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 1-4 of 4 results.

A132033 Product{0<=k<=floor(log_9(n)), floor(n/9^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 36, 38, 40, 42, 44, 46, 48, 50, 52, 81, 84, 87, 90, 93, 96, 99, 102, 105, 144, 148, 152, 156, 160, 164, 168, 172, 176, 225, 230, 235, 240, 245, 250, 255, 260, 265, 324, 330, 336, 342, 348, 354, 360, 366, 372
Offset: 1

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-9 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(85)=floor(85/9^0)*floor(85/9^1)*floor(85/9^2)=85*9*1=765; a(88)=792 since 88=107(base-9) and so a(88)=107*10*1(base-9)=88*9*1=792.

Crossrefs

Cf. A048651, A132025, A132037, A000217.
For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132032(p=8), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

  • Mathematica
    Table[Product[Floor[n/9^k],{k,0,Floor[Log[9,n]]}],{n,62}] (* James C. McMahon, Mar 03 2025 *)

Formula

Recurrence: a(n)=n*a(floor(n/9)); a(n*9^m)=n^m*9^(m(m+1)/2)*a(n).
a(k*9^m)=k^(m+1)*9^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_9(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_9(n)))/9^((1+floor(log_9(n)))*floor(log_9(n))/2); equality holds for n=k*9^m, 0=0. b(n) can also be written n^(1+floor(log_9(n)))/9^A000217(floor(log_9(n))).
Also: a(n)<=3^(1/4)*n^((1+log_9(n))/2)=1.316074013...*9^A000217(log_9(n)), equality holds for n=3*9^m, m>=0.
a(n)>c*b(n), where c=0.4689451783670236932832800... (see constant A132024).
Also: a(n)>c*2^((1-log_9(2))/2)*n^((1+log_9(n))/2)=0.4689451783670...*1.267747616...*9^A000217(log_9(n)).
lim inf a(n)/b(n)=0.4689451783670236932832800..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_9(n))/2)=0.4689451783670236932832800...*sqrt(2)/2^log_9(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_9(n))/2)=3^(1/4)=1.316074013..., for n-->oo.
lim inf a(n)/a(n+1)=0.4689451783670236932832800... for n-->oo (see constant A132025).

A054897 a(n) = Sum_{k>0} floor(n/8^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12
Offset: 0

Views

Author

Henry Bottomley, May 23 2000

Keywords

Comments

Different from the highest power of 8 dividing n!, A090617.

Examples

			a(100) = 13.
a(10^3) = 141.
a(10^4) = 1427.
a(10^5) = 14284.
a(10^6) = 142855.
a(10^7) = 1428569.
a(10^8) = 14285710.
a(10^9) = 142857138.

Links

Crossrefs

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A054899, A067080, A098844, A132032.

Programs

  • Magma
    m:=8;
function a(n) // a = A054897
  if n eq 0 then return n;
  else return a(Floor(n/m)) + Floor(n/m);
  end if;
end function;
[a(n): n in [0..103]]; // G. C. Greubel, Apr 28 2023
  • Mathematica
    Table[t=0; p=8; While[s=Floor[n/p]; t=t+s; s>0, p *= 8]; t, {n,0,100}]
  • Python
    def A054897(n): return (n-sum(int(d) for d in oct(n)[2:]))//7 # Chai Wah Wu, Jul 09 2022
  • SageMath
    m=8 # a = A054897
def a(n): return 0 if (n==0) else a(n//m) + (n//m)
[a(n) for n in range(104)] # G. C. Greubel, Apr 28 2023

Formula

a(n) = floor(n/8) + floor(n/64) + floor(n/512) + floor(n/4096) + ....
a(n) = (n - A053829(n))/7.
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence:
a(n) = floor(n/8) + a(floor(n/8));
a(8*n) = n + a(n);
a(n*8^m) = n*(8^m-1)/7 + a(n).
a(k*8^m) = k*(8^m-1)/7, for 0 <= k < 8, m >= 0.
Asymptotic behavior:
a(n) = n/7 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/7; equality holds for powers of 8.
a(n) >= (n-7)/7 - floor(log_8(n)); equality holds for n=8^m-1, m>0.
lim inf (n/7 - a(n)) = 1/7, for n -> oo.
lim sup (n/7 - log_8(n) - a(n)) = 0, for n -> oo.
lim sup (a(n+1) - a(n) - log_8(n)) = 0, for n -> oo.
G.f.: g(x) = ( Sum_{k>0} x^(8^k)/(1-x^(8^k)) )/(1-x). (End)
Partial sums of A244413. - R. J. Mathar, Jul 08 2021

Extensions

Examples added by Hieronymus Fischer, Jun 06 2012

A132024 Decimal expansion of Product_{k>=0} (1-1/(2*8^k)).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.46456888368647639098...

Crossrefs

Cf. A048651, A098844, A067080, A132019, A132026, A132032, A132036, A000217, A013730.

Programs

  • Mathematica
    RealDigits[QPochhammer[1/2,1/8],10,120][[1]] (* Harvey P. Dale, May 23 2011 *)
  • PARI
    prodinf(k=0, 1 - 1/(2*8^k)) \\ Amiram Eldar, May 09 2023

Formula

Equals lim inf_{n->oo} Product_{k=0..floor(log_8(n))} floor(n/8^k)*8^k/n.
Equals lim inf_{n->oo} A132032(n)/n^(1+floor(log_8(n)))*8^(1/2*(1+floor(log_8(n)))*floor(log_8(n))).
Equals lim inf_{n->oo} A132032(n)/n^(1+floor(log_8(n)))*8^A000217(floor(log_8(n))).
Equals (1/2)*exp(-Sum_{n>0} 8^(-n)*Sum_{k|n} 1/(k*2^k)).
Equals lim inf_{n->oo} A132032(n)/A132032(n+1).
Equals Product_{n>=0} (1 - 1/A013730(n)). - Amiram Eldar, May 09 2023

Extensions

Name corrected by Amiram Eldar, May 09 2023

A054900 a(n) = Sum_{j >= 1} floor(n/16^j).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6
Offset: 0

Views

Author

Henry Bottomley, May 23 2000

Keywords

Links

Crossrefs

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A053836, A054897, A054899, A067080, A098844, A132032.

Programs

  • Magma
    m:=16;
function a(n) // a = A054900, m = 16
  if n eq 0 then return 0;
  else return a(Floor(n/m)) + Floor(n/m);
  end if; end function;
[a(n): n in [0..127]]; // G. C. Greubel, Apr 28 2023
  • Mathematica
    a[n_, m_]:= If[n==0, 0, a[Floor[n/m], m] +Floor[n/m]];
Table[a[n, 16], {n,0,127}] (* G. C. Greubel, Apr 28 2023 *)
  • SageMath
    m=16 # a = A054900
def a(n): return 0 if (n==0) else a(n//m) + (n//m)
[a(n) for n in range(128)] # G. C. Greubel, Apr 28 2023

Formula

a(n) = (n - A053836(n))/15.
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence:
a(n) = a(floor(n/16)) + floor(n/16).
a(16*n) = a(n) + n.
a(n*16^m) = a(n) + n*(16^m - 1)/15.
a(k*16^m) = k*(16^m - 1)/15, for 0 <= k < 16, m>=0.
Asymptotic behavior:
a(n) = n/15 + O(log(n)).
a(n+1) - a(n) = O(log(n)) (this follows from the inequalities below).
a(n) <= (n-1)/15; equality holds for powers of 16.
a(n) >= (n-15)/15 - floor(log_16(n)); equality holds for n = 16^m - 1, m > 0.
Limits:
lim inf (n/15 - a(n)) = 1/15, for n --> oo.
lim sup (n/15 - log_16(n) - a(n)) = 0, for n --> oo.
lim sup (a(n+1) - a(n) - log_16(n)) = 0, for n --> oo.
Series:
G.f.: (1/(1-x))*Sum_{k > 0} x^(16^k)/(1-x^(16^k)). (End)
Showing 1-4 of 4 results.

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