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A123098
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Multiplicative encoding of triangle formed by reading Pascal's triangle mod 2 (A047999).
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9
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2, 6, 10, 210, 22, 858, 1870, 9699690, 46, 4002, 7130, 160660290, 20746, 1008940218, 2569288370, 32589158477190044730, 118, 21594, 39530, 3595293030, 94754, 17808161514, 44788794490, 7074421030108255253430, 263258, 141108130806, 281595235990, 296987147493893719182390, 944729501606
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OFFSET
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0,1
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COMMENTS
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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
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LINKS
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FORMULA
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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).
(End)
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EXAMPLE
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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.
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)
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)
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MAPLE
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f:=n->mul(ithprime(k+1)^(binomial(n, k) mod 2), k=0..n);
[seq(f(n), n=0..40)];
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MATHEMATICA
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a[n_] := Product[Prime[k+1]^Mod[Binomial[n, k], 2], {k, 0, n}];
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PROG
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(Python)
from operator import mul
from functools import reduce
from sympy import prime
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
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CROSSREFS
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Cf. A000040, A000120, A001316, A001317, A007188, A007318, A007913, A047999, A019565, A255484, A329329.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Changed offset from 1 to 0, corresponding changes to formulas and examples from Antti Karttunen, Sep 18 2016
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STATUS
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approved
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