[go: up one dir, main page]

login
Search: a145535 -id:a145535
     Sort: relevance | references | number | modified | created      Format: long | short | data
a(n) = number of numbers removed in the n-th step of Eratosthenes's sieve for 10^2.
+10
11
OFFSET
1,1
COMMENTS
Number of steps in Eratosthenes's sieve for 10^n is A122121(n).
Number of primes less than 10^2 is equal to 10^2 - (sum all of numbers in this sequence) - 1 = A006880(2).
MATHEMATICA
f3[k_Integer?Positive, i_Integer?Positive] := Module[{f, m, r, p}, p = Transpose[{r = Range[2, i], Prime[r]}]; f[x_] := Catch[Fold[If[Mod[x, #2[[2]]] == 0, Throw[m[ #2[[1]]] = m[ #2[[1]]] + 1], #1] &, If[Mod[x, 2] == 0, Throw[m[1] = m[1] + 1]], p]]; Table[m[n] = -1, {n, i}]; f /@ Range[k]; Table[m[n], {n, i}]]; nn = 2; kk = PrimePi[Sqrt[10^nn]]; t3 = f3[10^nn, kk] (*Bob Hanlon (hanlonr(AT)cox.net) *)
KEYWORD
fini,nonn
AUTHOR
Artur Jasinski with assistance from Bob Hanlon (hanlonr(AT)cox.net), Oct 14 2008
STATUS
approved
Natural numbers of the form (n!-2)/2.
+10
10
0, 2, 11, 59, 359, 2519, 20159, 181439, 1814399, 19958399, 239500799, 3113510399, 43589145599, 653837183999, 10461394943999, 177843714047999, 3201186852863999, 60822550204415999, 1216451004088319999
OFFSET
2,2
COMMENTS
Natural numbers of the form (n!-m)/m:
for m=1 n!-1 see A033312;
for m=2 (n!-2)/2 see A139172;
for m=3 (n!-3)/3 see A139173;
for m=4 (n!-4)/4 see A139174;
for m=5 (n!-5)/5 see A139175;
for m=6 (n!-6)/6 see A139176;
for m=7 (n!-7)/7 see A139177;
for m=8 (n!-8)/8 see A139183;
for m=9 (n!-9)/9 see A139184;
for m=10 (n!-10)/10 see A139185.
From Artur Jasinski, Oct 14 2008: (Start)
a(n) = Number of numbers removed in first step of Eratosthenes's sieve for n!
a(5)=A145532(1), a(6)=A145533(1), a(7)=A145534(1), a(8)=A145535(1), a(9)=A145536(1), a(10)=A145537(1). (End)
Generally, for n >= m, the formula a(n) = n*(a(n-1) + 1) - 1 applies to all natural numbers of the form (n!-m)/m, m >= 2. - Bob Selcoe, Mar 28 2015
LINKS
FORMULA
a(n) = Sum_{k=1..floor(n/2)} s(n,n-2*k), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 07 2012
a(n) = n*(a(n-1) + 1) - 1. - Bob Selcoe, Mar 28 2015
MATHEMATICA
Table[(n! - 2)/2, {n, 2, 20}]
PROG
(Magma) [(Factorial(n)-2)/2: n in [2..25]]; // Vincenzo Librandi, Jul 20 2011
(PARI) a(n)=n!/2-1 \\ Charles R Greathouse IV, Apr 07 2012
KEYWORD
nonn,easy
AUTHOR
Artur Jasinski, Apr 11 2008
STATUS
approved
Number of composites removed in each step of the Sieve of Eratosthenes for 10^10.
+10
3
4999999999, 1666666666, 666666666, 380952380, 207792207, 159840159, 112828348, 95013343, 74358271, 56409724, 50950713, 41311372, 36273411, 33742734, 30153115, 26170720, 23065826, 21931483, 19640105, 18256894, 17506397, 15954848, 14993294, 13813524, 12531256
OFFSET
1,1
COMMENTS
a(n) = the number of composites <= 10^10 for which the n-th prime is the least prime factor.
pi(sqrt(10^10)) = the number of terms of this sequence.
The sum of a(n) for n = 1..3401 = A000720(10^10) + A065855(10^10).
LINKS
EXAMPLE
a(1) = 10^10 \ 2 - 1.
a(2) = 10^10 \ 3 - 10^10 \ (2*3) - 1.
a(3) = 10^10 \ 5 - 10^10 \ (2*5) - 10^10 \ (3*5) + 10^10 \ (2*3*5) - 1.
a(4) = 10^10 \ 7 - 10^10 \ (2*7) - 10^10 \ (3*7) - 10^10 \ (5*7) + 10^10 \ (2*3*7) + 10^10 \ (2*5*7) + 10^10 \ (3*5*7) - 10^10 \ (2*3*5*7) - 1.
KEYWORD
nonn,fini
AUTHOR
Eric F. O'Brien, Jul 31 2013
STATUS
approved
a(n) = number of numbers removed in step n of Eratosthenes's sieve for 2^6.
+10
2
OFFSET
1,1
COMMENTS
Number of steps in Eratosthenes's sieve for 2^n is A060967(n).
Number of primes less than 2^6 is equal to 2^6 - (sum all of numbers in this sequence) - 1 = A007053(6).
MATHEMATICA
f3[k_Integer?Positive, i_Integer?Positive] := Module[{f, m, r, p}, p = Transpose[{r = Range[2, i], Prime[r]}]; f[x_] := Catch[Fold[If[Mod[x, #2[[2]]] == 0, Throw[m[ #2[[1]]] = m[ #2[[1]]] + 1], #1] &, If[Mod[x, 2] == 0, Throw[m[1] = m[1] + 1]], p]]; Table[m[n] = -1, {n, i}]; f /@ Range[k]; Table[m[n], {n, i}]]; nn = 6; kk = PrimePi[Sqrt[2^nn]]; t3 = f3[2^nn, kk] (* Bob Hanlon (hanlonr(AT)cox.net) *)
KEYWORD
fini,full,nonn
AUTHOR
Artur Jasinski with assistance from Bob Hanlon (hanlonr(AT)cox.net), Oct 14 2008
STATUS
approved

Search completed in 0.007 seconds