A005009 -id:A005009 - cp's OEIS frontend

A173787 Triangle read by rows: T(n,k) = 2^n - 2^k, 0 <= k <= n.

0, 1, 0, 3, 2, 0, 7, 6, 4, 0, 15, 14, 12, 8, 0, 31, 30, 28, 24, 16, 0, 63, 62, 60, 56, 48, 32, 0, 127, 126, 124, 120, 112, 96, 64, 0, 255, 254, 252, 248, 240, 224, 192, 128, 0, 511, 510, 508, 504, 496, 480, 448, 384, 256, 0, 1023, 1022, 1020, 1016, 1008, 992, 960, 896, 768, 512, 0

Offset: 0

Reinhard Zumkeller, Feb 28 2010

			Triangle begins as:
   0;
   1,  0;
   3,  2,  0;
   7,  6,  4,  0;
  15, 14, 12,  8,  0;
  31, 30, 28, 24, 16, 0;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

[2^n -2^k: k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 13 2021

Table[2^n -2^k, {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 13 2021 *)

flatten([[2^n -2^k for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 13 2021

A000120(T(n,k)) = A025581(n,k).
Row sums give A000337.
Central terms give A020522.
T(2*n+1, n) = A006516(n+1).
T(2*n+3, n+2) = A059153(n).
T(n, k) = A140513(n,k) - A173786(n,k), 0 <= k <= n.
T(n, k) = A173786(n,k) - A059268(n+1,k+1), 0 < k <= n.
T(2*n, 2*k) = T(n,k) * A173786(n,k), 0 <= k <= n.
T(n, 0) = A000225(n).
T(n, 1) = A000918(n) for n>0.
T(n, 2) = A028399(n) for n>1.
T(n, 3) = A159741(n-3) for n>3.
T(n, 4) = A175164(n-4) for n>4.
T(n, 5) = A175165(n-5) for n>5.
T(n, 6) = A175166(n-6) for n>6.
T(n, n-4) = A110286(n-4) for n>3.
T(n, n-3) = A005009(n-3) for n>2.
T(n, n-2) = A007283(n-2) for n>1.
T(n, n-1) = A000079(n-1) for n>0.
T(n, n) = A000004(n).

A110286 a(n) = 15*2^n.

Original entry on oeis.org

15, 30, 60, 120, 240, 480, 960, 1920, 3840, 7680, 15360, 30720, 61440, 122880, 245760, 491520, 983040, 1966080, 3932160, 7864320, 15728640, 31457280, 62914560, 125829120, 251658240, 503316480, 1006632960, 2013265920, 4026531840, 8053063680, 16106127360

Offset: 0

Alexandre Wajnberg, Sep 07 2005

The first differences are the sequence itself. Doubling the terms gives the same sequence (beginning one step further).

Vincenzo Librandi, Table of n, a(n) for n = 0..235
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A007283, A020714, A005009, A005010, A005015, A005029.

Cf. A000079, A051916, A173787.

[15*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

15*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[2#&,15,30] (* Harvey P. Dale, Oct 19 2014 *)

a(n)=15<Charles R Greathouse IV, Oct 07 2015

G.f.: 15/(1-2x). - Philippe Deléham, Nov 23 2008
a(n) = A000079(n)*15 = A007283(n)*5 = A020714(n)*3. - Omar E. Pol, Dec 17 2008
a(n) = A173787(n+4,n). - Reinhard Zumkeller, Feb 28 2010
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
a(n) = 2*a(n-1) (with a(0)=15). - Vincenzo Librandi, Dec 26 2010
E.g.f.: 15*exp(2*x). - Stefano Spezia, May 15 2021

Edited by Omar E. Pol, Dec 16 2008

A030067 The "Semi-Fibonacci sequence": a(1) = 1; a(n) = a(n/2) (n even); a(n) = a(n-1) + a(n-2) (n odd).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 1, 6, 3, 9, 2, 11, 5, 16, 1, 17, 6, 23, 3, 26, 9, 35, 2, 37, 11, 48, 5, 53, 16, 69, 1, 70, 17, 87, 6, 93, 23, 116, 3, 119, 26, 145, 9, 154, 35, 189, 2, 191, 37, 228, 11, 239, 48, 287, 5, 292, 53, 345, 16, 361, 69, 430, 1, 431, 70, 501, 17, 518, 87, 605, 6, 611, 93

Offset: 1

David W. Wilson

This is the "semi-Fibonacci sequence". The distinct numbers that appear are called "semi-Fibonacci numbers", and are given in A030068.
a(2n+1) >= a(2n-1) + 1 is monotonically increasing. a(2n)/n can be arbitrarily small, as a(2^n) = 1. There are probably an infinite number of primes in the sequence. - Jonathan Vos Post, Mar 28 2006
From Robert G. Wilson v, Jan 17 2014: (Start)
Positions where k occurs:
   k: sequence
   -:-----------------------------
   1: A000079;
   2: 3*A000079 = A007283;
   3: 5*A000079 = A020714;
   4: none in the first 10^6 terms;
   5: 7*A000079 = A005009;
   6: 9*A000079 = A005010;
   7: none in the first 10^6 terms;
   8: none in the first 10^6 terms;
   9: 11*A000079 = A005015;
  10: none in the first 10^6 terms;
  11: 13*A000079 = A005029;
  12: none in the first 10^6 terms;
(End)
Any integer N which occurs in this sequence first occurs as an odd-indexed term a(2k-1) = A030068(k-1), and thereafter at indices (2k-1)*2^j, j=1,2,3,... (Both of these statements follow immediately from the definition of even-indexed terms.) No N can occur a second time as an odd-indexed term: This follows from the definition of these terms, a(2n+1) = a(2n) + a(2n-1) = a(2n-1) + a(n), which shows that the subsequence of odd-indexed terms (A030068) is strictly increasing, and therefore equal to the range (or: set) of the semi-Fibonacci numbers. - M. F. Hasler, Mar 24 2017
The lines in the logarithmic scatterplot of the sequence corresponds to sets of indices with the same 2-adic valuation. - Rémy Sigrist, Nov 27 2017
Define the partition subsum polynomial of an integer partition m of n where m = (m_1, m_2, ...m_k) by ps(m,x) = Product_{i=1..k} (1+x^m_i). Expanding ps(m,x) gives 1+a_1 x+a_2 x^2+...+a_n x^n, where a_j is the number of ways to form the subsum j from the parts of m. Then the number of partitions m of n for which ps(m,x) has no repeated root is a(n). - George Beck, Nov 07 2018

			a(1) = 1 by definition.
a(2) = a(1) = 1.
a(3) = 1 + 1 = 2.
a(4) = a(2) = 1.
a(5) = 2 + 1 = 3.
a(6) = a(3) = 2.
a(7) = 3 + 2 = 5.
a(8) = a(4) = 1.
a(9) = 5 + 1 = 6.
a(10) = a(5) = 3.

T. D. Noe, Table of n, a(n) for n = 1..10000
Abdulaziz M. Alanazi, Augustine O. Munagi and Darlison Nyirenda, Power Partitions and Semi-m-Fibonacci Partitions, arXiv:1910.09482 [math.CO], 2019.
George E. Andrews, Binary and Semi-Fibonacci Partitions, Journal of Ramanujan Society of Mathematics and Mathematics Sciences, honoring A.K. Agarwal's 70th birthday, 7:1(2019), 01-06.
Cristina Ballantine and George Beck, Partitions enumerated by self-similar sequences, arXiv:2303.11493 [math.CO], 2023.
George Beck, Semi-Fibonacci Partitions
Rémy Sigrist, Colored logarithmic scatterplot of the first 10000 terms (where the color is function of the 2-adic valuation of n)

Cf. A000045, A030068, A074364, A129092, A129100.

See A109671 for a variant.

import Data.List (transpose)
a030067 n = a030067_list !! (n-1)
a030067_list = concat $ transpose [scanl (+) 1 a030067_list, a030067_list]
-- Reinhard Zumkeller, Jul 21 2013, Jul 07 2013

f:=proc(n) option remember; if n=1 then RETURN(1) elif n mod 2 = 0 then RETURN(f(n/2)) else RETURN(f(n-1)+f(n-2)); fi; end;

semiFibo[1] = 1; semiFibo[n_?EvenQ] := semiFibo[n] = semiFibo[n/2]; semiFibo[n_?OddQ] := semiFibo[n] = semiFibo[n - 1] + semiFibo[n - 2]; Table[semiFibo[n], {n, 80}] (* Jean-François Alcover, Aug 19 2013 *)

a(n) = if(n==1, 1, if(n%2 == 0, a(n/2), a(n-1) + a(n-2)));
vector(100, n, a(n)) \\ Altug Alkan, Oct 12 2015

a=[1]; [a.append(a[-2]+a[-1] if n%2 else a[n//2-1]) for n in range(2, 75)]
print(a) # Michael S. Branicky, Jul 07 2022

Theorem: a(2n+1) - a(2n-1) = a(n). Proof: a(2n+1) - a(2n-1) = a(2n) + a(2n-1) - a(2n-2) - a(2n-3) = a(n) - a(n-1) + a(n-1) (induction) = a(n). - N. J. A. Sloane, May 02 2010
a(2^n - 1) = A129092(n) for n >= 1, where A129092 forms the row sums and column 0 of triangle A129100, which is defined by the nice property that column 0 of matrix power A129100^(2^k) = column k of A129100 for k > 0. - Paul D. Hanna, Dec 03 2008
G.f. g(x) satisfies (1-x^2) g(x) = (1+x-x^2) g(x^2) + x. - Robert Israel, Mar 23 2017

A028313 Elements in the 5-Pascal triangle (by row).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 6, 6, 1, 1, 7, 12, 7, 1, 1, 8, 19, 19, 8, 1, 1, 9, 27, 38, 27, 9, 1, 1, 10, 36, 65, 65, 36, 10, 1, 1, 11, 46, 101, 130, 101, 46, 11, 1, 1, 12, 57, 147, 231, 231, 147, 57, 12, 1, 1, 13, 69, 204, 378, 462, 378, 204, 69, 13, 1, 1, 14, 82, 273, 582, 840, 840, 582, 273, 82, 14, 1

Offset: 0

Mohammad K. Azarian

			Triangle begins as:
  1;
  1,  1;
  1,  5,  1;
  1,  6,  6,   1;
  1,  7, 12,   7,   1;
  1,  8, 19,  19,   8,   1;
  1,  9, 27,  38,  27,   9,   1;
  1, 10, 36,  65,  65,  36,  10,  1;
  1, 11, 46, 101, 130, 101,  46, 11,  1;
  1, 12, 57, 147, 231, 231, 147, 57, 12,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000007, A005009, A010892, A022112, A028275, A028314, A028315.
Cf. A028316, A028317, A028318, A028319, A028320, A028321, A028322.
Cf. A028323, A028324, A028325, A029653, A051472, A051936, A051937.
Cf. A178915.

[n le 1 select 1 else Binomial(n,k) +3*Binomial(n-2,k-1): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 05 2024

Table[If[n<2, 1, Binomial[n,k] +3*Binomial[n-2,k-1]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 05 2024 *)

def A028313(n,k): return 1 if n<2 else binomial(n,k) + 3*binomial(n-2,k-1)
flatten([[A028313(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 05 2024

From Ralf Stephan, Jan 31 2005: (Start)
T(n, k) = C(n, k) + 3*C(n-2, k-1), with T(0, k) = T(1, k) = 1.
G.f.: (1 + 3*x^2*y)/(1 - x*(1+y)). (End)
From G. C. Greubel, Jan 05 2024: (Start)
T(n, n-k) = T(n, k).
T(n, n-1) = n + 3*(1 - [n=1]) = A178915(n+3), n >= 1.
T(n, n-2) = A051936(n+2), n >= 2.
T(n, n-3) = A051937(n+1), n >= 3.
T(2*n, n) = A028322(n).
Sum_{k=0..n} T(n, k) = A005009(n-2) - (3/4)*[n=0] - (3/2)*[n=1].
Sum_{k=0..n} (-1)^k * T(n, k) = A000007(n) - 3*[n=2].
Sum_{k=0..floor(n/2)} T(n-k, k) = A022112(n-2) + 3*([n=0] - [n=1]).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = 4*A010892(n) - 3*([n=0] + [n=1]). (End)

More terms from Sam Alexander (pink2001x(AT)hotmail.com)

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Wolfdieter Lang, First 10 rows and array of figurate numbers

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
Cf. A007318, A093564 (d=7), A228196, A228576.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

a093563 n k = a093563_tabl !! n !! k
a093563_row n = a093563_tabl !! n
a093563_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
-- Reinhard Zumkeller, Aug 31 2014

lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)

from math import comb, isqrt
def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A118416 Triangle read by rows: T(n,k) = (2k-1)2^(n-1), 0 < k <= n.

Original entry on oeis.org

1, 2, 6, 4, 12, 20, 8, 24, 40, 56, 16, 48, 80, 112, 144, 32, 96, 160, 224, 288, 352, 64, 192, 320, 448, 576, 704, 832, 128, 384, 640, 896, 1152, 1408, 1664, 1920, 256, 768, 1280, 1792, 2304, 2816, 3328, 3840, 4352, 512, 1536, 2560, 3584, 4608, 5632, 6656, 7680

Offset: 1

Reinhard Zumkeller, Apr 27 2006

Row sums give A014477: Sum_{k=1..n} T(n,k) = A014477(n-1);
central terms give A118415; T(2*k-1,k) = A058962(k-1);
T(n,1) = A000079(n-1);
T(n,2) = A007283(n-1) for n > 1;
T(n,3) = A020714(n-1) for n > 2;
T(n,4) = A005009(n-1) for n > 3;
T(n,5) = A005010(n-1) for n > 4;
T(n,n-1) = A118417(n-1) for n > 1;
T(n,n) = A014480(n-1) = A118413(n,n);
A001511(T(n,k)) = A002024(n,k);
A003602(T(n,k)) = A002260(n,k).
The alternating row sums, Sum_{k=1..n} (-1)^(k+1)*T(n,k), are: (a) in odd rows, the central term, T(n,(n+1)/2) = A058962((n-1)/2); (b) in even rows, the negation of the average of the two central terms, -(T(2n,n) + T(2n,+1))/2 = -A018215(m/2). The absolute values of the alternating row sums give the plain row means, Sum_{k=1..n} T(n,k)/n; the alternating sign row means are (-2)^(n-1). - Gregory Gerard Wojnar, Feb 10 2024

			Triangle begins:
   1;
   2,   6;
   4,  12,  20;
   8,  24,  40,  56;
  16,  48,  80, 112, 144;
  32,  96, 160, 224, 288, 352;
  64, 192, 320, 448, 576, 704, 832;

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened

Cf. A118413, A117303, A054582.

a118416 n k = a118416_tabl !! (n-1) !! (k-1)
a118416_row 1 = [1]
a118416_row n = (map (* 2) $ a118416_row (n-1)) ++ [a014480 (n-1)]
a118416_tabl = map a118416_row [1..]
-- Reinhard Zumkeller, Jan 22 2012

A118416 := proc(n,k) 2^(n-1)*(2*k-1) ; end proc: # R. J. Mathar, Sep 04 2011

Flatten[Table[(2k-1)2^(n-1),{n,10},{k,n}]] (* Harvey P. Dale, Aug 26 2014 *)

from math import isqrt
def A118416(n): return (a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))*(1-a)+(n<<1)-1<Chai Wah Wu, Jun 20 2025

T(n,k) = 2*T(n-1,k), 1 <= k < n; T(n,n) = A014480(n-1).

A002042 a(n) = 7*4^n.

Original entry on oeis.org

7, 28, 112, 448, 1792, 7168, 28672, 114688, 458752, 1835008, 7340032, 29360128, 117440512, 469762048, 1879048192, 7516192768, 30064771072, 120259084288, 481036337152, 1924145348608, 7696581394432, 30786325577728, 123145302310912, 492581209243648

Offset: 0

N. J. A. Sloane

Subsequence of A000069, the odious numbers. - Reinhard Zumkeller, Aug 26 2007
A rectangular prism with edge lengths 2^n, 2^(n+1) and 2^(n+2) has a surface area 2* (2^n*2^(n+1) + 2^(n+1)*2^(n+2) + 2^n*2^(n+2)) which equals 4*a(n). - J. M. Bergot, Aug 07 2013
x = A306472(n) and y = a(n) satisfy the Lebesgue-Ramanujan-Nagell equation x^2 + 3^(6*n+1) = 4*y^3 (see Theorem 2.1 in Chakraborty, Hoque and Sharma). - Stefano Spezia, Feb 18 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..500
K. Chakraborty, A. Hoque, R. Sharma, Complete solutions of certain Lebesgue-Ramanujan-Nagell type equations, arXiv:1812.11874 [math.NT], 2018.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (4).

First differences of A083597. Bisection of A005009.

Cf. A306472 (37*27^n), A009971 (27^n), A000302 (4^n), A000290 (n^2), A000578 (n^3).

[7*4^n: n in [0..30]]; // Vincenzo Librandi, May 31 2011

7*4^Range[0, 100] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
CoefficientList[Series[7/(1-4x), {x, 0, 33}], x] (* Vincenzo Librandi, Jun 25 2015 *)
NestList[4#&,7,30] (* Harvey P. Dale, Mar 19 2021 *)

a(n)=7<<(2*n) \\ Charles R Greathouse IV, Apr 17 2012

[7*4^n for n in (0..30)] # G. C. Greubel, Feb 18 2019

From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 4*a(n-1), n > 0, with a(0) = 7.
G.f.: 7/(1-4*x). (End)
a(n) = 7*A000302(n). - Michel Marcus, Jun 24 2015
E.g.f.: 7*exp(4*x). - G. C. Greubel, Feb 18 2019

A047345 Numbers that are congruent to {0, 4} mod 7.

Original entry on oeis.org

0, 4, 7, 11, 14, 18, 21, 25, 28, 32, 35, 39, 42, 46, 49, 53, 56, 60, 63, 67, 70, 74, 77, 81, 84, 88, 91, 95, 98, 102, 105, 109, 112, 116, 119, 123, 126, 130, 133, 137, 140, 144, 147, 151, 154, 158, 161, 165, 168, 172, 175, 179, 182, 186, 189, 193

Offset: 1

N. J. A. Sloane

Nonnegative k such that k or 5*k + 1 is divisible by 7. - Bruno Berselli, Feb 13 2018

Maximum number of 2's possible in an infinite Minesweeper grid with n mines. The pattern of mines (x) that generates these 2's looks like "...xx.xx.xx...". - Dmitry Kamenetsky, Apr 14 2018

David Lovler, Table of n, a(n) for n = 1..1000
Wikipedia, Minesweeper.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A030123.

Cf. A005009, A030308.

A047345:=n->ceil(7*(n-1)/2); seq(A047345(n), n=1..100); # Wesley Ivan Hurt, Mar 31 2014

Table[Ceiling[7 (n - 1)/2], {n, 100}] (* Wesley Ivan Hurt, Mar 31 2014 *)

forstep(n=0,200,[4,3],print1(n", ")) \\ Charles R Greathouse IV, Oct 17 2011

a(n) = ceiling(7*(n-1)/2).
a(n) = 7*n - a(n-1) - 10 for n>1, a(1)=0. - Vincenzo Librandi, Aug 05 2010
From R. J. Mathar, Oct 08 2011: (Start)
a(n) = 7*n/2 - 13/4 + (-1)^n/4.
G.f.: x^2*(4 + 3*x) / ((1 + x)*(x - 1)^2). (End)
a(n+1) = Sum_{k>=0} A030308(n,k)*b(k), with b(0) = 4, b(k) = A005009(k-1) = 7*2^(k-1), and k>0. - Philippe Deléham, Oct 17 2011.
a(n) = 4*(n - 1) - floor((n - 1)/2). - Wesley Ivan Hurt, Jun 14 2013
a(n) = 2*(n - 1) + floor((3*n - 2 - (n mod 2))/2). - Wesley Ivan Hurt, Mar 31 2014
E.g.f.: 3 + ((14*x - 13)*exp(x) + exp(-x))/4. - David Lovler, Aug 31 2022

A062092 a(n) = 2*a(n-1) - (-1)^n for n > 0, a(0)=2.

Original entry on oeis.org

2, 5, 9, 19, 37, 75, 149, 299, 597, 1195, 2389, 4779, 9557, 19115, 38229, 76459, 152917, 305835, 611669, 1223339, 2446677, 4893355, 9786709, 19573419, 39146837, 78293675, 156587349, 313174699, 626349397, 1252698795, 2505397589

Offset: 0

Amarnath Murthy, Jun 16 2001

Let A be the Hessenberg matrix of order n, defined by: A[1,j] = A[i,i] = 1, A[i,i-1] = -1, and A[i,j] = 0 otherwise. Then, for n>=1, a(n-1) = charpoly(A,3). - Milan Janjic, Jan 24 2010

T. Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.

Harry J. Smith, Table of n, a(n) for n = 0..200
Petro Kosobutskyy, The Collatz problem as a reverse n->0 problem on a graph tree formed from theta*2^n Jacobsthal-type numbers, arXiv:2306.14635 [math.GM], 2023.
Petro Kosobutskyy and Dariia Rebot, Collatz conjecture 3n+/-1 as a Newton binomial problem, Comp. Des. Sys. Theor. Prac., Lviv Nat'l Polytech. Univ. (Ukraine 2023) Vol. 5, No. 1, 137-145. See p. 140.
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A000032, A001045, A002487, A005009, A078008, A154879, A201630.

Cf. A171160 (first differences).

[(7*2^n-(-1)^n)/3: n in [0..40]]; // G. C. Greubel, Apr 04 2023

LinearRecurrence[{1, 2}, {2, 5}, 40] (* Jean-François Alcover, Aug 02 2021 *)

a(n) = (7*2^n - (-1)^n)/3; \\ Harry J. Smith, Aug 01 2009

[(7*2^n-(-1)^n)/3 for n in range(41)] # G. C. Greubel, Apr 04 2023

a(n) = a(n-1) + 2*a(n-2).
a(n) = (7*2^n - (-1)^n)/3.
a(n) = 2^(n+1) + A001045(n).
A002487(a(n)) = A000032(n+1).
G.f.: (2+3*x)/(1-x-2*x^2).
E.g.f.: (7*exp(2*x) - exp(-x))/3.
a(n) = Sum_{j=0..2} A001045(n-j) (sum of 3 consecutive elements of the Jacobsthal sequence). - Alexander Adamchuk, May 16 2006
From Paul Curtz, Jun 03 2022: (Start)
a(n) = A001045(n+3) - A078008(n).
a(n) = A078008(n+3) - A001045(n).
a(n) = A005009(n-1) - a(n-1) for n >= 1.
a(n) = a(n-2) + A005009(n-2) for n >= 2.
a(n) = A154879(n-2) + 3*A201630(n-2) for n >= 2. (End)

More terms from Jason Earls, Jun 18 2001

Additional comments from Michael Somos, Jun 24 2002

A070875 Binary expansion is 1x100...0 where x = 0 or 1.

Original entry on oeis.org

5, 7, 10, 14, 20, 28, 40, 56, 80, 112, 160, 224, 320, 448, 640, 896, 1280, 1792, 2560, 3584, 5120, 7168, 10240, 14336, 20480, 28672, 40960, 57344, 81920, 114688, 163840, 229376, 327680, 458752, 655360, 917504, 1310720, 1835008, 2621440

Offset: 0

N. J. A. Sloane, May 19 2002

A 2-automatic sequence. - Charles R Greathouse IV, Sep 24 2012
Third row in array A228405. - Richard R. Forberg, Sep 06 2013
Conjecture: a(n) = -1 +  positions of the ones in A309019(n+2) - A002487(n+2). - George Beck, Mar 26 2022
Consecutive integers for which the number of its proper nondivisors of the form 2^k (k > 0) is 2; proper nondivisors are defined in A173540 (5 has two such nondivisors: 2 and 4, 7 has 2 and 4, 10 has 4 and 8, 14 has 4 and 8, 20 has 8 and 16,...). - Lechoslaw Ratajczak, Dec 17 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2).
Index entries for 2-automatic sequences.

Cf. A070876, A123760, A063920.

[n le 2 select 2*n+3 else 2*Self(n-2): n in [1..39]]; // Bruno Berselli, Mar 01 2011

Flatten@ NestList[ 2# &, {5, 7}, 19] (* Or *)
CoefficientList[ Series[(5 + 7 x)/(1 - 2 x^2), {x, 0, 38}], x] (* Robert G. Wilson v, May 20 2002 *)

a(n)=if(n%2,7,5)<<(n\2) \\ Charles R Greathouse IV, Sep 24 2012

A093873(a(n)) = 2. - Reinhard Zumkeller, Oct 13 2006
For n>1, a(n+1) = a(n) + A000010(a(n)). - Stefan Steinerberger, Dec 20 2007
From Bruno Berselli, Mar 01 2011: (Start)
G.f.: (5+7*x)/(1-2*x^2).
a(n) = (6-(-1)^n)*2^((2*n+(-1)^n-1)/4). Therefore: a(n) = 5*2^(n/2) for n even, otherwise a(n) = 7*2^((n-1)/2).
a(n) = 2*a(n-2) for n>1. (End)
a(n+1) = A063757(n) + 6. - Philippe Deléham, Apr 13 2013
a(n) = sqrt(2*a(n-1) - (-2)^(n-1)). - Richard R. Forberg, Sep 06 2013
a(n+3) = a(n+2)*a(n+1)/a(n). - Richard R. Forberg, Sep 06 2013
For n>1, a(n) = 2*phi(a(n)) + phi(phi(a(n))). - Ivan Neretin, Feb 28 2016
a(2n) = A020714(n), a(2n+1) = A005009(n); for n>0. - Yosu Yurramendi, Jun 01 2016
From Ilya Gutkovskiy, Jun 02 2016: (Start)
E.g.f.: 7*sinh(sqrt(2)*x)/sqrt(2) + 5*cosh(sqrt(2)*x).
a(n) = 2^((n-3)/2)*(5*sqrt(2)*(1 + (-1)^n) + 7*(1 - (-1)^n)). (End)
Sum_{n>=0} 1/a(n) = 24/35. - Amiram Eldar, Mar 28 2022

Extended by Robert G. Wilson v, May 20 2002

cp's OEIS Frontend

A173787 Triangle read by rows: T(n,k) = 2^n - 2^k, 0 <= k <= n.

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A110286 a(n) = 15*2^n.

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A030067 The "Semi-Fibonacci sequence": a(1) = 1; a(n) = a(n/2) (n even); a(n) = a(n-1) + a(n-2) (n odd).

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A028313 Elements in the 5-Pascal triangle (by row).

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A093563 (6,1)-Pascal triangle.

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A118416 Triangle read by rows: T(n,k) = (2*k-1)*2^(n-1), 0 < k <= n.

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A118416 Triangle read by rows: T(n,k) = (2k-1)2^(n-1), 0 < k <= n.