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 1-4 of 4 results.

A277018 Numbers n for which A277017(n) = 0; range of A277022 sorted into ascending order.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 16, 17, 18, 20, 21, 22, 24, 25, 26, 28, 32, 33, 34, 36, 37, 38, 40, 41, 42, 44, 45, 48, 49, 50, 52, 53, 54, 56, 57, 58, 60, 64, 65, 66, 68, 69, 70, 72, 73, 74, 76, 77, 80, 81, 82, 84, 85, 86, 88, 89, 90, 92, 96, 97, 98, 100, 101, 102, 104, 105, 106, 108, 109, 112, 113, 114, 116, 117, 118, 120
Offset: 0

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Author

Antti Karttunen, Sep 26 2016

Keywords

Comments

Numbers such that no run of 1-bits has length >= A000040(1 + the total number of 0-bits anywhere right of that run in the binary expansion of n).
Indexing starts from zero as a(0) = 0 is a special case in this sequence.

Examples

			60 ("111100" in binary, A007088) is present as 4 < prime(2+1) = 5.

Links

Crossrefs

Complement: A277019.
Positions of zeros in A277017.
Sequence A277022 sorted into ascending order.
Cf. A000040, A005940, A007088, A129251.
Differs from its subsequence of A277008 for the first time at n=41, where a(41)=60, a value which is missing from A277008.

Formula

Other identitities:
A129251(A005940(1+a(n))) = 0 for all n.

A277007 Number of maximal runs of 1-bits (in binary expansion of n) such that the length of run > 1 + the total number of zeros anywhere right of that run.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0
Offset: 0

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Author

Antti Karttunen, Sep 25 2016

Keywords

Examples

			For n=3, "11" in binary, the only maximal run of 1-bits is of length 2, and 2 > 0+1 (where 0 is the total number of zeros to the right of it), thus a(3) = 1.
For n=59, "111011" in binary, both the length of run "11" at the least significant end exceeds the limit (see case n=3 above), and also the length of run "111" exceeds 1 + the total number of 0's to the right of it, thus a(59) = 1+1 = 2.
For n=60, "111100" in binary, the length of only run of 1's is 4, and 4 > 2+1, thus a(60) = 1.
For n=118, "1110110" in binary, the length of rightmost run of 1-bits is 2, but that is not > 1+1 (one more than the number of 0-bits right to it). Also, the length of the leftmost run of 1-bits is 3, but that is not > 1+2, thus a(118) = 0.
For n=246, "11110110" in binary, the rightmost run of 1-bits does not contribute, but the leftmost run of 1-bits has now length 4, which is more than 1+2 (where 2 is the total number of 0-bits right of it), thus a(246) = 1.

Links

Crossrefs

Cf. A277008 (positions of zeros), A277009 (of nonzeros).
Cf. A005940, A277011, A277012, A276077.
Differs from the similar entry A277017 for the first time at n=60, where a(60)=1, while A277017(60)=0.

Programs

  • Scheme
    ;; Standalone iterative implementation:
(define (A277007 n) (let loop ((e 0) (n n) (z 0) (r 0)) (cond ((zero? n) (+ e (if (> r (+ 1 z)) 1 0))) ((even? n) (loop (+ e (if (> r (+ 1 z)) 1 0)) (/ n 2) (+ 1 z) 0)) (else (loop e (/ (- n 1) 2) z (+ 1 r))))))

Formula

a(n) = A276077(A005940(1+n)).

A277019 Numbers not in range of A277022.

Original entry on oeis.org

3, 7, 11, 14, 15, 19, 23, 27, 29, 30, 31, 35, 39, 43, 46, 47, 51, 55, 59, 61, 62, 63, 67, 71, 75, 78, 79, 83, 87, 91, 93, 94, 95, 99, 103, 107, 110, 111, 115, 119, 123, 124, 125, 126, 127, 131, 135, 139, 142, 143, 147, 151, 155, 157, 158, 159, 163, 167, 171, 174, 175, 179, 183, 187, 189, 190, 191, 195, 199, 203, 206, 207, 211, 215
Offset: 1

Views

Author

Antti Karttunen, Sep 26 2016

Keywords

Comments

Numbers such that at least one run of 1-bits in their binary expansion has length >= A000040(1 + the total number of 0-bits anywhere right of that run).
Numbers n for which A277017(n) > 0.
Numbers n for which A129251(A005940(1+n)) > 0.

Examples

			3 ("11" in binary, A007088) is present as the length of that only run of 1's is 2, and 2 >= prime(1+0), where 0 is the total number of 0's to the right of that run.
60 ("111100" in binary) is NOT present as 4 < 5 = prime(2+1).

Links

Crossrefs

Complement: A277018.
Positions of nonzeros in A277017.
Cf. A000040, A005940, A129251.
Subsequence of A277009 from which this differs for the first time at n=20, where a(20)=61, skipping the value 60 present in A277009.

A277021 Left inverse of A277022.

Original entry on oeis.org

0, 1, 2, 2, 6, 3, 4, 3, 30, 7, 8, 4, 12, 5, 6, 4, 210, 31, 32, 8, 36, 9, 10, 5, 60, 13, 14, 6, 18, 7, 8, 5, 2310, 211, 212, 32, 216, 33, 34, 9, 240, 37, 38, 10, 42, 11, 12, 6, 420, 61, 62, 14, 66, 15, 16, 7, 90, 19, 20, 8, 24, 9, 10, 6, 30030, 2311, 2312, 212, 2316, 213, 214, 33, 2340, 217, 218, 34, 222, 35, 36, 10, 2520, 241, 242
Offset: 0

Views

Author

Antti Karttunen, Sep 26 2016

Keywords

Links

Crossrefs

Left inverse of A277022.
Cf. A005940, A276085.
Cf. also A277017.

Programs

  • Python
    from sympy import primorial, primepi, prime, factorint, floor, log
def a002110(n): return 1 if n<1 else primorial(n)
def a276085(n):
    f=factorint(n)
    return sum([f[i]*a002110(primepi(i) - 1) for i in f])
def A(n): return n - 2**int(floor(log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a(n): return a276085(b(n - 1))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 22 2017
  • Scheme
    (define (A277021 n) (let loop ((s 0) (n n) (r 0) (i 1) (pr 1)) (cond ((zero? n) (+ s (* r pr))) ((even? n) (loop (+ s (* r pr)) (/ n 2) 0 (+ 1 i) (* (A000040 i) pr))) (else (loop s (/ (- n 1) 2) (+ 1 r) i pr)))))

Formula

a(n) = A276085(A005940(1+n)).
Other identities. For all n >= 0:
a(A277022(n)) = n.
Showing 1-4 of 4 results.

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