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.
a(1)=59 because in first step we removed all numbers divisible by 2 (=60) with the exception of the first one, i.e., 2. a(2)=19 because the number of numbers divisible by 3 and not divisible by 2 is 20 and we remove all with the exception of the first one, i.e., 3.
A145537:=Array([seq(0,j=1..291)]): lim:=10!: p:=Array([seq(ithprime(j),j=1..291)]): for n from 4 to lim do if(isprime(n))then n:=n+1: fi: for k from 1 to 291 do if(n mod p[k] = 0)then A145537[k]:=A145537[k]+1: break: fi: od: od: seq(A145537[j],j=1..291); # Nathaniel Johnston, Jun 23 2011
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 = 10; kk = PrimePi[Sqrt[nn! ]]; t3 = f3[nn!, kk] (* Bob Hanlon (hanlonr(AT)cox.net) *)
a(1)=359 because in the first step we remove all numbers divisible by 2 (= 360) with the exception of the first one, i.e., 2. a(2)=119 because the number of numbers divisible by 3 and not divisible by 2 is 120 and we remove all such numbers with the exception of the first one, 3.
A145533 := {$(1..6!)}: for n from 1 do p:=ithprime(n): r:=0: lim:=6!/p: for k from 2 to lim do if(member(k*p,A145533))then r:=r+1: fi: A145533 := A145533 minus {k*p}: od: printf("%d, ", r): if(r=0)then break: fi: od: # Nathaniel Johnston, Jun 23 2011
{m1, m2, m3, m4, m5, m6, m7, m8, m9} = {-1, -1, -1, -1, -1, -1, -1, -1, -1};
Do[If[Mod[n, 2] == 0, m1 = m1 + 1,
If[Mod[n, 3] == 0, m2 = m2 + 1,
If[Mod[n, 5] == 0, m3 = m3 + 1,
If[Mod[n, 7] == 0, m4 = m4 + 1,
If[Mod[n, 11] == 0, m5 = m5 + 1,
If[Mod[n, 13] == 0, m6 = m6 + 1,
If[Mod[n, 17] == 0, m7 = m7 + 1,
If[Mod[n, 19] == 0, m8 = m8 + 1,
If[Mod[n, 23] == 0, m9 = m9 + 1]]]]]]]]], {n, 1, 6!}];
Print[{m1, m2, m3, m4, m5, m6, m7, m8, m9}] (* Artur Jasinski *)
A145534 := {$(1..7!)}: for n from 1 do p:=ithprime(n): r:=0: lim:=7!/p: for k from 2 to lim do if(member(k*p,A145534))then r:=r+1: fi: A145534 := A145534 minus {k*p}: od: printf("%d, ", r): if(r=0)then break: fi: od: # Nathaniel Johnston, Jun 23 2011
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 = 7; kk = PrimePi[Sqrt[nn! ]]; t3 = f3[nn!, kk] (* Bob Hanlon (hanlonr(AT)cox.net) *)
A145535 := {$(1..8!)}: for n from 1 do p:=ithprime(n): r:=0: lim:=8!/p: for k from 2 to lim do if(member(k*p,A145535))then r:=r+1: fi: A145535 := A145535 minus {k*p}: od: printf("%d, ", r): if(r=0)then break: fi: od: # Nathaniel Johnston, Jun 23 2011
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 = 8; kk = PrimePi[Sqrt[nn! ]]; t3 = f3[nn!, kk] (* Bob Hanlon (hanlonr(AT)cox.net) *)
A145536:=Array([seq(0,j=1..110)]): lim:=9!: p:=Array([seq(ithprime(j),j=1..110)]): for n from 4 to lim do if(isprime(n))then n:=n+1: fi: for k from 1 to 110 do if(n mod p[k] = 0)then A145536[k]:=A145536[k]+1: break: fi: od: od: seq(A145536[j],j=1..110); # Nathaniel Johnston, Jun 23 2011
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 = 9; kk = PrimePi[Sqrt[nn! ]]; t3 = f3[nn!, kk] (* Bob Hanlon (hanlonr(AT)cox.net) *)
For n = 2, prime(n) = 3, a(n) = 1666666: 3 divides 10^7 3333333 times. 6 is the common multiple of 2 and 3, thus 10^7 \ 6 multiples of 3 (1666666) have already been eliminated by a(1). 3333333 less 1666666 = 1666667, less 1 because 3 itself is not eliminated. Thus a(2) = 3333333 - 1666666 - 1 = 1666666.
For n = 3, prime(n) = 5, a(n) = 6666666: 5 divides 10^8 20000000 times. 10 is the least common multiple of 2 (prime(1)) and 5 and 15 is the least common multiple of 3 (prime(2)) and 5; thus [10^8 / 10] multiples of 5 and [10^8 / 15] multiples of 5 have already been eliminated by a(1) and a(2), and thereby respectively reduce a(3) by 10000000 and 6666666 offset by [10^8 / 30] multiples of 5 which would otherwise excessively reduce a(3) by 3333333 because 30 is the least common multiple of 2, 3 and 5. a(3) is further reduced by 1 as 5 itself is not eliminated.
a(1) = 10^9 \ 2 - 1. a(2) = 10^9 \ 3 - 10^9 \ (2*3) - 1 a(3) = 10^9 \ 5 - 10^9 \ (2*5) - 10^9 \ (3*5) + 10^9 \ (2*3*5) - 1 a(4) = 10^9 \ 7 - 10^9 \ (2*7) - 10^9 \ (3*7) - 10^9 \ (5*7) + 10^9 \ (2*3*7) + 10^9 \ (2*5*7) + 10^9 \ (3*5*7) - 10^9 \ (2*3*5*7) - 1.
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.
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