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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A248732 Number of partitions of 6^n.

Original entry on oeis.org

1, 11, 17977, 15285151248481, 1398703012615213588677365804960180341, 3173477897288016617984809197028065610087051214582584606785402878333070481745149246796102615681
Offset: 0

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Crossrefs

Programs

  • Magma
    [NumberOfPartitions(6^n): n in [0..6]]; // Vincenzo Librandi, Oct 13 2014
  • Mathematica
    Table[ PartitionsP[ 6^n], {n, 0, 5}]

Formula

A248732 = A000041 o A000400. \\ M. F. Hasler, Oct 16 2014

A248733 Number of digits in the decimal expansion of the number of partitions of 6^n.

Original entry on oeis.org

1, 2, 5, 14, 37, 94, 236, 584, 1437, 3529, 8654, 21210, 51966, 127302, 311840, 763864, 1871094, 4583243
Offset: 0

Views

Author

Robert G. Wilson v, Oct 12 2014

Keywords

Crossrefs

Programs

  • Magma
    [Floor(Log(10,(NumberOfPartitions(6^n))))+1: n in [0..7]]; // Vincenzo Librandi, Oct 13 2014
    
  • Mathematica
    f[n_] := Floor[ Log[10, PartitionsP[ 6^n]] + 1]; Table[ f@n, {n, 0, 17}]
  • PARI
    a(n) = #Str(numbpart(6^n)); \\ Michel Marcus, Oct 16 2014
    
  • Python
    from sympy import npartitions
    from gmpy2 import digits
    def A248733(n): return len(digits(npartitions(6**n))) # Chai Wah Wu, Jul 15 2024

Formula

A248733 = A055642 o A000041 o A000400. \\ M. F. Hasler, Oct 16 2014

A013607 a(n) = 6^n-prevprime(6^n).

Original entry on oeis.org

1, 5, 5, 5, 17, 7, 17, 7, 7, 7, 59, 19, 17, 13, 7, 19, 49, 13, 19, 7, 23, 35, 19, 89, 17, 119, 29, 55, 5, 19, 5, 59, 85, 5, 17, 35, 59, 83, 49, 115, 19, 23, 43, 109, 103, 7, 23, 19, 7, 245, 43, 13, 5, 55, 89, 83, 143, 53, 53, 383, 7, 55, 113, 37, 5, 23, 143, 115
Offset: 1

Views

Author

James Kilfiger (mapdn(AT)csv.warwick.ac.uk)

Keywords

Crossrefs

Cf. A000400 (6^n), A104091 (largest prime <= 6^n).

Programs

  • Maple
    seq(6^i-prevprime(6^i),i=1..100);
  • Mathematica
    pp[n_]:=Module[{n6=6^n},n6-NextPrime[n6,-1]]; Array[pp,70] (* Harvey P. Dale, May 04 2011 *)
    #-NextPrime[#,-1]&/@(6^Range[70]) (* Harvey P. Dale, May 28 2023 *)

Formula

a(n) = A000400(n) - A104091(n). - Michel Marcus, Jan 13 2020

A013612 Triangle of coefficients in expansion of (1+5x)^n.

Original entry on oeis.org

1, 1, 5, 1, 10, 25, 1, 15, 75, 125, 1, 20, 150, 500, 625, 1, 25, 250, 1250, 3125, 3125, 1, 30, 375, 2500, 9375, 18750, 15625, 1, 35, 525, 4375, 21875, 65625, 109375, 78125, 1, 40, 700, 7000, 43750, 175000, 437500, 625000, 390625, 1, 45, 900, 10500, 78750, 393750, 1312500, 2812500, 3515625, 1953125
Offset: 0

Views

Author

Keywords

Comments

Mirror image of A038243. - Zerinvary Lajos, Nov 25 2007
T(n,k) equals the number of n-length words on {0,1,...,5} having n-k zeros. - Milan Janjic, Jul 24 2015

Programs

  • Maple
    T:= n-> (p-> seq(coeff(p, x, k), k=0..n))((1+5*x)^n):
    seq(T(n), n=0..10);  # Alois P. Heinz, Jun 10 2014
  • Mathematica
    row[n_] := CoefficientList[(1 + 5x)^n, x]; Table[row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Feb 13 2016 *)

Formula

G.f.: 1 / [1 - x(1+5y)].
T(n,k) = 5^k*C(n,k) = Sum_{i=n-k..n} C(i,n-k) *C(n,i) *4^(n-i). Row sums are 6^n = A000400(n). - Mircea Merca, Apr 28 2012

A014830 a(1)=1; for n > 1, a(n) = 7*a(n-1) + n.

Original entry on oeis.org

1, 9, 66, 466, 3267, 22875, 160132, 1120932, 7846533, 54925741, 384480198, 2691361398, 18839529799, 131876708607, 923136960264, 6461958721864, 45233711053065, 316635977371473, 2216451841600330, 15515162891202330, 108606140238416331, 760242981668914339, 5321700871682400396
Offset: 1

Views

Author

Keywords

Examples

			For n=5, a(5) = 1*15 + 6*20 + 6^2*15 + 6^3*6 + 6^4*1 = 3267. - _Bruno Berselli_, Nov 13 2015
		

Crossrefs

Row n=7 of A126885.

Programs

  • Maple
    a:=n->sum((7^(n-j)-1)/6,j=0..n): seq(a(n), n=1..19); # Zerinvary Lajos, Jan 15 2007
  • Mathematica
    a[1] = 1; a[n_] := 7*a[n-1]+n; Table[a[n], {n, 10}] (* Zak Seidov, Feb 06 2011 *)
    LinearRecurrence[{9, -15, 7}, {1, 9, 66}, 30] (* Harvey P. Dale, Jul 22 2013 *)
  • PARI
    Vec(x/((1 - x)^2*(1 - 7*x)) + O(x^25)) \\ Colin Barker, Jun 03 2020

Formula

a(n) = (7^(n+1) - 6*n - 7)/36. - Rolf Pleisch, Oct 19 2010
a(1)=1, a(2)=9, a(3)=66; for n > 3, a(n) = 9*a(n-1) - 15*a(n-2) + 7*a(n-3). - Harvey P. Dale, Jul 22 2013
a(n) = Sum_{i=0..n-1} 6^i*binomial(n+1,n-1-i). - Bruno Berselli, Nov 13 2015
G.f.: x/((1 - x)^2*(1 - 7*x)). - Colin Barker, Jun 03 2020
E.g.f.: exp(x)*(7*exp(6*x) - 6*x - 7)/36. - Elmo R. Oliveira, Mar 29 2025

A036292 a(n) = n*6^n.

Original entry on oeis.org

0, 6, 72, 648, 5184, 38880, 279936, 1959552, 13436928, 90699264, 604661760, 3990767616, 26121388032, 169789022208, 1097098297344, 7052774768640, 45137758519296, 287753210560512, 1828079220031488, 11577835060199424, 73123168801259520, 460675963447934976
Offset: 0

Views

Author

Keywords

Crossrefs

Cf. A000400.

Programs

Formula

a(n) = n*A000400(n). - Michel Marcus, Nov 09 2013
From Vincenzo Librandi, Apr 10 2016: (Start)
G.f.: 6*x/(1-6*x)^2.
a(n) = 12*a(n-1) - 36*a(n-2) for n > 1. (End)
From Amiram Eldar, Jul 20 2020: (Start)
Sum_{n>=1} 1/a(n) = log(6/5).
Sum_{n>=1} (-1)^(n+1)/a(n) = log(7/6). (End)
E.g.f.: 6*x*exp(6*x). - Elmo R. Oliveira, Sep 09 2024

A165424 a(1) = 1, a(2) = 6, a(n) = product of the previous terms for n >= 3.

Original entry on oeis.org

1, 6, 6, 36, 1296, 1679616, 2821109907456, 7958661109946400884391936, 63340286662973277706162286946811886609896461828096
Offset: 1

Views

Author

Jaroslav Krizek, Sep 17 2009

Keywords

Crossrefs

See A173501 for another version. - N. J. A. Sloane, Feb 23 2010

Programs

  • Mathematica
    a[1]:= 1; a[2]:= 6; a[n_]:= Product[a[j], {j,1,n-1}]; Table[a[n],{n,1, 12}] (* G. C. Greubel, Oct 19 2018 *)
  • PARI
    {a(n) = if(n==1, 1, if(n==2, 6, prod(j=1,n-1, a(j))))};
    for(n=1,10, print1(a(n), ", ")) \\ G. C. Greubel, Oct 19 2018

Formula

a(1) = 1, a(2) = 6, a(n) = Product_{i=1..n-1} a(i), n >= 3.
a(1) = 1, a(2) = 6, a(n) = A000400(2^(n-3)) = 6^(2^(n-3)), n >= 3.
a(1) = 1, a(2) = 6, a(3) = 6, a(n) = (a(n-1))^2, n >= 4.

A173501 a(1) = 2, a(2) = 3, a(n) = 6^(2^(n-3)) for n >= 3.

Original entry on oeis.org

2, 3, 6, 36, 1296, 1679616, 2821109907456, 7958661109946400884391936, 63340286662973277706162286946811886609896461828096
Offset: 1

Views

Author

Giovanni Teofilatto, Feb 20 2010

Keywords

Comments

Except for first three terms, a(n) = (a(n-1))^2.
Essentially the same as A165424. - R. J. Mathar, Feb 21 2010

Crossrefs

Programs

  • Mathematica
    Table[If[n<3, n+1, 6^(2^(n-3))], {n,10}] (* G. C. Greubel, Apr 25 2021 *)
  • PARI
    a(n) = if(n<=2, n+1, 6^(2^(n-3))) \\ Felix Fröhlich, Apr 25 2021
  • SageMath
    [2,3]+[6^(2^(n-3)) for n in (3..10)] # G. C. Greubel, Apr 25 2021
    

Formula

a(n) = Product_{i=1..n-1} a(i) with a(1) = 2 and a(2) = 3.
a(n) = 6^(2^(n-3)) = A000400(A000079(n-3)) for n>2.

Extensions

Definition corrected by Glenn Tesler, Aug 19 2017

A175457 (Digit sum of 6^n) mod n.

Original entry on oeis.org

0, 1, 0, 2, 2, 3, 1, 4, 0, 6, 1, 9, 10, 12, 3, 6, 4, 0, 6, 12, 18, 19, 3, 18, 15, 21, 18, 6, 19, 27, 6, 30, 27, 33, 30, 18, 24, 21, 6, 11, 3, 27, 24, 21, 27, 42, 21, 9, 15, 21, 12, 15, 4, 9, 52, 50, 27, 55, 48, 3, 24, 39, 36, 60, 10, 27, 33, 3, 36, 42, 66, 54, 33, 57, 63, 51, 75, 36, 51, 39, 45, 60, 1, 27, 15, 21, 27, 15, 48, 0, 69, 84, 27, 51, 39, 54, 94, 66, 45, 42
Offset: 1

Views

Author

N. J. A. Sloane, Dec 03 2010

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Mod[Total[IntegerDigits[6^n]],n],{n,120}] (* Harvey P. Dale, May 24 2016 *)

Extensions

Corrected by N. J. A. Sloane, May 23 2016 at the suggestion of Harvey P. Dale.

A202821 Position of 6^n among 3-smooth numbers A003586.

Original entry on oeis.org

1, 5, 14, 26, 43, 64, 89, 119, 153, 191, 233, 279, 330, 385, 444, 507, 575, 646, 722, 802, 886, 975, 1067, 1164, 1266, 1371, 1481, 1595, 1713, 1835, 1961, 2092, 2227, 2366, 2509, 2657, 2809, 2965, 3125, 3289, 3458, 3630, 3807, 3989, 4174, 4364, 4558, 4756
Offset: 0

Views

Author

Zak Seidov, Dec 25 2011

Keywords

Examples

			a(0) = 1 because A003586(1) = 6^0 = 1.
a(1) = 5 because A003586(5) = 6^1 = 6.
a(2) = 14 because A003586(14) = 6^2 = 36.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[Floor[Log[3, 6^n/2^i]] + 1, {i, 0, Log2[6^n]}]; Array[a, 50, 0] (* Amiram Eldar, Jul 15 2023 *)
  • Python
    # uses imports/function in A372401
    print(list(islice(A372401gen(p=3), 1000))) # Michael S. Branicky, Jun 06 2024
    
  • Python
    from sympy import integer_log
    def A202821(n): return 1+n*(n+1)+sum((m:=3**i).bit_length()+((1<Chai Wah Wu, Oct 22 2024

Formula

A003586(a(n)) = 6^n, for n >= 0.
a(n) ~ (log(6))^2/(log(3)*log(4))*n^2 = 2.1079...*n^2.
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