OFFSET
0,1
COMMENTS
This is to A047999 "Triangle formed by reading Pascal's triangle mod 2" as A007188 "Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i)" is to A007318 "Pascal's triangle read by rows." a(2^n - 1) = primorial(2^n) = A002110(A000079(n)). In row(n) the primes with exponent 1 form row(n) of a Sierpinski sieve, so this sequence is a kind of Gödelization of a Sierpinski sieve.
All terms are divisible by 2 and the n-th term, a(n-1), is also divisible by prime(n). This sequence appears as first column of the square array A255483; its second column A276804 is very similar, with all prime factors shifted to the net larger prime (cf. A003961). - M. F. Hasler, Sep 17 2016
a(n) is the n-th power of 6 in the ring defined in A329329. - Peter Munn, Jan 04 2020
LINKS
Chai Wah Wu, Table of n, a(n) for n = 0..510
C. Cobeli, A. Zaharescu, A game with divisors and absolute differences of exponents, Journal of Difference Equations and Applications, Vol. 20, #11 (2014) pp. 1489-1501, DOI: 10.1080/10236198.2014.940337. Also available as arXiv:1411.1334 [math.NT], 2014.
FORMULA
a(n) = Product_{i=0..n} p(i+1)^(C(n,i) mod 2).
a(n) = Product_{i=0..n} p(i+1)^T(n,i), where T(n,i) are as in A047999 and where Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
From Antti Karttunen, Sep 18 2016: (Start)
(End)
a(0) = 2, and for n > 0, a(n) = A329329(a(n-1), 6). - Peter Munn, Jan 04 2020
EXAMPLE
a(0) = 2^T(0,0) = 2^1 = 2.
a(1) = 2^T(1,0) * 3^T(1,1) = 2^1 * 3^1 = 6.
a(2) = 2^T(2,0) * 3^T(2,1) * 5^T(2,2) = 2^1 * 3^0 * 5^1 = 10.
a(3) = 2^T(3,0) * 3^T(3,1) * 5^T(3,2) * 7^T(3,3) = 2^1 * 3^1 * 5^1 * 7^1 = 210.
a(4) = 2^1 * 3^0 * 5^0 * 7^0 * 11^1 = 22.
a(5) = 2^1 * 3^1 * 5^0 * 7^0 * 11^1 * 13^1 = 858.
a(6) = 2^1 * 3^0 * 5^1 * 7^0 * 11^1 * 13^0 * 17^1 = 1870.
a(7) = 2^1 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 * 19^1 = 9699690.
a(8) = 2^1 * 3^0 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 = 46.
a(9) = 2^1 * 3^1 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^1 = 4002.
a(10) = 2^1 * 3^0 * 5^1 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^0 * 31^1 = 7130.
a(11) = 2^1 * 3^1 * 5^1 * 7^1 * 11^0 * 13^0 * 17^0 * 19^0 * 23^1 * 29^1 * 31^1 * 37^1 = 160660290.
a(12) = 2^1 * 3^0 * 5^0 * 7^0 * 11^1 * 13^0 * 17^0 * 19^0 * 23^1 * 29^0 * 31^0 * 37^0 * 41^1 = 20746.
From N. J. A. Sloane, Feb 28 2015: (Start)
Factorizations of initial terms, from Cobeli-Zaharescu paper:
2 = 2
6 = 2*3
10 = 2*5
210 = 2*3*5*7
22 = 2*11
858 = 2*3*11*13
1870 = 2*5*11*17
9699690 = 2*3*5*7*11*13*17*19
46 = 2*23
4002 = 2*3*23*29
7130 = 2*5*23*31
160660290 = 2*3*5*7*23*29*31*37
20746 = 2*11*23*41
1008940218 = 2*3*11*13*23*29*41*43
2569288370 = 2*5*11*17*23*31*41*47
32589158477190044730 = 2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53
... (End)
From Jon E. Schoenfield, Jun 09 2019: (Start)
n | Factorization of a(n)
---+-----------------------------------------------
0 | 2
1 | 2* 3
2 | 2 * 5
3 | 2* 3* 5* 7
4 | 2 *11
5 | 2* 3 *11*13
6 | 2 * 5 *11 *17
7 | 2* 3* 5* 7*11*13*17*19
8 | 2 *23
9 | 2* 3 *23*29
10 | 2 * 5 *23 *31
11 | 2* 3* 5* 7 *23*29*31*37
12 | 2 *11 *23 *41
13 | 2* 3 *11*13 *23*29 *41*43
14 | 2 * 5 *11 *17 *23 *31 *41 *47
15 | 2* 3* 5* 7*11*13*17*19*23*29*31*37*41*43*47*53
... (End)
MAPLE
f:=n->mul(ithprime(k+1)^(binomial(n, k) mod 2), k=0..n);
[seq(f(n), n=0..40)];
MATHEMATICA
a[n_] := Product[Prime[k+1]^Mod[Binomial[n, k], 2], {k, 0, n}];
Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Oct 01 2018, from Maple *)
PROG
(Python)
from operator import mul
from functools import reduce
from sympy import prime
def A123098(n):
return reduce(mul, (1 if ~(n-1) & k else prime(k+1) for k in range(n))) # Chai Wah Wu, Feb 08 2016
(PARI) a(n) = prod (k=0, n, if (binomial(n, k)%2, prime(k+1), 1)) \\ Rémy Sigrist, Jun 09 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Vos Post, Nov 05 2006
EXTENSIONS
Further terms from N. J. A. Sloane, Feb 28 2015
Changed offset from 1 to 0, corresponding changes to formulas and examples from Antti Karttunen, Sep 18 2016
STATUS
approved