

A174842


Irregular triangle T(i,n) giving the number of elements of Zp having multiplicative order di, the ith divisor of p1, where p is the nth prime.


2



1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 2, 6, 6, 1, 1, 10, 10, 1, 1, 2, 6, 6, 12, 1, 1, 2, 4, 2, 4, 8, 8, 1, 1, 2, 2, 2, 6, 4, 6, 12, 1, 1, 2, 4, 4, 4, 8, 16, 1, 1, 2, 2, 6, 6, 12, 12, 1, 1, 22, 22, 1, 1, 2, 12, 12, 24, 1, 1, 28, 28, 1, 1, 2, 2, 4, 2, 4, 4, 8, 8
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OFFSET

1,6


COMMENTS

The divisors of p1 are assumed to be in increasing order. The first row, for prime 2, has only one term. All other rows begin with two 1s and end with phi(p1). There are tau(p1), the number of divisors of p1, terms in each row. The sum of the terms in each row is p1. When p is a prime of the form 4k1, then the last two terms in the row are equal. When p is a prime of the form 4k+1, then the last two terms in the row have a ratio of 2.


LINKS

T. D. Noe, Rows n=1..500 of triangle, flattened
Eric W. Weisstein, MathWorld: Modulo Multiplication Group


FORMULA

T(i,n) = phi(di), where di is the ith divisor of prime(n)1.


EXAMPLE

For prime p=17, the 7th prime, the multiplicative order of the numbers 1 to p1 is 1, 8, 16, 4, 16, 16, 16, 8, 8, 16, 16, 16, 4, 16, 8, 2. There is one 1, one 2, two 4's, four 8's, and eight 16's. Hence row 7 is 1, 1, 2, 4, 8.


MATHEMATICA

Flatten[Table[EulerPhi[Divisors[p1]], {p, Prime[Range[100]]}]]


CROSSREFS

A008328 (tau(p1)), A008330 (phi(p1)), A174843 (divisors of p1)
Sequence in context: A106476 A101566 A176653 * A156074 A051287 A278218
Adjacent sequences: A174839 A174840 A174841 * A174843 A174844 A174845


KEYWORD

nonn,tabf


AUTHOR

T. D. Noe, Mar 30 2010


STATUS

approved



