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A124010 Triangle in which first row is 0, n-th row (n>1) lists the exponents of distinct prime factors ("ordered prime signature") in the prime factorization of n. 451

%I #63 Apr 09 2022 14:05:13

%S 0,1,1,2,1,1,1,1,3,2,1,1,1,2,1,1,1,1,1,1,4,1,1,2,1,2,1,1,1,1,1,1,3,1,

%T 2,1,1,3,2,1,1,1,1,1,1,5,1,1,1,1,1,1,2,2,1,1,1,1,1,3,1,1,1,1,1,1,2,1,

%U 2,1,1,1,1,4,1,2,1,2,1,1,2,1,1,1,3,1,1,3,1,1,1,1,1,1,2,1,1,1,1,1,2,1,6,1,1,1,1,1,1,2,1,1,1,1,1,1,1,3,2,1

%N Triangle in which first row is 0, n-th row (n>1) lists the exponents of distinct prime factors ("ordered prime signature") in the prime factorization of n.

%C A001222(n) = Sum(T(n,k), 1 <= k <= A001221(n)); A005361(n) = Product(T(n,k), 1 <= k <= A001221(n)), n>1; A051903(n) = Max(T(n,k): 1 <= k <= A001221(n)); A051904(n) = Min(T(n,k), 1 <= k <= A001221(n)); A067029(n) = T(n,1); A071178(n) = T(n,A001221(n)); A064372(n)=Sum(A064372(T(n,k)), 1 <= k <= A001221(n)). - _Reinhard Zumkeller_, Aug 27 2011

%C Any finite sequence of natural numbers appears as consecutive terms. - _Paul Tek_, Apr 27 2013

%C For n > 1: n-th row = n-th row of A067255 without zeros. - _Reinhard Zumkeller_, Jun 11 2013

%C Most often the prime signature is given as a sorted representative of the multiset of the nonzero exponents, either in increasing order, which yields A118914, or, most commonly, in decreasing order, which yields A212171. - _M. F. Hasler_, Oct 12 2018

%H Reinhard Zumkeller, <a href="/A124010/b124010.txt">Rows n = 1..10000 of triangle, flattened</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>

%F n = Product_k A027748(n,k)^a(n,k).

%e Initial values of exponents are:

%e 1, [0]

%e 2, [1]

%e 3, [1]

%e 4, [2]

%e 5, [1]

%e 6, [1, 1]

%e 7, [1]

%e 8, [3]

%e 9, [2]

%e 10, [1, 1]

%e 11, [1]

%e 12, [2, 1]

%e 13, [1]

%e 14, [1, 1]

%e 15, [1, 1]

%e 16, [4]

%e 17, [1]

%e 18, [1, 2]

%e 19, [1]

%e 20, [2, 1]

%e ...

%p expts:=proc(n) local t1,t2,t3,t4,i; if n=1 then RETURN([0]); fi; if isprime(n) then RETURN([1]); fi; t1:=ifactor(n); if nops(factorset(n))=1 then RETURN([op(2,t1)]); fi; t2:=nops(t1); t3:=[]; for i from 1 to t2 do t4:=op(i,t1); if nops(t4) = 1 then t3:=[op(t3),1]; else t3:=[op(t3),op(2,t4)]; fi; od; RETURN(t3); end; # _N. J. A. Sloane_, Dec 20 2007

%t row[1] = {0}; row[n_] := FactorInteger[n][[All, 2]] // Flatten; Table[row[n], {n, 1, 80}] // Flatten (* _Jean-François Alcover_, Aug 19 2013 *)

%o (Haskell)

%o a124010 n k = a124010_tabf !! (n-1) !! (k-1)

%o a124010_row 1 = [0]

%o a124010_row n = f n a000040_list where

%o f 1 _ = []

%o f u (p:ps) = h u 0 where

%o h v e | m == 0 = h v' (e + 1)

%o | m /= 0 = if e > 0 then e : f v ps else f v ps

%o where (v',m) = divMod v p

%o a124010_tabf = map a124010_row [1..]

%o -- _Reinhard Zumkeller_, Jun 12 2013, Aug 27 2011

%o (PARI) print1(0); for(n=2,50, f=factor(n)[,2]; for(i=1,#f,print1(", "f[i]))) \\ _Charles R Greathouse IV_, Nov 07 2014

%o (PARI) A124010_row(n)=if(n,factor(n)[,2]~,[0]) \\ _M. F. Hasler_, Oct 12 2018

%o (Python)

%o from sympy import factorint

%o def a(n):

%o f=factorint(n)

%o return [0] if n==1 else [f[i] for i in f]

%o for n in range(1, 21): print(a(n)) # _Indranil Ghosh_, May 16 2017

%Y Cf. A027748, A001221 (row lengths, n>1), A001222 (row sums), A027746, A020639, A064372, A067029 (first column).

%Y Sorted rows: A118914, A212171.

%K easy,nonn,tabf

%O 1,4

%A _Franklin T. Adams-Watters_, Nov 01 2006

%E Name edited by _M. F. Hasler_, Apr 08 2022

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Last modified April 24 22:17 EDT 2024. Contains 371964 sequences. (Running on oeis4.)