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A170818 a(n) = product of primes (with multiplicity) of form 4k+1 that divide n. 6
1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 1, 1, 13, 1, 5, 1, 17, 1, 1, 5, 1, 1, 1, 1, 25, 13, 1, 1, 29, 5, 1, 1, 1, 17, 5, 1, 37, 1, 13, 5, 41, 1, 1, 1, 5, 1, 1, 1, 1, 25, 17, 13, 53, 1, 5, 1, 1, 29, 1, 5, 61, 1, 1, 1, 65, 1, 1, 17, 1, 5, 1, 1, 73, 37, 25, 1, 1, 13, 1, 5, 1, 41, 1, 1, 85, 1, 29, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

COMMENTS

Completely multiplicative with a(p) = p if p = 4k+1 and a(p) = 1 otherwise. - Tom Edgar, Mar 05 2015

LINKS

Alois P. Heinz, Table of n, a(n) for n = 1..10000

A. Tripathi, On Pythagorean triples containing a fixed integer, Fib. Q., 46/47 (2008/2009), 331-340.

Index to divisibility sequences

FORMULA

a(n) = n/A072438(n). - Michel Marcus, Mar 05 2015

MAPLE

a:= n-> mul(`if`(irem(i[1], 4)=1, i[1]^i[2], 1), i=ifactors(n)[2]):

seq(a(n), n=1..100);  # Alois P. Heinz, Jun 09 2014

PROG

(PARI) a(n)=my(f=factor(n)); prod(i=1, #f~, if(f[i, 1]%4>1, 1, f[i, 1])^f[i, 2]) \\ Charles R Greathouse IV, Jun 28 2015

(Python)

from sympy import factorint

from operator import mul

def a072438(n):

    f = factorint(n)

    return 1 if n == 1 else reduce(mul, [1 if i%4==1 else i**f[i] for i in f])

def a(n): return n/a072438(n) # Indranil Ghosh, May 08 2017

CROSSREFS

Cf. A170817-A170819, A097706, A083025, A072438, A286361.

Sequence in context: A060904 A135469 A170817 * A046622 A170825 A140214

Adjacent sequences:  A170815 A170816 A170817 * A170819 A170820 A170821

KEYWORD

nonn,mult

AUTHOR

N. J. A. Sloane, Dec 22 2009

STATUS

approved

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Last modified August 20 11:43 EDT 2017. Contains 290835 sequences.