A275735 - OEIS

%I #32 Apr 03 2022 13:49:21

%S 1,2,2,4,3,6,2,4,4,8,6,12,3,6,6,12,9,18,5,10,10,20,15,30,2,4,4,8,6,12,

%T 4,8,8,16,12,24,6,12,12,24,18,36,10,20,20,40,30,60,3,6,6,12,9,18,6,12,

%U 12,24,18,36,9,18,18,36,27,54,15,30,30,60,45,90,5,10,10,20,15,30,10,20,20,40,30,60,15,30,30,60,45,90,25,50,50,100,75

%N Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

%C These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of times a nonzero digit k occurs in the factorial base representation of n. See the examples.

%H Antti Karttunen, <a href="/A275735/b275735.txt">Table of n, a(n) for n = 0..40320</a>

%H <a href="/index/Fa#facbase">Index entries for sequences related to factorial base representation</a>

%F a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

%F Other identities and observations. For all n >= 0:

%F a(n) = A275734(A225901(n)).

%F A001221(a(n)) = A275806(n).

%F A001222(a(n)) = A060130(n).

%F A048675(a(n)) = A275729(n).

%F A051903(a(n)) = A264990(n).

%F A008683(a(A265349(n))) = -1 or +1 for all n >= 0.

%F A008683(a(A265350(n))) = 0 for all n >= 1.

%F From _Antti Karttunen_, Apr 03 2022: (Start)

%F A342001(a(n)) = A351954(n).

%F a(n) = A181819(A276076(n)).

%F (End)

%e For n = 0 whose factorial base representation (A007623) is also 0, there are no nonzero digits at all, thus there cannot be any prime present in the encoding, and a(0) = 1.

%e For n = 1 there is just one 1, thus a(1) = prime(1) = 2.

%e For n = 2 ("10", there is just one 1-digit, thus a(2) = prime(1) = 2.

%e For n = 3 ("11") there are two 1-digits, thus a(3) = prime(1)^2 = 4.

%e For n = 18 ("300") there is just one 3, thus a(18) = prime(3) = 5.

%e For n = 19 ("301") there is one 1 and one 3, thus a(19) = prime(1)*prime(3) = 2*5 = 10.

%e For n = 141 ("10311") there are three 1's and one 3, thus a(141) = prime(1)^3 * prime(3) = 2^3 * 5^1 = 40.

%o (Scheme, with memoization-macro definec)

%o (definec (A275735 n) (if (zero? n) 1 (* (A000079 (A257511 n)) (A003961 (A275735 (A257684 n))))))

%o (Python)

%o from sympy import prime

%o from operator import mul

%o import collections

%o def a007623(n, p=2): return n if n<p else a007623(n//p, p+1)*10 + n%p

%o def a(n):

%o     y=collections.Counter(map(int, list(str(a007623(n)).replace("0", "")))).most_common()

%o     return 1 if n==0 else reduce(mul, [prime(y[i][0])**y[i][1] for i in range(len(y))])

%o print([a(n) for n in range(101)]) # _Indranil Ghosh_, Jun 19 2017

%o (PARI)

%o A276076(n) = { my(i=0,m=1,f=1,nextf); while((n>0),i=i+1; nextf = (i+1)*f; if((n%nextf),m*=(prime(i)^((n%nextf)/f));n-=(n%nextf));f=nextf); m; };

%o A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2])));

%o A275735(n) = A181819(A276076(n)); \\ _Antti Karttunen_, Apr 03 2022

%Y Cf. A000079, A001221, A001222, A003961, A007623, A008683, A181819, A225901, A257511, A257684, A265349, A265350, A264990, A275729, A275806, A351954.

%Y Cf. also A275725, A275733, A275734 for other such prime factorization encodings of A060117/A060118-related polynomials, and also A276076.

%Y Differs from A227154 for the first time at n=18, where a(18) = 5, while A227154(18) = 4.

%K nonn,base,look

%O 0,2

%A _Antti Karttunen_, Aug 09 2016