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A318510
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Completely multiplicative with a(prime(k)) = A002487(prime(k+1)).
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2
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1, 2, 3, 4, 3, 6, 5, 8, 9, 6, 5, 12, 5, 10, 9, 16, 7, 18, 7, 12, 15, 10, 7, 24, 9, 10, 27, 20, 5, 18, 11, 32, 15, 14, 15, 36, 11, 14, 15, 24, 13, 30, 9, 20, 27, 14, 13, 48, 25, 18, 21, 20, 11, 54, 15, 40, 21, 10, 9, 36, 11, 22, 45, 64, 15, 30, 13, 28, 21, 30, 15, 72, 13, 22, 27, 28, 25, 30, 19, 48, 81, 26, 17, 60, 21, 18, 15, 40
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OFFSET
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1,2
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COMMENTS
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Provided that the conjecture given in A261179 holds, then for all n >= 1, A007814(a(n)) = A007814(n), i.e., then the sequence preserves the 2-adic valuation of n.
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LINKS
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FORMULA
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PROG
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(PARI)
A002487(n) = { my(a=1, b=0); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); (b); }; \\ From A002487
A318510(n) = { my(f=factor(n)); for (i=1, #f~, f[i, 1] = A002487(prime(1+primepi(f[i, 1])))); factorback(f); };
(Python)
from math import prod
from functools import reduce
from sympy import factorint, nextprime
def A318510(n): return prod(sum(reduce(lambda x, y:(x[0], x[0]+x[1]) if int(y) else (x[0]+x[1], x[1]), bin(nextprime(p))[-1:2:-1], (1, 0)))**e for p, e in factorint(n).items()) # Chai Wah Wu, May 18 2023
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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