A103889 -id:A103889 - cp's OEIS frontend

A114113 a(n) = sum{k=1 to n} (A114112(k)). (For n>=2, a(n) = sum{k=1 to n} (A014681(k)) =sum{k=1 to n} (A103889(k)).).

1, 3, 7, 10, 16, 21, 29, 36, 46, 55, 67, 78, 92, 105, 121, 136, 154, 171, 191, 210, 232, 253, 277, 300, 326, 351, 379, 406, 436, 465, 497, 528, 562, 595, 631, 666, 704, 741, 781, 820, 862, 903, 947, 990, 1036, 1081, 1129, 1176, 1226, 1275, 1327, 1378, 1432

Offset: 1

Leroy Quet, Nov 13 2005

a(n) is not divisible by (A114112(n+1)).

Sequence is A130883 union A014105 - {0,2}. - Anthony Hernandez, Sep 08 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A014105, A014681, A026035, A084265, A103889, A114112.

I:=[1,3,7,10,16]; [n le 5 select I[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..60]]; // Vincenzo Librandi, Mar 13 2018

Join[{1}, LinearRecurrence[{2, 0, -2, 1}, {3, 7, 10, 16}, 52]] (* Jean-François Alcover, Sep 22 2017 *)
CoefficientList[Series[(1 + x + x^2 -2 x^3 + x^4)/((1 + x) (1 - x)^3), {x, 0, 60}], x] (* Vincenzo Librandi, Mar 13 2018 *)

def A114113(n): return 1 if n == 1 else (m:=n//2)*(n+1) + (n+1-m)*(n-2*m) # Chai Wah Wu, May 24 2022

a(1)=1. a(2n) = n*(2n+1). a(2n+1) = 2n^2 +3n +2.
From R. J. Mathar, Nov 04 2008: (Start)
a(n) = A026035(n+1) - A026035(n), n>1.
G.f.: x(1+x+x^2-2x^3+x^4)/((1+x)(1-x)^3).
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4), n>5. (End)
This is (essentially) 1 + A084265, - N. J. A. Sloane, Mar 12 2018

More terms from R. J. Mathar, Aug 31 2007

A289805 p-INVERT of A103889, where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 9, 36, 153, 624, 2584, 10632, 43865, 180774, 745347, 3072528, 12666854, 52218790, 215273737, 887468000, 3658604277, 15082652352, 62178493132, 256331858332, 1056732372729, 4356396740786, 17959318086575, 74037587378784, 305221185520298, 1258279413185322

Offset: 0

Clark Kimberling, Aug 12 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).

See A289780 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,2,-8,8,-2,1)

Cf. A103889, A289780.

z = 60; s = x*(x^2 - x + 2)/((x - 1)^2*(1 + x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A103889 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289805 *)
LinearRecurrence[{4,2,-8,8,-2,1},{2,9,36,153,624,2584},30] (* Harvey P. Dale, Jun 30 2021 *)

G.f.: (-2 - x + 4 x^2 - 7 x^3 + 4 x^4 - 2 x^5)/(-1 + 4 x + 2 x^2 - 8 x^3 + 8 x^4 - 2 x^5 + x^6).

a(n) = 4*a(n-1) + 2*a(n-2) - 8*a(n-3) + 8*a(n-4) - 2*a(n-5) + a(n-6).

A203196 (n-1)-st elementary symmetric function of the first n terms of (2,1,4,3,6,5,8,7,...)=A103889.

Original entry on oeis.org

1, 3, 14, 50, 324, 1764, 14832, 109584, 1136160, 10628640, 131172480, 1486442880, 21289201920, 283465647360, 4622628648960, 70734282393600, 1294139872972800, 22376988058521600, 453942134876160000, 8752948036761600000

Offset: 1

Clark Kimberling, Dec 30 2011

nonn

a(2n)=A000254(2n) for n=1,2,3...

Cf. A103889.

f[k_] := If[Mod[k, 2] == 0, k - 1, k + 1]
t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 14}]  (* A203196 *)

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

Original entry on oeis.org

0, 2, 1, 5, 2, 8, 3, 11, 4, 14, 5, 17, 6, 20, 7, 23, 8, 26, 9, 29, 10, 32, 11, 35, 12, 38, 13, 41, 14, 44, 15, 47, 16, 50, 17, 53, 18, 56, 19, 59, 20, 62, 21, 65, 22, 68, 23, 71, 24, 74, 25, 77, 26, 80, 27, 83, 28, 86, 29, 89, 30, 92, 31, 95, 32, 98, 33, 101, 34, 104

Offset: 0

Mohammad K. Azarian

This is the function usually denoted by T(n) in the literature on the 3x+1 problem. See A006370 for further references and links.
Intertwining of sequence A016789 '2,5,8,11,... ("add 3")' and the nonnegative integers.
a(n) = log_2(A076936(n)). - Amarnath Murthy, Oct 19 2002
The average value of a(0), ..., a(n-1) is A004526(n). - Amarnath Murthy, Oct 19 2002
Partial sums are A093353. - Paul Barry, Mar 31 2008
Absolute first differences are essentially in A014681 and A103889. - R. J. Mathar, Apr 05 2008
Only terms of A016789 occur twice, at positions given by sequences A005408 (odd numbers) and A016957 (6n+4): (1,4), (3,10), (5,16), (7,22), ... - Antti Karttunen, Jul 28 2017
a(n) represents the unique congruence class modulo 2n+1 that is represented an odd number of times in any 2n+1 consecutive oblong numbers (A002378). This property relates to Jim Singh's 2018 formula, as n^2 + n is a relevant oblong number. - Peter Munn, Jan 29 2022

			a(3) = -3*(-1) - 2*1 - 1*(-1) - 0*1 + 1*(-1) + 2*1 + 3*(-1) + 4*1 + 5*(-1) + 6*1 = 5. - _Bruno Berselli_, Dec 14 2015

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.

N. J. A. Sloane (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..100000
C. J. Everett, Iteration of the number-theoretic function f(2n) = n, f(2n + 1) = 3n + 2, Advances in Mathematics, Volume 25, Issue 1, July 1977, Pages 42-45.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
Keenan Monks, Kenneth G. Monks, Kenneth M. Monks, and Maria Monks, Strongly sufficient sets and the distribution of arithmetic sequences in the 3x+1 graph, arXiv:1204.3904v1 [math.DS], Apr 17 2012.
R. Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A002378, A004526, A076936, A139391, A016116, A126241, A060412, A060413, A006370, A070168 (iterations), A005408, A016957, A064455, A153285, A008619, A193356.
Bisections: A001477, A016789.
Cf. A093353.

a014682 n = if r > 0 then div (3 * n + 1) 2 else n'
            where (n', r) = divMod n 2
-- Reinhard Zumkeller, Oct 03 2014

[IsOdd(n) select (3*n+1)/2 else n/2: n in [0..52]]; // Vincenzo Librandi, Sep 28 2018

T:=proc(n) if n mod 2 = 0 then n/2 else (3*n+1)/2; fi; end; # N. J. A. Sloane, Jan 31 2011
A076936 := proc(n) option remember ; local apr,ifr,me,i,a ; if n <=2 then n^2 ; else apr := mul(A076936(i),i=1..n-1) ; ifr := ifactors(apr)[2] ; me := -1 ; for i from 1 to nops(ifr) do me := max(me, op(2,op(i,ifr))) ; od ; me := me+ n-(me mod n) ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; if a = A076936(n-1) then me := me+n ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; fi ; RETURN(a) ; fi ; end: A014682 := proc(n) log[2](A076936(n)) ; end: for n from 1 to 85 do printf("%d, ",A014682(n)) ; od ; # R. J. Mathar, Mar 20 2007

Collatz[n_?OddQ] := (3n + 1)/2; Collatz[n_?EvenQ] := n/2; Table[Collatz[n], {n, 0, 79}] (* Alonso del Arte, Apr 21 2011 *)
LinearRecurrence[{0, 2, 0, -1}, {0, 2, 1, 5}, 70] (* Jean-François Alcover, Sep 23 2017 *)
Table[If[OddQ[n], (3 n + 1) / 2, n / 2], {n, 0, 60}] (* Vincenzo Librandi, Sep 28 2018 *)

a(n)=if(n%2,3*n+1,n)/2 \\ Charles R Greathouse IV, Sep 02 2015

a(n)=if(n<2,2*n,(n^2-n-1)%(2*n+1)) \\ Jim Singh, Sep 28 2018

def a(n): return n//2 if n%2==0 else (3*n + 1)//2
print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 29 2017

From Paul Barry, Mar 31 2008: (Start)
G.f.: x*(2 + x + x^2)/(1-x^2)^2.
a(n) = (4*n+1)/4 - (2*n+1)*(-1)^n/4. (End)
a(n) = -a(n-1) + a(n-2) + a(n-3) + 4. - John W. Layman
For n > 1 this is the image of n under the modified "3x+1" map (cf. A006370): n -> n/2 if n is even, n -> (3*n+1)/2 if n is odd. - Benoit Cloitre, May 12 2002
O.g.f.: x*(2+x+x^2)/((-1+x)^2*(1+x)^2). - R. J. Mathar, Apr 05 2008
a(n) = 5/4 + (1/2)*((-1)^n)*n + (3/4)*(-1)^n + n. - Alexander R. Povolotsky, Apr 05 2008
a(n) = Sum_{i=-n..2*n} i*(-1)^i. - Bruno Berselli, Dec 14 2015
a(n) = Sum_{k=0..n-1} Sum_{i=0..k} C(i,k) + (-1)^k. - Wesley Ivan Hurt, Sep 20 2017
a(n) = (n^2-n-1) mod (2*n+1) for n > 1. - Jim Singh, Sep 26 2018
The above formula can be rewritten to show a pattern: a(n) = (n*(n+1)) mod (n+(n+1)). - Peter Munn, Jan 29 2022
Binary: a(n) = (n shift left (n AND 1)) - (n shift right 1) = A109043(n) - A004526(n). - Rudi B. Stranden, Jun 15 2021
From Rudi B. Stranden, Mar 21 2022: (Start)
a(n) = A064455(n+1) - 1, relating the number ON cells in row n of cellular automaton rule 54.
a(n) = 2*n - A071045(n).
(End)
E.g.f.: (1 + x)*sinh(x)/2 + 3*x*cosh(x)/2 = ((4*x+1)*e^x + (2*x-1)*e^(-x))/4. - Rénald Simonetto, Oct 20 2022
a(n) = n*(n mod 2) + ceiling(n/2) = A193356(n) + A008619(n+1). - Jonathan Shadrach Gilbert, Mar 12 2023
a(n) = 2*a(n-2) - a(n-4) for n > 3. - Chai Wah Wu, Apr 17 2024

Edited by N. J. A. Sloane, Apr 26 2008, at the suggestion of Artur Jasinski

Edited by N. J. A. Sloane, Jan 31 2011

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 14, 47, 156, 517, 1714, 5684, 18851, 62520, 207349, 687676, 2280686, 7563923, 25085844, 83197513, 275925586, 915110636, 3034975799, 10065534960, 33382471801, 110713382644, 367182309614, 1217764693607, 4038731742156, 13394504020957, 44423039068114

Offset: 0

Clark Kimberling, Aug 10 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).
Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).
Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
t(A000012) = t(1,1,1,1,1,1,1,...)    = A001906
t(A000290) = t(1,4,9,16,25,36,...)   = A289779
t(A000027) = t(1,2,3,4,5,6,7,8,...)  = A289780
t(A000045) = t(1,2,3,5,8,13,21,...)  = A289781
t(A000032) = t(2,1,3,4,7,11,14,...)  = A289782
t(A000244) = t(1,3,9,27,81,243,...)  = A289783
t(A000302) = t(1,4,16,64,256,...)    = A289784
t(A000351) = t(1,5,25,125,625,...)   = A289785
t(A005408) = t(1,3,5,7,9,11,13,...)  = A289786
t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
t(A016777) = t(1,4,7,10,13,16,...)   = A289789
t(A016789) = t(2,5,8,11,14,17,...)   = A289790
t(A008585) = t(3,6,9,12,15,18,...)   = A289795
t(A000217) = t(1,3,6,10,15,21,...)   = A289797
t(A000225) = t(1,3,7,15,31,63,...)   = A289798
t(A000578) = t(1,8,27,64,625,...)    = A289799
t(A000984) = t(1,2,6,20,70,252,...)  = A289800
t(A000292) = t(1,4,10,20,35,56,...)  = A289801
t(A002620) = t(1,2,4,6,9,12,16,...)  = A289802
t(A001906) = t(1,3,8,21,55,144,...)  = A289803
t(A001519) = t(1,1,2,5,13,34,...)    = A289804
t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
t(A008619) = t(1,1,2,2,3,3,4,4,...)  = A289806
t(A080513) = t(1,2,2,3,3,4,4,5,...)  = A289807
t(A133622) = t(1,2,1,3,1,4,1,5,...)  = A289809
t(A000108) = t(1,1,2,5,14,42,...)    = A081696
t(A081696) = t(1,1,3,9,29,97,...)    = A289810
t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
t(A175676) = t(1,0,0,2,0,0,3,0,...)  = A289844
t(A079977) = t(1,0,1,0,2,0,3,...)    = A289845
t(A059841) = t(1,0,1,0,1,0,1,...)    = A289846
t(A000040) = t(2,3,5,7,11,13,...)    = A289847
t(A008578) = t(1,2,3,5,7,11,13,...)  = A289828
t(A000142) = t(1!, 2!, 3!, 4!, ...)  = A289924
t(A000201) = t(1,3,4,6,8,9,11,...)   = A289925
t(A001950) = t(2,5,7,10,13,15,...)   = A289926
t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
t(A000045*) = t(0,1,1,2,3,5,...)     = A289975 (* indicates prepended 0's)
t(A000045*) = t(0,0,1,1,2,3,5,...)   = A289976
t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
t(A290990*) = t(0,1,2,3,4,5,...)     = A290990
t(A290990*) = t(0,0,1,2,3,4,5,...)   = A290991
t(A290990*) = t(0,0,01,2,3,4,5,...)  = A290992

			Example 1:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
- p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
t(s) = (1,3,8,21,...) = A001906.
***
Example 2:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
S(x) =  x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
- p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
t(s) = (1,4,14,47,...) = A289780.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -7, 5, -1)

Cf. A000027.

P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # Muniru A Asiru, Sep 03 2017

z = 60; s = x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289780 *)

x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ Altug Alkan, Aug 13 2017

G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).

a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).

A014681 Fix 0; exchange even and odd numbers.

Original entry on oeis.org

0, 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14, 13, 16, 15, 18, 17, 20, 19, 22, 21, 24, 23, 26, 25, 28, 27, 30, 29, 32, 31, 34, 33, 36, 35, 38, 37, 40, 39, 42, 41, 44, 43, 46, 45, 48, 47, 50, 49, 52, 51, 54, 53, 56, 55, 58, 57, 60, 59, 62, 61, 64, 63, 66, 65, 68, 67, 70

Offset: 0

Mohammad K. Azarian

A self-inverse permutation of the nonnegative numbers.
If we ignore the first term 0, then this can be obtained as: a(n) is the smallest number different from n, not occurring earlier and coprime to n. - Amarnath Murthy, Apr 16 2003 [Corrected by Alois P. Heinz, May 06 2015]
a(0)=0, a(1)=2, then repeatedly subtract 1 and then add 3. - Jon Perry, Aug 12 2014
The biggest term of the pair [a(n), a(n+1)] is always even. This is the lexicographically first sequence with this property starting with a(1) = 0 and always extented with the smallest integer not yet present. - Eric Angelini, Feb 20 2017

Derek Orr, Table of n, a(n) for n = 0..10000
Index entries for sequences that are permutations of the natural numbers
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Composing this permutation with A065190 gives A065164.
Equals 1 + A004442.
Cf. A103889.

Table[n - (-1)^n, {n, 1, 60}]
Join[{0},LinearRecurrence[{1, 1, -1},{2, 1, 4},69]] (* Ray Chandler, Sep 03 2015 *)

a(n)=n - (-1)^n \\ Charles R Greathouse IV, May 06 2015

G.f.: x*(2-x+x^2)/((1-x)*(1-x^2)). - N. J. A. Sloane
a(n) = n - (-1)^n = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 2. - Henry Bottomley, Mar 29 2000
a(0) = 0; a(2m+1) = 2m+2; for m > 0 a(2m) = 2m - 1. - George E. Antoniou, Dec 04 2001
a(n) = n - (-1)^n + 0^n for n >= 0. - Bruno Berselli, Nov 16 2010
E.g.f.: 1 + (x - 1)*cosh(x) + (1 + x)*sinh(x). - Stefano Spezia, Sep 02 2022

A339399 Pairwise listing of the partitions of k into two parts (s,t), with 0 < s <= t ordered by increasing values of s and where k = 2,3,... .

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 2, 1, 4, 2, 3, 1, 5, 2, 4, 3, 3, 1, 6, 2, 5, 3, 4, 1, 7, 2, 6, 3, 5, 4, 4, 1, 8, 2, 7, 3, 6, 4, 5, 1, 9, 2, 8, 3, 7, 4, 6, 5, 5, 1, 10, 2, 9, 3, 8, 4, 7, 5, 6, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6, 6, 1, 12, 2, 11, 3, 10, 4, 9, 5, 8, 6, 7, 1, 13, 2, 12, 3, 11

Offset: 1

Wesley Ivan Hurt, Dec 02 2020

a(n-1) and a(n) are the lesser and greater of a twin prime pair if and only if a(n) = a(n-1) + 2 where a(n-1) and a(n) are prime.

			                                                                     [1,9]
                                                     [1,7]   [1,8]   [2,8]
                                     [1,5]   [1,6]   [2,6]   [2,7]   [3,7]
                     [1,3]   [1,4]   [2,4]   [2,5]   [3,5]   [3,6]   [4,6]
     [1,1]   [1,2]   [2,2]   [2,3]   [3,3]   [3,4]   [4,4]   [4,5]   [5,5]
   k   2       3       4       5       6       7       8       9      10
  --------------------------------------------------------------------------
   k   Nondecreasing partitions of k
  --------------------------------------------------------------------------
   2   1,1
   3   1,2
   4   1,3,2,2
   5   1,4,2,3
   6   1,5,2,4,3,3
   7   1,6,2,5,3,4
   8   1,7,2,6,3,5,4,4
   9   1,8,2,7,3,6,4,5
  10   1,9,2,8,3,7,4,6,5,5
  ...

Index entries for sequences related to partitions

Cf. A103889, A339443.

Bisections: A122197 (odd), A199474 (even).

t[n_] := Flatten[Reverse /@ IntegerPartitions[n, {2}]]; Array[t, 14, 2] // Flatten (* Amiram Eldar, Dec 03 2020 *)
Table[(1 + (-1)^n) (1 + Floor[Sqrt[2 n - 1 - (-1)^n]])/2 - ((2 n + 1 - (-1)^n)/2 - 2 Sum[Floor[(k + 1)/2], {k, -1 + Floor[Sqrt[2 n - 2 - (-1)^n]]}]) (-1)^n/2, {n, 100}] (* Wesley Ivan Hurt, Dec 04 2020 *)

row(n) = vector(n\2, i, [i, n-i]);
tabf(nn) = for (n=2, nn, print(row(n))); \\ Michel Marcus, Dec 03 2020

a(n) = (1+(-1)^n)*(1+floor(sqrt(2*n-1-(-1)^n)))/2-((2*n+1-(-1)^n)/2-2 *Sum_{k=1..floor(sqrt(2*n-2-(-1)^n)-1)} floor((k+1)/2))*(-1)^n/2.

a(n) = A339443(A103889(n)). - Wesley Ivan Hurt, May 09 2021

A114285 Expansion of g.f. (1 - 3x)/((1 - x)(1 - x^2)).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Nov 20 2005

Diagonal sums of A114284.

Bérénice Delcroix-Oger and Clément Dupont, Lie-operads and operadic modules from poset cohomology, arXiv:2505.06094 [math.CO], 2025. See p. 28.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A103889, A114284.

[(-1)^n-n : n in [0..70]]; // Wesley Ivan Hurt, Sep 06 2015

I:=[1,-2,-1]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..70]]; // Vincenzo Librandi, Sep 07 2015

A114285:=n->(-1)^n-n: seq(A114285(n), n=0..70); # Wesley Ivan Hurt, Sep 06 2015

Table[(-1)^n-n, {n,0,70}] (* Wesley Ivan Hurt, Sep 06 2015 *)
CoefficientList[Series[(1 - 3 x)/((1 - x) (1 - x^2)), {x, 0, 70}] ,x] (* Vincenzo Librandi, Sep 07 2015 *)
LinearRecurrence[{1,1,-1},{1,-2,-1},70] (* Harvey P. Dale, Jul 24 2019 *)

a(n) = Sum_{k=0..floor(n/2)} 3*0^(n-2k) - 2.
a(n) = 3*(1 + (-1)^n)/2 - 2*floor((n+2)/2).
a(n) = - A103889(n). - R. J. Mathar, Apr 06 2008
From Wesley Ivan Hurt, Sep 06 2015: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3), n > 2.
a(n) = (-1)^n - n. (End)
E.g.f.: exp(-x) - x*exp(x). - Stefano Spezia, May 03 2023

A359855 Array read by antidiagonals: T(n,k) is the number of Hamiltonian cycles in the stacked prism graph P_n X C_k, n >= 1, k >= 2.

Original entry on oeis.org

1, 1, 4, 1, 3, 4, 1, 6, 6, 4, 1, 5, 22, 12, 4, 1, 8, 30, 82, 24, 4, 1, 7, 86, 160, 306, 48, 4, 1, 10, 126, 776, 850, 1142, 96, 4, 1, 9, 318, 1484, 7010, 4520, 4262, 192, 4, 1, 12, 510, 6114, 18452, 63674, 24040, 15906, 384, 4, 1, 11, 1182, 12348, 126426, 229698, 578090, 127860, 59362, 768, 4

Offset: 1

Andrew Howroyd, Feb 18 2025

The case for P_n X C_2 is determined using a double edge for C_2.

			Array begins:
=========================================================
n\k | 2   3     4      5       6        7          8 ...
----+---------------------------------------------------
  1 | 1   1     1      1       1        1          1 ...
  2 | 4   3     6      5       8        7         10 ...
  3 | 4   6    22     30      86      126        318 ...
  4 | 4  12    82    160     776     1484       6114 ...
  5 | 4  24   306    850    7010    18452     126426 ...
  6 | 4  48  1142   4520   63674   229698    2588218 ...
  7 | 4  96  4262  24040  578090  2861964   53055038 ...
  8 | 4 192 15906 127860 5247824 35663964 1087362018 ...
   ...

Eric Weisstein's World of Mathematics, Hamiltonian Cycle.
Eric Weisstein's World of Mathematics, Stacked Prism Graph.

Columns 2..12 are A123932(n-1), A003945(n-1), A003699, A003731, A180582, A180583, A180584, A180585, A180586, A180587, A180588.
Rows 1..2 are A000012, A103889(n+1).
Cf. A222196 (order of recurrences), A222197 (main diagonal), A270273, A321172.

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

Original entry on oeis.org

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

Offset: 1

Leroy Quet, Nov 13 2005

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

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

Vincenzo Librandi, Table of n, a(n) for n = 1..2000.
ProofWiki, Cesàro mean.
Wikipedia, Ernesto Cesàro.
Wikipédia, Lemme de Cesàro (in French).
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

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

a[1] = 1; a[n_] := a[n] = Block[{k = 1, s, t = Table[ a[i], {i, n - 1}]}, s = Plus @@ t; While[ Position[t, k] != {} || Mod[s, k] == 0, k++ ]; k]; Array[a, 72] (* Robert G. Wilson v, Nov 18 2005 *)

a(n) = if (n<=2, n, if (n%2, n+1, n-1)); \\ Michel Marcus, May 16 2022

def A114112(n): return n + (0 if n <= 2 else -1+2*(n%2)) # Chai Wah Wu, May 24 2022

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

More terms from Robert G. Wilson v, Nov 18 2005

Entry edited (with simpler definition) by N. J. A. Sloane, Mar 12 2018

Showing 1-10 of 32 results. Next

cp's OEIS Frontend

A114113 a(n) = sum{k=1 to n} (A114112(k)). (For n>=2, a(n) = sum{k=1 to n} (A014681(k)) =sum{k=1 to n} (A103889(k)).).

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A289805 p-INVERT of A103889, where p(S) = 1 - S - S^2.

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A203196 (n-1)-st elementary symmetric function of the first n terms of (2,1,4,3,6,5,8,7,...)=A103889.

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A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

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A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

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A014681 Fix 0; exchange even and odd numbers.

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A339399 Pairwise listing of the partitions of k into two parts (s,t), with 0 < s <= t ordered by increasing values of s and where k = 2,3,... .

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A114285 Expansion of g.f. (1 - 3x)/((1 - x)(1 - x^2)).