A054861 -id:A054861 - cp's OEIS frontend

A085604 T(n,k) = highest power of prime(k) dividing n!, read by rows.

0, 1, 0, 1, 1, 0, 3, 1, 0, 0, 3, 1, 1, 0, 0, 4, 2, 1, 0, 0, 0, 4, 2, 1, 1, 0, 0, 0, 7, 2, 1, 1, 0, 0, 0, 0, 7, 4, 1, 1, 0, 0, 0, 0, 0, 8, 4, 2, 1, 0, 0, 0, 0, 0, 0, 8, 4, 2, 1, 1, 0, 0, 0, 0, 0, 0, 10, 5, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 10, 5, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 11, 5, 2, 2, 1, 1, 0, 0, 0

Offset: 1

Reinhard Zumkeller, Jul 07 2003

T(n,1) = A011371(n); T(n,2) = A054861(n) for n>1;
T(n,k) = number of occurrences of prime(k) as factor in numbers <= n (with repetitions);
Sum{T(n,k): 1<=k<=n} = A022559(n);
T(n, A000720(n)) = 1; T(n,k) = 0, A000720(n)T(n,k) = A115627(n,k) for n > 1 and k=1..A000720(n). - Reinhard Zumkeller, Nov 01 2013

			0;
1,0;
1,1,0;
3,1,0,0;
3,1,1,0,0;
4,2,1,0,0,0;
4,2,1,1,0,0,0;
7,2,1,1,0,0,0,0;
7,4,1,1,0,0,0,0,0;
8,4,2,1,0,0,0,0,0,0;

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Cf. A141809, A115627, A000142.

a085604 n k = a085604_tabl !! (n-2) !! (k-1)
a085604_row 1 = [0]
a085604_row n = a115627_row n ++ (take $ a062298 $ fromIntegral n) [0,0..]
a085604_tabl = map a085604_row [1..]
-- Reinhard Zumkeller, Nov 01 2013

T[n_, k_] := Module[{p = Prime[k], jm}, jm = Floor[Log[p, n]]; Sum[Quotient[n, p^j], {j, 1, jm}]];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 19 2021 *)

A090617 Exponent of highest power of 8 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

			a(8)=2 since 8! = 40320 = 8^2 * 630.

Cf. A011371, A054861, A090616, A027868, A054896.

Table[IntegerExponent[n!,8],{n,0,80}] (* Harvey P. Dale, Mar 21 2013 *)

a(n) = valuation(n!, 8); \\ Michel Marcus, Jul 10 2022

a(n) = A090622(n, 8) = floor(A011371(n)/3) = floor((floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ...)/3).

a(n) = A244413(n!) . - R. J. Mathar, Jul 08 2021

A064458 Highest power of 11 dividing n!.

Original entry on oeis.org

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

Offset: 0

Robert G. Wilson v, Oct 03 2001

Harry J. Smith, Table of n, a(n) for n = 0..1000

Cf. A011371, A054861, A054899.

Table[t = 0; p = 11; While[s = Floor[n/p]; t = t + s; s > 0, p *= 11]; t, {n, 0, 100} ]
IntegerExponent[Range[0,110]!,11] (* Harvey P. Dale, Aug 07 2017 *)

{ for (n=0, 1000, a=0; p=11; while (s = n\p, a+=s; p*=11); write("b064458.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 14 2009

a(n) = valuation(n!, 11); \\ Michel Marcus, Apr 07 2016

a(n) = floor[n/11] + floor[n/121] + floor[n/1331] + floor[n/14641] + ....

a(n) = (n-A053831(n))/10. [R. J. Mathar, Oct 17 2010]

A065039 If n in base 10 is d_1 d_2 ... d_k then a(n) = d_1 + d_1d_2 + d_1d_2d_3 + ... + d_1...d_k.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70

Offset: 0

Santi Spadaro, Nov 04 2001

a(n) = (D(n) - sod(n))/9, for n >= 1, with sod(n) the sum of digits of n, and with D(n) any of the 10 numbers given in base 10 representation by d_(nod(n)-1) d_(nod(n)-2) ... d_0 b_0, where nod(n) is the number of digits of n = d_(nod(n)-1) d_(nod(n)-2) ... d_0 in base 10, and b_0 from {0, 1, ..., 9}. E.g., D(1234) stands for any number from {12340, 12341, ..., 12349}. This corresponds the well known (and easy to prove) rule that any number after subtraction of its sum of digits is divisible by 9. In this subtraction any of the last digit b_0 leads to the same result. Some mathematical tricks are based on this rule. See the Gardner reference. - Wolfdieter Lang, May 04 2010

			a(1234)=1370 because 1+12+123+1234=1370.
With repunits: a(1234) = 4*1 + 3*11 + 2*111 + 1*1111 = 1370. - _Wolfdieter Lang_, May 04 2010

M. Gardner, Mathematische Zaubereien, Dumont, 2004, p. 39. German translation of: Mathematics, Magic and Mystery, Dover, 1956. [From Wolfdieter Lang, May 04 2010]

Harry J. Smith, Table of n, a(n) for n = 0..1000

Complement of A065438.

Cf. A067079, A067080, A067082, A054899, A054861, A067080, A098844, A132027, A005187.

import Data.List (inits)
a065039 n = sum $ map read $ tail $ inits $ show n
-- Reinhard Zumkeller, Mar 31 2011

A065039 := proc(n) local d,m: d:=convert(n,base,10): m:=nops(d): return add(op(convert(d[(m-k+1)..m], base, 10, 10^m)),k=1..m): end: seq(A065039(n),n=0..64); # Nathaniel Johnston, Jun 27 2011

a[n_] := Apply[Plus, Table[FromDigits[Take[IntegerDigits[n], k]], {k, 1, Length[IntegerDigits[n]]}]]
Table[d = IntegerDigits[n]; rd = 0; While[ Length[d] > 0, rd = rd + FromDigits[d]; d = Drop[d, -1]]; rd, {n, 0, 75} ]
f[n_] := Plus @@ NestList[ Quotient[ #, 10] &, n, Max[1, Floor@ Log[10, n]]]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 29 2010 *)
Array[Total[Table[FromDigits[Take[IntegerDigits[#],x]],{x, IntegerLength[ #]}]]&,100,0](* Harvey P. Dale, Jan 02 2016 *)

{ for (n=0, 1000, a=0; k=n; until (k==0, a+=k; k\=10); write("b065039.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 04 2009

a(n) = sum( k>=0, floor(n/10^ k)) = n+A054899(n). - Benoit Cloitre, Aug 03 2002
From Hieronymus Fischer, Aug 14 2007: (Start)
a(10*n)=10*n+a(n); a(n*10^m)=10*n*(10^m-1)/9+a(n).
a(k*10^m)=k*(10^(m+1)-1)/2, 0<=k<10, m>=0.
a(n)=10/9*n+O(log(n)), a(n+1)-a(n)=O(log(n)); this follows from the inequalities below.
a(n)<=(10*n-1)/9; equality holds for powers of 10.
a(n)>=(10*n-9)/9-floor(log_10(n)); equality holds for n=10^m-1, m>0.
lim inf (10*n/9-a(n))=1/9, for n-->oo.
lim sup (10*n/9-log_10(n)-a(n))=0, for n-->oo.
lim sup (a(n+1)-a(n)-log_10(n))=1, for n-->oo.
G.f.: sum{k>=0, x^(10^k)/(1-x^(10^k))}/(1-x).
(End)
a(n) = sum(d_(k)*RU(k+1),k=0..nod(n)-1), with the notation nod(n)and d_k given in a comment above, and RU(k)is the repunit (10^k-1)/9 (k times 1). - Wolfdieter Lang, May 04 2010

A090620 Highest power of 13 dividing n!.

Original entry on oeis.org

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, 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, 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, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8

Offset: 0

Henry Bottomley, Dec 06 2003

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)

Cf. A011371, A054861.

a:= proc(n) option remember; `if`(n=0, 0,
      a(n-1)+padic[ordp](n, 13))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Jun 20 2020

IntegerExponent[Range[0,110]!,13] (* Harvey P. Dale, Aug 22 2011 *)
FoldList[Plus, 0, IntegerExponent[Range[100], 13]] (* T. D. Noe, Apr 10 2012 *)

a(n)=my(t);while(n,t+=n\=13);t \\ Charles R Greathouse IV, Aug 06 2012

a(n) = A090622(n, 13) = A090623(n, 13) = [n/13]+[n/169]+[n/2197]+...

a(n) = n/12 + O(log n). - Charles R Greathouse IV, Aug 06 2012

A090623 Triangle of T(n,k) = [n/k] + [n/k^2] + [n/k^3] + [n/k^4] + ... for n, k > 1.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 4, 2, 1, 1, 1, 4, 2, 1, 1, 1, 1, 7, 2, 2, 1, 1, 1, 1, 7, 4, 2, 1, 1, 1, 1, 1, 8, 4, 2, 2, 1, 1, 1, 1, 1, 8, 4, 2, 2, 1, 1, 1, 1, 1, 1, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 11, 5, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 11, 6, 3, 3, 2, 2, 1, 1, 1

Offset: 2

Henry Bottomley, Dec 06 2003

			Rows start:
  1;
  1,1;
  3,1,1;
  3,1,1,1;
  4,2,1,1,1;
  4,2,1,1,1,1;
  7,2,2,1,1,1,1;
  7,4,2,1,1,1,1,1;
  8,4,2,2,1,1,1,1,1;
  ...

Zhuorui He, Table of n, a(n) for n = 2..11326
Wenguang Zhai, On the prime power factorization of n!, Journal of Number Theory, Volume 129, Issue 8, August 2009, Pages 1820-1836.

Columns include A011371, A054861, A054893, A027868, A054895, A054896, A054897, A054898, A054899, A064458, A064459, A090620, A054900.
Row sums: A078632.
Cf. A240236.

A090623[n_, k_] := Quotient[n - DigitSum[n, k], k - 1];
Table[A090623[n, k], {n, 2, 15}, {k, 2, n}] (* Paolo Xausa, Sep 02 2025 *)

T(n,k) = {my(s = 0, j = 1); while(p=n\k^j, s += p; j++); s;} \\ Michel Marcus, Feb 02 2016

T(n,k) = (n - sumdigits(n,k))/(k-1) \\ Zhuorui He, Aug 25 2025

For p prime, T(n, p) = A090622(n, p) is the number of times that p is a factor of n!.

T(n,k) = (n - A240236(n, k))/(k - 1). - Zhuorui He, Aug 25 2025

a(41) onward corrected by Zhuorui He, Aug 25 2025

A261231 a(n) = number of steps to reach 0 when starting from k = n and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..19683

Cf. A000244, A053735, A054861, A261232, A261233, A261234, A261235.

Cf. also A071542.

from sympy.ntheory.factor_ import digits
def a054861(n): return (n - sum(digits(n, 3)[1:]))/2
def a(n): return 0 if n==0 else 1 + a(2*a054861(n)) # Indranil Ghosh, May 22 2017

a(0) = 0; for n >= 1, a(n) = 1 + a(2*A054861(n)). [Note that A054861(n) = (n - A053735(n))/2, where A053735(n) = sum of digits of n, when written in base 3.]

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

Henry Bottomley, May 23 2000

nonn

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

			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.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.

Cf. A054899, A067080, A098844, A132032.

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

Table[t=0; p=8; While[s=Floor[n/p]; t=t+s; s>0, p *= 8]; t, {n,0,100}]

def A054897(n): return (n-sum(int(d) for d in oct(n)[2:]))//7 # Chai Wah Wu, Jul 09 2022

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

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

Examples added by Hieronymus Fischer, Jun 06 2012

A054898 a(n) = Sum_{k>0} floor(n/9^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11

Offset: 0

Henry Bottomley, May 23 2000

nonn

Different from the highest power of 9 dividing n!, A090618.

			a(100)=12.
a(10^3)=124.
a(10^4)=1248.
a(10^5)=12498.
a(10^6)=124996.
a(10^7)=1249997.
a(10^8)=12499996.
a(10^9)=124999997.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.

Cf. A054899, A067080, A098844, A132033.

Table[t = 0; p = 9; While[s = Floor[n/p]; t = t + s; s > 0, p *= 9]; t, {n, 0, 100} ]
Table[Sum[Floor[n/9^k],{k,n}],{n,0,100}] (* Harvey P. Dale, Jul 10 2024 *)

a(n) = floor(n/9) + floor(n/81) + floor(n/729) + floor(n/6561) + ....
a(n) = (n-A053830(n))/8.
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence:
a(n) = floor(n/9) + a(floor(n/9));
a(9*n) = n + a(n);
a(n*9^m) = n*(9^m-1)/8 + a(n).
a(k*9^m) = k*(9^m-1)/8, for 0<=k<9, m>=0.
Asymptotic behavior:
a(n) = n/8 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/8; equality holds for powers of 9.
a(n) >= (n-8)/8 - floor(log_9(n)); equality holds for n=9^m-1, m>0.
lim inf (n/8 - a(n)) =1/8, for n-->oo.
lim sup (n/8 - log_9(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_9(n)) = 0, for n-->oo.
G.f.: g(x) = sum{k>0, x^(9^k)/(1-x^(9^k))}/(1-x). (End)

Examples added by Hieronymus Fischer, Jun 06 2012

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cp's OEIS Frontend

A054893 a(n) = Sum_{j > 0} floor(n/4^j).

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A085604 T(n,k) = highest power of prime(k) dividing n!, read by rows.

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A090617 Exponent of highest power of 8 dividing n!.

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A064458 Highest power of 11 dividing n!.

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A065039 If n in base 10 is d_1 d_2 ... d_k then a(n) = d_1 + d_1d_2 + d_1d_2d_3 + ... + d_1...d_k.

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A090620 Highest power of 13 dividing n!.

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A090623 Triangle of T(n,k) = [n/k] + [n/k^2] + [n/k^3] + [n/k^4] + ... for n, k > 1.

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