A098844 -id:A098844 - cp's OEIS frontend

A132263 Product{0<=k<=floor(log_11(n)), floor(n/11^k)}, n>=1.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 275, 280, 285, 290, 295, 300, 305, 310

Offset: 1

Hieronymus Fischer, Aug 20 2007

If n is written in base-11 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).

			a(50)=floor(50/11^0)*floor(50/11^1)=50*4=200; a(63)=315 since 63=58(base-11) and so a(63)=58*5(base-11)=63*5=315.

Cf. A048651, A132019-A132026, A132265, A132267, 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)-A132033(p=9), A067080(p=10), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Recurrence: a(n)=n*a(floor(n/11)); a(n*11^m)=n^m*11^(m(m+1)/2)*a(n).

a(k*11^m)=k^(m+1)*11^(m(m+1)/2), for 0

Asymptotic behavior: a(n)=O(n^((1+log_11(n))/2)); this follows from the inequalities below.

a(n)<=b(n), where b(n)=n^(1+floor(log_11(n)))/p^((1+floor(log_11(n)))*floor(log_11(n))/2); equality holds for n=k*11^m, 0=0. b(n) can also be written n^(1+floor(log_11(n)))/11^A000217(floor(log_11(n))).

Also: a(n)<=3^((1-log_11(3))/2)*n^((1+log_11(n))/2)=1.346673852...^((1-log_11(3))/2)*11^A000217(log_11(n)), equality holds for n=3*11^m, m>=0.

a(n)>c*b(n), where c=0.4751041275076031053975644472... (see constant A132265).

Also: a(n)>c*(sqrt(2)/2^log_11(sqrt(2)))*n^((1+log_11(n))/2)=0.607848303...*11^00217(log_11(n)).

lim inf a(n)/b(n)=0.4751041275076031053975644472..., for n-->oo.

lim sup a(n)/b(n)=1, for n-->oo.

lim inf a(n)/n^((1+log_p(n))/2)=0.4751041275076031...*sqrt(2)/2^log_11(sqrt(2)), for n-->oo.

lim sup a(n)/n^((1+log_p(n))/2)=sqrt(3)/3^log_11(sqrt(3))=1.346673852..., for n-->oo.

lim inf a(n)/a(n+1)=0.4751041275076031053975644472... for n-->oo (see constant A132265).

A132264 Product{0<=k<=floor(log_12(n)), floor(n/12^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, 18, 19, 20, 21, 22, 23, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232, 236, 300, 305, 310, 315

Offset: 1

Hieronymus Fischer, Aug 20 2007

If n is written in base-12 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).

			a(50)=floor(50/12^0)*floor(50/12^1)=50*4=200.
a(65)=325 since 65=55(base-12) and so a(65)=55*5(base-12)=65*5=325.

Cf. A048651, A132019-A132026, A132265, A132267, A000217, A002378.
For the product of terms floor(n/p^k) for p=2 to p=11 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

The following formulas are given for a general parameter p considering the product of terms floor(n/p^k) for 0<=k<=floor(log_p(n)), where p=12 for this sequence.
Recurrence: a(n)=n*a(floor(n/p)); a(n*p^m)=n^m*p^(m(m+1)/2)*a(n).
a(k*p^m)=k^(m+1)*p^(m(m+1)/2), for 0Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_p(n)))/p^((1+floor(log_p(n)))*floor(log_p(n))/2); equality holds for n=k*p^m, 0=0. b(n) can also be written n^(1+floor(log_p(n)))/p^A000217(floor(log_p(n))).
Also: a(n)<=q^((1-log_p(q))/2)*n^((1+log_p(n))/2)=q^((1-log_p(q))/2)*p^A000217(log_p(n)), equality holds for n=q*p^m, m>=0, where q=floor(sqrt(p)+1/2). Also, equality holds for n=(q+1)*p^m, provided p is a A002378-number (in this case we have p=q*(q+1) and so q^((1-log_p(q))/2)=(q+1)^((1-log_p(q+1))/2)).
a(n)>c*b(n), where c=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... (for p=12 see constant A132265).
Also: a(n)>c*(sqrt(2)/2^log_p(sqrt(2)))*n^((1+log_p(n))/2)=0.612870619...*p^A000217(log_p(n)), (p=12).
lim inf a(n)/b(n)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380198334286365820..., for n-->oo (for p=12 see constant A132265).
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_p(n))/2)=(sqrt(2)/2^log_p(sqrt(2)))*product{k>0, 1-1/(2*p^k)}=0.612870619..., for n-->oo, (p=12).
lim sup a(n)/n^((1+log_p(n))/2)=sqrt(q)/q^log_p(sqrt(q))=1.358593737..., for n-->oo, (p=12, q=round(sqrt(p))=3).
lim inf a(n)/a(n+1)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... for n-->oo (for p=12 see constant A132265).

A132269 a(n) = Product_{k>=0} (1 + floor(n/2^k)).

Original entry on oeis.org

1, 2, 6, 8, 30, 36, 56, 64, 270, 300, 396, 432, 728, 784, 960, 1024, 4590, 4860, 5700, 6000, 8316, 8712, 9936, 10368, 18200, 18928, 21168, 21952, 27840, 28800, 31744, 32768, 151470, 156060, 170100, 174960, 210900, 216600, 234000, 240000, 340956, 349272, 374616

Offset: 0

Hieronymus Fischer, Aug 20 2007

nonn

If n is written in base 2 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 (1+d(m)d(m-1)d(m-2)...d(2)d(1)d(0))*(1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).
From Gary W. Adamson, Aug 25 2016: (Start)
Given the following production matrix M =
  1, 0, 0, 0, 0, ...
  2, 0, 0, 0, 0, ...
  0, 3, 0, 0, 0, ...
  0, 4, 0, 0, 0, ...
  0, 0, 5, 0, 0, ...
  0, 0, 6, 0, 0, ...
  0, 0, 0, 7, 0, ...
  ...
the sequence is the left-shifted vector as lim_{n->infinity} M^n. (End)

			a(10) = (1 + floor(10/2^0))*(1 + floor(10/2^1))*(1 + floor(10/2^2))*(1 + floor(10/2^3)) = 11*6*3*2 = 396;
a(17) = 4860 since 17 = 10001_2 and so a(17) = (1+10001_2)*(1+1000_2)*(1+100_2)*(1+10_2)*(1+1) = 18*9*5*3*2 = 4860.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A048651, A081845, A132270, A132271(p=10), A132272, A132327(p=3), A132328.
For formulas regarding a general parameter p (i.e., terms 1+floor(n/p^k)) see A132271.
For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

[1] cat [n le 1 select 2 else (1+n)*Self(Floor(n/2)): n in [1..50]]; // Vincenzo Librandi, Aug 26 2016

f:= proc(n) option remember; (1+n)*procname(floor(n/2)) end proc:
f(0):= 1:
map(f, [$0..100]); # Robert Israel, Aug 26 2016

Table[Product[1 + Floor[2 n/2^k], {k, 2 n}], {n, 0, 42}] (* or *)
Table[Function[w, Times @@ Map[1 + FromDigits[PadRight[w, #], 2] &, Range@ Length@ w]]@ IntegerDigits[n, 2], {n, 0, 42}] (* Michael De Vlieger, Aug 26 2016 *)

Recurrence: a(n)=(1+n)*a(floor(n/2)); a(2n)=(1+2n)*a(n); a(n*2^m) = (Product_{k=1..m} (1 + n*2^k))*a(n).
a(2^m-1) = 2^(m*(m+1)/2), a(2^m) = 2^(m*(m+1)/2)*Product_{k=0..m} (1 + 1/2^k), m>=1.
a(n) = A132270(2n) = (1+n)*A132270(n).
Asymptotic behavior: a(n) = O(n^((1+log_2(n))/2)); this follows from the inequalities below.
a(n) <= A098844(n)*Product_{k=0..floor(log_2(n))} (1 + 1/2^k).
a(n) >= A098844(n)/Product_{k=1..floor(log_2(n))} (1 - 1/2^k).
a(n) < c*n^((1+log_2(n))/2) = c*2^A000217(log_2(n)), where c = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627... (see constant A081845).
a(n) > n^((1+log_2(n))/2) = 2^A000217(log_2(n)),
lim sup a(n)/A098844(n) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627..., for n->oo (see constant A081845).
lim inf a(n)/A098844(n) = 1/Product_{k>=1} (1 - 1/2^k) = 1/0.288788095086602421..., for n->oo (see constant A048651).
lim inf a(n)/n^((1+log_2(n))/2) = 1, for n->oo.
lim sup a(n)/n^((1+log_2(n))/2) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627..., for n->oo (see constant A081845).
lim inf a(n+1)/a(n) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627... for n->oo (see constant A081845).
G.f. g(x) satisfies g(x) = (1+2x)*g(x^2) + 2*x^2*(1+x)*g'(x^2). - Robert Israel, Aug 26 2016

A132327 a(n) = Product{k>=0} (1 + floor(n/3^k)).

Original entry on oeis.org

1, 2, 3, 8, 10, 12, 21, 24, 27, 80, 88, 96, 130, 140, 150, 192, 204, 216, 399, 420, 441, 528, 552, 576, 675, 702, 729, 2240, 2320, 2400, 2728, 2816, 2904, 3264, 3360, 3456, 4810, 4940, 5070, 5600, 5740, 5880, 6450, 6600, 6750, 8832, 9024, 9216, 9996, 10200

Offset: 0

Hieronymus Fischer, Aug 20 2007

If n is written in base-3 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 (1+d(m)d(m-1)d(m-2)...d(2)d(1)d(0))*(1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).

			a(12)=(1+floor(12/3^0))*(1+floor(12/3^1))*(1+floor(12/3^2))=13*5*2=130; a(20)=441 since 20=202(base-3) and so
a(20)=(1+202)*(1+20)*(1+2)(base-3)=21*7*3=441.

Cf. A100220, A132027, A132038, A132264, A132269(for p=2), A132271(for p=10).
For formulas regarding a general parameter p (i.e. terms 1+floor(n/p^k)) see A132271.
For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Table[Product[1+Floor[n/3^k],{k,0,n}],{n,0,49}] (* James C. McMahon, Mar 07 2025 *)

Recurrence: a(n)=(1+n)*a(floor(n/3)); a(3n)=(1+3n)*a(n); a(n*3^m)=product{1<=k<=m, 1+n*3^k}*a(n).
a(k*3^m-j)=(k*3^m-j+1)*3^m*p^(m(m-1)/2), for 0=1, a(3^m)=3^(m(m+1)/2)*product{0<=k<=m, 1+1/3^k}, m>=1.
a(n)=A132328(3*n)=(1+n)*A132328(n).
Asymptotic behavior: a(n)=O(n^((1+log_3(n))/2)); this follows from the inequalities below.
a(n)<=A132027(n)*product{0<=k<=floor(log_3(n)), 1+1/3^k}.
a(n)>=A132027(n)/product{1<=k<=floor(log_3(n)), 1-1/3^k}.
a(n)A000217(log_3(n)), where c=product{k>=0, 1+1/p^k}=3.12986803713402307587769821345767... (see constant A132323).
a(n)>n^((1+log_3(n))/2)=3^A000217(log_3(n)).
lim sup a(n)/A132027(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf a(n)/A132027(n)=1/product{k>0, 1-1/3^k}=1/0.560126077927948944969792243314140014..., for n-->oo (see constant A100220).
lim inf a(n)/n^((1+log_3(n))/2)=1, for n-->oo.
lim sup a(n)/n^((1+log_3(n))/2)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf a(n+1)/a(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... for n-->oo (see constant A132323).

A054896 a(n) = Sum_{k>0} floor(n/7^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13

Offset: 0

Henry Bottomley, May 23 2000

nonn

Exponent of the highest power of 7 dividing n!.

			  a(10^0) = 0.
  a(10^1) = 1.
  a(10^3) = 16.
  a(10^3) = 164.
  a(10^4) = 1665.
  a(10^5) = 16662.
  a(10^6) = 166664.
  a(10^7) = 1666661.
  a(10^8) = 16666662.
  a(10^9) = 166666661

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

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A053828, A054893, A054895, A054899, A067080, A098844, A132031, A214411.
Cf. A000142, A214411.

function A054896(n)
  if n eq 0 then return n;
  else return A054896(Floor(n/7)) + Floor(n/7);
  end if; return A054896;
end function;
[A054896(n): n in [0..100]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=7; While[s=Floor[n/p]; t=t+s; s>0, p *= 7]; t, {n,0,100}]

a(n)=(n-sumdigits(n, 7))\6 \\ Alan Michael Gómez Calderón, Oct 08 2024

def A054896(n):
    if (n==0): return 0
    else: return A054896(n//7) + (n//7)
[A054896(n) for n in range(101)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/7) + floor(n/49) + floor(n/343) + floor(n/2401) + ...
a(n) = (n - A053828(n))/6.
From Hieronymus Fischer, Aug 14 2007: (Start)
a(n) = a(floor(n/7)) + floor(n/7).
a(7*n) = n + a(n).
a(n*7^m) = a(n) + n*(7^m-1)/6.
a(k*7^m) = k*(7^m-1)/6, for 0 <= k < 7, m >= 0.
Asymptotic behavior:
a(n) = n/6 + O(log(n)).
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/6; equality holds for powers of 7.
a(n) >= (n-6)/6 - floor(log_7(n)); equality holds for n=7^m-1, m>0. -
lim inf (n/6 - a(n)) = 1/6, for n-->oo.
lim sup (n/6 - log_7(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_7(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(7^k)/(1-x^(7^k)). (End)
Partial sums of A214411. - R. J. Mathar, Jul 08 2021
a(n) = A214411(A000142(n)). - Michel Marcus, Oct 07 2024

Examples added by Hieronymus Fischer, Jun 06 2012

A132035 Decimal expansion of Product_{k>0} (1-1/7^k).

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Aug 14 2007

			0.8367954070890378710...

Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.

Cf. A027875, A048651, A100220, A132023, A132026, A132038, A098844, A067080, A000203.

digits = 103; NProduct[1-1/7^k, {k, 1, Infinity}, NProductFactors -> 200, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
RealDigits[QPochhammer[1/7], 10, 100][[1]] (* Amiram Eldar, May 09 2023 *)

prodinf(k=1, 1 - 1/(7^k)) \\ Amiram Eldar, May 09 2023

Equals exp(-Sum_{n>0} sigma_1(n)/(n*7^n)) = exp(-Sum_{n>0} A000203(n)/(n*7^n)).
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(7)) * exp(log(7)/24 - Pi^2/(6*log(7))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(7))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027875(n). (End)

A054895 a(n) = Sum_{k>0} floor(n/6^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16

Offset: 0

Henry Bottomley, May 23 2000

nonn

Different from the highest power of 6 dividing n! (cf. A054861). - Hieronymus Fischer, Aug 14 2007

Partial sums of A122841. - Hieronymus Fischer, Jun 06 2012

			  a(10^0) = 0.
  a(10^1) = 1.
  a(10^2) = 18.
  a(10^3) = 197.
  a(10^4) = 1997.
  a(10^5) = 19996.
  a(10^6) = 199995.
  a(10^7) = 1999995.
  a(10^8) = 19999994.
  a(10^9) = 199999993.

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

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

Cf. A053827, A054861, A054899, A067080, A098844, A122841, A132030.

a054895 n = a054895_list !! n
a054895_list = scanl (+) 0 a122841_list
-- Reinhard Zumkeller, Nov 10 2013

function A054895(n)
  if n eq 0 then return n;
  else return A054895(Floor(n/6)) + Floor(n/6);
  end if; return A054895;
end function;
[A054895(n): n in [0..100]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=6; While[s=Floor[n/p]; t=t+s; s>0, p *= 6]; t, {n,0,100}]

def A054895(n):
    if (n==0): return 0
    else: return A054895(n//6) + (n//6)
[A054895(n) for n in range(104)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/6) + floor(n/36) + floor(n/216) + floor(n/1296) + ...
a(n) = (n - A053827(n))/5.
From Hieronymus Fischer, Aug 14 2007: (Start)
a(n) = a(floor(n/6)) + floor(n/6).
a(6*n) = n + a(n).
a(n*6^m) = n*(6^m-1)/5 + a(n).
a(k*6^m) = k*(6^m-1)/5, for 0 <= k < 6, m >= 0.
Asymptotic behavior:
a(n) = (n/5) + O(log(n)).
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/5; equality holds for powers of 6.
a(n) >= ((n-5)/5) - floor(log_6(n)); equality holds for n=6^m-1, m>0.
lim inf (n/5 - a(n)) = 1/5, for n-->oo.
lim sup (n/5 - log_6(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_6(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(6^k)/(1-x^(6^k)). (End)

An incorrect formula was deleted by N. J. A. Sloane, Nov 18 2008

Examples added by Hieronymus Fischer, Jun 06 2012

A132020 Decimal expansion of Product_{k>=0} (1 - 1/(2*4^k)).

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Aug 14 2007

This is the limiting probability that a large random symmetric binary matrix is nonsingular (cf. A086812, A048651). In other words, equals Lim_{n->oo} A086812(n)/A006125(n+1).- H. Tracy Hall, Sep 07 2024

			0.41942244179510759770995610770297425223395323439266674908044991663177...

John E. Cremona and R. W. K. Odoni, Some density results for negative Pell equations; an application of graph theory, Journal of the London Mathematical Society s2-39:1 (1989), pp. 16-28.

Cf. A004171, A048651, A098844, A067080, A132019, A132026, A132028, A100221, A000217, A081845.

evalf(1+sum((-1)^n*2^(n*(n-1)/2)/product(2^k-1, k=1..n), n=1..infinity), 120); # Robert FERREOL, Feb 23 2020

RealDigits[ Product[1 - 1/(2*4^i), {i, 0, 175}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
RealDigits[QPochhammer[1/2, 1/4], 10, 105][[1]] (* Jean-François Alcover, Nov 18 2015 *)

prodinf(k=0,1-1.>>(2*k+1)) \\ Charles R Greathouse IV, Nov 16 2012

Equals lim inf_{n->oo} Product_{k=0..floor(log_4(n))} floor(n/4^k)*4^k/n.
Equals lim inf_{n->oo} A132028(n)/n^(1+floor(log_4(n)))*4^((1/2)*(1+floor(log_4(n)))*floor(log_4(n))).
Equals lim inf_{n->oo} A132028(n)/n^(1+floor(log_4(n)))*4^A000217(floor(log_4(n))).
Equals (1/2)*exp(-Sum_{n>0} (4^(-n)*(Sum_{k|n} 1/(k*2^k)))).
Equals lim inf_{n->oo} A132028(n)/A132028(n+1).
Equals Product_{k>0} (1-1/(2^k+1)). - Robert G. Wilson v, May 25 2011
From Robert FERREOL, Feb 23 2020: (Start)
Equals Product_{k>0} (1 + 1/2^k)^(-1) = 2/A081845.
Equals 1 + Sum_{n>=1} (-1)^n*2^(n*(n-1)/2)/((2-1)*(2^2-1)*...*(2^n-1)). (End)
From Peter Bala, Jan 15 2021: (Start)
Constant C = Sum_{n >= 0} 2^n/Product_{k = 1..n} (1 - 4^k).
Faster converging series:
2*C = (1/2)*Sum_{n >= 0} 2^(-n)/Product_{k = 1..n} (1 - 4^k);
(2^4)*C = 7*Sum_{n >= 0} 2^(-3*n)/Product_{k = 1..n} (1 - 4^k);
(2^9)*C = 7*31*Sum_{n >= 0} 2^(-5*n)/Product_{k = 1..n} (1 - 4^k), and so on.
Slower converging series:
C = -Sum_{n >= 0} 2^(3*n)/Product_{k = 1..n} (1 - 4^k);
7*C = Sum_{n >= 0} 2^(5*n)/Product_{k = 1..n} (1 - 4^k);
7*31*C = -Sum_{n >= 0} 2^(7*n)/Product_{k = 1..n} (1 - 4^k), and so on. (End)
Equals Product_{n>=0} (1 - 1/A004171(n)). - Amiram Eldar, May 09 2023

Name corrected by Charles R Greathouse IV, Nov 16 2012

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

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, 12, 12, 13, 13, 13, 13, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24

Offset: 0

Henry Bottomley, May 23 2000

Different from highest power of 4 dividing n! (see A090616).

			  a(10^0) = 0.
  a(10^1) = 2.
  a(10^2) = 32.
  a(10^3) = 330.
  a(10^4) = 3331.
  a(10^5) = 33330.
  a(10^6) = 333330.
  a(10^7) = 3333329.
  a(10^8) = 33333328.
  a(10^9) = 333333326.

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

Cf. A053737, A235127 (first differences).

Cf. A011371, A027868, A054861, A054899, A067080, A090616, A098844, A132028.

function A054893(n)
  if n eq 0 then return n;
  else return A054893(Floor(n/4)) + Floor(n/4);
  end if; return A054893;
end function;
[A054893(n): n in [0..103]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=4; While[s=Floor[n/p]; t=t+s; s>0, p *= 4]; t, {n,0,100}]
Table[Total[Floor/@(n/NestList[4#&,4,6])],{n,0,80}] (* Harvey P. Dale, Jun 12 2022 *)

a(n) = (n - sumdigits(n,4))/3; \\ Kevin Ryde, Jan 08 2024

def A054893(n):
    if (n==0): return 0
    else: return A054893(n//4) + (n//4)
[A054893(n) for n in range(104)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/4) + floor(n/16) + floor(n/64) + floor(n/256) + ...
a(n) = (n - A053737(n))/3.
From Hieronymus Fischer, Sep 15 2007: (Start)
a(n) = a(floor(n/4)) + floor(n/4).
a(4*n) = a(n) + n.
a(n*4^m) = a(n) + n*(4^m-1)/3.
a(k*4^m) = k*(4^m-1)/3, for 0 <= k < 4, m >= 0.
Asymptotic behavior:
a(n) = n/3 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/3; equality holds true for powers of 4.
a(n) >= (n-3)/3 - floor(log_4(n)); equality holds true for n = 4^m - 1, m>0. lim inf (n/3 - a(n)) = 1/3, for n-->oo.
lim sup (n/3 - log_4(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_4(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(4^k)/(1-x^(4^k)). (End)
Partial sums of A235127. - R. J. Mathar, Jul 08 2021

Edited by Hieronymus Fischer, Sep 15 2007

Examples added by Hieronymus Fischer, Jun 06 2012

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

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cp's OEIS Frontend

A132263 Product{0<=k<=floor(log_11(n)), floor(n/11^k)}, n>=1.

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A132264 Product{0<=k<=floor(log_12(n)), floor(n/12^k)}, n>=1.

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A132269 a(n) = Product_{k>=0} (1 + floor(n/2^k)).

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A132327 a(n) = Product{k>=0} (1 + floor(n/3^k)).

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A054896 a(n) = Sum_{k>0} floor(n/7^k).

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A132035 Decimal expansion of Product_{k>0} (1-1/7^k).

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A054895 a(n) = Sum_{k>0} floor(n/6^k).

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