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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A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

Original entry on oeis.org

1, 5, 23, 107, 497, 2309, 10727, 49835, 231521, 1075589, 4996919, 23214443, 107848529, 501037445, 2327695367, 10813893803, 50238661313, 233396326661, 1084301290583, 5037394142315, 23402480441009, 108722104190981, 505095858086951, 2346549744920747
Offset: 0

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Author

R. H. Hardin, Dec 27 2006

Keywords

Comments

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.
From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.
The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)
Inverse binomial transform of A154244. (End)
Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010
Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Crossrefs

Cf. 5 symbol differing by two or less A126392, one or less A057960.
Cf. Red king sequences side squares [numerical value A[5]]: A086347 [495], A179598 [239], A126473 [367], A123347 [335], A179602 [95], A154964 [31], A015448 [327], A152187 [27], A003947 [325], A108981 [11], A007483 [2]. - Johannes W. Meijer, Aug 01 2010
Cf. A055099.

Programs

  • Maple
    with(LinearAlgebra): nmax:=19; m:=2; A[5]:= [1,0,1,1,0,1,1,1,1]: A:=Matrix([[0,1,0,1,1,0,0,0,0],[1,0,1,1,1,1,0,0,0],[0,1,0,0,1,1,0,0,0],[1,1,0,0,1,0,1,1,0],A[5],[0,1,1,0,1,0,0,1,1],[0,0,0,1,1,0,0,1,0],[0,0,0,1,1,1,1,0,1],[0,0,0,0,1,1,0,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Aug 01 2010
    # second Maple program:
    a:= n-> (M-> M[1,2]+M[2,2])(<<0|1>, <3|4>>^n):
    seq(a(n), n=0..24);  # Alois P. Heinz, Jun 28 2021
  • Mathematica
    LinearRecurrence[{4, 3}, {1, 5}, 24] (* Jean-François Alcover, Dec 10 2024 *)
  • PARI
    a(n)=([0,1; 3,4]^n*[1;5])[1,1] \\ Charles R Greathouse IV, May 10 2016

Formula

From Johannes W. Meijer, Aug 01 2010: (Start)
G.f.: (1+x)/(1-4*x-3*x^2).
a(n) = 4*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = ((1+3/sqrt(7))/2)*(A)^(-n) + ((1-3/sqrt(7))/2)*(B)^(-n) with A = (-2 + sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7)-A108851(n))
(End)
a(n) = A015330(n)+A015330(n+1). - R. J. Mathar, May 09 2023

Extensions

Edited by Johannes W. Meijer, Aug 10 2010

A374848 Obverse convolution A000045**A000045; see Comments.

Original entry on oeis.org

0, 1, 2, 16, 162, 3600, 147456, 12320100, 2058386904, 701841817600, 488286500625000, 696425232679321600, 2038348954317776486400, 12259459134020160144810000, 151596002479762016373851690400, 3855806813438155578522841251840000
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2024

Keywords

Comments

The obverse convolution of sequences
s = (s(0), s(1), ...) and t = (t(0), t(1), ...)
is introduced here as the sequence s**t given by
s**t(n) = (s(0)+t(n)) * (s(1)+t(n-1)) * ... * (s(n)+t(0)).
Swapping * and + in the representation s(0)*t(n) + s(1)*t(n-1) + ... + s(n)*t(0)
of ordinary convolution yields s**t.
If x is an indeterminate or real (or complex) variable, then for every sequence t of real (or complex) numbers, s**t is a sequence of polynomials p(n) in x, and the zeros of p(n) are the numbers -t(0), -t(1), ..., -t(n).
Following are abbreviations in the guide below for triples (s, t, s**t):
F = (0,1,1,2,3,5,...) = A000045, Fibonacci numbers
L = (2,1,3,4,7,11,...) = A000032, Lucas numbers
P = (2,3,5,7,11,...) = A000040, primes
T = (1,3,6,10,15,...) = A000217, triangular numbers
C = (1,2,6,20,70, ...) = A000984, central binomial coefficients
LW = (1,3,4,6,8,9,...) = A000201, lower Wythoff sequence
UW = (2,5,7,10,13,...) = A001950, upper Wythoff sequence
[ ] = floor
In the guide below, sequences s**t are identified with index numbers Axxxxxx; in some cases, s**t and Axxxxxx differ in one or two initial terms.
Table 1. s = A000012 = (1,1,1,1...) = (1);
t = A000012; 1 s**t = A000079; 2^(n+1)
t = A000027; n s**t = A000142; (n+1)!
t = A000040, P s**t = A054640
t = A000040, P (1/3) s**t = A374852
t = A000079, 2^n s**t = A028361
t = A000079, 2^n (1/3) s**t = A028362
t = A000045, F s**t = A082480
t = A000032, L s**t = A374890
t = A000201, LW s**t = A374860
t = A001950, UW s**t = A374864
t = A005408, 2*n+1 s**t = A000165, 2^n*n!
t = A016777, 3*n+1 s**t = A008544
t = A016789, 3*n+2 s**t = A032031
t = A000142, n! s**t = A217757
t = A000051, 2^n+1 s**t = A139486
t = A000225, 2^n-1 s**t = A006125
t = A032766, [3*n/2] s**t = A111394
t = A034472, 3^n+1 s**t = A153280
t = A024023, 3^n-1 s**t = A047656
t = A000217, T s**t = A128814
t = A000984, C s**t = A374891
t = A279019, n^2-n s**t = A130032
t = A004526, 1+[n/2] s**t = A010551
t = A002264, 1+[n/3] s**t = A264557
t = A002265, 1+[n/4] s**t = A264635
Sequences (c)**L, for c=2..4: A374656 to A374661
Sequences (c)**F, for c=2..6: A374662, A374662, A374982 to A374855
The obverse convolutions listed in Table 1 are, trivially, divisibility sequences. Likewise, if s = (-1,-1,-1,...) instead of s = (1,1,1,...), then s**t is a divisibility sequence for every choice of t; e.g. if s = (-1,-1,-1,...) and t = A279019, then s**t = A130031.
Table 2. s = A000027 = (0,1,2,3,4,5,...) = (n);
t = A000027, n s**t = A007778, n^(n+1)
t = A000290, n^2 s**t = A374881
t = A000040, P s**t = A374853
t = A000045, F s**t = A374857
t = A000032, L s**t = A374858
t = A000079, 2^n s**t = A374859
t = A000201, LW s**t = A374861
t = A005408, 2*n+1 s**t = A000407, (2*n+1)! / n!
t = A016777, 3*n+1 s**t = A113551
t = A016789, 3*n+2 s**t = A374866
t = A000142, n! s**t = A374871
t = A032766, [3*n/2] s**t = A374879
t = A000217, T s**t = A374892
t = A000984, C s**t = A374893
t = A038608, n*(-1)^n s**t = A374894
Table 3. s = A000290 = (0,1,4,9,16,...) = (n^2);
t = A000290, n^2 s**t = A323540
t = A002522, n^2+1 s**t = A374884
t = A000217, T s**t = A374885
t = A000578, n^3 s**t = A374886
t = A000079, 2^n s**t = A374887
t = A000225, 2^n-1 s**t = A374888
t = A005408, 2*n+1 s**t = A374889
t = A000045, F s**t = A374890
Table 4. s = t;
s = t = A000012, 1 s**s = A000079; 2^(n+1)
s = t = A000027, n s**s = A007778, n^(n+1)
s = t = A000290, n^2 s**s = A323540
s = t = A000045, F s**s = this sequence
s = t = A000032, L s**s = A374850
s = t = A000079, 2^n s**s = A369673
s = t = A000244, 3^n s**s = A369674
s = t = A000040, P s**s = A374851
s = t = A000201, LW s**s = A374862
s = t = A005408, 2*n+1 s**s = A062971
s = t = A016777, 3*n+1 s**s = A374877
s = t = A016789, 3*n+2 s**s = A374878
s = t = A032766, [3*n/2] s**s = A374880
s = t = A000217, T s**s = A375050
s = t = A005563, n^2-1 s**s = A375051
s = t = A279019, n^2-n s**s = A375056
s = t = A002398, n^2+n s**s = A375058
s = t = A002061, n^2+n+1 s**s = A375059
If n = 2*k+1, then s**s(n) is a square; specifically,
s**s(n) = ((s(0)+s(n))*(s(1)+s(n-1))*...*(s(k)+s(k+1)))^2.
If n = 2*k, then s**s(n) has the form 2*s(k)*m^2, where m is an integer.
Table 5. Others
s = A000201, LW t = A001950, UW s**t = A374863
s = A000045, F t = A000032, L s**t = A374865
s = A005843, 2*n t = A005408, 2*n+1 s**t = A085528, (2*n+1)^(n+1)
s = A016777, 3*n+1 t = A016789, 3*n+2 s**t = A091482
s = A005408, 2*n+1 t = A000045, F s**t = A374867
s = A005408, 2*n+1 t = A000032, L s**t = A374868
s = A005408, 2*n+1 t = A000079, 2^n s**t = A374869
s = A000027, n t = A000142, n! s**t = A374871
s = A005408, 2*n+1 t = A000142, n! s**t = A374872
s = A000079, 2^n t = A000142, n! s**t = A374874
s = A000142, n! t = A000045, F s**t = A374875
s = A000142, n! t = A000032, L s**t = A374876
s = A005408, 2*n+1 t = A016777, 3*n+1 s**t = A352601
s = A005408, 2*n+1 t = A016789, 3*n+2 s**t = A064352
Table 6. Arrays of coefficients of s(x)**t(x), where s(x) and t(x) are polynomials
s(x) t(x) s(x)**t(x)
n x A132393
n^2 x A269944
x+1 x+1 A038220
x+2 x+2 A038244
x x+3 A038220
nx x+1 A094638
1 x^2+x+1 A336996
n^2 x x+1 A375041
n^2 x 2x+1 A375042
n^2 x x+2 A375043
2^n x x+1 A375044
2^n 2x+1 A375045
2^n x+2 A375046
x+1 F(n) A375047
x+1 x+F(n) A375048
x+F(n) x+F(n) A375049

Examples

			a(0) = 0 + 0 = 0
a(1) = (0+1) * (1+0) = 1
a(2) = (0+1) * (1+1) * (1+0) = 2
a(3) = (0+2) * (1+1) * (1+1) * (2+0) = 16
As noted above, a(2*k+1) is a square for k>=0. The first 5 squares are 1, 16, 3600, 12320100, 701841817600, with corresponding square roots 1, 4, 60, 3510, 837760.
If n = 2*k, then s**s(n) has the form 2*F(k)*m^2, where m is an integer and F(k) is the k-th Fibonacci number; e.g., a(6) = 2*F(3)*(192)^2.
		

Crossrefs

Programs

  • Maple
    a:= n-> (F-> mul(F(n-j)+F(j), j=0..n))(combinat[fibonacci]):
    seq(a(n), n=0..15);  # Alois P. Heinz, Aug 02 2024
  • Mathematica
    s[n_] := Fibonacci[n]; t[n_] := Fibonacci[n];
    u[n_] := Product[s[k] + t[n - k], {k, 0, n}];
    Table[u[n], {n, 0, 20}]
  • PARI
    a(n)=prod(k=0, n, fibonacci(k) + fibonacci(n-k)) \\ Andrew Howroyd, Jul 31 2024

Formula

a(n) ~ c * phi^(3*n^2/4 + n) / 5^((n+1)/2), where c = QPochhammer(-1, 1/phi^2)^2/2 if n is even and c = phi^(1/4) * QPochhammer(-phi, 1/phi^2)^2 / (phi + 1)^2 if n is odd, and phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 01 2024

A026150 a(0) = a(1) = 1; a(n+2) = 2*a(n+1) + 2*a(n).

Original entry on oeis.org

1, 1, 4, 10, 28, 76, 208, 568, 1552, 4240, 11584, 31648, 86464, 236224, 645376, 1763200, 4817152, 13160704, 35955712, 98232832, 268377088, 733219840, 2003193856, 5472827392, 14952042496, 40849739776
Offset: 0

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Author

Keywords

Comments

a(n+1)/A002605(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
a(n+1)/a(n) converges to 1 + sqrt(3) = 2.732050807568877293.... - Philippe Deléham, Jul 03 2005
Binomial transform of expansion of cosh(sqrt(3)x) (A000244 with interpolated zeros); inverse binomial transform of A001075. - Philippe Deléham, Jul 04 2005
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 3 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(3). - Cino Hilliard, Sep 25 2005
Inverse binomial transform of A001075: (1, 2, 7, 26, 97, 362, ...). - Gary W. Adamson, Nov 23 2007
Starting (1, 4, 10, 28, 76, ...), the sequence is the binomial transform of [1, 3, 3, 9, 9, 27, 27, 81, 81, ...], and inverse binomial transform of A001834: (1, 5, 19, 71, 265, ...). - Gary W. Adamson, Nov 30 2007
[1, 3; 1, 1]^n * [1,0] = [a(n), A002605(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = a(n) + A002605(n)*(sqrt(3)). - Gary W. Adamson, Mar 21 2008
Equals right border of triangle A143908. Also, starting (1, 4, 10, 28, ...) = row sums of triangle A143908 and INVERT transform of (1, 3, 3, 3, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A002605 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 1, 1, 24, 1, 48, 1, 3, 24, 10, 1, 12, 48, 24, 1,144, 3,180, 24, ... - R. J. Mathar, Aug 10 2012
(1 + sqrt(3))^n = a(n) + A002605(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
a(n) is also the number of solutions for cyclic three-dimensional stable matching instances with master preference lists of size n (Escamocher and O'Sullivan 2018). - Guillaume Escamocher, Jun 15 2018
Starting from a(1), first differences of A005665. - Ivan N. Ianakiev, Nov 22 2019
Number of 3-permutations of n elements avoiding the patterns 231, 312. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

Examples

			G.f. = 1 + x + 4*x^2 + 10*x^3 + 28*x^4 + 76*x^5 + 208*x^6 + 568*x^7 + ...
		

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

First differences of A002605.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Haskell
    a026150 n = a026150_list !! n
    a026150_list = 1 : 1 : map (* 2) (zipWith (+) a026150_list (tail
    a026150_list))
    -- Reinhard Zumkeller, Oct 15 2011
    
  • Magma
    [n le 2 select 1 else 2*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018
  • Maple
    with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL2,ZL2,ZL2), b=ZL1], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n)/3, n=2..27); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    Expand[Table[((1 + Sqrt[3])^n + (1 - Sqrt[3])^n)/(2), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
    LinearRecurrence[{2, 2}, {1, 1}, 30] (* T. D. Noe, Mar 25 2011 *)
    Round@Table[LucasL[n, Sqrt[2]] 2^(n/2 - 1), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
  • Maxima
    a(n) := if n<=1 then 1 else 2*a(n-1)+2*a(n-2);
    makelist(a(n),n,0,20); /* Emanuele Munarini, Apr 14 2017 */
    
  • PARI
    {a(n) = if( n<0, 0, real((1 + quadgen(12))^n))};
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2; it = recur_gen2(1,1,2,2); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [lucas_number2(n,2,-2)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = (1/2)*((1 + sqrt(3))^n + (1 - sqrt(3))^n). - Benoit Cloitre, Oct 28 2002
G.f.: (1 - x)/(1 - 2*x - 2*x^2).
a(n) = a(n-1) + A083337(n-1). A083337(n)/a(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
From Paul Barry, May 15 2003: (Start)
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*3^k;
E.g.f.: exp(x)*cosh(sqrt(3)x). (End)
a(n) = Sum_{k=0..n} A098158(n,k)*3^(n - k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1, 1; 3, 1]^n. (1 + sqrt(3))^n = a(n) + A083337(n)/(sqrt(3)). - Gary W. Adamson, Mar 12 2008
a(n) = A080040(n)/2. - Philippe Deléham, Nov 19 2008
If p[1] = 1, and p[i] = 3, (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j + 1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
a(n) = 2 * A052945(n-1). - Vladimir Joseph Stephan Orlovsky, Mar 24 2011
a(n) = round((1 + sqrt(3))^n/2) for n > 0. - Bruno Berselli, Feb 04 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k - 1)/(x*(3*k + 2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
a(n) = (-sqrt(2)*i)^n*T(n,sqrt(2)*i/2), with i = sqrt(-1) and the Chebyshev T-polynomials (A053120). - Wolfdieter Lang, Feb 10 2018

A048883 a(n) = 3^wt(n), where wt(n) = A000120(n).

Original entry on oeis.org

1, 3, 3, 9, 3, 9, 9, 27, 3, 9, 9, 27, 9, 27, 27, 81, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81, 81, 243, 9, 27, 27, 81, 27, 81, 81, 243, 27, 81, 81, 243, 81, 243, 243, 729, 3, 9, 9, 27, 9, 27, 27, 81, 9, 27, 27, 81, 27, 81
Offset: 0

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Author

Keywords

Comments

Or, a(n)=number of 1's ("live" cells) at stage n of a 2-dimensional cellular automata evolving by the rule: 1 if NE+NW+S=1, else 0.
This is the odd-rule cellular automaton defined by OddRule 013 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015
Or, start with S=[1]; replace S by [S, 3*S]; repeat ad infinitum.
Fixed point of the morphism 1 -> 13, 3 -> 39, 9 -> 9(27), ... = 3^k -> 3^k 3^(k+1), ... starting from a(0) = 1; 1 -> 13 -> 1339 -> = 1339399(27) -> 1339399(27)399(27)9(27)(27)(81) -> ..., . - Robert G. Wilson v, Jan 24 2006
Equals row sums of triangle A166453 (the square of Sierpiński's gasket, A047999). - Gary W. Adamson, Oct 13 2009
First bisection of A169697=1,5,3,19,3,. a(2n+2)+a(2n+3)=12,12,36,=12*A147610 ? Distribution of terms (in A000244): A011782=1,A000079 for first array, A000079 for second. - Paul Curtz, Apr 20 2010
a(A000225(n)) = A000244(n) and a(m) != A000244(n) for m < A000225(n). - Reinhard Zumkeller, Nov 14 2011
This sequence pertains to phenotype Punnett square mathematics. Start with X=1. Each hybrid cross involves the equation X:3X. Therefore, the ratio in the first (mono) hybrid cross is X=1:3X=3(1) or 3; or 3:1. When you move up to the next hybridization level, replace the previous cross ratio with X. X now represents 2 numbers-1:3. Therefore, the ratio in the second (di) hybrid cross is X=(1:3):3X=[3(1):3(3)] or (3:9). Put it together and you get 1:3:3:9. Each time you move up a hybridization level, replace the previous ratio with X, and use the same equation-X:3X to get its ratio. - John Michael Feuk, Dec 10 2011
Number of odd values in the n-th layer of Pascal's tetrahedron (see A268240). - Caden Le, Mar 03 2025
a(x*y) <= a(x)^A000120(y). - Joe Amos, Mar 28 2025

Examples

			From _Omar E. Pol_, Jun 07 2009: (Start)
Triangle begins:
  1;
  3;
  3,9;
  3,9,9,27;
  3,9,9,27,9,27,27,81;
  3,9,9,27,9,27,27,81,9,27,27,81,27,81,81,243;
  3,9,9,27,9,27,27,81,9,27,27,81,27,81,81,243,9,27,27,81,27,81,81,243,27,...
Or
  1;
  3,3;
  9,3,9,9;
  27,3,9,9,27,9,27,27;
  81,3,9,9,27,9,27,27,81,9,27,27,81,27,81,81;
  243,3,9,9,27,9,27,27,81,9,27,27,81,27,81,81,243,9,27,27,81,27,81,81,243,27...
(End)
		

Crossrefs

For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
A generalization of A001316. Cf. A102376.
Partial sums give A130665. - David Applegate, Jun 11 2009

Programs

  • Haskell
    a048883 = a000244 . a000120  -- Reinhard Zumkeller, Nov 14 2011
  • Mathematica
    Nest[ Join[#, 3#] &, {1}, 6] (* Robert G. Wilson v, Jan 24 2006 and modified Jul 27 2014*)
    a[n_] := 3^DigitCount[n, 2, 1]; Array[a, 80, 0] (* Jean-François Alcover, Nov 15 2017 *)
  • PARI
    a(n)=n=binary(n);3^sum(i=1,#n,n[i])
    

Formula

a(n) = Product_{k=0..log_2(n)} 3^b(n,k), where b(n,k) = coefficient of 2^k in binary expansion of n (offset 0). - Paul D. Hanna
a(n) = 3*a(n/2) if n is even, otherwise a(n) = a((n+1)/2).
G.f.: Product_{k>=0} (1+3*x^(2^k)). The generalization k^A000120 has generating function (1 + kx)*(1 + kx^2)*(1 + kx^4)*...
a(n+1) = Sum_{i=0..n} (binomial(n, i) mod 2) * Sum_{j=0..i} (binomial(i, j) mod 2). - Benoit Cloitre, Nov 16 2003
a(0)=1, a(n) = 3*a(n-A053644(n)) for n > 0. - Joe Slater, Jan 31 2016
G.f. A(x) satisfies: A(x) = (1 + 3*x) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019

Extensions

Corrected by Ralf Stephan, Jun 19 2003
Entry revised by N. J. A. Sloane, May 30 2009
Offset changed to 0, Jun 11 2009

A087207 A binary representation of the primes that divide a number, shown in decimal.

Original entry on oeis.org

0, 1, 2, 1, 4, 3, 8, 1, 2, 5, 16, 3, 32, 9, 6, 1, 64, 3, 128, 5, 10, 17, 256, 3, 4, 33, 2, 9, 512, 7, 1024, 1, 18, 65, 12, 3, 2048, 129, 34, 5, 4096, 11, 8192, 17, 6, 257, 16384, 3, 8, 5, 66, 33, 32768, 3, 20, 9, 130, 513, 65536, 7, 131072, 1025, 10, 1, 36, 19, 262144, 65, 258
Offset: 1

Views

Author

Mitch Cervinka (puritan(AT)planetkc.com), Oct 26 2003

Keywords

Comments

The binary representation of a(n) shows which prime numbers divide n, but not the multiplicities. a(2)=1, a(3)=10, a(4)=1, a(5)=100, a(6)=11, a(10)=101, a(30)=111, etc.
For n > 1, a(n) gives the (one-based) index of the column where n is located in array A285321. A008479 gives the other index. - Antti Karttunen, Apr 17 2017
From Antti Karttunen, Jun 18 & 20 2017: (Start)
A268335 gives all n such that a(n) = A248663(n); the squarefree numbers (A005117) are all the n such that a(n) = A285330(n) = A048675(n).
For all n > 1 for which the value of A285331(n) is well-defined, we have A285331(a(n)) <= floor(A285331(n)/2), because then n is included in the binary tree A285332 and a(n) is one of its ancestors (in that tree), and thus must be at least one step nearer to its root than n itself.
Conjecture: Starting at any n and iterating the map n -> a(n), we will always reach 0 (see A288569). This conjecture is equivalent to the conjecture that at any n that is neither a prime nor a power of two, we will eventually hit a prime number (which then becomes a power of two in the next iteration). If this conjecture is false then sequence A285332 cannot be a permutation of natural numbers. On the other hand, if the conjecture is true, then A285332 must be a permutation of natural numbers, because all primes and powers of 2 occur in definite positions in that tree. This conjecture also implies the conjectures made in A019565 and A285320 that essentially claim that there are neither finite nor infinite cycles in A019565.
If there are any 2-cycles in this sequence, then both terms of the cycle should be present in A286611 and the larger one should be present in A286612.
(End)
Binary rank of the distinct prime indices of n, where the binary rank of an integer partition y is given by Sum_i 2^(y_i-1). For all prime indices (with multiplicity) we have A048675. - Gus Wiseman, May 25 2024

Examples

			a(38) = 129 because 38 = 2*19 = prime(1)*prime(8) and 129 = 2^0 + 2^7 (in binary 10000001).
a(140) = 13, binary 1101 because 140 is divisible by the first, third and fourth primes and 2^(1-1) + 2^(3-1) + 2^(4-1) = 13.
		

Crossrefs

For partial sums see A288566.
Sequences with related definitions: A007947, A008472, A027748, A048675, A248663, A276379 (same sequence shown in base 2), A288569, A289271, A297404.
Cf. A286608 (numbers n for which a(n) < n), A286609 (n for which a(n) > n), and also A286611, A286612.
A003986, A003961, A059896 are used to express relationship between terms of this sequence.
Related to A267116 via A225546.
Positions of particular values are: A000079\{1} (1), A000244\{1} (2), A033845 (3), A000351\{1} (4), A033846 (5), A033849 (6), A143207 (7), A000420\{1} (8), A033847 (9), A033850 (10), A033851 (12), A147576 (14), A147571 (15), A001020\{1} (16), A033848 (17).
A048675 gives binary rank of prime indices.
A061395 gives greatest prime index, least A055396.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.
Binary indices (listed A048793):
- length A000120, complement A023416
- min A001511, opposite A000012
- sum A029931, product A096111
- max A029837 or A070939, opposite A070940
- complement A368494, sum A359400
- opposite complement A371571, sum A359359
- opposite A371572, sum A230877

Programs

  • Haskell
    a087207 = sum . map ((2 ^) . (subtract 1) . a049084) . a027748_row
    -- Reinhard Zumkeller, Jul 16 2013
    
  • Mathematica
    a[n_] := Total[ 2^(PrimePi /@ FactorInteger[n][[All, 1]] - 1)]; a[1] = 0; Table[a[n], {n, 1, 69}] (* Jean-François Alcover, Dec 12 2011 *)
  • PARI
    a(n) = {if (n==1, 0, my(f=factor(n), v = []); forprime(p=2, vecmax(f[,1]), v = concat(v, vecsearch(f[,1], p)!=0);); fromdigits(Vecrev(v), 2));} \\ Michel Marcus, Jun 05 2017
    
  • PARI
    A087207(n)=vecsum(apply(p->1<M. F. Hasler, Jun 23 2017
    
  • Python
    from sympy import factorint, primepi
    def a(n):
        return sum(2**primepi(i - 1) for i in factorint(n))
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 06 2017
    
  • Scheme
    (definec (A087207 n) (if (= 1 n) 0 (+ (A000079 (+ -1 (A055396 n))) (A087207 (A028234 n))))) ;; This uses memoization-macro definec
    (define (A087207 n) (A048675 (A007947 n))) ;; Needs code from A007947 and A048675. - Antti Karttunen, Jun 19 2017

Formula

Additive with a(p^e) = 2^(i-1) where p is the i-th prime. - Vladeta Jovovic, Oct 29 2003
a(n) gives the m such that A019565(m) = A007947(n). - Naohiro Nomoto, Oct 30 2003
A000120(a(n)) = A001221(n); a(n) = Sum(2^(A049084(p)-1): p prime-factor of n). - Reinhard Zumkeller, Nov 30 2003
G.f.: Sum_{k>=1} 2^(k-1)*x^prime(k)/(1-x^prime(k)). - Franklin T. Adams-Watters, Sep 01 2009
From Antti Karttunen, Apr 17 2017, Jun 19 2017 & Dec 06 2018: (Start)
a(n) = A048675(A007947(n)).
a(1) = 0; for n > 1, a(n) = 2^(A055396(n)-1) + a(A028234(n)).
A000035(a(n)) = 1 - A000035(n). [a(n) and n are of opposite parity.]
A248663(n) <= a(n) <= A048675(n). [XOR-, OR- and +-variants.]
a(A293214(n)) = A218403(n).
a(A293442(n)) = A267116(n).
A069010(a(n)) = A287170(n).
A007088(a(n)) = A276379(n).
A038374(a(n)) = A300820(n) for n >= 1.
(End)
From Peter Munn, Jan 08 2020: (Start)
a(A059896(n,k)) = a(n) OR a(k) = A003986(a(n), a(k)).
a(A003961(n)) = 2*a(n).
a(n^2) = a(n).
a(n) = A267116(A225546(n)).
a(A225546(n)) = A267116(n).
(End)

Extensions

More terms from Don Reble, Ray Chandler and Naohiro Nomoto, Oct 28 2003
Name clarified by Antti Karttunen, Jun 18 2017

A015441 Generalized Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159
Offset: 0

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Author

Keywords

Comments

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

Examples

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...
		

Crossrefs

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
  • Maple
    A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014
  • Mathematica
    a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
    CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)
  • PARI
    {a(n) = (3^n - (-2)^n) / 5};
    
  • Sage
    [lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

A000149 a(n) = floor(e^n).

Original entry on oeis.org

1, 2, 7, 20, 54, 148, 403, 1096, 2980, 8103, 22026, 59874, 162754, 442413, 1202604, 3269017, 8886110, 24154952, 65659969, 178482300, 485165195, 1318815734, 3584912846, 9744803446, 26489122129, 72004899337, 195729609428, 532048240601, 1446257064291
Offset: 0

Views

Author

Keywords

Comments

A000079(n) <= a(n) <= A000244(n); for n > 0: A064780(n) = a(n+1) - a(n). - Reinhard Zumkeller, Mar 17 2015
Satisfies Benford's law [Whyman et al., 2016]. - N. J. A. Sloane, Feb 12 2017

References

  • Federal Works Agency, Work Projects Administration for the City of NY, Tables of the Exponential Function. National Bureau of Standards, Washington, DC, 1939.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 230.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisection: A116472.
Cf. A001113, A003619, A000079, A000244, A064780 (first differences, apart from initial term).
Cf. A000227 (round e^n), A001671 (ceiling e^n).

Programs

  • Haskell
    a000149 = floor . (exp 1 ^)
    a000149_list = let e = exp 1 in map floor $ iterate (* e) 1
    -- Reinhard Zumkeller, Mar 17 2015
    
  • Mathematica
    a[n_]:=Floor[E^n]; (* Vladimir Joseph Stephan Orlovsky, Dec 12 2008 *)
    Floor[E^Range[0,30]] (* Harvey P. Dale, Apr 01 2012 *)
  • PARI
    a(n) = floor(exp(n)); \\ Arkadiusz Wesolowski, Nov 26 2011
    
  • PARI
    apply( A000149(n)=exp(n)\1, [0..30]) \\ An error message will say so if default(realprecision) must be increased, for large n. - M. F. Hasler, May 27 2018
    
  • Python
    from sympy import floor, E
    def a(n):  return floor(E**n)
    print([a(n) for n in range(29)]) # Michael S. Branicky, Jul 20 2021

Formula

a(n)^(1/n) converges to e because |1-a(n)/e^n|=|e^n-a(n)|/e^n < e^(-n) and so a(n)^(1/n)=(e^n*(1+o(1)))^(1/n)=e*(1+o(1)). - Hieronymus Fischer, Jan 22 2006

A087436 Number of odd prime factors of n, counted with repetitions.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 2, 0, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 0, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 2, 2, 1, 1, 3, 2, 1, 2, 1, 1, 2, 1, 1, 3, 0, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 2, 1, 1, 4, 1, 1, 2, 2, 1, 2, 1, 1, 3, 2, 1, 2, 1, 2, 1, 1, 2, 3, 2, 1, 2
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 03 2003

Keywords

Comments

Number of parts larger than 1 in the partition with Heinz number n. The Heinz number of an integer partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). Example: a(9) = 2 because the partition with Heinz number 9 (=3*3) is [2,2]. - Emeric Deutsch, Oct 02 2015
Totally additive because both A001222 and A007814 are. a(2) = 0, and a(p) = 1 for odd primes p, a(m*n) = a(m)+a(n) for m, n > 1. - Antti Karttunen, Jul 10 2020

Examples

			a(9) = 2 because 9 = 3*3 has 2 odd prime factors. - _Emeric Deutsch_, Oct 02 2015
		

Crossrefs

Cf. A000244 (the first occurrence of each n, and also the positions of records).

Programs

  • Maple
    seq(bigomega(n) - padic[ordp](n, 2), n=1..102); # Peter Luschny, Dec 06 2017
  • Mathematica
    Join[{0},Table[Length[Select[Flatten[Table[#[[1]],{#[[2]]}]&/@ FactorInteger[ n]],OddQ]],{n,2,110}]] (* Harvey P. Dale, Feb 01 2013 *)
  • PARI
    a(n) = bigomega(n) - valuation(n, 2); \\ Michel Marcus, Sep 10 2019
    
  • PARI
    A087436(n) = (bigomega(n>>valuation(n,2))); \\ Antti Karttunen, Jul 10 2020

Formula

a(n) = A001222(n) - A007814(n).
a(n) = A001222(A000265(n)). - Antti Karttunen, Jul 10 2020
Sum_{k=1..n} a(k) = n * (log(log(n)) + B_2 - 1) + O(n/log(n)), where B_2 = A083342. - Amiram Eldar, May 16 2025

A009964 Powers of 20.

Original entry on oeis.org

1, 20, 400, 8000, 160000, 3200000, 64000000, 1280000000, 25600000000, 512000000000, 10240000000000, 204800000000000, 4096000000000000, 81920000000000000, 1638400000000000000, 32768000000000000000
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequences E(1, 20), L(1, 20), P(1, 20), T(1, 20). Essentially same as Pisot sequences E(20, 400), L(20, 400), P(20, 400), T(20, 400). See A008776 for definitions of Pisot sequences.
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 20-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
a(n) gives the number of small cubes in the n-th iteration of the Menger sponge fractal. - Felix Fröhlich, Jul 09 2016
Equivalently, the number of vertices in the n-Menger sponge graph.

Crossrefs

Cf. A291066 (edge count).
Cf. A291066, A083233, and A332705 on the surface area of the n-Menger sponge graph.

Programs

Formula

G.f.: 1/(1-20*x).
E.g.f.: exp(20*x).
a(n) = A159991(n)/A000244(n). - Reinhard Zumkeller, May 02 2009
From Vincenzo Librandi, Nov 21 2010: (Start)
a(n) = 20^n.
a(n) = 20*a(n-1) for n > 0, a(0) = 1. (End)
a(n) = A000079(n)*A011557(n) = A000302(n)*A000351(n). - Felix Fröhlich, Jul 09 2016

A122803 Powers of -2: a(n) = (-2)^n.

Original entry on oeis.org

1, -2, 4, -8, 16, -32, 64, -128, 256, -512, 1024, -2048, 4096, -8192, 16384, -32768, 65536, -131072, 262144, -524288, 1048576, -2097152, 4194304, -8388608, 16777216, -33554432, 67108864, -134217728, 268435456, -536870912, 1073741824, -2147483648, 4294967296, -8589934592, 17179869184
Offset: 0

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The number -2 can be used as a base of numeration (see the Weisstein link). - Alonso del Arte, Mar 30 2014
Contribution from M. F. Hasler, Oct 21 2014: (Start)
This is the inverse binomial transform of A033999 = n->(-1)^n, and the binomial transform of A033999*A000244 = n->(-3)^n, see also A141413.
Prefixed with one 0, i.e., (0,1,-2,4,...) = -A033999*A131577, it is the binomial transform of (0, 1, -4, 13, -40, 121,...) = -A033999*A003462, and inverse binomial transform of (0,1,0,1,0,1,...) = A000035.
Prefixed with two 0's, i.e., (0,0,1,-2,4,-8,...), it is the binomial transform of (0,0,1,-5,18,-58,179,-543,...) (cf. A000340) and inverse binomial transform of (0,0,1,1,2,2,3,3,...) = A004526. (End)
Prefixed with three 0's, this is the inverse binomial difference of (0, 0, 0, 1, 2, 4, 6, 9, 12, 16,...) = concat(0, A002620), which has as successive differences (0, 0, 1, 1, 2, 2,...) = A004526, then (0, 1, 0, 1,...) = A000035, then (1, -1, 1, -1,...) = A033999, and then (-2)^k*A033999 with k=1,2,3,... - Paul Curtz, Oct 16 2014, edited by M. F. Hasler, Oct 21 2014
Stirling-Bernoulli transform of triangular numbers: 1, 3, 6, 10, 15, 21, 28, ... - Philippe Deléham, May 25 2015

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Programs

Formula

a(n) = (-2)^n = (-1)^n * 2^n.
a(n) = -2*a(n-1), n > 0; a(0) = 1. G.f.: 1/(1+2x). - Philippe Deléham, Nov 19 2008
Sum_{n >= 0} 1/a(n) = 2/3. - Jaume Oliver Lafont, Mar 01 2009
E.g.f.: 1/exp(2*x). - Arkadiusz Wesolowski, Aug 13 2012
a(n) = Sum_{k = 0..n} (-2)^(n-k)*binomial(n, k)*A030195(n+1). - R. J. Mathar, Oct 15 2012
G.f.: 1/(1+2x). A122803 = A033999 * A000079. - M. F. Hasler, Oct 21 2014
a(n) = Sum_{k = 0..n} A163626(n,k)*A000217(k+1). - Philippe Deléham, May 25 2015
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