A093353 -id:A093353 - cp's OEIS frontend

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

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

A079326 a(n) = the largest number m such that if m monominoes are removed from an n X n square then an L-tromino must remain.

Original entry on oeis.org

1, 2, 7, 9, 17, 20, 31, 35, 49, 54, 71, 77, 97, 104, 127, 135, 161, 170, 199, 209, 241, 252, 287, 299, 337, 350, 391, 405, 449, 464, 511, 527, 577, 594, 647, 665, 721, 740, 799, 819, 881, 902, 967, 989, 1057, 1080, 1151, 1175, 1249, 1274, 1351, 1377, 1457

Offset: 2

Mambetov Timur (timur_teufel(AT)mail.ru), Feb 13 2003

			a(3)=2 because if a middle row of 3 monominoes are removed from the 3 X 3, no L remains.

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Frobenius number for k successive numbers: A028387 (k=2), this sequence (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Cf. A093353, A104519.

Table[FrobeniusNumber[{a, a + 1, a + 2}], {a, 2, 54}] (* Zak Seidov, Jan 08 2015 *)

a(n) = (n^2)/2 - 1 (n even), (n^2-n)/2 - 1 (n odd).
a(n) = A204557(n-1) / (n-1). - Reinhard Zumkeller, Jan 18 2012
From Bruno Berselli, Jan 18 2011: (Start)
G.f.: x^2*(1+x+3*x^2-x^4)/((1+x)^2*(1-x)^3).
a(n) = n*(2*n+(-1)^n-1)/4 - 1.
a(n) = A105638(-n+2). (End)

Edited by Don Reble, May 28 2007

A226141 Sum of the squared parts of the partitions of n into exactly two parts.

Original entry on oeis.org

0, 2, 5, 18, 30, 64, 91, 156, 204, 310, 385, 542, 650, 868, 1015, 1304, 1496, 1866, 2109, 2570, 2870, 3432, 3795, 4468, 4900, 5694, 6201, 7126, 7714, 8780, 9455, 10672, 11440, 12818, 13685, 15234, 16206, 17936, 19019, 20940, 22140, 24262, 25585, 27918, 29370, 31924, 33511

Offset: 1

Wesley Ivan Hurt, May 27 2013

			a(5) = 30; 5 has exactly 2 partitions into two parts, (4,1) and (3,2). Squaring the parts and adding, we get: 1^2 + 2^2 + 3^2 + 4^2 = 30.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), this sequence (k=2), A294270 (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).

Cf. A000330, A308422.

[n*(8*n^2 - 9*n + 4)/24 + (-1)^n*n^2/8 : n in [1..80]]; // Wesley Ivan Hurt, Jun 22 2024

a:=n->sum(i^2 + (n-i)^2, i=1..floor(n/2)); seq((a(k), k=1..40);

Array[Sum[i^2 + (# - i)^2, {i, Floor[#/2]}] &, 39] (* Michael De Vlieger, Jan 23 2018 *)
LinearRecurrence[{1,3,-3,-3,3,1,-1},{0,2,5,18,30,64,91},50] (* Harvey P. Dale, Jul 23 2019 *)

a(n) = Sum_{i=1..floor(n/2)} (i^2 + (n-i)^2).
a(n) = n*(8*n^2 - 9*n + 4)/24 + (-1)^n*n^2/8. - Giovanni Resta, May 29 2013
G.f.: x^2*(2+3*x+7*x^2+3*x^3+x^4) / ( (1+x)^3*(x-1)^4 ). - R. J. Mathar, Jun 07 2013
a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7). - Wesley Ivan Hurt, Jun 22 2024
a(n) = A000330(n) - A308422(n). - Wesley Ivan Hurt, Jul 16 2025

A235988 Sum of the partition parts of 3n into 3 parts.

Original entry on oeis.org

3, 18, 63, 144, 285, 486, 777, 1152, 1647, 2250, 3003, 3888, 4953, 6174, 7605, 9216, 11067, 13122, 15447, 18000, 20853, 23958, 27393, 31104, 35175, 39546, 44307, 49392, 54897, 60750, 67053, 73728, 80883, 88434, 96495, 104976, 113997, 123462, 133497, 144000

Offset: 1

Wesley Ivan Hurt, Jan 17 2014

			a(2) = 18; 3(2) = 6 has 3 partitions into 3 parts: (4, 1, 1), (3, 2, 1), and (2, 2, 2). The sum of the parts is 18.
Figure 1: The partitions of 3n into 3 parts for n = 1, 2, 3, ...
                                               13 + 1 + 1
                                               12 + 2 + 1
                                               11 + 3 + 1
                                               10 + 4 + 1
                                                9 + 5 + 1
                                                8 + 6 + 1
                                                7 + 7 + 1
                                   10 + 1 + 1  11 + 2 + 2
                                    9 + 2 + 1  10 + 3 + 2
                                    8 + 3 + 1   9 + 4 + 2
                                    7 + 4 + 1   8 + 5 + 2
                                    6 + 5 + 1   7 + 6 + 2
                        7 + 1 + 1   8 + 2 + 2   9 + 3 + 3
                        6 + 2 + 1   7 + 3 + 2   8 + 4 + 3
                        5 + 3 + 1   6 + 4 + 2   7 + 5 + 3
                        4 + 4 + 1   5 + 5 + 2   6 + 6 + 3
            4 + 1 + 1   5 + 2 + 2   6 + 3 + 3   7 + 4 + 4
            3 + 2 + 1   4 + 3 + 2   5 + 4 + 3   6 + 5 + 4
1 + 1 + 1   2 + 2 + 2   3 + 3 + 3   4 + 4 + 4   5 + 5 + 5
   3(1)        3(2)        3(3)        3(4)        3(5)     ..    3n
------------------------------------------------------------------------
    3           18          63         144         285      ..   a(n)
- _Wesley Ivan Hurt_, Sep 07 2019

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A001651, A077043, A093353.

[3*n^3-3*n*Floor(n^2/4): n in [1..100]]; // Wesley Ivan Hurt, Nov 01 2015

[3*n*(1-(-1)^n+6*n^2)/8: n in [1..40]]; // Vincenzo Librandi, Nov 18 2015

A235988:=n->3*n^3 - 3*n*floor(n^2/4); seq(A235988(n), n=1..100);

Table[3 n^3 - 3 n*Floor[n^2/4], {n, 100}] (* or *) CoefficientList[ Series[3*x*(x^4 + 4*x^3 + 8*x^2 + 4*x + 1)/((x - 1)^4*(x + 1)^2), {x, 0, 30}], x]
LinearRecurrence[{2,1,-4,1,2,-1},{3,18,63,144,285,486},40] (* Harvey P. Dale, May 17 2018 *)

a(n)=3*n^3 - n^2\4*3*n \\ Charles R Greathouse IV, Oct 07 2015

x='x+O('x^50); Vec(3*x*(x^4+4*x^3+8*x^2+4*x+1)/((x-1)^4*(x+1)^2)) \\ Altug Alkan, Nov 01 2015

a(n) = 3*n^3 - 3*n*floor(n^2/4).
a(n) = 3n * A077043(n).
a(n) = a(n-1) + 3*A077043(n-1) + A001651(n) + A093353(3n-2).
From Colin Barker, Jan 18 2014: (Start)
a(n) = (3*n*(1-(-1)^n+6*n^2))/8.
G.f.: 3*x*(x^4+4*x^3+8*x^2+4*x+1) / ((x-1)^4*(x+1)^2). (End)
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n > 6. - Wesley Ivan Hurt, Nov 15 2015
E.g.f.: 3*x*((4 + 9*x + 3*x^2)*cosh(x) + 3*(1 + 3*x + x^2)*sinh(x))/4. - Stefano Spezia, Feb 09 2023

a(165) in b-file corrected by Andrew Howroyd, Feb 21 2018

A294270 Sum of the cubes of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 9, 44, 100, 252, 441, 848, 1296, 2150, 3025, 4572, 6084, 8624, 11025, 14912, 18496, 24138, 29241, 37100, 44100, 54692, 64009, 77904, 90000, 107822, 123201, 145628, 164836, 192600, 216225, 250112, 278784, 319634, 354025, 402732, 443556, 501068, 549081

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

a(n) is a square when n is odd. In fact: a(2*k+1) = (2*k^2 + k)^2; a(2*k) = k^2*(4*k^2 - 3*k + 1), where (2*k)^2 > 4*k^2 - 3*k + 1 > (2*k - 1)^2 for k>0. - Bruno Berselli, Nov 20 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), this sequence (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).

[0] cat &cat[[k^2*(4*k^2-3*k+1),k^2*(2*k+1)^2]: k in [1..20]]; // Bruno Berselli, Nov 22 2017

Table[Sum[i^3 + (n - i)^3, {i, Floor[n/2]}], {n, 80}]

concat(0, Vec(x^2*(2 + 7*x + 27*x^2 + 28*x^3 + 24*x^4 + 7*x^5 + x^6) / ((1 - x)^5*(1 + x)^4) + O(x^40))) \\ Colin Barker, Nov 20 2017

a(n) = Sum_{i=1..floor(n/2)} i^3 + (n-i)^3.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 7*x + 27*x^2 + 28*x^3 + 24*x^4 + 7*x^5 + x^6) / ((1 - x)^5*(1 + x)^4).
a(n) = n^2*(4*n^2 - 7*n + 4 + n*(-1)^n)/16.
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) - 6*a(n-4) + 6*a(n-5) + 4*a(n-6) - 4*a(n-7) - a(n-8) + a(n-9) for n>9. (End)

A331702 Number of distinct intersections among all circles that can be constructed on vertices of an n-sided regular polygon, using only a compass.

Original entry on oeis.org

0, 2, 6, 40, 55, 145, 238, 584, 612, 1350, 1804, 2401, 3523, 5180, 6150, 9312, 11101, 13645, 17746, 22300, 25998, 33462, 39514, 43993, 55225, 66976, 74088, 88956, 102109, 111841, 133672, 155808, 170940, 198798, 220150, 243937, 275983, 313728, 338208, 382480, 419143, 448561, 507658

Offset: 1

Matej Veselovac, Jan 25 2020

nonn

Sequence counts intersections among all distinct circles such that: A circle is defined by a pair of distinct points of a regular n-sided polygon. First point is the center of the circle, while the distance between the points defines the radius of the circle.
It seems one additional intersection exists at the center of the polygon if and only if n is a multiple of 6. From this and n symmetries of the n-sided regular polygon, it would follow that n divides either a(n) or a(n)-1, depending on whether n is a multiple of 6.
A093353(n-1) gives the number of unique circles whose intersections a(n) counts.
From Scott R. Shannon, Dec 15 2022 (Start)
The values for n which lead to all vertices, other than those defining the n-sided regular polygon, being simple start 2, 3, 4, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, ... . These are all prime values except for the prime squares 4 and 25 which also appear. It is likely all primes appear although what other values lead to only simple vertices is unknown. (End)

			a(1)=0, we need at least two points to define a radius and a center.
a(2)=2, 2 circles constructed on segment endpoints intersect at 2 points.
a(3)=6, 3 circles on vertices of a triangle intersect at 6 distinct points.
a(4)=40, 8 circles can be constructed on vertices of a square and intersect at 40 distinct points.
a(5)=55, 10 circles can be constructed on vertices of a pentagon and intersect at 55 distinct points.

Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 3.
Scott R. Shannon, Image for n = 4.
Scott R. Shannon, Image for n = 5.
Scott R. Shannon, Image for n = 6.
Scott R. Shannon, Image for n = 7.
Scott R. Shannon, Image for n = 8.
Scott R. Shannon, Image for n = 9.
Scott R. Shannon, Image for n = 10.
Scott R. Shannon, Image for n = 11.
Scott R. Shannon, Image for n = 12.
Scott R. Shannon, Image for n = 18.
Scott R. Shannon, Image for n = 25.
N. J. A. Sloane, Illustration for A331702(4) = 40. Shows the planar graph. Annotated version of an illustration in the Math StackEchange link.
Math StackExchange, Intersections of circles drawn on vertices of regular polygons, 2020.

Cf. A093353, A359046 (regions), A359047 (edges), A359061 (k-gons), A358746.

n = Slider(2, 10, 1);
C = Unique(RemoveUndefined(Flatten(Sequence(Sequence(Circle(Point({cos((2v Pi) / n), sin((2v Pi) / n)}), 2sin((c Pi) / n)), c, 1, floor(n / 2)), v, 1, n))));
I = Unique(RemoveUndefined(Flatten(Sequence(Sequence(Intersect(Element(C, i), Element(C, j)), j, 1, Length(C)), i, 1, Length(C)))));
a_n = Length(I);

a(24)-a(30) from Giovanni Resta, Mar 27 2020

a(31)-a(43) from Scott R. Shannon, Dec 14 2022

A227906 Coins left after packing heart patterns (fixed orientation) into n X n coins.

Original entry on oeis.org

2, 4, 4, 9, 6, 13, 8, 17, 10, 21, 12, 25, 14, 29, 16, 33, 18, 37, 20, 41, 22, 45, 24, 49, 26, 53, 28, 57, 30, 61, 32, 65, 34, 69, 36, 73, 38, 77, 40, 81, 42, 85, 44, 89, 46, 93, 48, 97, 50, 101, 52, 105, 54, 109, 56, 113, 58, 117, 60, 121

Offset: 2

Kival Ngaokrajang, Oct 19 2013

On the Japanese TV show "Tsuki no Koibito", a girl told her boyfriend that she saw a heart in 4 coins. Actually there are a total of 6 distinct patterns appearing in 2 X 2 coins in which each pattern consists of a part of the perimeter of each coin and forms a continuous area.

a(n) is the number of coins left after packing fixed orientation heart patterns (type 4c2s1: 4-curve cover 2 coins and symmetry) into n X n coins. The total number of hearts is A093005 and the number of voids left is A093353. See illustration in links.

Kival Ngaokrajang, Illustration of initial terms (U)
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A008795, A230370 (3-curves), A074148, A229093, A229154 (4-curves), A001399, A230267, A230276 (5-curves), A229593, A228949, A229598 (6-curves).

With[{nn=60},Join[{2,4},Riffle[Range[4,nn,2],Range[9,2nn+1,4]]]] (* Harvey P. Dale, Feb 11 2015 *)

Vec(-x^2*(x^5-x^3-4*x-2)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Oct 30 2013

From Colin Barker, Oct 30 2013: (Start)
a(n) = (-1 + (-1)^n - (-3 + (-1)^n)*n)/2 for n>3.
a(n) = n for n>3 and even.
a(n) = 2*n-1 for n > 3 and odd.
a(n) = 2*a(n-2) - a(n-4) for n>7.
G.f.: -x^2*(x^5-x^3-4*x-2) / ((x-1)^2*(x+1)^2).(End)

A294271 Sum of the fourth powers of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 17, 114, 354, 1060, 2275, 4932, 8772, 15958, 25333, 41270, 60710, 91672, 127687, 182408, 243848, 333930, 432345, 572666, 722666, 931788, 1151403, 1451980, 1763020, 2182206, 2610621, 3180478, 3756718, 4514624, 5273999, 6263056, 7246096, 8515538, 9768353

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,5,-5,-10,10,10,-10,-5,5,1,-1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), this sequence (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).

[(n*(-16 + 160*n^2 + 15*(-15 + (-1)^n)*n^3 + 96*n^4))/480 : n in [1..50]]; // Wesley Ivan Hurt, Jul 12 2025

Table[Sum[i^4 + (n - i)^4, {i, Floor[n/2]}], {n, 60}]
Table[Total[Flatten[IntegerPartitions[n,{2}]]^4],{n,40}] (* Harvey P. Dale, Mar 01 2019 *)

concat(0, Vec(x^2*(2 + 15*x + 87*x^2 + 165*x^3 + 241*x^4 + 165*x^5 + 77*x^6 + 15*x^7 + x^8) / ((1 - x)^6*(1 + x)^5) + O(x^40))) \\ Colin Barker, Nov 20 2017

a(n) = sum(i=1, n\2, i^4 + (n-i)^4); \\ Michel Marcus, Nov 20 2017

a(n) = Sum_{i=1..floor(n/2)} i^4 + (n-i)^4.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 15*x + 87*x^2 + 165*x^3 + 241*x^4 + 165*x^5 + 77*x^6 + 15*x^7 + x^8) / ((1 - x)^6*(1 + x)^5).
a(n) = (1/480)*(n*(-16 + 160*n^2 + 15*(-15 + (-1)^n)*n^3 + 96*n^4)).
a(n) = a(n-1) + 5*a(n-2) - 5*a(n-3) - 10*a(n-4) + 10*a(n-5) + 10*a(n-6) - 10*a(n-7) - 5*a(n-8) + 5*a(n-9) + a(n-10) - a(n-11) for n>11.
(End)

A294272 Sum of the fifth powers of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 33, 308, 1300, 4668, 12201, 30032, 61776, 123950, 220825, 389652, 630708, 1018808, 1539825, 2331968, 3347776, 4826682, 6657201, 9233300, 12333300, 16578452, 21571033, 28256208, 35970000, 46106918, 57617001, 72503732, 89176276, 110446800, 133987425

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-15,15,20,-20,-15,15,6,-6,-1,1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), A294271 (k=4), this sequence (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).

[(n^2*(-16 + 80*n^2 + 3*(-31 + (-1)^n)*n^3 + 32*n^4))/192 : n in [1..50]]; // Wesley Ivan Hurt, Jul 12 2025

Table[Sum[i^5 + (n - i)^5, {i, Floor[n/2]}], {n, 50}]
Table[Total[Flatten[IntegerPartitions[n,{2}]]^5],{n,35}] (* or *) LinearRecurrence[{1,6,-6,-15,15,20,-20,-15,15,6,-6,-1,1},{0,2,33,308,1300,4668,12201,30032,61776,123950,220825,389652,630708},40] (* Harvey P. Dale, Jun 07 2025 *)

concat(0, Vec(x^2*(2 + 31*x + 263*x^2 + 806*x^3 + 1748*x^4 + 2046*x^5 + 1708*x^6 + 806*x^7 + 238*x^8 + 31*x^9 + x^10) / ((1 - x)^7*(1 + x)^6) + O(x^40))) \\ Colin Barker, Nov 20 2017

a(n) = Sum_{i=1..floor(n/2)} i^5 + (n-i)^5.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 31*x + 263*x^2 + 806*x^3 + 1748*x^4 + 2046*x^5 + 1708*x^6 + 806*x^7 + 238*x^8 + 31*x^9 + x^10) / ((1 - x)^7*(1 + x)^6).
a(n) = (1/192)*(n^2*(-16 + 80*n^2 + 3*(-31 + (-1)^n)*n^3 + 32*n^4)).
a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - 15*a(n-4) + 15*a(n-5) + 20*a(n-6) - 20*a(n-7) - 15*a(n-8) + 15*a(n-9) + 6*a(n-10) - 6*a(n-11) - a(n-12) + a(n-13) for n>13.
(End)

A294273 Sum of the sixth powers of the parts in the partitions of n into two parts.

Original entry on oeis.org

0, 2, 65, 858, 4890, 21244, 67171, 188916, 446964, 994030, 1978405, 3796622, 6735950, 11680408, 19092295, 30745064, 47260136, 71929146, 105409929, 153455810, 216455810, 303993492, 415601835, 566623708, 754740700, 1003708134, 1307797101, 1702747126

Offset: 1

Wesley Ivan Hurt, Oct 26 2017

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for sequences related to partitions
Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-21,21,35,-35,-35,35,21,-21,-7,7,1,-1).

Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), A294270 (k=3), A294271 (k=4), A294272 (k=5), this sequence (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).

[(n/42 - n^3/6 + n^5/2 + 1/128*(-63 + (-1)^n)*n^6 + n^7/7) : n in [1..50]]; // Wesley Ivan Hurt, Jul 12 2025

Table[Sum[i^6 + (n - i)^6, {i, Floor[n/2]}], {n, 50}]

concat(0, Vec(x^2*(2 + 63*x + 779*x^2 + 3591*x^3 + 10845*x^4 + 19026*x^5 + 23850*x^6 + 19026*x^7 + 10600*x^8 + 3591*x^9 + 723*x^10 + 63*x^11 + x^12) / ((1 - x)^8*(1 + x)^7) + O(x^40))) \\ Colin Barker, Nov 20 2017

a(n) = Sum_{i=1..floor(n/2)} i^6 + (n-i)^6.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 63*x + 779*x^2 + 3591*x^3 + 10845*x^4 + 19026*x^5 + 23850*x^6 + 19026*x^7 + 10600*x^8 + 3591*x^9 + 723*x^10 + 63*x^11 + x^12) / ((1 - x)^8*(1 + x)^7).
a(n) = (n/42 - n^3/6 + n^5/2 + 1/128*(-63 + (-1)^n)*n^6 + n^7/7).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - 21*a(n-4) + 21*a(n-5) + 35*a(n-6) - 35*a(n-7) - 35*a(n-8) + 35*a(n-9) + 21*a(n-10) - 21*a(n-11) - 7*a(n-12) + 7*a(n-13) + a(n-14) - a(n-15) for n>15.
(End)

cp's OEIS Frontend

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

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A079326 a(n) = the largest number m such that if m monominoes are removed from an n X n square then an L-tromino must remain.

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A226141 Sum of the squared parts of the partitions of n into exactly two parts.

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A235988 Sum of the partition parts of 3n into 3 parts.

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A294270 Sum of the cubes of the parts in the partitions of n into two parts.

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A331702 Number of distinct intersections among all circles that can be constructed on vertices of an n-sided regular polygon, using only a compass.

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