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.
(* first run the Mmca in A240751, then *) records[s_List] := Block[{k = 1, lmt = 1 + Length@s, lst = {}, mx = 0}, While[k < lmt, If[s[[k]] > mx, mx = s[[k]]; AppendTo[lst, mx]]; k++]; lst]; records[ Array[f, 250000]]
n=7: 7! = 5040 = 2*2*2*2*3*3*5*7; three different exponents arise: 4, 2 and 1; a(7)=3.
n=7: { floor(7/p) : p <= 7, p prime } = {3,2,1}. So, its cardinality is 3. - _Md Rahil Miraj_, Apr 05 2024
ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ep[x_] := Table[Part[ffi[x], 2*w], {w, 1, lf[x]}] Table[Length[Union[ep[w! ]]], {w, 1, 100}]
Table[Length[Union[Last/@If[n==1,{},FactorInteger[n!]]]],{n,30}] (* Gus Wiseman, May 15 2019 *)
a(n) = #Set(factor(n!)[, 2]); \\ Michel Marcus, Sep 05 2017
Let n = 4. A050376(4)=5. For k = 2, 3, 4, 5, 6, we have the following factorizations over distinct terms of A050376: 2! = 2, 3! = 2*3, 4! = 2*3*4, 5! = 2*3*4*5, 6! = 5*9*16. Only the last factorization begins with 5. So a(4) = 6. From _David A. Corneth_, Jun 19 2025: (Start) a(6) = 130. Once we checked that a(6) is > 125 we try 126. The minimal factor of k! into distinct products must be A050376(6) = 8. For 126! we have the 5-adic valuation of 31 so the minimal factor is at most 5. To get rid of the 5 we try the next candidate > 126 that is a multiple of 5. This is 130. We can just skip 127, 128 and 129 altogether. It turns out this smallest factor for 130! is 8 giving the value for a(6). (End)
\\ See Corneth link
\\ See Eldar link
81 is in the sequence, since 9! = 2*4*5*7*16*81.
Factorization of 5! over distinct terms of A050376 is 5! = 2*3*4*5. Thus 5 is the smallest k such that such a factorization contains 3 primes: 2,3,5. So a(3)=5.
f[p_, e_] := Mod[e, 2]; b[1] = 0; b[n_] := Plus @@ (f @@@ FactorInteger[n]); m = 56; v = Table[0, {m}]; c = 0; p = 1; n = 2; While[c < m, p *= n; i = b[p]; If[i <= m && v[[i]] == 0, c++; v[[i]] = n]; n++]; v (* Amiram Eldar, Sep 17 2019 *)
nbp(n) = {f = factor(n); sum (i=1, #f~, f[i,2] % 2);}
a(n) = {k = 1; while(nbp(k!) != n, k++); k;} \\ Michel Marcus, Apr 27 2014
For k=2,3,4,5,6, we have the following factorizations of k! over distinct terms of A050376: 2!=2, 3!=2*3, 4!=2*3*4, 5!=2*3*4*5, 6!=5*9*16. Therefore, a(1)=4, a(2)=6.
f[n_] := DigitCount[n, 2, 1] - Mod[n, 2]; nb[n_] := Total@(f/@ FactorInteger[n][[;;,2]]); a[n_] := (k=1; While[nb[k!] < n, k++]; k); Array[a, 60] (* Amiram Eldar, Dec 16 2018 from the PARI code *)
nb(n) = {my(f = factor(n)); sum(k=1, #f~, hammingweight(f[k,2]) - (f[k,2] % 2));}
a(n) = {my(k=1); while (nb(k!) < n, k++); k;} \\ Michel Marcus, Dec 16 2018
Factorization of 4! over distinct terms of A050376 is 4! = 2*3*4. This factorization contains only one A050376-nonprime. So a(4)=1.
b[n_] := 2^(-1 + Position[Reverse@IntegerDigits[n, 2], ?(# == 1 &)]) // Flatten; a[n] := Module[{np = PrimePi[n]}, v = Table[0, {np}]; Do[p = Prime[k]; Do[v[[k]] += IntegerExponent[j, p], {j, 2, n}], {k, 1, np}]; Length[Select[(b /@ v) // Flatten, # > 1 &]]]; Array[a, 73, 2] (* Amiram Eldar, Sep 17 2019 *)
a(n)={my(f=factor(n!)[,2]); sum(i=1, #f~, hammingweight(f[i]>>1))} \\ Andrew Howroyd, Sep 17 2019
0!=1, 1!=1; further we have the following factorizations of k! over distinct terms of A050376 for k = 2,3,4,5,6: 2!=2, 3!=2*3, 4!=2*3*4, 5!=2*3*4*5, 6!=5*9*16. Thus, in the sense of the factorizations being considered, 6! is relatively prime to 0!,1!,2!,3!, and 4!. So a(6)=5.
b[n_] := 2^(-1 + Position[Reverse@IntegerDigits[n, 2], ?(# == 1 &)]) // Flatten; infp[n] := Module[{np = PrimePi[n]}, v = Table[0, {np}]; Do[p = Prime[k]; Do[v[[k]] += IntegerExponent[j, p], {j, 2, n}], {k, 1, np}]; (Prime /@ Range[np])^(b /@ v) // Flatten]; infCoprimeQ[x_, y_] := Intersection[infp[x], infp[y]] == {}; a[n_] := Length @ Select[Range[0, n], infCoprimeQ[n, #] & ]; Array[a, 87, 0] (* Amiram Eldar, Sep 17 2019 *)
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