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.

A026353 a(n) = sum of the numbers between the two n's in A026350.

Original entry on oeis.org

0, 4, 8, 17, 29, 38, 55, 67, 89, 114, 131, 161, 194, 216, 254, 279, 322, 368, 398, 449, 482, 538, 597, 635, 699, 766, 809, 881, 927, 1004, 1084, 1135, 1220, 1308, 1364, 1457, 1516, 1614, 1715, 1779, 1885, 1952, 2063, 2177, 2249
Offset: 2

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Keywords

A026351 a(n) = floor(n*phi) + 1, where phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 4, 5, 7, 9, 10, 12, 13, 15, 17, 18, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 47, 49, 51, 52, 54, 56, 57, 59, 60, 62, 64, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89, 91, 93, 94, 96, 98, 99
Offset: 0

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Comments

a(n)=least k such that s(k)=n, where s=A026350.
a(n)=position of n-th 1 in A096270.
From Wolfdieter Lang, Jun 27 2011: (Start)
a(n) = A(n)+1, with Wythoff sequence A(n)=A000201(n), n>=1, and A(0)=0.
a(n) = -floor(-n*phi). Recall that floor(-x) = -(floor(x)+1) if x is not integer and -floor(x) otherwise.
An exhaustive and disjoint decomposition of the integers is given by the following two Wythoff sequences A' and B: A'(0):=-1 (not 0), A'(-n):=-a(n)=-(A(n)+1), n>=1, A'(n) = A(n), n>=1, and B(-n):=-(B(n)+1)= -A026352(n), n>=1, with B(n)=A001950(n), n>=1, and B(0)=0.
(End)
Where odd terms in A060142 occur: A060142(a(n)) = A219608(n). - Reinhard Zumkeller, Nov 26 2012

Crossrefs

Essentially same as A004956. Cf. A000201.
Complement of A026352.
Cf. A283733 (partial sums).

Programs

  • Haskell
    import Data.List (findIndices)
    a026351 n = a026351_list !! n
    a026351_list = findIndices odd a060142_list
    -- Reinhard Zumkeller, Nov 26 2012
    
  • Mathematica
    Table[Floor[n*GoldenRatio] + 1, {n, 0, 100}] (* T. D. Noe, Apr 15 2011 *)
  • Python
    from math import isqrt
    def A026351(n): return (n+isqrt(5*n**2)>>1)+1 # Chai Wah Wu, Aug 17 2022

A026352 a(n) = floor(n*tau) + n + 1 where tau is the golden ratio A001622.

Original entry on oeis.org

1, 3, 6, 8, 11, 14, 16, 19, 21, 24, 27, 29, 32, 35, 37, 40, 42, 45, 48, 50, 53, 55, 58, 61, 63, 66, 69, 71, 74, 76, 79, 82, 84, 87, 90, 92, 95, 97, 100, 103, 105, 108, 110, 113, 116, 118, 121, 124, 126, 129, 131, 134, 137, 139, 142, 144
Offset: 0

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Author

Clark Kimberling, Dec 11 1999

Keywords

Comments

a(n) = greatest k such that s(k) = n+1, where s = A026350.
Indices at which blocks (0;1) occur in infinite Fibonacci word; i.e., n such that A005614(n)=0 and A005614(n+1)=1. - Benoit Cloitre, Nov 15 2003
Except for the first term, these are the numbers whose lazy Fibonacci representation (see A095791) includes both 1 and 2; thus this sequence is a subsequence of the lower Wythoff sequence, A000201. - Clark Kimberling, Jun 10 2004 [A-number typo corrected by Nathan Fox, May 03 2014]
a(n) = n-th number k whose lazy Fibonacci representation (as in A095791) has more summands than that of k-1. - Clark Kimberling, Jun 12 2004
a(n) = position of n-th 0 in A096270. - Clark Kimberling, Apr 22 2011
Maximum number of chips in a pile created at each step in the game described by Roland Schroeder in his comment at A000201. (From Allan C. Wechsler via Seqfan.)
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,3)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+3. - Michael De Vlieger, Jul 28 2025

Crossrefs

Essentially same as A004957.
Subsequence of A000201.
Complement of A026351.

Programs

Formula

a(n) = A026351(n)+n. - R. J. Mathar, Jun 24 2025

A026272 a(n) = smallest k such that k=a(n-k-1) is the only appearance of k so far; if there is no such k, then a(n) = least positive integer that has not yet appeared.

Original entry on oeis.org

1, 2, 1, 3, 2, 4, 5, 3, 6, 7, 4, 8, 5, 9, 10, 6, 11, 7, 12, 13, 8, 14, 15, 9, 16, 10, 17, 18, 11, 19, 20, 12, 21, 13, 22, 23, 14, 24, 15, 25, 26, 16, 27, 28, 17, 29, 18, 30, 31, 19, 32, 20, 33, 34, 21, 35, 36, 22, 37, 23, 38, 39, 24, 40, 41, 25
Offset: 1

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From Daniel Joyce, Apr 13 2001: (Start)
This sequence displays every positive integer exactly twice and the gap between the two occurrences of n contains exactly n other values. The first occurrence of n precedes the first occurrence of n+1.
Also related to the Wythoff array (A035513) and the Para-Fibonacci sequence (A035513) where every positive integer is displayed exactly once in the whole array. Take any integer n in A026272 and let C = number of terms from the beginning of the sequence to the second occurrence of n. Then C = (2nd term after n in the applicable sequence for n in A035513).
Also in the second occurrence of n in A026272, let N=n ( - one term) = (first term value after n in the applicable sequence for n in A035513). In this format the second occurrence of n in A026272 will produce in A035513, n itself and two of the succeeding terms of n in the Wythoff array where every positive integer can only be displayed once.
In A026272 if |a(n)-a(n+1)| > 10 then phi ~ a(n)/|a(n)-a(n+1)|. When n -> infinity it will converge to phi. (End)
Or, put a copy of n in A000027 n places further along! - Zak Seidov, May 24 2008
Another version would prefix this sequence with two leading 0's (see the Angelini reference). If we use this form and write down the indices of the two 0's, the two 1's, the two 2's, the two 3's, etc., then we get A072061. - Jacques ALARDET, Jul 26 2008

References

  • Eric Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.

Crossrefs

Programs

  • Mathematica
    s=Range[1000];n=0;Do[n++;s=Insert[s,n,Position[s,n][[1]]+n+1],{500}];A026272=Take[s,1000] (* Zak Seidov, May 24 2008 *)
  • PARI
    A026272=apply(t->t-1,A026242[3..-1]) \\ Use vecextract(A026242,"3..") in PARI versions < 2.7. - M. F. Hasler, Sep 17 2014
    
  • Python
    from collections import Counter
    from itertools import count, islice
    def agen(): # generator of terms
        aset, alst, k, mink, counts = set(), [0], 0, 1, Counter()
        for n in count(1):
            for k in range(1, len(alst)-1):
                if k == alst[n-k-1] and counts[alst[n-k-1]] == 1: an = k; break
            else: an = mink
            yield an; aset.add(an); alst.append(an); counts.update([an])
            while mink in aset: mink += 1
    print(list(islice(agen(), 66))) # Michael S. Branicky, Jun 27 2022

Formula

a(n) = A026242(n+2) - 1 = A026350(n+3) - 2 = A026354(n+4) - 3.

Extensions

Edited by Max Alekseyev, May 31 2011
Showing 1-4 of 4 results.