cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A104414 Number of prime factors, with multiplicity, of the heptanacci numbers A066178.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 5, 6, 1, 2, 6, 3, 7, 6, 9, 8, 1, 4, 2, 5, 6, 5, 8, 9, 2, 3, 10, 6, 7, 7, 16, 10, 4, 2, 7, 5, 6, 9, 12, 10, 4, 3, 6, 4, 9, 8, 14, 12, 2, 3, 7, 6, 11, 8, 7, 10, 5, 5, 12, 6, 7, 9, 12, 11, 3, 4, 3, 6, 7, 5, 6, 11, 4, 2, 9, 4, 7, 9, 14, 8, 4, 3
Offset: 0

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Author

Jonathan Vos Post, Mar 06 2005

Keywords

Comments

Prime heptanacci numbers: a(2) = 2, a(8) = 127, a(16) = 31489, ... Semiprime heptanacci numbers: a(4) = 4 = 2^2, a(9) = 253 = 11 * 23, a(18) = 124946 = 2 * 62473, a(24) = 7805695 = 5 * 1561139.

Examples

			a(0)=a(1)=0 because the first two nonzero heptanacci numbers are both 1, which has zero prime divisors.
a(2)=1 because the 3rd nonzero heptanacci number is 2, a prime, with only one prime divisor.
a(3)=2 because the 4th nonzero pentanacci number is 4 = 2^2 which has (with multiplicity) 2 prime divisors (which happen to be equal).
a(4)=3 because the 5th nonzero heptanacci number is 8 = 2^3.
a(12)= 7 because A066178(12) = 2000 = 2^4 * 5^3 which has seven prime factors (four of the 2, three of them 5).
		

Crossrefs

Programs

  • Mathematica
    PrimeOmega[#]&/@LinearRecurrence[{1,1,1,1,1,1,1},{1,1,2,4,8,16,32},100] (* Harvey P. Dale, Oct 08 2015 *)

Formula

a(n) = A001222(A066178(n)). a(n) = bigomega(A066178(n)).

Extensions

More terms from Harvey P. Dale, Oct 08 2015

A104415 Number of prime factors, with multiplicity, of the nonzero octanacci numbers A079262.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 5, 6, 7, 3, 1, 4, 5, 6, 6, 11, 7, 9, 1, 4, 3, 6, 8, 7, 8, 11, 10, 2, 2, 8, 4, 9, 7, 11, 11, 12, 3, 2, 4, 5, 6, 9, 10, 11, 12, 2, 4, 10, 5, 10, 9, 17, 12, 10, 4, 4, 4, 9, 11, 8, 8, 12, 12, 4, 4, 10, 11, 9, 11, 15, 13, 9, 5, 6, 5, 9, 6, 9, 9
Offset: 1

Views

Author

Jonathan Vos Post, Mar 06 2005

Keywords

Examples

			a(0)=a(1)=0 because the first two nonzero octanacci numbers are both 1, which has zero prime divisors.
a(2)=1 because the 3rd nonzero octanacci number is 2, a prime, with only one prime divisor.
a(3)=2 because the 4th nonzero octanacci number is 4 = 2^2 which has (with multiplicity) 2 prime divisors (which happen to be equal).
a(4)=3 because the 5th nonzero octanacci number is 8 = 2^3.
a(10)=3 because A079262(10) = 255 = 3 * 5 * 17 which has 3 prime factors.
		

Crossrefs

Formula

a(n) = A001222(A079262(n+6)).

Extensions

Offset corrected and more terms added by Amiram Eldar, Sep 08 2019

A104418 Number of prime factors, with multiplicity, of the nonzero 9-acci numbers.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 6, 3, 5, 7, 9, 9, 11, 9, 3, 2, 2, 8, 7, 7, 7, 10, 11, 10, 3, 2, 7, 8, 11, 7, 12, 13, 15, 11, 3, 2, 6, 7, 7, 10, 9, 12, 12, 13, 5, 2, 5, 8, 8, 7, 13, 12, 10, 12, 6, 3, 3, 6, 12, 11, 12, 10, 12, 12, 2, 6, 12, 8, 11, 9, 14, 13, 13, 13, 7, 2
Offset: 1

Views

Author

Jonathan Vos Post, Mar 06 2005

Keywords

Comments

Prime 9-acci numbers: b(3) = 2, b(12) = 1021, ... Semiprime 9-acci numbers: b(4) = 4 = 2^2, b(11) = 511 = 7 * 73, b(22) = 1035269 = 47 * 22027, b(23) = 2068498 = 2 * 1034249, b(32) = 1049716729 = 1051 * 998779 b(42) = 1064366053385 = 5 * 212873210677, b(52) = 1079219816432629 = 28669 * 37644138841, b(71) = 555323195719171835391 = 3 * 185107731906390611797, b(82) = 1125036467745713090813969 = 37 * 30406391020154407859837.

Examples

			a(1)=a(2)=0 because the first two nonzero 9-acci numbers are both 1, which has zero prime divisors.
a(3)=1 because the 3rd nonzero 9-acci number is 2, a prime, with only one prime divisor.
a(4)=2 because the 4th nonzero 9-acci number is 4 = 2^2 which has (with multiplicity) 2 prime divisors (which happen to be equal).
a(5)=3 because the 5th nonzero 9-acci number is 8 = 2^3.
a(13) = 6 because b(13) = 2040 = 2^3 * 3 * 5 * 17 so has 6 prime factors (2 with multiplicity 3 and 3, 5 and 17 once each).
		

Crossrefs

Formula

a(n) = A001222(A104144(n+7)).
Showing 1-3 of 3 results.