A020714 -id:A020714 - cp's OEIS frontend

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).

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

A095666 Pascal (1,4) triangle.

Original entry on oeis.org

4, 1, 4, 1, 5, 4, 1, 6, 9, 4, 1, 7, 15, 13, 4, 1, 8, 22, 28, 17, 4, 1, 9, 30, 50, 45, 21, 4, 1, 10, 39, 80, 95, 66, 25, 4, 1, 11, 49, 119, 175, 161, 91, 29, 4, 1, 12, 60, 168, 294, 336, 252, 120, 33, 4, 1, 13, 72, 228, 462, 630, 588, 372, 153, 37, 4, 1, 14, 85, 300, 690, 1092

Offset: 0

Wolfdieter Lang, Jun 11 2004

This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. 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) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - Reinhard Zumkeller, Apr 08 2012

			Triangle begins:
  [4];
  [1,4];
  [1,5,4];
  [1,6,9,4];
  [1,7,15,13,4];
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A020714(n-1), n >= 1, 4 if n=0.
Alternating row sums are [4, -3, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+4), A055999(n+1), A060488(n+3), A095667-71, A095819.

a095666 n k = a095666_tabl !! n !! k
a095666_row n = a095666_tabl !! n
a095666_tabl = [4] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]
-- Reinhard Zumkeller, Apr 08 2012

a(n,k):=(1+3*k/n)*binomial(n,k) # Mircea Merca, Apr 08 2012

A095666[n_, k_] := If[n == k,  4, (3*k/n + 1)*Binomial[n, k]];
Table[A095666[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1 + 3*k/n)*binomial(n,k). - Mircea Merca, Apr 08 2012

A093561 (4,1) Pascal triangle.

Original entry on oeis.org

1, 4, 1, 4, 5, 1, 4, 9, 6, 1, 4, 13, 15, 7, 1, 4, 17, 28, 22, 8, 1, 4, 21, 45, 50, 30, 9, 1, 4, 25, 66, 95, 80, 39, 10, 1, 4, 29, 91, 161, 175, 119, 49, 11, 1, 4, 33, 120, 252, 336, 294, 168, 60, 12, 1, 4, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1, 4, 41, 190, 525, 960, 1218

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
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+3*z)/(1-(1+x)*z).
The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. 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
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

			Triangle begins
  [1];
  [4, 1];
  [4, 5, 1];
  [4, 9, 6, 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
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.
Cf. A007318, A093562 (d=5), A228196, A228576.

a093561 n k = a093561_tabl !! n !! k
a093561_row n = a093561_tabl !! n
a093561_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093561(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+3*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*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)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 3*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)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*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

A094958 Numbers of the form 2^k or 5*2^k.

Original entry on oeis.org

1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, 2560, 4096, 5120, 8192, 10240, 16384, 20480, 32768, 40960, 65536, 81920, 131072, 163840, 262144, 327680, 524288, 655360, 1048576, 1310720, 2097152

Offset: 1

Ralf Stephan, Jun 01 2004

The subset {a(1),...,a(2k)} together with a(2k+2) is the set of proper divisors of 5*2^k.
For a(n)>4: number of vertices of complete graphs that can be properly edge-colored in such a way that the edges can be partitioned into edge disjoint multicolored isomorphic spanning trees.
(Editor's note: The following 3 comments are equivalent.)
From Wouter Meeussen, Apr 10 2005: This appears to be the same sequence as "Numbers n such that n^2 is not the sum of three nonzero squares". Don Reble and Paul Pollack respond: Yes, that is correct.
Also numbers k such that k^2=a^2+b^2+c^2 has no solutions in the positive integers a, b and c. - Wouter Meeussen, Apr 20 2005
The only natural numbers which cannot be the lengths of an interior diagonal of a cuboid with natural edges. - Michael Somos, Mar 02 2004

Wacław Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, p. 101, MR2002669.

Gregory Constantine, Multicolored isomorphic spanning trees in complete graphs, Discrete Mathematics and Theoretical Computer Science, Vol. 5 (2002), pp. 121-126.
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A029744, A029745.
Union of A000079 and A020714.
Complement of A005767.

With[{c=2^Range[0,30]},Union[Join[c,5c]]] (* Harvey P. Dale, Jul 15 2012 *)

def A094958(n): return 1<>1)+1 if n&1 else 5<<((n>>1)-2) # Chai Wah Wu, Feb 14 2025

a(1)=1, a(2)=2, a(3)=4, for n>=0, a(2n+3) = 4*2^n, a(2n+4) = 5*2^n.
Recurrence: for n>4, a(n) = 2a(n-2).
G.f.: x*(1+x)*(1+x+x^2)/(1-2x^2).
Sum_{n>=1} 1/a(n) = 12/5. - Amiram Eldar, Jan 21 2022

Edited by T. D. Noe and M. F. Hasler, Nov 12 2010

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).

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

A091629 Product of digits associated with A091628(n). Essentially the same as A007283.

Original entry on oeis.org

6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472

Offset: 1

Enoch Haga, Jan 24 2004

Sequence arising in Farideh Firoozbakht's solution to Prime Puzzle 251 - 23 is the only pointer prime (A089823) not containing digit "1".

The monotonic increasing value of successive product of digits strongly suggests that in successive n the digit 1 must be present.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Carlos Rivera, Puzzle 251, Pointer primes, The Prime Puzzles and Problems Connection.
Index entries for linear recurrences with constant coefficients, signature (2).

Sequences of the form (2*m+1)*2^n: A000079 (m=0), A007283 (m=1), A020714 (m=2), A005009 (m=3), A005010 (m=4), A005015 (m=5), A005029 (m=6), A110286 (m=7), A110287 (m=8), A110288 (m=9), A175805 (m=10), A248646 (m=11), A164161 (m=12), A175806 (m=13), A257548 (m=15).
Cf. A089823, A091628, A091630, A091631, A091632.
Similar to A003945, A042950, A058764, A087009, A081808.

[3*2^n : n in [1..40]]; // Wesley Ivan Hurt, Jul 17 2020

3*2^Range[1, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)

[3*2^n for n in range(1,51)] # G. C. Greubel, Jan 05 2023

a(n) = 3 * 2^n = product of digits of A091628(n).
From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 6*2^(n-1).
a(n) = 2*a(n-1), with a(1) = 6.
G.f.: 6*x/(1-2*x). (End)
E.g.f.: 3*(exp(2*x) - 1). - G. C. Greubel, Jan 05 2023

Edited and extended by Ray Chandler, Feb 07 2004

A334101 Numbers of the form q*(2^k), where q is one of the Fermat primes and k >= 0; Numbers n for which A329697(n) == 1.

Original entry on oeis.org

3, 5, 6, 10, 12, 17, 20, 24, 34, 40, 48, 68, 80, 96, 136, 160, 192, 257, 272, 320, 384, 514, 544, 640, 768, 1028, 1088, 1280, 1536, 2056, 2176, 2560, 3072, 4112, 4352, 5120, 6144, 8224, 8704, 10240, 12288, 16448, 17408, 20480, 24576, 32896, 34816, 40960, 49152, 65537, 65792, 69632, 81920, 98304, 131074, 131584, 139264

Offset: 1

Antti Karttunen, Apr 14 2020

nonn

Numbers k that themselves are not powers of two, but for which A171462(k) = k-A052126(k) is [a power of 2].
Numbers k such that A000265(k) is in A019434.
Squares of these numbers can be found (as a subset) in A334102, and the cubes (as a subset) in A334103.

Row 1 of A334100.
Cf. A000120, A000265, A052126, A171462, A209229, A329697, A334102, A334103.
Cf. A019434 (primes present), A007283, A020714, A110287 (other subsequences).
Subsequence of A018900.

A000265(n) = (n>>valuation(n,2));
isA019434(n) = ((n>2)&&isprime(n)&&!bitand(n-2,n-1));
isA334101(n) = isA019434(A000265(n));

A052126(n) = if(1==n,n,n/vecmax(factor(n)[, 1]));
A209229(n) = (n && !bitand(n,n-1));
isA334101(n) = ((!A209229(n))&&A209229(n-A052126(n)));

For all n, A000120(a(n)) = 2.

A084215 Expansion of g.f.: (1+x^2)/(1-2*x).

Original entry on oeis.org

1, 2, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240

Offset: 0

Paul Barry, May 19 2003

Associated with a math magic problem.

Elements are the sums of consecutive pairs of elements of A084214.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A020714, A060816, A084214.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+x^2)/(1-2*x))); // G. C. Greubel, Oct 08 2018

Join[{1, 2, a = 5}, Table[a = 2*a, {n,0,40}]] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
Table[Int[2^(n-2)*5],{n,0,40}] (* Taher Jamshidi, Sep 15 2012 *)
CoefficientList[Series[(1 + x^2)/(1 - 2 x), {x, 0, 30}], x] (* G. C. Greubel, Oct 08 2018 *)

x='x+O('x^30); Vec((1+x^2)/(1-2*x)) \\ G. C. Greubel, Oct 08 2018

a(n) = Sum_{k=0..n} 2^(n-k)*binomial(1, k/2)*(1+(-1)^k)/2. - Paul Barry, Oct 15 2004
a(n) = A020714(n-2), n > 1. - R. J. Mathar, Dec 19 2008
From Gary W. Adamson, Aug 26 2011: (Start)
a(n) is the sum of top row terms of M^n, M is an infinite square production matrix as follows:
  1, 1, 0, 0, 0, 0, ...
  1, 1, 1, 0, 0, 0, ...
  0, 0, 0, 0, 0, 0, ...
  0, 0, 0, 0, 0, 0, ...
  ...
E.g.: a(4) = 20 = (8 + 8 + 4) since the top row of M^4 = (8, 8, 4, 0, 0, 0, ...). (End)
a(n) = floor(2^(n-2)*5). - Taher Jamshidi, Sep 15 2012
a(n) = 2*a(n-1) for n >= 3, a(0) = 1, a(1) = 2, a(2) = 5. - Philippe Deléham, Mar 13 2013
E.g.f.: (5*exp(2*x) - 2*x - 1)/4. - Stefano Spezia, Feb 20 2023

A257113 a(1) = 2, a(2) = 3; thereafter a(n) is the sum of all the previous terms.

Original entry on oeis.org

2, 3, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240

Offset: 1

Giovanni Teofilatto, Apr 24 2015

Except for first three terms, a(n) is 10 times 2^(n-4).

These values comprise the tile values used in the "fives" variant of the game 2048, including 1 as the zeroth term. - Michael De Vlieger, Jul 18 2018

Colin Barker, Table of n, a(n) for n = 1..1000
E. R. Berlekamp, A contribution to mathematical psychometrics, Unpublished Bell Labs Memorandum, Feb 08 1968 [Annotated scanned copy]
David Eppstein, Making Change in 2048, arXiv:1804.07396 [cs.DM], 2018.
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A000079, A020714. Essenitally the same as A084215.

[2,3] cat [5*2^n: n in [0..35]]; // Vincenzo Librandi, May 15 2015

t = {2, 3}; For[k = 3, k <= 27, k++, AppendTo[t, Total@ t]]; t (* Michael De Vlieger, May 14 2015 *)
Join[{2, 3}, Table[5 2^n, {n, 0, 40}]] (* Vincenzo Librandi, May 15 2015 *)
Join[{2,3},NestList[2#&,5,40]] (* Harvey P. Dale, Apr 06 2018 *)

a(n) = if(n<3, n+1, 5*2^(n-3)); \\ Altug Alkan, Jul 18 2018

Vec(x*(1 - x)*(2 + x) / (1 - 2*x) + O(x^40)) \\ Colin Barker, Nov 17 2018

a(n) = ceil(5*2^(n-3)) \\ Alan Michael Gómez Calderón, Mar 30 2022

a(n) = A020714(n-3) for n>2.
a(n) = A146523(n-2) for n>2. - R. J. Mathar, May 14 2015
G.f.: x*(1 - x)*(2 + x) / (1 - 2*x). - Colin Barker, Nov 17 2018

cp's OEIS Frontend

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