A158302 -id:A158302 - cp's OEIS frontend

A086893 a(n) is the index of F(n+1) at the unique occurrence of the ordered pair of reversed consecutive terms (F(n+1),F(n)) in Stern's diatomic sequence A002487, where F(k) denotes the k-th term of the Fibonacci sequence A000045.

1, 3, 5, 13, 21, 53, 85, 213, 341, 853, 1365, 3413, 5461, 13653, 21845, 54613, 87381, 218453, 349525, 873813, 1398101, 3495253, 5592405, 13981013, 22369621, 55924053, 89478485, 223696213, 357913941, 894784853, 1431655765, 3579139413

Offset: 1

John W. Layman, Sep 18 2003

If the Fibonacci pairs are kept in the natural order (F(n),F(n+1)), it appears that the first term of the pair occurs in A002487 at the index given by A061547(n).
Equals row sums of triangle A177954. - Gary W. Adamson, May 15 2010
Starting at n=3, begin subtracting from (2^(n-1)-1)/2^(n-1): 3/4 - 1/2 = 1/4 with 1+4=5=a(3); 7/8 - 1/4 = 5/8 with 5+8=13=a(4); 15/16 - 5/8 = 5/16 with 5+16=21= a(5); 31/32 - 5/16 = 21/32 with 21+32=53=a(6); 63/64 - 21/32 = 21/64 with 21+64=85=a(7) and so on. For n odd in the first fraction (2^(n-1)-1)/2^(n-1), the result approaches 1/3, and for n even in the first fraction, the result approaches 2/3. - J. M. Bergot, May 08 2015
Also, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017

			A002487 begins 0,1,1,2,1,3,2,... with offset 0. Thus a(1)=1 since (F(2),F(1)) = (1,1) occurs at term 1 of A002487. Similarly, a(2)=3 and a(3)=5, since (F(3),F(2))=(2,1) occurs at term 3 and (F(4),F(3))=(3,2) at term 5 of A002487.

A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata..., Fig. 12.

Cf. A000045, A002487, A061547, A077954, A283641, A372286, A372352.
Interleaving of A002450\{0} and A072197.
Positive terms of A096773 in ascending order.
Partial sums of A158302.

[2^(n-1)*(3-(-1)^n/3)-1/3: n in [0..35]]; // Vincenzo Librandi, May 09 2015

f[n_] := Module[{a = 1, b = 0, m = n}, While[m > 0, If[OddQ@ m, b = a + b, a = a + b]; m = Floor[m/2]]; b]; a = Table[f[n], {n, 0, 10^6}]; b = Reverse /@ Partition[Map[Fibonacci, Range[Ceiling@ Log[GoldenRatio, Max@ a] + 1]], 2, 1]; Map[If[Length@ # > 0, #[[1, 1]] - 1, 0] &@ SequencePosition[a, #] &, b] (* Michael De Vlieger, Mar 15 2017, Version 10.1, after Jean-François Alcover at A002487 *)

a(n)=if(n%2,2^(n+1),2^(n+1)+2^(n-1))\3 \\ Charles R Greathouse IV, May 08 2015

def A086893(n): return (1<Chai Wah Wu, Apr 29 2024

It appears that a(n)=(4^((n+1)/2)-1)/3 if n is odd and a(n)=(a(n-1)+a(n+1))/2 if n is even.
G.f.: (1+2*x-2*x^2)/((1-x)*(1-4*x^2)); a(n) = 2^(n-1)(3-(-1)^n/3)-1/3 (offset 0); a(n) = Sum{k=0..n+1, 4^floor(k/2)/2} (offset 0); a(2n) = A002450(n+1) (offset 0); a(2n+1) = A072197(n) (offset 0). - Paul Barry, May 21 2004
a(n+2) = 4*a(n) + 1, a(1) = 1, a(2) = 3, n > 0. - Yosu Yurramendi, Mar 07 2017
a(n+1) = a(n) + A158302(n), a(1) = 1, n > 0. - Yosu Yurramendi, Mar 07 2017

More terms from Paul Barry, May 21 2004

A303325 T(n,k)=Number of nXk 0..1 arrays with every element equal to 0, 2, 4, 5 or 6 horizontally, diagonally or antidiagonally adjacent elements, with upper left element zero.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 8, 1, 8, 1, 8, 4, 2, 16, 1, 32, 1, 8, 1, 32, 1, 32, 4, 2, 16, 2, 64, 1, 128, 1, 12, 4, 32, 1, 128, 1, 128, 4, 10, 64, 3, 64, 2, 256, 1, 512, 1, 46, 25, 62, 3, 128, 1, 512, 1, 512, 4, 50, 368, 56, 204, 10, 256, 2, 1024, 1, 2048, 1, 204, 201, 758, 136, 744, 9, 512, 1

Offset: 1

R. H. Hardin, Apr 21 2018

Table starts
...1.1...1..1...1....1.....1.....1.......1........1.........1..........1
...2.2...8..8..32...32...128...128.....512......512......2048.......2048
...4.1...4..1...4....1.....4.....1.......4........1.........4..........1
...8.2...8..2..12...10....46....50.....204......290......1034.......1682
..16.1..16..4..64...25...368...201....2545.....1855.....21082......17922
..32.2..32..3..62...56...758...822...11950....15100....206189.....274746
..64.1..64..3.204..136..2956..3929...69328...130531...2005898....4227664
.128.2.128.10.744..531.15494.24759..629227..1489177..33766564...89782726
.256.1.256..9.900.1035.44101.97205.3531980.11297739.359263250.1265799068

			All solutions for n=5 k=4
..0..0..1..1. .0..0..0..1. .0..1..1..1. .0..1..0..1
..0..0..1..1. .0..0..0..1. .0..1..1..1. .0..1..0..1
..0..1..0..1. .0..1..0..1. .0..1..0..1. .0..1..0..1
..0..0..1..1. .0..0..0..1. .0..1..1..1. .0..1..0..1
..0..0..1..1. .0..0..0..1. .0..1..1..1. .0..1..0..1

R. H. Hardin, Table of n, a(n) for n = 1..242

Column 1 is A000079(n-1).
Column 3 is A000079(n-1) for n>2.
Row 2 is A158302(n+1).
Row 3 is A010685(n+8).

Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = a(n-2)
k=3: a(n) = 2*a(n-1) for n>3
k=4: [order 55] for n>57
k=5: [order 33] for n>35
Empirical for row n:
n=1: a(n) = a(n-1)
n=2: a(n) = 4*a(n-2)
n=3: a(n) = a(n-2)
n=4: [order 18] for n>19
n=5: [order 34] for n>35

A335949 a(n) = denominator(b_n(x)), where b_n(x) are the polynomials defined in A335947.

Original entry on oeis.org

1, 1, 12, 4, 240, 48, 1344, 192, 3840, 1280, 33792, 3072, 5591040, 430080, 245760, 49152, 16711680, 983040, 522977280, 27525120, 1211105280, 173015040, 1447034880, 62914560, 22900899840, 4580179968, 1409286144, 469762048, 116769423360, 4026531840, 7689065201664

Offset: 0

Peter Luschny, Jul 01 2020

nonn

The sequence can also be computed without reference to the Bernoulli polynomials (ultimately thanks to the von Staudt-Clausen theorem) by the method of Kellner and Sondow (2019). Compare the SageMath program.

Bernd C. Kellner and Jonathan Sondow, On Carmichael and polygonal numbers, Bernoulli polynomials, and sums of base-p digits, arXiv:1902.10672 [math.NT], 2019.

Cf. A335947/A335948, A144845, A158302.

def A335949(n):
    a = set(prime_divisors(n + 1)) - set([2])
    b = (
        p
        for p in prime_range(3, (n + 2) // (2 + n % 2))
        if not p.divides(n + 1) and sum((n + 1).digits(base=p)) >= p
    )
    p = list(a.union(set(b)))
    return 4 ^ (n // 2) * mul(p)
print([A335949(n) for n in range(31)])

a(n) = min {m | m*([x^k] b(n, x)) is an integer for all k = 0..n}.
The odd part of a(n) is squarefree (A000265).
a(n) and A144845(n) have the same odd prime factors.
a(n)/A144845(n) = 4^floor(n/2)/2 for n >= 1.
a(n)/rad(a(n)) = A158302(n+1), (rad=A007947).

A154388 Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,-1,0,0,0,0,0,0,0,...] DELTA [1,-1,-1,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Philippe Deléham, Jan 08 2009

			Triangle begins:
  1;
  0, 1;
  0, 1, 0;
  0, 0, 0, 1;
  0, 0, 0, 1, 0;
  0, 0, 0, 0, 0, 1; ...

Sum_{k=0..n} T(n,k)*x^(n-k) = A135528(n+1), A000012(n), A040001(n), A153284(n+1) for x = 0,1,2,3 respectively.
G.f.: (1+y*x+(y-y^2)*x^2)/(1-y^2*x^2). - Philippe Deléham, Dec 17 2011
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000012(n), A158302(n) for x = 0, 1, 2 respectively. - Philippe Deléham, Dec 17 2011

A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

Original entry on oeis.org

1, 1, 3, 1, 2, 13, 0, 1, 5, 3, -1, -1, 2, 29, 119, 0, -1, -1, 1, 31, 5, 1, 1, -1, -8, -1, 43, 253, 0, 1, 1, 4, -4, -1, 41, 7, -1, -1, -1, 4, 8, 4, -1, 29, 239, 0, -1, -1, -8, -4, 4, 8, 1, 31, 9, 5, 5, 7, -4, -116, -32, -116, -4, 7, 71, 665, 0

Offset: 0

Paul Curtz, Feb 03 2015

The difference table of Bernoulli(n,2) or B(n,2) = A164558(n)/A027642(n) is defined by placing the fractions in the upper row and calculating further rows as the differences of their preceding row:
1,       3/2, 13/6,     3, 119/30, ...
1/2,     2/3,  5/6, 29/30, ...
1/6,     1/6, 2/15, ...
0,     -1/30, ...
-1/30, ...
etc.
The first column is A164555(n)/A027642(n).
In particular, the sums of the antidiagonals
1                              = 1
1/2 + 3/2                      = 2
1/6 + 2/3 + 13/6               = 3
0   + 1/6 +  5/6 + 3           = 4
etc. are the positive natural numbers. (This is rewritten for Bernoulli(n,3) in A157809).
We also have for Bernoulli(.,2)
B(0,2) = 1
B(0,2) + 2*B(1,2) = 4
B(0,2) + 3*B(1,2) + 3*B(2,2) = 12
B(0,2) + 4*B(1,2) + 6*B(2,2) + 4*B(3,2) = 32
etc. with right hand sides provided by A001787.
More generally sum_{s=0..t-1} binomial(t,s)*Bernoulli(s,q) gives A027471(t) for q=3, A002697 for q=4 etc, by reading A104002 downwards the q-th column.

Cf. A027641, A027642, A074909, A085737, A085738, A104002, A157809, A157920, A157930, A157945, A157946, A157965, A164555, A164558, A190339, A158302, A181131 (numerators and denominators of the main diagonal).

nmax = 11; A164558 = Table[BernoulliB[n,2], {n, 0, nmax}]; D164558 = Table[ Differences[A164558, n], {n, 0, nmax}]; Table[ D164558[[n-k+1, k+1]] // Numerator, {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 04 2015 *)

cp's OEIS Frontend

A086893 a(n) is the index of F(n+1) at the unique occurrence of the ordered pair of reversed consecutive terms (F(n+1),F(n)) in Stern's diatomic sequence A002487, where F(k) denotes the k-th term of the Fibonacci sequence A000045.

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A303325 T(n,k)=Number of nXk 0..1 arrays with every element equal to 0, 2, 4, 5 or 6 horizontally, diagonally or antidiagonally adjacent elements, with upper left element zero.

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A335949 a(n) = denominator(b_n(x)), where b_n(x) are the polynomials defined in A335947.

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A154388 Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,-1,0,0,0,0,0,0,0,...] DELTA [1,-1,-1,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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Examples

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A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

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Programs

Mathematica