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.

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A218722 a(n) = (19^n-1)/18.

Original entry on oeis.org

0, 1, 20, 381, 7240, 137561, 2613660, 49659541, 943531280, 17927094321, 340614792100, 6471681049901, 122961939948120, 2336276859014281, 44389260321271340, 843395946104155461, 16024522975978953760, 304465936543600121441
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 19 (A001029); q-integers for q=19: diagonal k=1 in triangle A022183.
Partial sums are in A014903. Also, the sequence is related to A014936 by A014936(n) = n*a(n)-sum(a(i), i=0..n-1) for n>0. - Bruno Berselli, Nov 06 2012

Crossrefs

Programs

Formula

a(n) = floor(19^n/18).
G.f.: x/((1-x)*(1-19*x)). - Bruno Berselli, Nov 06 2012
a(n) = 20*a(n-1) - 19*a(n-2). - Vincenzo Librandi, Nov 07 2012
E.g.f.: exp(10*x)*sinh(9*x)/9. - Stefano Spezia, Mar 11 2023

A064108 a(n) = (20^n - 1)/19.

Original entry on oeis.org

0, 1, 21, 421, 8421, 168421, 3368421, 67368421, 1347368421, 26947368421, 538947368421, 10778947368421, 215578947368421, 4311578947368421, 86231578947368421, 1724631578947368421, 34492631578947368421, 689852631578947368421, 13797052631578947368421, 275941052631578947368421
Offset: 0

Views

Author

Jason Earls, Sep 17 2001

Keywords

Comments

Partial sums of powers of 20 (A009964), q-integers for q=20: diagonal k=1 in triangle A022184.
Partial sums are in A014904. Also, the sequence is related to A014937 by A014937(n) = n*a(n)-Sum_{i=0..n-1} a(i), for n>0. - Bruno Berselli, Nov 06 2012
For n >= 1, a(n) is the total number of holes in a certain box fractal (start with 20 boxes, 1 hole) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 28 2015

Examples

			From _N. J. A. Sloane_, Nov 04 2014: Can also be obtained by writing powers of 2 in a staggered array and adding them (cf. A249604). For example, a(9) is:
..........1
.........2
........4
.......8
.....16
....32
...64
.128
256
-----------
26947368421
		

Crossrefs

Programs

  • Maple
    a:=n->sum(20^(n-j), j=0..n): seq(a(n), n=0..15); # Zerinvary Lajos, Feb 11 2007
  • Mathematica
    (20^Range[20]-1)/19 (* or *) NestList[20#+1&,1,20] (* Harvey P. Dale, Oct 04 2012 *)
  • Maxima
    A064108(n):=(20^n-1)/19$ makelist(A064108(n),n,1,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    for (n=0, 100, write("b064108.txt", n, " ", (20^n - 1)/19))  \\ Harry J. Smith, Sep 07 2009
    
  • PARI
    A064108(n)=20^n\19  \\ M. F. Hasler, Nov 04 2012
    
  • Sage
    [gaussian_binomial(n,1,20) for n in range(1,17)] # Zerinvary Lajos, May 29 2009
    

Formula

a(n) = 20*a(n-1) + 1, with a(0)=0. - Vincenzo Librandi, Aug 07 2010
a(0)=0, a(1)=1, a(n) = 21*a(n-1) - 20*a(n-2). - Harvey P. Dale, Oct 04 2012
a(n) = floor(20^n/19). - M. F. Hasler, Nov 04 2012
G.f.: x/((1 - x)*(1 - 20*x)). - Bruno Berselli, Nov 06 2012
E.g.f.: exp(x)*(exp(19*x) - 1)/19. - Stefano Spezia, Mar 23 2023

Extensions

Edited and extended to offset 0 by M. F. Hasler, Nov 04 2012

A135518 Generalized repunits in base 15.

Original entry on oeis.org

1, 16, 241, 3616, 54241, 813616, 12204241, 183063616, 2745954241, 41189313616, 617839704241, 9267595563616, 139013933454241, 2085209001813616, 31278135027204241, 469172025408063616, 7037580381120954241, 105563705716814313616, 1583455585752214704241
Offset: 1

Views

Author

Julien Peter Benney (jpbenney(AT)gmail.com), Feb 19 2008

Keywords

Comments

Primes in this sequence are given in A006033.
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=15, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Partial sums are in A014898. Also, the sequence is related to A014930 by A014930(n) = n*a(n) - Sum_{i=1..n-1}( a(i) ). - Bruno Berselli, Nov 06 2012

Examples

			For n=4, a(4) = 15^3+15^2+15^1+1 = 3375+225+15+1 = 3616.
For n=6, a(6) = 1*6 + 14*15 + 14^2*20 + 14^3*15 + 14^4*6 + 14^5*1 = 813616. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[PadRight[{},n,1],15],{n,20}] (* or *) LinearRecurrence[{16,-15},{1,16},20] (* Harvey P. Dale, Jul 08 2013 *)
  • Maxima
    A135518(n):=(15^n-1)/14$ makelist(A135518(n),n,1,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=(15^n-1)/14 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Python
    def a(n): return int('1'*n, 15)
    print([a(n) for n in range(1, 20)]) # Michael S. Branicky, Jan 16 2021
  • Sage
    [gaussian_binomial(n,1,15) for n in range(1,15)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [(15^n-1)/14 for n in (1..30)] # Bruno Berselli, Nov 12 2015
    

Formula

a(n) = (15^n - 1)/14.
a(n) = 15*a(n-1) + 1 with n>1, a(1)=1. - Vincenzo Librandi, Aug 03 2010
G.f.: x/((1-x)*(1-15*x)). - Bruno Berselli, Nov 07 2012
a(1)=1, a(2)=16; for n>2, a(n) = 16*a(n-1) - 15*a(n-2). - Harvey P. Dale, Jul 08 2013
a(n) = Sum_{i=0...n-1} 14^i*binomial(n,n-1-i). - Bruno Berselli, Nov 12 2015
E.g.f.: (1/14)*(exp(15*x) - exp(x)). - G. C. Greubel, Oct 17 2016

A135519 Generalized repunits in base 14.

Original entry on oeis.org

1, 15, 211, 2955, 41371, 579195, 8108731, 113522235, 1589311291, 22250358075, 311505013051, 4361070182715, 61054982558011, 854769755812155, 11966776581370171, 167534872139182395, 2345488209948553531
Offset: 1

Views

Author

Julien Peter Benney (jpbenney(AT)gmail.com), Feb 19 2008

Keywords

Comments

Primes are given in A006032.
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=14, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010

Examples

			a(4) = 2955 because (14^4-1)/13 = 38416/13 = 2955.
For n=6, a(6) = 1*6 + 13*15 + 169*20 + 2197*15 + 28561*6 + 371293*1 = 579195. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[PadRight[{}, n, 1], 14], {n, 20}] (* or *) LinearRecurrence[{15, -14}, {1, 15}, 20] (* Harvey P. Dale, Aug 29 2016 *)
  • Maxima
    A135519(n):=(14^n-1)/13$ makelist(A135519(n),n,1,30); /* Martin Ettl, Nov 05 2012 */
  • Sage
    [gaussian_binomial(n,1,14) for n in range(1,15)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [(14^n-1)/13 for n in (1..30)] # Bruno Berselli, Nov 12 2015
    

Formula

a(n) = (14^n - 1)/13.
a(n) = 14*a(n-1) + 1 for n>1, a(1)=1. - Vincenzo Librandi, Aug 03 2010
a(n) = Sum_{i=0..n-1} 13^i*binomial(n,n-1-i). - Bruno Berselli, Nov 12 2015
From G. C. Greubel, Oct 17 2016: (Start)
G.f.: x/((1-x)*(1-14*x)).
E.g.f.: (1/13)*(exp(14*x) - exp(x)). (End)

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264
Offset: 0

Views

Author

Keywords

Comments

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Crossrefs

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011
    
  • Mathematica
    LinearRecurrence[{2,-10}, {0,1}, 50]
  • PARI
    a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016
    
  • SageMath
    [lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

Formula

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A218724 a(n) = (21^n - 1)/20.

Original entry on oeis.org

0, 1, 22, 463, 9724, 204205, 4288306, 90054427, 1891142968, 39714002329, 833994048910, 17513875027111, 367791375569332, 7723618886955973, 162195996626075434, 3406115929147584115, 71528434512099266416, 1502097124754084594737, 31544039619835776489478
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 21 (A009965); q-integers for q=21: diagonal k=1 in triangle A022185.
Partial sums are in A014905. Also, the sequence is related to A014938 by A014938(n) = n*a(n) - Sum_{i=0..n-1} a(i) for n > 0. - Bruno Berselli, Nov 06 2012
For n >= 1, 4*a(n) is the total number of holes in a certain box fractal (start with 21 boxes, 4 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 27 2015

Crossrefs

Programs

Formula

a(n) = floor(21^n/20).
G.f.: x/((1-x)*(1-21*x)). - Bruno Berselli, Nov 06 2012
a(n) = 22*a(n-1) - 21*a(n-2). - Vincenzo Librandi, Nov 07 2012
a(n) = 21*a(n-1) + 1. - Kival Ngaokrajang, Jan 27 2015
a(n) = a(n-1) + 21^(n-1), n >= 1, a(0) = 0. - Wolfdieter Lang, Feb 02 2015
E.g.f.: exp(11*x)*sinh(10*x)/10. - Elmo R. Oliveira, Aug 29 2024

A218734 a(n) = (31^n - 1)/30.

Original entry on oeis.org

0, 1, 32, 993, 30784, 954305, 29583456, 917087137, 28429701248, 881320738689, 27320942899360, 846949229880161, 26255426126284992, 813918209914834753, 25231464507359877344, 782175399728156197665, 24247437391572842127616, 751670559138758105956097
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 31 (A009975).

Crossrefs

Programs

  • Magma
    [n le 2 select n-1 else 32*Self(n-1)-31*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012
    
  • Mathematica
    LinearRecurrence[{32, -31}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)
  • Maxima
    A218734(n):=(31^n-1)/30$
    makelist(A218734(n),n,0,30); /* Martin Ettl, Nov 07 2012 */
  • PARI
    a(n)=31^n\30
    

Formula

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 31*x)).
a(n) = 32*a(n-1) - 31*a(n-2) for n > 1.
a(n) = floor(31^n/30). (End)
E.g.f.: exp(16*x)*sinh(15*x)/15. - Stefano Spezia, Mar 11 2023

A132469 a(n) = (2^(5*n) - 1)/31.

Original entry on oeis.org

0, 1, 33, 1057, 33825, 1082401, 34636833, 1108378657, 35468117025, 1134979744801, 36319351833633, 1162219258676257, 37191016277640225, 1190112520884487201, 38083600668303590433, 1218675221385714893857, 38997607084342876603425, 1247923426698972051309601
Offset: 0

Views

Author

A.K. Devaraj, Aug 22 2007

Keywords

Comments

Partial sums of powers of 32 (A009976), a.k.a. q-numbers for q=32. - M. F. Hasler, Nov 05 2012

References

  • A. K. Devaraj, "Minimum Universal Exponent Generalisation of Fermat's Theorem", in ISSN #1550-3747, Proceedings of Hawaii Intl Conference on Statistics, Mathematics & Related Fields, 2004.

Crossrefs

Programs

Formula

a(n) = (32^n - 1)/31 = floor(32^n/31) = Sum_{k=0..n} 32^k. - M. F. Hasler, Nov 05 2012
G.f.: x/((1 - x)*(1 - 32*x)). - Bruno Berselli, Nov 06 2012
E.g.f.: exp(x)*(exp(31*x) - 1)/31. - Stefano Spezia, Mar 23 2023

Extensions

Edited and extended by Robert G. Wilson v, Aug 22 2007
Edited and extended to offset 0 by M. F. Hasler, Nov 05 2012

A218721 a(n) = (18^n-1)/17.

Original entry on oeis.org

0, 1, 19, 343, 6175, 111151, 2000719, 36012943, 648232975, 11668193551, 210027483919, 3780494710543, 68048904789775, 1224880286215951, 22047845151887119, 396861212733968143, 7143501829211426575, 128583032925805678351
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 18 (A001027), q-integers for q=18: diagonal k=1 in triangle A022182.
Partial sums are in A014901. Also, the sequence is related to A014935 by A014935(n) = n*a(n) - Sum_{i=0..n-1} a(i), for n>0. - Bruno Berselli, Nov 06 2012
From Bernard Schott, May 06 2017: (Start)
Except for 0, 1 and 19, all terms are Brazilian repunits numbers in base 18, and so belong to A125134. From n = 3 to n = 8286, all terms are composite. See link "Generalized repunit primes".
As explained in the extensions of A128164, a(25667) = (18^25667 - 1)/17 would be (is) the smallest prime in base 18. (End)

Examples

			a(3) = (18^3 - 1)/17 = 343 = 7 * 49; a(6) = (18^6 - 1)/17 = 2000719 = 931 * 2149. - _Bernard Schott_, May 01 2017
		

Crossrefs

Programs

Formula

a(n) = floor(18^n/17).
G.f.: x/((1-x)*(1-18*x)). - Bruno Berselli, Nov 06 2012
a(n) = 19*a(n-1) - 18*a(n-2). - Vincenzo Librandi, Nov 07 2012
E.g.f.: exp(x)*(exp(17*x) - 1)/17. - Stefano Spezia, Mar 11 2023

A218753 a(n) = (49^n - 1)/48.

Original entry on oeis.org

0, 1, 50, 2451, 120100, 5884901, 288360150, 14129647351, 692352720200, 33925283289801, 1662338881200250, 81454605178812251, 3991275653761800300, 195572507034328214701, 9583052844682082520350, 469569589389422043497151, 23008909880081680131360400
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 49 (A087752).

Crossrefs

Programs

Formula

G.f.: x/((1-x)*(1-49*x)). - Vincenzo Librandi, Nov 08 2012
a(n) = 50*a(n-1) - 49*a(n-2) with a(0)=0, a(1)=1. - Vincenzo Librandi, Nov 08 2012
a(n) = 49*a(n-1) + 1 with a(0)=0. - Vincenzo Librandi, Nov 08 2012
a(n) = floor(49^n/48). - Vincenzo Librandi, Nov 08 2012
E.g.f.: exp(25*x)*sinh(24*x)/24. - Elmo R. Oliveira, Aug 27 2024
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