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
A067240
If n = Product_{i} p_i^e_i, a(n) = Sum_{i} (p_i - 1)*p_i^(e_i - 1).
Original entry on oeis.org
0, 1, 2, 2, 4, 3, 6, 4, 6, 5, 10, 4, 12, 7, 6, 8, 16, 7, 18, 6, 8, 11, 22, 6, 20, 13, 18, 8, 28, 7, 30, 16, 12, 17, 10, 8, 36, 19, 14, 8, 40, 9, 42, 12, 10, 23, 46, 10, 42, 21, 18, 14, 52, 19, 14, 10, 20, 29, 58, 8, 60, 31, 12, 32, 16, 13, 66, 18, 24, 11, 70, 10, 72, 37, 22, 20, 16, 15, 78, 12, 54, 41, 82, 10, 20, 43, 30, 14, 88, 11, 18, 24, 32, 47, 22, 18, 96, 43, 16, 22
Offset: 1
-
a067240 1 = 0
a067240 n = sum $ map a000010 $ a141809_row $ toInteger n
-- Reinhard Zumkeller, Jun 13 2012
-
with(numtheory); A067240 := proc(n) local e,j; e := ifactors(n)[2]: add((e[j][1]-1)*e[j][1]^(e[j][2]-1),j=1..nops(e)); end;
-
a[n_] := Total[ EulerPhi[ Power @@ #] & /@ FactorInteger[n]]; a[1] = 0; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jun 22 2012, after T. D. Noe *)
-
A067240(n)=
{
local(f=factor(n), r=0, p, e);
for (i=1, matsize(f)[1],
p=f[i,1]; e=f[i,2];
r += (p-1)*p^(e-1);
);
return(r);
} /* Joerg Arndt, Jun 10 2011 */
-
a(n)=my(f=factor(n)); sum(i=1,#f~, (f[i,1]-1)*f[i,1]^(f[i,2]-1)) \\ Charles R Greathouse IV, Sep 10 2015
-
from sympy import factorint
a = lambda n: sum((p**(e-1))*(p-1) for p,e in factorint(n).items() if e > 0) # Darío Clavijo, Feb 15 2024
A141810
Irregular table: Row n (of A001221(n) terms, for n>=2) consists of the largest powers that divides n of each distinct prime that divides n. Terms are arranged by the sizes of the prime-powers. Row 1 = (1).
Original entry on oeis.org
1, 2, 3, 4, 5, 2, 3, 7, 8, 9, 2, 5, 11, 3, 4, 13, 2, 7, 3, 5, 16, 17, 2, 9, 19, 4, 5, 3, 7, 2, 11, 23, 3, 8, 25, 2, 13, 27, 4, 7, 29, 2, 3, 5, 31, 32, 3, 11, 2, 17, 5, 7, 4, 9, 37, 2, 19, 3, 13, 5, 8, 41, 2, 3, 7, 43, 4, 11, 5, 9, 2, 23, 47, 3, 16, 49, 2, 25, 3, 17, 4, 13, 53, 2, 27, 5, 11, 7, 8, 3
Offset: 1
60 has the prime factorization of 2^2 * 3^1 * 5^1. So row 60 is, arranged by the sizes of the prime-powers, (3,4,5).
A080736
Multiplicative function defined by a(1)=1, a(2)=0, a(2^r) = phi(2^r) (r>1), a(p^r) = phi(p^r) (p odd prime, r>=1), where phi is Euler's function A000010.
Original entry on oeis.org
1, 0, 2, 2, 4, 0, 6, 4, 6, 0, 10, 4, 12, 0, 8, 8, 16, 0, 18, 8, 12, 0, 22, 8, 20, 0, 18, 12, 28, 0, 30, 16, 20, 0, 24, 12, 36, 0, 24, 16, 40, 0, 42, 20, 24, 0, 46, 16, 42, 0, 32, 24, 52, 0, 40, 24, 36, 0, 58, 16, 60, 0, 36, 32, 48, 0, 66, 32, 44, 0, 70, 24, 72, 0, 40, 36, 60, 0, 78, 32
Offset: 1
-
a080736 n = if n `mod` 4 == 2 then 0 else a000010 n
-- Reinhard Zumkeller, Jun 13 2012, Jun 11 2012
-
a[n_] := If[Mod[n, 4] == 2, 0, EulerPhi[n]]; Array[a, 100] (* Amiram Eldar, Nov 02 2023 *)
-
{for(n=1,81,f=factor(n); print1(if(n==1,1,if(f[1,1]==2&&f[1,2]==1,0,prod(j=1,matsize(f)[1],eulerphi(f[j,1]^f[j,2])))),","))}
A140831
Numbers in whose canonical prime factorization the powers of the primes do not form an increasing sequence.
Original entry on oeis.org
12, 24, 40, 45, 48, 56, 60, 63, 80, 84, 90, 96, 112, 120, 126, 132, 135, 144, 156, 160, 168, 175, 176, 180, 189, 192, 204, 208, 224, 228, 240, 252, 264, 270, 275, 276, 280, 288, 297, 300, 312, 315, 320, 325, 336, 348, 350, 351, 352, 360, 372, 378, 384, 405
Offset: 1
The prime factorization of 90 is, when arranged by size of the distinct primes, 2^1 * 3^2 * 5^1. Since 3^2 is > 5^1, even though 5 > 3, 90 is in the sequence.
-
fQ[n_] := Block[{f = First@# ^ Last@# & /@ FactorInteger@n}, f != Sort@f]; Select[ Range@ 407, fQ@# &] (* Robert G. Wilson v, Aug 31 2008 *)
-
is(n) = { my(f = factor(n)); for(i = 1, #f~-1, if(f[i,1]^f[i,2] > f[i+1,1]^f[i+1,2], return(1) ) ); 0 } \\ David A. Corneth, Jun 16 2022
A333235
a(n) is the product of indices of unitary prime power divisors of n.
Original entry on oeis.org
1, 1, 2, 3, 4, 2, 5, 6, 7, 4, 8, 6, 9, 5, 8, 10, 11, 7, 12, 12, 10, 8, 13, 12, 14, 9, 15, 15, 16, 8, 17, 18, 16, 11, 20, 21, 19, 12, 18, 24, 20, 10, 21, 24, 28, 13, 22, 20, 23, 14, 22, 27, 24, 15, 32, 30, 24, 16, 25, 24, 26, 17, 35, 27, 36, 16, 28, 33, 26, 20
Offset: 1
a(600) = a(2^3 * 3 * 5^2) = a(A246655(6) * A246655(2) * A246655(14)) = 6 * 2 * 14 = 168.
-
N:= 1000: # for a(1)..a(N)
R:= NULL: p:= 2:
while p < N do
R:= R, seq(p^k,k=1..ilog[p](N));
p:= nextprime(p);
od:
L:= sort([R]):
f:= proc(n) local F, t;
F:= ifactors(n)[2];
mul(ListTools:-BinarySearch(L,t[1]^t[2]),t=F)
end proc:
map(f, [$1..N]); # Robert Israel, Feb 11 2021
-
PrimePowerPi[n_] := Sum[Boole[PrimePowerQ[k]], {k, 1, n}]; a[1] = 1; a[n_] := Times @@ (PrimePowerPi[#[[1]]^#[[2]]] & /@ FactorInteger[n]); Table[a[n], {n, 1, 70}]
-
apply( {A333235(n)=vecprod([A322981(f[1]^f[2])|f<-factor(n)~])}, [1..99]) \\ M. F. Hasler, Jun 16 2021
A304233
If n = Product (p_j^k_j) then a(n) = min{p_j^k_j}*max{p_j^k_j}.
Original entry on oeis.org
1, 4, 9, 16, 25, 6, 49, 64, 81, 10, 121, 12, 169, 14, 15, 256, 289, 18, 361, 20, 21, 22, 529, 24, 625, 26, 729, 28, 841, 10, 961, 1024, 33, 34, 35, 36, 1369, 38, 39, 40, 1681, 14, 1849, 44, 45, 46, 2209, 48, 2401, 50, 51, 52, 2809, 54, 55, 56, 57, 58, 3481, 15, 3721, 62, 63, 4096, 65
Offset: 1
a(60) = 15 because 60 = 2^2*3*5, min{2^2,3,5} = 3, max{2^2,3,5} = 5 and 3*5 = 15.
A059401
Numbers that contain as proper substrings every maximal prime power dividing them.
Original entry on oeis.org
1197, 2510, 2570, 5210, 5230, 5290, 12590, 14673, 15230, 20530, 21530, 22510, 23510, 23570, 24590, 25030, 25210, 25310, 25390, 25430, 25490, 25510, 25570, 25790, 25910, 25930, 26570, 26590, 27530, 28510, 28570, 29530, 29570, 32510
Offset: 1
1197 = 9 * 7 * 19 and all of these are substrings.
Equivalent sequence for primes instead of maximal prime powers:
A050694.
-
import Data.List (isInfixOf)
a059401 n = a059401_list !! (n-1)
a059401_list = filter (\x -> a010055 x == 0 &&
all (`isInfixOf` show x) (map show $ a141809_row x)) [1..]
-- Reinhard Zumkeller, Dec 16 2013
-
psmppQ[n_]:=Module[{pp=#[[1]]^#[[2]]&/@FactorInteger[n], idn= IntegerDigits[ n]}, Length[pp]>1&&And@@Table[ SequenceCount[ idn, IntegerDigits[pp[[i]]]]>0,{i,Length[pp]}]]; Select[Range[ 33000], psmppQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 04 2016 *)
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