A016933 -id:A016933 - cp's OEIS frontend

A016936 a(n) = (6*n + 2)^4.

16, 4096, 38416, 160000, 456976, 1048576, 2085136, 3748096, 6250000, 9834496, 14776336, 21381376, 29986576, 40960000, 54700816, 71639296, 92236816, 116985856, 146410000, 181063936, 221533456, 268435456, 322417936, 384160000, 454371856, 533794816, 623201296

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A016780, A016933, A016934, A016935.

[(6*n+2)^4: n in [0..30]]; // Vincenzo Librandi, May 04 2011

(6*Range[0,30]+2)^4 (* or *) LinearRecurrence[{5,-10,10,-5,1},{16,4096,38416,160000,456976},30] (* Harvey P. Dale, Aug 22 2012 *)

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Aug 22 2012
From Amiram Eldar, Mar 29 2022: (Start)
a(n) = A016933(n)^4 = A016934(n)^2.
a(n) = 16*A016780(n).
Sum_{n>=0} 1/a(n) = PolyGamma(3, 1/3)/7776. (End)

A138750 a(n) = ceiling(n/2) if n == 2 (mod 3), a(n) = 2n otherwise.

Original entry on oeis.org

0, 2, 1, 6, 8, 3, 12, 14, 4, 18, 20, 6, 24, 26, 7, 30, 32, 9, 36, 38, 10, 42, 44, 12, 48, 50, 13, 54, 56, 15, 60, 62, 16, 66, 68, 18, 72, 74, 19, 78, 80, 21, 84, 86, 22, 90, 92, 24, 96, 98, 25, 102, 104, 27, 108, 110, 28, 114, 116, 30, 120, 122, 31, 126, 128, 33, 132, 134, 34

Offset: 0

M. F. Hasler, Mar 28 2008

This map is inspired by A124123, which hides in fact a variation of the Collatz problem, defined on the set of primes and working mod 3 instead of mod 2. See A138751 for more information.
The use of ceiling() is here equivalent to round().
The main reason for defining this function is to write A124123 as complement of A007918(A138750(A000040)), and to express the recursion function occurring there in terms of this map.
It might have been more natural to define this map as a(n) = 2n if n == 1 (mod 3), a(n) = ceiling(n/2) otherwise, which is equivalent for all primes > 3 (which are either == 1 or == 2 (mod 3)) and would have "better" properties regarding the analysis of orbits of all integers under this map.
However, for the prime n=3 it does make a difference, and in order to reproduce the map occurring in A124123, we had to adopt the present convention.

			a(0) = 2*0 = 0, a(1) = 2*1 = 2, a(3) = 2*3 = 6, a(4) = 2*4 = 8, ... since these indices are not congruent to 2 (mod 3).
a(2) = ceiling(2/2) = 1, a(5) = ceiling(5/2) = 3, a(8) = ceiling(8/2) = 4, a(11) = ceiling(11/2) = 6, ... since these indices are congruent to 2 (mod 3).

Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,0,1,0,0,-1).
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A001281, A124123, A138751, A138752, A138753, A008588 (trisection), A016933 (trisection), A032766 (trisection)

Table[If[Mod[n,3]==2,Ceiling[n/2],2n],{n,0,70}] (* or *) LinearRecurrence[{0,0,1,0,0,1,0,0,-1},{0,2,1,6,8,3,12,14,4},70] (* Harvey P. Dale, Nov 20 2013 *)

A138750(n) = if( n%3==2, ceil(n/2), 2*n )

G.f.: x*(2 + x + 6*x^2 + 6*x^3 + 2*x^4 + 6*x^5 + 4*x^6) / ( (1+x)*(x^2-x+1)*(x-1)^2*(1+x+x^2)^2 ). - R. J. Mathar, Oct 16 2013
a(n) = a(n-3) + a(n-6) - a(n-9); a(0)=0, a(1)=2, a(2)=1, a(3)=6, a(4)=8, a(5)=3, a(6)=12, a(7)=14, a(8)=4. - Harvey P. Dale, Nov 20 2013
Sum_{n>=1} (-1)^n/a(n) = log(3)/2 - log(2)/3 = log(27/4)/6. - Amiram Eldar, Jul 26 2024

A157176 a(n+1) = a(n - n mod 2) + a(n - n mod 3), a(0) = 1.

Original entry on oeis.org

1, 2, 2, 3, 5, 8, 8, 16, 16, 24, 40, 64, 64, 128, 128, 192, 320, 512, 512, 1024, 1024, 1536, 2560, 4096, 4096, 8192, 8192, 12288, 20480, 32768, 32768, 65536, 65536, 98304, 163840, 262144, 262144, 524288, 524288, 786432, 1310720, 2097152, 2097152, 4194304, 4194304

Offset: 0

Reinhard Zumkeller, Feb 24 2009

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,8).

Cf. A001018, A008588, A013730, A016921, A016933, A016945, A016957, A016969, A067412, A103333.

LinearRecurrence[{0,0,0,0,0,8},{1, 2, 2, 3, 5, 8},45] (* Stefano Spezia, May 29 2024 *)

a(n+6) = 8*a(n).
a(6*k) = 8^k; a(A008588(n))=A001018(n);
a(6*k+1) = a(6*k+2) = 2*8^k; a(A016921(n))=a(A016933(n))=A013730(n);
a(6*k+3) = 3*8^k; a(A016945(n))=A103333(n+1);
a(6*k+4) = 5*8^k; a(A016957(n))=A067412(n+1);
a(6*k+5) = 8^(k+1); a(A016969(n))=A001018(n+1).
G.f.: (1 + 2*x + 2*x^2 + 3*x^3 + 5*x^4 + 8*x^5)/((1 - 2*x^2)*(1 + 2*x^2 + 4*x^4)). - Stefano Spezia, May 29 2024

a(43)-a(44) from Stefano Spezia, May 29 2024

A016937 a(n) = (6*n + 2)^5.

Original entry on oeis.org

32, 32768, 537824, 3200000, 11881376, 33554432, 79235168, 164916224, 312500000, 550731776, 916132832, 1453933568, 2219006624, 3276800000, 4704270176, 6590815232, 9039207968, 12166529024, 16105100000, 21003416576, 27027081632, 34359738368, 43204003424, 53782400000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A016781, A016933, A016934, A016935, A016936.

[(6*n+2)^5: n in [0..30]]; // Vincenzo Librandi, May 04 2011

(6*Range[0,20]+2)^5 (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{32,32768,537824,3200000,11881376,33554432},20] (* Harvey P. Dale, Dec 13 2012 *)

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Harvey P. Dale, Dec 13 2012
From Amiram Eldar, Mar 29 2022: (Start)
a(n) = A016933(n)^5.
a(n) = 32*A016781(n).
Sum_{n>=0} 1/a(n) = Pi^5/(11664*sqrt(3)) + 121*zeta(5)/7776. (End)

A016938 a(n) = (6*n + 2)^6.

Original entry on oeis.org

64, 262144, 7529536, 64000000, 308915776, 1073741824, 3010936384, 7256313856, 15625000000, 30840979456, 56800235584, 98867482624, 164206490176, 262144000000, 404567235136, 606355001344, 885842380864, 1265319018496, 1771561000000, 2436396322816, 3297303959104

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A016782, A016933, A016934, A016935, A016936, A016937.

[(6*n+2)^6: n in [0..30]]; // Vincenzo Librandi, May 04 2011

(6Range[0,20]+2)^6  (* Harvey P. Dale, Apr 01 2011 *)

From Amiram Eldar, Mar 29 2022: (Start)
a(n) = A016933(n)^6 = A016934(n)^3 = A016935(n)^2.
a(n) = 64*A016782(n).
Sum_{n>=0} 1/a(n) = PolyGamma(5, 1/3)/5598720. (End)

A134494 a(n) = Fibonacci(6n+2).

Original entry on oeis.org

1, 21, 377, 6765, 121393, 2178309, 39088169, 701408733, 12586269025, 225851433717, 4052739537881, 72723460248141, 1304969544928657, 23416728348467685, 420196140727489673, 7540113804746346429, 135301852344706746049, 2427893228399975082453

Offset: 0

Artur Jasinski, Oct 28 2007

Colin Barker, Table of n, a(n) for n = 0..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A001519, A015448, A014445, A033887-A033891, A049310, A049660, A099100, A102312, A103134, A134490 - A134504.

[Fibonacci(6*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

seq( combinat[fibonacci](6*n+2),n=0..10) ; # R. J. Mathar, Apr 17 2011

Table[Fibonacci[6n+2], {n, 0, 30}]
Table[ChebyshevU[3*n, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+2) \\ Charles R Greathouse IV, Jun 11 2015

Vec((1+3*x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: ( 1+3*x ) / ( 1-18*x+x^2 ).
a(n) = 3*A049660(n)+A049660(n+1). (End)
a(n) = A000045(A016933(n)). - Michel Marcus, Nov 07 2013
a(n) = ((5-3*sqrt(5)+(5+3*sqrt(5))*(9+4*sqrt(5))^(2*n)))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n, 3) = S(n,18) + 3*S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Index in definition corrected by T. D. Noe, Joerg Arndt, Apr 17 2011

A365442 Partial sums of A365412.

Original entry on oeis.org

3, 18, 42, 84, 126, 189, 249, 333, 426, 546, 642, 768, 882, 1068, 1200, 1368, 1539, 1749, 1965, 2175, 2361, 2616, 2820, 3156, 3378, 3678, 3918, 4212, 4536, 4908, 5244, 5580, 5874, 6339, 6651, 7029, 7359, 7863, 8295, 8715, 9114, 9594, 9978, 10566, 11046, 11604, 12024, 12528

Offset: 0

Omar E. Pol, Sep 07 2023

Partial sums of the sum of the divisors of the numbers of the form 6*k + 2, k >= 0.
Consider a spiral similar to the spiral described in A239660 but instead of having four quadrants on the square grid the new spiral has six wedges on the triangular grid. A "diamond" formed by two adjacent triangles has area 1. a(n) is the total number of diamonds (or the total area) in the second wedge after n + 1 turns. The interesting fact is that for n >> 1 the geometric pattern in the second wedge of the spiral is very similar to the geometric pattern of the fourth wedge but it is different from the other wedges. Note that the six wedge spiral shows more and better geometric patterns than the four quadrants spiral.
The graph is very close to the graph of A365444 (see the Links section).

Michael De Vlieger, Table of n, a(n) for n = 0..10000
OEIS Plot 2, Plot pairs of A365442 and A365444
Omar E. Pol, Plot 6. Area of the spiral in the six wedges

Sequences of the same family are A363161, this sequence, A383403, A365444, A383405, A365446.

Cf. A000203, A016933, A237593, A239660, A365412.

Accumulate[Table[DivisorSigma[1, 6*n + 2], {n, 0, 50}]] (* Amiram Eldar, Sep 08 2023 *)

a(n) = sum(k=0, n, sigma(6*k+2)); \\ Michel Marcus, Sep 09 2023

a(n) = (5*Pi^2/9) * n^2 + O(n*log(n)). - Amiram Eldar, Sep 08 2023

A016939 a(n) = (6n+2)^7.

Original entry on oeis.org

128, 2097152, 105413504, 1280000000, 8031810176, 34359738368, 114415582592, 319277809664, 781250000000, 1727094849536, 3521614606208, 6722988818432, 12151280273024, 20971520000000, 34792782221696, 55784660123648, 86812553324672, 131593177923584, 194871710000000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A016783, A016933, A016934, A016935, A016936, A016937, A016938.

[(6*n+2)^7: n in [0..30]]; // Vincenzo Librandi, May 04 2011

Table[(6n+2)^7,{n,0,100}] (* Mohammad K. Azarian, Jun 15 2016 *)

a(n) = 128*A016783(n). - R. J. Mathar, May 07 2008
G.f.: 128*(1 + 16376*x + 692499*x^2 + 3870352*x^3 + 4890287*x^4 + 1475736*x^5 + 77101*x^6 + 128*x^7)/(1 - x)^8. - Ilya Gutkovskiy, Jun 16 2016
From Amiram Eldar, Mar 29 2022: (Start)
a(n) = A016933(n)^7.
Sum_{n>=0} 1/a(n) = 7*Pi^7/(3149280*sqrt(3)) + 1093*zeta(7)/279936. (End)

A112414 a(n) = 3*n + 7.

Original entry on oeis.org

7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 130, 133, 136, 139, 142, 145, 148, 151, 154, 157, 160, 163, 166, 169, 172, 175, 178, 181, 184

Offset: 0

Zerinvary Lajos, Dec 09 2005

At least the first 2 million terms from a(1) on coincide with the corresponding terms of A086822(n+2). - R. J. Mathar, Aug 15 2008

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

Cf. A016933, A059845.

3Range[0,70]+7 (* or *) LinearRecurrence[{2,-1},{7,10},70] (* Harvey P. Dale, May 12 2019 *)

a(n)=3*n+7 \\ Charles R Greathouse IV, Jul 10 2016

From Elmo R. Oliveira, Oct 30 2024: (Start)
G.f.: (7 - 4*x)/(1 - x)^2.
E.g.f.: (7 + 3*x)*exp(x).
a(n) = 2*a(n-1) - a(n-2) for n >= 2.
a(n) = A016933(n+2)/2 = A059845(n+1) - A059845(n). (End)

Better definition from T. D. Noe, Nov 30 2006

A334715 A(n,k) = !n + [n > 0] * (k * n!), where !n = A000166(n) is subfactorial of n and [] is an Iverson bracket; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 3, 5, 8, 9, 1, 4, 7, 14, 33, 44, 1, 5, 9, 20, 57, 164, 265, 1, 6, 11, 26, 81, 284, 985, 1854, 1, 7, 13, 32, 105, 404, 1705, 6894, 14833, 1, 8, 15, 38, 129, 524, 2425, 11934, 55153, 133496, 1, 9, 17, 44, 153, 644, 3145, 16974, 95473, 496376, 1334961

Offset: 0

Alois P. Heinz, May 08 2020

			Square array A(n,k) begins:
     1,    1,     1,     1,     1,     1,     1,     1, ...
     0,    1,     2,     3,     4,     5,     6,     7, ...
     1,    3,     5,     7,     9,    11,    13,    15, ...
     2,    8,    14,    20,    26,    32,    38,    44, ...
     9,   33,    57,    81,   105,   129,   153,   177, ...
    44,  164,   284,   404,   524,   644,   764,   884, ...
   265,  985,  1705,  2425,  3145,  3865,  4585,  5305, ...
  1854, 6894, 11934, 16974, 22014, 27054, 32094, 37134, ...
  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Wikipedia, Derangement
Wikipedia, Iverson bracket

Columns k=0-3 give: A000166, A001120, A110043, A110149.
Rows n=0-3 give: A000012, A001477, A005408, A016933.
Main diagonal gives A334716.
Cf. A000142.

A:= proc(n, k) option remember; `if`(n<2,
      (k-1)*n+1, n*A(n-1, k)+(-1)^n)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..10);

A[n_, k_] := Subfactorial[n] + Boole[n>0] k n!;
Table[A[n, d-n], {d, 0, 10}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 11 2021 *)

E.g.f. of column k: (k*exp(x)*x+1)*exp(-x)/(1-x).
A(n,k) = A000166(n) + [n > 0] * (k * n!).
A(n,k) = (k-1)*n + 1 if n<2, A(n,k) = n*A(n-1, k) + (-1)^n if n>=2.

cp's OEIS Frontend

A016936 a(n) = (6*n + 2)^4.

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A138750 a(n) = ceiling(n/2) if n == 2 (mod 3), a(n) = 2n otherwise.

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A157176 a(n+1) = a(n - n mod 2) + a(n - n mod 3), a(0) = 1.

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A016937 a(n) = (6*n + 2)^5.

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A016938 a(n) = (6*n + 2)^6.

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A134494 a(n) = Fibonacci(6n+2).

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A365442 Partial sums of A365412.

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A016939 a(n) = (6n+2)^7.

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