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.

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A379410 a(n) = n + floor(n*r/t) + floor(n*s/t), where r = u^(1/4), s = u^(1/2), t = u^(3/4), u = golden ratio (A001622).

Original entry on oeis.org

1, 4, 7, 10, 12, 15, 18, 21, 23, 25, 28, 31, 34, 37, 39, 42, 45, 47, 49, 52, 55, 58, 61, 63, 66, 69, 71, 74, 76, 79, 82, 85, 87, 90, 93, 95, 98, 100, 103, 106, 109, 112, 114, 117, 119, 122, 124, 127, 130, 133, 136, 138, 140, 143, 146, 149, 151, 154, 157, 160, 162, 164, 167, 170
Offset: 1

Views

Author

Clark Kimberling, Jan 21 2025

Keywords

Comments

This sequence, A379408, and A379409 partition the positive integers.

Crossrefs

Cf. A001622, A184908, A379408, A379409.

Programs

  • Mathematica
    u = (1 + 5^(1/2))/2;
r = u^(1/4); s = u^(1/2); t = u^(3/4);
Table[n + Floor[n*s/r] + Floor[n*t/r], {n, 1, 120}]  (* A379408 *)
Table[n + Floor[n*r/s] + Floor[n*t/s], {n, 1, 120}]  (* A379409 *)
Table[n + Floor[n*r/t] + Floor[n*s/t], {n, 1, 120}]  (* A379410 *)

Formula

a(n) = n + floor(n/r^2) + floor(n/r), where r = u^(1/4), u = golden ratio.

Extensions

Values corrected by R. J. Mathar, Jan 28 2025

A184907 Let S_n be the set of the integers having alternating bit sum equal to n. There are a(n) primes among the smallest 2n+1 numbers of S_n.

Original entry on oeis.org

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

Views

Author

Washington Bomfim, Jan 25 2011

Keywords

Comments

The smallest 2n+1 numbers in the set S_n are given by (4^(n+1)-1)/3 - (-2)^k for k in 0..2*n. - Andrew Howroyd, Dec 15 2024

Examples

			The smallest 2n+1 = 5 numbers of the set S_2 of the integers having alternating bit sum 2, are 5, 17, 20, 23, and 29, so a(2)=4.

Links

Crossrefs

Cf. A065359, A065084, A002450, A184908.

Programs

  • PARI
    II()={i = (4^n - 1)/3 - 2^(2*n-2) + 2^(2*n); if(isprime(i),an++)};
III()={w = 2^(2*n-2); for(j=1, n-1, i += w; w /= 4; i -= w; if(isprime(i), an++;))};
IV()={i+=3; if(isprime(i), an++); w=2; for(j=1, n-1, i -= w; w *= 4; i+=w; if(isprime(i),an++))};
print1("0, 1, 4, ");for(n=3,80, an=0; II(); III(); IV(); print1(an,", ")) \\ Washington Bomfim, Jan 25 2011
  • PARI
    a(n)={my(m=(4^(n+1)-1)/3); sum(k=0, 2*n, isprime(bitxor(m,1<Andrew Howroyd, Dec 15 2024

Formula

a(n) = 0 for n == 0 (mod 3). - Andrew Howroyd, Dec 15 2024
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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