, 1550842085, 6221613541
Offset: 0

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Keywords

Comments

The coefficients of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^n*A157784(n,k). - R. J. Mathar, Mar 12 2013

Examples

			Triangle begins:
  1;
  1,    1;
  1,    5,       1;
  1,   21,      21,        1;
  1,   85,     357,       85,        1;
  1,  341,    5797,     5797,      341,       1;
  1, 1365,   93093,   376805,    93093,    1365,    1;
  1, 5461, 1490853, 24208613, 24208613, 1490853, 5461, 1;
		

References

  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 698.

Crossrefs

Cf. A006118 (row sums), A002450 (k=1), A006105 (k=2), A006106 (k=3).

Programs

  • Maple
    A022168 := proc(n,m)
            A027637(n)/A027637(n-m)/A027637(m) ;
    end proc: # R. J. Mathar, Nov 14 2011
  • Mathematica
    gaussianBinom[n_, k_, q_] := Product[q^i - 1, {i, n}]/Product[q^j - 1, {j, n - k}]/Product[q^l - 1, {l, k}]; Column[Table[gaussianBinom[n, k, 4], {n, 0, 8}, {k, 0, n}], Center] (* Alonso del Arte, Nov 14 2011 *)
    Table[QBinomial[n,k,4], {n,0,10}, {k,0,n}]//Flatten (* or *) q:= 4; T[n_, 0]:= 1; T[n_,n_]:= 1; T[n_,k_]:= T[n,k] = If[k < 0 || n < k, 0, T[n-1, k -1] +q^k*T[n-1,k]]; Table[T[n,k], {n,0,10}, {k,0,n}] // Flatten  (* G. C. Greubel, May 27 2018 *)
  • PARI
    {q=4; T(n,k) = if(k==0,1, if (k==n, 1, if (k<0 || nG. C. Greubel, May 27 2018

Formula

T(n,k) = T(n-1,k-1) + q^k * T(n-1,k). - Peter A. Lawrence, Jul 13 2017
G.f. of column k: x^k * exp( Sum_{j>=1} f((k+1)*j)/f(j) * x^j/j ), where f(j) = 4^j - 1. - Seiichi Manyama, May 09 2025

A027871 a(n) = Product_{i=1..n} (3^i - 1).

Original entry on oeis.org

1, 2, 16, 416, 33280, 8053760, 5863137280, 12816818094080, 84078326697164800, 1654829626053597593600, 97714379759212830706892800, 17309711516825516108403231948800
Offset: 0

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Comments

2*(10)^m|a(n) where 4*m <= n <= 4*m+3 for m >= 1. - G. C. Greubel, Nov 20 2015
Given probability p = 1/3^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, 1-a(n)/A047656(n+1) is the probability that the outcome has occurred at or before the n-th iteration. The limiting ratio is 1-A100220 ~ 0.4398739. - Bob Selcoe, Mar 01 2016

Crossrefs

Cf. A005329 (q=2), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Programs

Formula

a(n) ~ c * 3^(n*(n+1)/2), where c = A100220 = Product_{k>=1} (1-1/3^k) = 0.560126077927948944969792243314140014379736333798... . - Vaclav Kotesovec, Nov 21 2015
a(n) = 3^(binomial(n+1,2))*(1/3;1/3){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
a(n) = Product_{i=1..n} A024023(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 3^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 3^k*x). - Ilya Gutkovskiy, May 22 2017
From Amiram Eldar, Feb 19 2022: (Start)
Sum_{n>=0} 1/a(n) = A132324.
Sum_{n>=0} (-1)^n/a(n) = A100220. (End)

A027878 a(n) = Product_{i=1..n} (10^i - 1).

Original entry on oeis.org

1, 9, 891, 890109, 8900199891, 890011088900109, 890010198889020099891, 8900101098880002109889900109, 890010100987899112108987901010099891, 890010100097889011121088788901111989989900109
Offset: 0

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Keywords

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027879 (q=11), A027880 (q=12).

Programs

  • Magma
    [1] cat [&*[10^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015
  • Mathematica
    Table[Product[10^i-1,{i,n}],{n,0,10}] (* Harvey P. Dale, Aug 15 2011 *)
    Abs@QPochhammer[10, 10, Range[0, 30]] (* G. C. Greubel, Nov 24 2015 *)
  • PARI
    a(n) = prod(k=1, n, 10^k - 1) \\ Altug Alkan, Nov 25 2015
    

Formula

a(n) ~ c * 10^(n*(n+1)/2), where c = Product_{k>=1} (1-1/10^k) = A132038 = 0.890010099998999000000100009999999989999900000000... . - Vaclav Kotesovec, Nov 21 2015
3^n*(11)^(floor(n/2)) divides a(n) for n>=0. - G. C. Greubel, Nov 24 2015
Equals 10^(binomial(n+1,2))*(1/10;1/10){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
G.f.: Sum_{n>=0} 10^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 10^k*x). - Ilya Gutkovskiy, May 22 2017
From Amiram Eldar, May 07 2023: (Start)
Sum_{n>=0} 1/a(n) = A132326.
Sum_{n>=0} (-1)^n/a(n) = A132038. (End)

A027872 a(n) = Product_{i=1..n} (5^i - 1).

Original entry on oeis.org

1, 4, 96, 11904, 7428096, 23205371904, 362560730628096, 28324694519589371904, 11064305472020078810628096, 21609960560733744406929189371904, 211034749490954911990173458030810628096
Offset: 0

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Comments

Given probability p = 1/5^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, 1 - a(n)/A109345(n+1) is the probability that the outcome has occurred at or before the n-th iteration. The limiting ratio is 1-A100222 ~ 0.2396672. - Bob Selcoe, Mar 01 2016

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Programs

Formula

4^n|a(n) for n >= 1. - G. C. Greubel, Nov 21 2015
a(n) ~ c * 5^(n*(n+1)/2), where c = Product_{k>=1} (1-1/5^k) = A100222 . - Vaclav Kotesovec, Nov 21 2015
a(n) = 5^(binomial(n+1,2))*(1/5; 1/5){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 23 2015
a(n) = Product_{i=1..n} A024049(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 5^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 5^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A100222. - Amiram Eldar, May 07 2023

A027873 a(n) = Product_{i=1..n} (6^i - 1).

Original entry on oeis.org

1, 5, 175, 37625, 48724375, 378832015625, 17674407688984375, 4947685316415841015625, 8310206472731792807458984375, 83747726219216824716765369541015625
Offset: 0

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Keywords

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).
Cf. A132034.

Programs

Formula

5^n|a(n) for n>=0. - G. C. Greubel, Nov 20 2015
a(n) ~ c * 6^(n*(n+1)/2), where c = Product_{k>=1} (1-1/6^k) = A132034 = 0.805687728162164940923750215496298968917997628693... . - Vaclav Kotesovec, Nov 21 2015
a(n) = 6^(binomial(n+1,2))*(1/6;1/6){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
a(n) = Product_{i=1..n} A024062(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 6^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 6^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132034. - Amiram Eldar, May 07 2023

A027875 a(n) = Product_{i=1..n} (7^i - 1).

Original entry on oeis.org

1, 6, 288, 98496, 236390400, 3972777062400, 467389275837235200, 384914699001548351078400, 2218956256804125934296760320000, 89542886518308517126993353029713920000
Offset: 0

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Keywords

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).
Cf. A132035.

Programs

Formula

2*(10)^(2m)|a(n) where 4*m <= n <= 4*m+3, for m >= 1. - G. C. Greubel, Nov 20 2015
a(n) ~ c * 7^(n*(n+1)/2), where c = Product_{k>=1} (1-1/7^k) = A132035 = 0.836795407089037871026729798146136241352436435876... . - Vaclav Kotesovec, Nov 21 2015
a(n) = 7^(binomial(n+1,2))*(1/7;1/7){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
a(n) = Product_{i=1..n} A024075(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 7^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 7^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132035. - Amiram Eldar, May 07 2023

A027876 a(n) = Product_{i=1..n} (8^i - 1).

Original entry on oeis.org

1, 7, 441, 225351, 922812345, 30237792108615, 7926625536728661945, 16623330670976050126618695, 278893192683059452825059069034425, 37432410397693271164043156886536608251975
Offset: 0

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Author

Keywords

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).
Cf. A132036.

Programs

Formula

a(n) ~ c * 8^(n*(n+1)/2), where c = Product_{k>=1} (1-1/8^k) = A132036 = 0.859405994400702866200758580064418894909484979588... . - Vaclav Kotesovec, Nov 21 2015
7^n | a(n). - G. C. Greubel, Nov 21 2015
It appears that 7^m | a(n) iff 7^m | (7n)!. - Robert Israel, Dec 24 2015
a(n) = 8^(binomial(n+1,2))*(1/8;1/8){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
G.f. g(x) satisfies (1+x) g(x) = 1 + 8 x g(8x). - Robert Israel, Dec 24 2015
a(n) = Product_{i=1..n} A024088(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 8^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 8^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132036. - Amiram Eldar, May 07 2023

A027877 a(n) = Product_{i=1..n} (9^i - 1).

Original entry on oeis.org

1, 8, 640, 465920, 3056435200, 180476385689600, 95912370410881024000, 458745798479390789599232000, 19747501938318761090457052119040000, 7650586837724400321220283274999910891520000
Offset: 0

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Author

Keywords

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027878 (q=10), A027879 (q=11), A027880 (q=12).
Cf. A132037.

Programs

  • Magma
    [1] cat [&*[ 9^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015
    
  • Mathematica
    Abs@QPochhammer[9, 9, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)
  • PARI
    a(n) = prod(i=1, n, 9^i-1); \\ Altug Alkan, Dec 24 2015

Formula

a(n) ~ c * 3^(n*(n+1)), where c = Product_{k>=1} (1-1/9^k) = A132037 = 0.876560354035964205836019838417862010106635101174... . - Vaclav Kotesovec, Nov 21 2015
From - G. C. Greubel, Dec 24 2015: (Start)
8^n * 10^(floor(n/2))|a(n), for n>=0.
a(n) = 9^(binomial(n+1,2))*(1/9;1/9){n}, where (a;q){n} is the q-Pochhammer symbol. (End)
a(n) = Product_{i=1..n} A024101(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 9^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 9^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132037. - Amiram Eldar, May 07 2023
Showing 1-10 of 22 results. Next