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 1-6 of 6 results.

A354280 a(n) is the numerator of Cesàro means sequence c(n) of A237420 when the denominator is A141310(n).

Original entry on oeis.org

0, 0, 2, 1, 6, 2, 12, 3, 20, 4, 30, 5, 42, 6, 56, 7, 72, 8, 90, 9, 110, 10, 132, 11, 156, 12, 182, 13, 210, 14, 240, 15, 272, 16, 306, 17, 342, 18, 380, 19, 420, 20, 462, 21, 506, 22, 552, 23, 600, 24, 650, 25, 702, 26, 756, 27, 812, 28, 870, 29, 930, 30, 992, 31, 1056, 32, 1122, 33, 1190
Offset: 0

Views

Author

Bernard Schott, May 22 2022

Keywords

Comments

So, we get c(n) = a(n) / A141310(n) for n >= 0 (see Formula and Example section).
Cesàro mean theorem: when the series u(n) has a limit (finite or infinite) in the usual sense, then c(n) = (u(0)+...+u(n))/(n+1) has the same Cesàro limit, but the converse is false.
A237420 is such a counterexample in the case of an infinite limit.
Proof: A237420 is not convergent in the usual sense because a(2n+1) = 0, while a(2n) -> oo when n -> oo. Now, the successive arithmetic means c(n) of the first n terms of the sequence are 0/1, 0/2, 2/3, 2/4, 6/5, 6/6, 12/7, 12/8, 20/9, 20/10, ... so c(2n)= (n*(n+1))/(2*n+1) ~ n/2 and c(2n+1) = n/2, hence the Cesàro limit is infinity because c(n) -> oo as n -> oo (Arnaudiès et al.), QED.
The first few irreducible fractions c(n) are in the last row of the Example section. The differences between row 4 and last row exist only when n = 4*k+1, k>0, then respectively c(n) = 2k/2 = k/1.
This sequence consists of the oblong numbers (A002378) interlaced with the natural numbers (A001477)
Note that A033999 is a counterexample in the case of a finite Cesàro limit.
Also, the converse of the Cesàro mean theorem is true iff u(n) is monotonic.

Examples

			Table with the first few terms:
       Indices n         :   0,   1,   2,   3,   4,   5,    6,   7,    8,   9, ...
       A237420(n)        :   0,   0,   2,   0,   4,   0,    6,   0,    8,   0, ...
      Partial sums       :   0,   0,   2,   2,   6,   6,   12,  12,   20,  20, ...
    Cesàro means c(n)    : 0/1, 0/2, 2/3, 1/2, 6/5, 2/2, 12/7, 3/2, 20/9, 4/2, ...
      Numerator a(n)     :   0,   0,   2,   1,   6,   2,   12,   3,   20,   4, ...
Denominator A141310(n)   :   1,   2,   3,   2,   5,   2,    7,   2,    9,   2, ...
Irreducible Cesàro mean  : 0/1, 0/2, 2/3, 1/2, 6/5, 1/1, 12/7, 3/2, 20/9, 2/1, ...
		

References

  • J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Crossrefs

Cf. A001477, A002378, A033999, A141310 (denominators), A237420.

Programs

  • Mathematica
    m = 50; Accumulate[Table[If[OddQ[n], 0, n], {n, 0, 2*m - 1}]] * Flatten[Table[{2*n - 1, 2}, {n, 1, m}]] / Range[2*m] (* Amiram Eldar, Jun 05 2022 *)
  • PARI
    c(n) = sum(k=0, n, if (k%2, 0, k))/(n+1);
    f(n) = if(n%2, 2, 1+n); \\ A141310
    a(n) = c(n)*f(n); \\ Michel Marcus, Jun 06 2022

Formula

a(n) = (A141310(n)/(n+1)) * Sum_{k=0..n} A237420(k).
For n >= 0, a(2n) = n*(n+1) = A002378(n), a(2n+1) = n = A001477(n).
G.f.: x^2*(2 + x - x^3)/(1 - x^2)^3. - Stefano Spezia, May 23 2022

A114112 a(1)=1, a(2)=2; thereafter a(n) = n+1 if n odd, n-1 if n even.

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14, 13, 16, 15, 18, 17, 20, 19, 22, 21, 24, 23, 26, 25, 28, 27, 30, 29, 32, 31, 34, 33, 36, 35, 38, 37, 40, 39, 42, 41, 44, 43, 46, 45, 48, 47, 50, 49, 52, 51, 54, 53, 56, 55, 58, 57, 60, 59, 62, 61, 64, 63, 66, 65, 68, 67, 70, 69, 72, 71
Offset: 1

Views

Author

Leroy Quet, Nov 13 2005

Keywords

Comments

a(1)=1; for n>1, a(n) is the smallest positive integer not occurring earlier in the sequence such that a(n) does not divide Sum_{k=1..n-1} a(k). - Leroy Quet, Nov 13 2005 (This was the original definition. A simple induction argument shows that this is the same as the present definition. - N. J. A. Sloane, Mar 12 2018)
Define b(1)=2; for n>1, b(n) is the smallest number not yet in the sequence which shares a prime factor with the sum of all preceding terms. Then a simple induction argument shows that the b(n) sequence is the same as the present sequence with the first term omitted. - David James Sycamore, Feb 26 2018
Here are the details of the two induction arguments (Start)
For a(n), let A(n) = a(1)+...+a(n). The claim is that for n>2 a(n)=n+1 if n odd, n-1 if n even.
The induction hypotheses are: for i
For b(n), the argument is very similar, except that the missing numbers when looking for b(n) are slightly different, and (setting B(n) = b(1)+...b(n)), we have B(2i)=(i+1)(2i+1), B(2i+1)=(i+2)(2i+1). - N. J. A. Sloane, Mar 14 2018
When sequence a(n) is increasing, then the Cesàro means sequence c(n) = (a(1)+...+a(n))/n is also increasing, but the converse is false. This sequence is a such an example where c(n) is increasing, while a(n) is not increasing (Arnaudiès et al.). See proof in A354008. - Bernard Schott, May 11 2022

References

  • J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Crossrefs

All of A014681, A103889, A113981, A114112, A114285 are essentially the same sequence. - N. J. A. Sloane, Mar 12 2018
Cf. A114113 (partial sums).
See A084265 for the partial sums of the b(n) sequence.
About Cesàro mean theorem: A033999, A141310, A237420, A354008.
Cf. A244009.

Programs

  • Mathematica
    a[1] = 1; a[n_] := a[n] = Block[{k = 1, s, t = Table[ a[i], {i, n - 1}]}, s = Plus @@ t; While[ Position[t, k] != {} || Mod[s, k] == 0, k++ ]; k]; Array[a, 72] (* Robert G. Wilson v, Nov 18 2005 *)
  • PARI
    a(n) = if (n<=2, n, if (n%2, n+1, n-1)); \\ Michel Marcus, May 16 2022
    
  • Python
    def A114112(n): return n + (0 if n <= 2 else -1+2*(n%2)) # Chai Wah Wu, May 24 2022

Formula

G.f.: x*(x^4-2*x^3+x^2+x+1)/((1-x)*(1-x^2)). - N. J. A. Sloane, Mar 12 2018
The g.f. for the b(n) sequence is x*(x^3-3*x^2+2*x+2)/((1-x)*(1-x^2)). - Conjectured (correctly) by Colin Barker, Mar 04 2018
E.g.f.: 1 - x + x^2/2 + (x - 1)*cosh(x) + (x + 1)*sinh(x). - Stefano Spezia, Sep 02 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - log(2) (A244009). - Amiram Eldar, Jun 29 2025

Extensions

More terms from Robert G. Wilson v, Nov 18 2005
Entry edited (with simpler definition) by N. J. A. Sloane, Mar 12 2018

A299483 Irregular triangle read by rows in which row n lists the odd divisors of n in increasing order together with the even divisors of n in decreasing order.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 2, 1, 5, 1, 3, 6, 2, 1, 7, 1, 8, 4, 2, 1, 3, 9, 1, 5, 10, 2, 1, 11, 1, 3, 12, 6, 4, 2, 1, 13, 1, 7, 14, 2, 1, 3, 5, 15, 1, 16, 8, 4, 2, 1, 17, 1, 3, 9, 18, 6, 2, 1, 19, 1, 5, 20, 10, 4, 2, 1, 3, 7, 21, 1, 11, 22, 2, 1, 23, 1, 3, 24, 12, 8, 6, 4, 2, 1, 5, 25, 1, 13, 26, 2, 1, 3, 9, 27
Offset: 1

Author

Omar E. Pol, Feb 11 2018

Keywords

Comments

Consider the diagram with overlapping periodic curves that appears in the Links section (figure 1). The number of curves that contain the point [n,0] equals the number of divisors of n. The simpler interpretation of the diagram is that the curve of diameter d represents the divisor d of n. Now here we introduce a new interpretation: the curve of diameter d that contains the point [n,0] represents the divisor c of n, where c = n/d. This model has the property that each odd quadrant centered at [n,0] contains the curves that represent the even divisors of n, and each even quadrant centered at [n,0] contains the curves that represent the odd divisors of n.
We can find the n-th row of the triangle as follows:
Consider only the semicircumferences that contain the point [n,0].
In the second quadrant from top to bottom we can see the curves that represent the odd divisors of n in increasing order. Also we can see these curves in the fourth quadrant from bottom to top.
Then, if n is an even number, in the first quadrant from bottom to top we can see the curves that represent the even divisors of n in decreasing order. Also we can see these curves in the third quadrant from top to bottom (see example).
Sequences of the same family are shown below:
-----------------------------------
Triangle Order of divisors of n
-----------------------------------
A299481 odd v t.w. even ^
This seq. odd ^ t.w. even v
A319844 even v t.w. odd ^
A319845 even ^ t.w. odd v
A319846 odd v t.w. even v
A319847 odd ^ t.w. even ^
A319848 even v t.w. odd v
A319849 even ^ t.w. odd ^
-----------------------------------
In the above table we have that:
"even v" means "even divisors of n in decreasing order".
"even ^" means "even divisors of n in increasing order".
"odd v" means "odd divisors of n in decreasing order".
"odd ^" means "odd divisors of n in increasing order".
"t.w." means "together with".

Examples

			Triangle begins:
1;
1,  2;
1,  3;
1,  4,  2;
1,  5;
1,  3,  6,  2;
1,  7;
1,  8,  4,  2;
1,  3,  9;
1,  5, 10,  2;
1, 11;
1,  3, 12,  6, 4, 2;
1, 13;
1,  7, 14,  2;
1,  3,  5, 15;
1, 16,  8,  4, 2;
1, 17;
1,  3,  9, 18, 6, 2;
1, 19;
1,  5, 20, 10, 4, 2;
1,  3,  7, 21;
1, 11, 22,  2;
1, 23;
1,  3, 24, 12, 8, 6, 4, 2;
1,  5, 25;
1, 13, 26,  2;
1,  3,  9, 27;
1,  7, 28, 14, 4, 2;
...
For n = 12 the divisors of 12 are [1, 2, 3, 4, 6, 12]. The odd divisors of 12 in increasing order are [1, 3], and the even divisors of 12 in decreasing order are [12, 6, 4, 2], so the 12th row of triangle is [1, 3, 12, 6, 4, 2].
On the other hand, consider the diagram that appears in the Links section (figure 1). Then consider only the semicircumferences that contain the point [12,0]. In the second quadrant, from top to bottom, we can see the curves with diameters [12, 4]. Also we can see these curves in the fourth quadrant from bottom to top. The associated numbers c = 12/d are [1, 3] respectively. These are the odd divisors of 12 in increasing order. Then, in the first quadrant, from bottom to top, we can see the curves with diameters [1, 2, 3, 6]. Also we can see these curves in the third quadrant from top the bottom. The associated numbers c = 12/d are [12, 6, 4, 2] respectively. These are the even divisors of n in decreasing order. Finally all numbers c obtained are [1, 3, 12, 6, 4, 2] equaling the 12th row of triangle.
		

Crossrefs

Row sums give A000203.
Row n has length A000005(n).
Column 1 gives A000012.
Right border gives A141310.

Programs

  • PARI
    row(n) = my(d=divisors(n)); concat(select(x->(x%2), d), Vecrev(select(x->!(x%2), d)));
    lista(nn) = {for (n=1, nn, my(r = row(n)); for (k=1, #r, print1(r[k], ", ")););} \\ Michel Marcus, Jan 17 2019

A319702 Filter sequence for sequences that are constant for all even terms >= 2.

Original entry on oeis.org

1, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 2, 11, 2, 12, 2, 13, 2, 14, 2, 15, 2, 16, 2, 17, 2, 18, 2, 19, 2, 20, 2, 21, 2, 22, 2, 23, 2, 24, 2, 25, 2, 26, 2, 27, 2, 28, 2, 29, 2, 30, 2, 31, 2, 32, 2, 33, 2, 34, 2, 35, 2, 36, 2, 37, 2, 38, 2, 39, 2, 40, 2, 41, 2, 42, 2, 43, 2, 44, 2, 45, 2, 46, 2, 47, 2, 48, 2, 49, 2, 50, 2, 51, 2, 52, 2, 53, 2, 54, 2, 55, 2, 56, 2, 57, 2, 58, 2, 59, 2, 60, 2, 61, 2
Offset: 1

Author

Antti Karttunen, Oct 02 2018

Keywords

Comments

Restricted growth sequence transform of A141310.
For n > 2, a(n-1) is the number of occurrences of n in A319840. - Stefano Spezia, Apr 07 2023

Crossrefs

Programs

  • PARI
    A319702(n) = if(n<=2, n, if(!(n%2), 2, (n+3)/2));

Formula

a(1) = 1, and for n > 1, if n is even, a(n) = 2, otherwise a(n) = (n+3)/2.
From Stefano Spezia, Apr 07 2023: (Start)
O.g.f.: x*(1 + 2*x + x^2 - 2*x^3 - x^4)/((1 - x)^2*(1 + x)^2).
E.g.f.: ((4 + x)*cosh(x) + 3*sinh(x) - 2*(2 + x))/2. (End)

A354008 Numerators of Cesàro means sequence of A114112.

Original entry on oeis.org

1, 3, 7, 5, 16, 7, 29, 9, 46, 11, 67, 13, 92, 15, 121, 17, 154, 19, 191, 21, 232, 23, 277, 25, 326, 27, 379, 29, 436, 31, 497, 33, 562, 35, 631, 37, 704, 39, 781, 41, 862, 43, 947, 45, 1036, 47, 1129, 49, 1226, 51, 1327, 53, 1432, 55, 1541, 57, 1654, 59, 1771, 61, 1892, 63, 2017, 65
Offset: 1

Author

Bernard Schott, May 13 2022

Keywords

Comments

This sequence lists the numerators of c(n) = (Sum_{k=1..n} A114112(k)) / n. The corresponding denominator is A141310(n-1) (see Example section).
When a sequence u(n) is increasing, then Cesàro means sequence c(n) = (u(1)+...+u(n))/n is also increasing, but the converse is false.
A114112 is such a counterexample.
Proof: A114112 is clearly not increasing; now, the successive arithmetic means c(n) of the first specific terms of the sequence are 1/1, 3/2, 7/3, 10/4, 16/5, 21/6, 29/7, ... so, if m >= 1, c(2m) = (2m+1)/2 and c(2m+1) = m+1 + 1/(2m+1), c(1) = 1. We get c(n) = a(n) / A141310(n-1) for n >= 1.
We have c(2m+1) - c(2m) = 1/(2m+1) + 1/2 > 0 and c(2m+2) - c(2m+1) = (2m-1) / (4m+2) > 0 when m >= 1; hence for m >= 1, c(2m) < c(2m+1) < c(2m+2), and also c(1) = 1 < c(2) = 3/2; QED.

Examples

			Table with the first few terms:
       Indices n        :  1,   2,   3,   4,    5,   6,    7,   8,    9,   10, ...
       A114112(n)       :  1,   2,   4,   3,    6,   5,    8,   7,   10,    9, ...
      Partial sums      :  1,   3,   7,  10,   16,  21,   29,  36,   46,   55, ...
    Cesàro means c(n)   :  1, 3/2, 7/3, 5/2, 16/5, 7/2, 29/7, 9/2, 46/9, 11/2, ...
      Numerator a(n)    :  1,   3,   7,   5,   16,   7,   29,   9,   46,   11, ...
Denominator A141310(n-1):  1,   2,   3,   2,    5,   2,    7,   2,    9,    2, ...
		

References

  • J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Crossrefs

Programs

  • Mathematica
    s[1] = 1; s[2] = 2; s[n_] := If[OddQ[n], n + 1, n - 1]; m = 100; Numerator[Accumulate[Array[s, m]]/Range[m]] (* Amiram Eldar, May 15 2022 *)
  • PARI
    f(n) = if (n<=2, n, if (n%2, n+1, n-1)); \\ A114112
    a(n) = numerator(sum(k=1, n, f(k))/n); \\ Michel Marcus, May 16 2022
    
  • Python
    from math import gcd
    def A354008(n): return 1 if n == 1 else (k:= (m:=n//2)*(n+1) + (n+1-m)*(n-2*m))//gcd(k,n) # Chai Wah Wu, May 24 2022

Formula

a(1) = 1, then for m >= 1: a(2m+1) = A130883(m+1) and a(2m) = A005408(m) = 2m+1.
G.f.: x*(1 + 3*x + 4*x^2 - 4*x^3 - 2*x^4 + x^5 + x^6)/(1 - x^2)^3. - Stefano Spezia, May 15 2022

A303226 Number of minimal total dominating sets in the n-gear graph.

Original entry on oeis.org

0, 6, 12, 6, 30, 30, 56, 110, 156, 306, 506, 870, 1560, 2652, 4692, 8190, 14280, 25122, 43890, 77006, 135056, 236682, 415380, 728462, 1278030, 2242506, 3934272, 6903756, 12113880, 21256710, 37301556, 65456190, 114864806, 201569006, 353722056, 620732310
Offset: 1

Author

Eric W. Weisstein, Apr 20 2018

Keywords

Comments

Sequence extrapolated to n=1 using recurrence. - Andrew Howroyd, Apr 20 2018

Crossrefs

Programs

  • Mathematica
    Table[RootSum[-1 - # + #^3 &, #^n &] + RootSum[-1 + # - 2 #^2 + #^3 &, #^n &] + 2 RootSum[-1 + #^2 + #^3 &, #^(n + 2) (1 + #) &], {n, 20}]
    LinearRecurrence[{1, 2, 1, -3, -1, -1, 0, 0, 1}, {0, 6, 12, 6, 30, 30, 56, 110, 156}, 20]
    CoefficientList[Series[-2 x (3 + 3 x - 9 x^2 - 3 x^3 - 3 x^4 + x^5 + 6 x^7)/(-1 + x + 2 x^2 + x^3 - 3 x^4 - x^5 - x^6 + x^9), {x, 0, 20}], x]
  • PARI
    concat([0], Vec(2*(3 + 3*x - 9*x^2 - 3*x^3 - 3*x^4 + x^5 + 6*x^7)/((1 - 2*x + x^2 - x^3)*(1 + x - x^3)*(1 - x^2 - x^3)) + O(x^40))) \\ Andrew Howroyd, Apr 20 2018

Formula

From Andrew Howroyd, Apr 20 2018: (Start)
a(n) = a(n-1) + 2*a(n-2) + a(n-3) - 3*a(n-4) - a(n-5) - a(n-6) + a(n-9) for n > 9.
G.f.: 2*x^2*(3 + 3*x - 9*x^2 - 3*x^3 - 3*x^4 + x^5 + 6*x^7)/((1 - 2*x + x^2 - x^3)*(1 + x - x^3)*(1 - x^2 - x^3)).
(End)

Extensions

a(1)-a(2) and terms a(11) and beyond from Andrew Howroyd, Apr 20 2018
Showing 1-6 of 6 results.