A060747 -id:A060747 - cp's OEIS frontend

A067998 a(n) = n^2 - 2*n.

0, -1, 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600, 2703, 2808, 2915, 3024, 3135, 3248, 3363

Offset: 0

George E. Antoniou, Feb 06 2002

a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n - n^2 and a(n) = -P_0(n). - Peter Luschny, Jul 08 2011
For n >= 3, the denominator of the probability of winning the prize by switching from the initial choice of doors in a generalized Monty Hall problem with n doors: After a prize has been placed behind exactly one of the n doors at random, a contestant chooses a door. Then the host, who knows where the prize is, deliberately opens exactly one unchosen door that does not hide the prize (picked at random by the host among such doors when there is a choice) and then gives the contestant an opportunity to switch to any other door not yet opened. The numerator of this probability is n-1 (incidentally, gcd(n-1, n*(n-2)) = 1). The probability of winning by switching minus the probability of winning by not switching is thus (n-1)/(n*(n-2)) - 1/n = 1/a(n), which approaches zero as n approaches infinity, but nevertheless makes the switching strategy better for every finite n >= 3. The winning probability is 2/3 from switching in the classic 3-door Monty Hall problem; we have 3/8 and 4/15, respectively, in the 4- and 5-door generalizations. (The above analysis was independent but is consistent with the even more general "N-doors" section of the Wikipedia article, other parts of which make clear the historical importance of wording this problem as carefully as possible. See also A122774.) - Rick L. Shepherd, May 31 2014, clarified Oct 29 2015
For n > 1, a(n) is the largest integer k such that k + n^2 is a multiple of k + n. - Derek Orr, Sep 04 2014

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Wikipedia, Monty Hall problem.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Essentially the same as A005563.
Cf. A060747 (first differences).
Cf. A000290.

a067998 n = n * (n - 2)
a067998_list = scanl (+) 0 [-1, 1 ..]
-- Reinhard Zumkeller, Aug 26 2013

[n^2-2*n : n in [0..50]]; // Wesley Ivan Hurt, Sep 04 2014

A067998:=n->n^2-2*n: seq(A067998(n), n=0..50); # Wesley Ivan Hurt, Sep 04 2014

Table[ n^2 - 2*n, {n, 0, 60} ] (* George E. Antoniou *)
LinearRecurrence[{3, -3, 1}, {0, -1, 0}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 23 2012 *)

PARI
```
a(n)=n^2-2*n;
```

concat(0, Vec(x*(3*x-1)/(1-x)^3 + O(x^100))) \\ Altug Alkan, Oct 30 2015

a(n) = A005563(n-2) = A005563(-n) = A000290(n-1)-1.
G.f.: x*(3*x-1)/(1-x)^3. - Paul Barry, Mar 27 2007
E.g.f.: exp(x)*(x^2-x). - Paul Barry, Mar 27 2007
a(n) = 2*n + a(n-1) - 3 (with a(0)=0). - Vincenzo Librandi, Aug 08 2010
From Amiram Eldar, Feb 17 2023: (Start)
Sum_{n>=3} 1/a(n) = 3/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = 1/4. (End)

Edited and extended by Robert G. Wilson v, Feb 08 2002

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55

Offset: 1

Roger L. Bagula, Aug 05 2008

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x)  (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
Carlton Gamer, David W. Roeder, and John J. Watkins, Trapezoidal Numbers, Mathematics Magazine 58:2 (1985), pp. 108-110.
L. E. Jeffery, Unit-primitive matrices
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

a141419 n k =  k * (2 * n - k + 1) `div` 2
a141419_row n = a141419_tabl !! (n-1)
a141419_tabl = map (scanl1 (+)) a004736_tabl
-- Reinhard Zumkeller, Aug 04 2014

a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018

T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10.  T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Simpler name by Stefano Spezia, Oct 14 2018

A140811 a(n) = 6*n^2 - 1.

Original entry on oeis.org

-1, 5, 23, 53, 95, 149, 215, 293, 383, 485, 599, 725, 863, 1013, 1175, 1349, 1535, 1733, 1943, 2165, 2399, 2645, 2903, 3173, 3455, 3749, 4055, 4373, 4703, 5045, 5399, 5765, 6143, 6533, 6935, 7349, 7775, 8213, 8663, 9125, 9599, 10085, 10583, 11093, 11615

Offset: 0

Paul Curtz, Jul 16 2008

Also: The numerators in the j=2 column of the array a(i,j) defined in A140825, where the columns j=0 and j=1 are represented by A000012 and A005408. This could be extended to column j=3: 1, -1, 9, 55, 161, ... The common feature of these sequences derived from a(i,j) is that their j-th differences are constant sequences defined by A091137(j).
a(n) is the set of all k such that 6*k + 6 is a perfect square. - Gary Detlefs, Mar 04 2010
The identity (6*n^2 - 1)^2 - (9*n^2 - 3)*(2*n)^2 = 1 can be written as a(n+1)^2 - A157872(n)*A005843(n+1)^2 = 1. - Vincenzo Librandi, Feb 05 2012
Apart from first term, sequence found by reading the line from 5, in the direction 5, 23, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Jul 18 2012
From Paul Curtz, Sep 17 2018: (Start)
Terms from center to right in the following spiral:
.
                    65--63--61--59
                    /             \
                  67  31--29--27  57
                  /   /         \   \
                69  33   9---7  25  55
                /   /   /     \   \   \
              71  35  11  -1===5==23==53==>
                  /   /   /   /   /   /
                37  13   1---3  21  51
                  \   \         /   /
                  39  15--17--19  49
                    \             /
                    41--43--45--47 (End)

P. Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Note 12, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969, 132 pages, pp. 28-36. CCSA, then CELAR. Now DGA Maitrise de l'Information 35131 Bruz.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Barred Stars.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A157872, A060747, A103115(n+1), A141417 (array).
Cf. A000012, A001318, A005408, A091137, A140825.
Cf. A003154, A032528, A033581, A016777, A017593.

[6*n^2 - 1: n in [0..50]]; // Vincenzo Librandi, Jun 02 2011

LinearRecurrence[{3,-3,1},{-1,5,23},40] (* Vincenzo Librandi, Feb 05 2012 *)
CoefficientList[Series[(1-8*x-5*x^2)/(x-1)^3 , {x, 0, 40}], x] (* Stefano Spezia, Sep 17 2018 *)

a(n)=6*n^2-1 \\ Charles R Greathouse IV, Jun 01 2011

a(n) = 2*a(n-1) - a(n-2) + 12.
First differences: a(n+1) - a(n) = A017593(n).
Second differences: A071593(n+1) - A071593(n) = 12.
G.f.: (1-8*x-5*x^2)/(x-1)^3. - Jaume Oliver Lafont, Aug 30 2009
From Vincenzo Librandi, Feb 05 2012: (Start)
a(n) = a(n-1) + 12*n - 6.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
a(n) = A033581(n) - 1. - Omar E. Pol, Jul 18 2012
a(n) = A032528(2*n) - 1. - Adriano Caroli, Jul 21 2013
For n > 0, a(n) = floor(3/(cosh(1/n) - 1)) = floor(1/(n*sinh(1/n) - 1)); for similar formulas for cosine and sine, see A033581. - Clark Kimberling, Oct 19 2014, corrected by M. F. Hasler, Oct 21 2014
a(-n) = a(n). - Paul Curtz, Sep 17 2018
From Amiram Eldar, Feb 04 2021: (Start)
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(6))*cot(Pi/sqrt(6)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(6))*csc(Pi/sqrt(6)) - 1)/2.
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(6))*csc(Pi/sqrt(6)).
Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(6))*sin(Pi/sqrt(3))/sqrt(2). (End)
a(n) = A003154(n+1) - 2*A016777(n). - Leo Tavares, May 13 2022
E.g.f.: exp(x)*(6*x^2 + 6*x - 1). - Elmo R. Oliveira, Jan 16 2025

Edited and extended by R. J. Mathar, Aug 06 2008

Better description Ray Chandler, Feb 03 2009

A141530 a(n) = 4n^3 - 6n^2 + 1.

Original entry on oeis.org

1, -1, 9, 55, 161, 351, 649, 1079, 1665, 2431, 3401, 4599, 6049, 7775, 9801, 12151, 14849, 17919, 21385, 25271, 29601, 34399, 39689, 45495, 51841, 58751, 66249, 74359, 83105, 92511, 102601, 113399, 124929, 137215, 150281, 164151, 178849, 194399, 210825, 228151

Offset: 0

Paul Curtz, Aug 12 2008

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A046092, A141047, A141417.

See Librandi's comment in A078371.

[4*n^3 -6*n^2 +1: n in [0..50]]; // G. C. Greubel, Mar 29 2021

A141530:= n-> 4*n^3 -6*n^2 +1; seq(A141530(n), n=0..50); # G. C. Greubel, Mar 29 2021

Array[4*#^3-6*#^2+1&,50,0] (* Vladimir Joseph Stephan Orlovsky, Nov 03 2009 *)
LinearRecurrence[{4,-6,4,-1},{1,-1,9,55},50] (* Harvey P. Dale, Nov 30 2011 *)

a(n)=4*n^3-6*n^2+1 \\ Charles R Greathouse IV, Oct 07 2015

def A141530(n): return (m:=(n<<1)-1)*(n*(m-1)-1) # Chai Wah Wu, Mar 11 2024

[4*n^3 -6*n^2 +1 for n in (0..50)] # G. C. Greubel, Mar 29 2021

a(n) = (2*n-1)*(2*n^2 - 2*n - 1) = A060747(n)*A132209(n-1), n > 1. - R. J. Mathar, Feb 22 2009
G.f.: (1 - 5*x + 19*x^2 + 9*x^3)/(1-x)^4. - Jaume Oliver Lafont, Aug 30 2009
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) with a(0)=1, a(1)=-1, a(2)=9, a(3)=55. - Harvey P. Dale, Nov 30 2011
E.g.f.: (1 - 2*x + 6*x^2 + 4*x^3)*exp(x). - G. C. Greubel, Mar 29 2021

Corrected, completed and edited, following an observation from Vincenzo Librandi, by M. F. Hasler, Feb 12 2009

Further edited by N. J. A. Sloane, Feb 13 2009

A048495 a(n) = (n-1)*2^n + 2.

Original entry on oeis.org

1, 2, 6, 18, 50, 130, 322, 770, 1794, 4098, 9218, 20482, 45058, 98306, 212994, 458754, 983042, 2097154, 4456450, 9437186, 19922946, 41943042, 88080386, 184549378, 385875970, 805306370, 1677721602, 3489660930, 7247757314

Offset: 0

Clark Kimberling

Binomial transform of 1 followed by the odd numbers (2n-1+2*0^n, or abs(A060747)). Binomial transform is A084643. - Paul Barry, Jun 09 2003
Total number of bits of all binary numbers less than 2^n (see example).
Total number of zero bits of all binary numbers less than 2^(n+1). - Olivier Gérard, Feb 25 2014.
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>2, 4>3} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element, and the fourth element is larger than the third element. - Sergey Kitaev, Dec 08 2020

			a(1)=2 : 0 1
a(2)=6 : 0 1 10 11
a(3)=18 : 0 1 10 11 100 101 110 111
a(4)=50 : 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111
...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alice L. L. Gao, Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao, Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
Index entries for linear recurrences with constant coefficients, signature (5,-8,4).

a(n) = T(1, n), array T given by A048494.

Cf. A002064, A060747, A084643.

[(n-1)*2^n + 2: n in [0..30]]; // Vincenzo Librandi, Sep 25 2011

A048495:=n->(n-1)*2^n + 2; seq(A048495(n), n=0..30); # Wesley Ivan Hurt, Jun 29 2014

f[n_]:=(n-1)2^n+2;Array[f,29,0] (* Robert G. Wilson v, Jun 29 2014 *)
LinearRecurrence[{5,-8,4},{1,2,6},30] (* Harvey P. Dale, Jan 23 2015 *)

a(n) = (n-1)*2^n + 2; \\ Joerg Arndt, Feb 25 2014

Vec(-(4*x^2-3*x+1)/((x-1)*(2*x-1)^2) + O(x^100)) \\ Colin Barker, Jun 29 2014

a(n) - 1 = Sum_{i=0..n-1} (n-i) * 2^(n-i-1) = n*2^(n-1) + (n-1)*2^(n-2) + (n-2)*2^(n-3) + ... + 1*(2^0). - Matthew Erbst (matt(AT)erbst.org), Apr 19 2006
a(n) = 2 * A002064(n-1), n >= 1. - Omar E. Pol, Sep 30 2012
a(n) = a(n-1) + (2^n - 2^(n-1)) * n = a(n-1) + n*2^(n-1). - Olivier Gérard, Feb 25 2014
G.f.: -(4*x^2-3*x+1) / ((x-1)*(2*x-1)^2). - Colin Barker, Jun 29 2014
E.g.f.: exp(x)*(2 + exp(x)*(2*x - 1)). - Stefano Spezia, Feb 14 2025

Better description from John W. Layman, May 04 1999

A141310 The odd numbers interlaced with the constant-2 sequence.

Original entry on oeis.org

1, 2, 3, 2, 5, 2, 7, 2, 9, 2, 11, 2, 13, 2, 15, 2, 17, 2, 19, 2, 21, 2, 23, 2, 25, 2, 27, 2, 29, 2, 31, 2, 33, 2, 35, 2, 37, 2, 39, 2, 41, 2, 43, 2, 45, 2, 47, 2, 49, 2, 51, 2, 53, 2, 55, 2, 57, 2, 59, 2, 61, 2, 63, 2, 65, 2, 67, 2, 69, 2, 71, 2, 73, 2, 75, 2, 77, 2, 79, 2, 81, 2, 83, 2, 85, 2, 87, 2, 89, 2, 91, 2, 93, 2, 95, 2, 97

Offset: 0

Paul Curtz, Aug 02 2008

Similarly, the principle of interlacing a sequence and its first differences leads from A000012 and its differences A000004 to A059841, or from A140811 and its first differences A017593 to a sequence -1, 6, 5, 18, ...
If n is even then a(n) = n + 1 ; otherwise a(n) = 2. - Wesley Ivan Hurt, Jun 05 2013
Denominators of floor((n+1)/2) / (n+1), n > 0. - Wesley Ivan Hurt, Jun 14 2013
a(n) is also the number of minimum total dominating sets in the (n+1)-gear graph for n>1. - Eric W. Weisstein, Apr 11 2018
a(n) is also the number of minimum total dominating sets in the (n+1)-sun graph for n>1. - Eric W. Weisstein, Sep 09 2021
Denominators of Cesàro means sequence of A114112, corresponding numerators are in A354008. -  Bernard Schott, May 14 2022
Also, denominators of Cesàro means sequence of A237420, corresponding numerators are in A354280. -  Bernard Schott, May 22 2022

Antti Karttunen, Table of n, a(n) for n = 0..16384
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Minimum Total Dominating Set.
Eric Weisstein's World of Mathematics, Sun Graph.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A000004, A000012, A000142, A005408, A008586, A017593, A059841, A060747, A109613, A140811, A211374, A319702 (rgs-transform).
Cf. A114112, A354008.
Cf. A237420, A354280.

a:= n-> n+1-(n-1)*(n mod 2): seq(a(n), n=0..96); # Wesley Ivan Hurt, Jun 05 2013

Flatten[Table[{2 n - 1, 2}, {n, 40}]] (* Alonso del Arte, Jun 15 2013 *)
Riffle[Range[1, 79, 2], 2] (* Alonso del Arte, Jun 14 2013 *)
Table[((-1)^n (n - 1) + n + 3)/2, {n, 0, 20}] (* Eric W. Weisstein, Apr 11 2018 *)
Table[Floor[(n + 1)/2]/(n + 1), {n, 0, 20}] // Denominator (* Eric W. Weisstein, Apr 11 2018 *)
LinearRecurrence[{0, 2, 0, -1}, {2, 3, 2, 5}, {0, 20}] (* Eric W. Weisstein, Apr 11 2018 *)
CoefficientList[Series[(1 + 2 x + x^2 - 2 x^3)/(-1 + x^2)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Apr 11 2018 *)

A141310(n) = if(n%2,2,1+n); \\ (for offset=0 version) - Antti Karttunen, Oct 02 2018

A141310off1(n) = if(n%2,n,2); \\ (for offset=1 version) - Antti Karttunen, Oct 02 2018

def A141310(n): return 2 if n % 2 else n + 1 # Chai Wah Wu, May 24 2022

a(2n) = A005408(n). a(2n+1) = 2.
First differences: a(n+1) - a(n) = (-1)^(n+1)*A109613(n-1), n > 0.
b(2n) = -A008586(n), and b(2n+1) = A060747(n), where b(n) = a(n+1) - 2*a(n).
a(n) = 2*a(n-2) - a(n-4). - R. J. Mathar, Feb 23 2009
G.f.: (1+2*x+x^2-2*x^3)/((x-1)^2*(1+x)^2). - R. J. Mathar, Feb 23 2009
From Wesley Ivan Hurt, Jun 05 2013: (Start)
a(n) = n + 1 - (n - 1)*(n mod 2).
a(n) = (n + 1) * (n - floor((n+1)/2))! / floor((n+1)/2)!.
a(n) = A000142(n+1) / A211374(n+1). (End)

Edited by R. J. Mathar, Feb 23 2009

Term a(45) corrected, and more terms added by Antti Karttunen, Oct 02 2018

A165747 a(n) = 1-2n.

Original entry on oeis.org

1, -1, -3, -5, -7, -9, -11, -13, -15, -17, -19, -21, -23, -25, -27, -29, -31, -33, -35, -37, -39, -41, -43, -45, -47, -49, -51, -53, -55, -57, -59, -61, -63, -65, -67, -69, -71, -73, -75, -77, -79, -81, -83, -85, -87, -89, -91, -93, -95, -97, -99, -101, -103

Offset: 0

Philippe Deléham, Sep 26 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Ângela Mestre, José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for linear recurrences with constant coefficients, signature (2, -1).

Cf. A060747

Table[1 - 2 n, {n, 0, 100}] (* G. C. Greubel, Apr 07 2016 *)

x='x+O('x^99); Vec((1-3*x)/(1-x)^2) \\ Altug Alkan, Apr 07 2016

a(n) = -A060747(n).
a(n) = 2*a(n-1) - a(n-2), a(0)= 1, a(1)= -1.
G.f.: (1-3x)/(1-x)^2.
a(n) = Sum_{k, 0<=k<=n} A112555(n,k)*(-2)^(n-k).
E.g.f.: (1-2*x)*exp(x). - G. C. Greubel, Apr 07 2016

A159964 a(n) = 2^n*(1-n).

Original entry on oeis.org

1, 0, -4, -16, -48, -128, -320, -768, -1792, -4096, -9216, -20480, -45056, -98304, -212992, -458752, -983040, -2097152, -4456448, -9437184, -19922944, -41943040, -88080384, -184549376, -385875968, -805306368, -1677721600, -3489660928

Offset: 0

Paul Barry, Apr 28 2009

Hankel transform of A124791. Binomial transform of -A060747.

{1} U A159964 is a composition of generating functions of A165747 and A000012, with H=G(F(x)) with F(x) for A000012 and G(x) for A165747. - Oboifeng Dira, Aug 29 2019

Oboifeng Dira, A Note on Composition and Recursion, Southeast Asian Bulletin of Mathematics (2017), Vol. 41, Issue 6, 849-853.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A058922, A060747, A124791.

LinearRecurrence[{4,-4},{1,0},30] (* Harvey P. Dale, May 02 2016 *)

G.f.: (1-4x)/(1-2x)^2.
a(n) = -A058922(n). - Jeffrey R. Goodwin, Nov 11 2011
E.g.f.: U(0) where U(k)=  1 - 2*x/(2 - 4/(2 - (k+1)/U(k+1))) ; (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 18 2012
a(n) = Sum_{k=0..n} (1-2k) * C(n,k). - Wesley Ivan Hurt, Sep 23 2017
From Amiram Eldar, Jan 13 2021: (Start)
Sum_{n>=2} 1/a(n) = -log(2)/2.
Sum_{n>=2} (-1)^n/a(n) = -log(3/2)/2. (End)

A270710 a(n) = 3n^2 + 2n - 1.

Original entry on oeis.org

-1, 4, 15, 32, 55, 84, 119, 160, 207, 260, 319, 384, 455, 532, 615, 704, 799, 900, 1007, 1120, 1239, 1364, 1495, 1632, 1775, 1924, 2079, 2240, 2407, 2580, 2759, 2944, 3135, 3332, 3535, 3744, 3959, 4180, 4407, 4640, 4879, 5124, 5375, 5632, 5895, 6164, 6439, 6720, 7007, 7300, 7599

Offset: 0

Ilya Gutkovskiy, Mar 22 2016

In general, the ordinary generating function for the values of quadratic polynomial p*n^2 + q*n + k, is (k + (p + q - 2*k)*x + (p - q + k)*x^2)/(1 - x)^3.
From Bruno Berselli, Mar 25 2016: (Start)
This sequence and A140676 provide all integer m such that 3*m + 4 is a square.
Numbers related to A135713 by A135713(n) = n*a(n) - Sum_{k=0..n-1} a(k).
After -1, second bisection of A184005. (End)

			a(0) = 3*0^2 + 2*0 - 1 = -1;
a(1) = 3*1^2 + 2*1 - 1 =  4;
a(2) = 3*2^2 + 2*2 - 1 = 15;
a(3) = 3*3^2 + 2*3 - 1 = 32, etc.

Ilya Gutkovskiy, Examples of the ordinary generating function for the values of quadratic polynomial.
Leo Tavares, Triple Diamond Illustration.
Eric Weisstein's World of Mathematics, Quadratic Polynomial.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033428, A045944, A056105, A056109, A060747, A135713, A140676, A184005.

Cf. A000290.

List([0..50], n -> 3*n^2+2*n-1); # Bruno Berselli, Feb 16 2018

[3*n^2+2*n-1: n in [0..50]]; // Bruno Berselli, Mar 25 2016

Table[3 n^2 + 2 n - 1, {n, 0, 50}]
LinearRecurrence[{3, -3, 1}, {-1, 4, 15}, 51]

makelist(3*n^2+2*n-1, n, 0, 50); /* Bruno Berselli, Mar 25 2016 */

Vec((-1 + 7*x)/(1 - x)^3 + O(x^60)) \\ Michel Marcus, Mar 22 2016

lista(nn) = {for(n=0, nn, print1(3*n^2 + 2*n - 1, ", ")); } \\ Altug Alkan, Mar 25 2016

vector(50, n, n--; 3*n^2+2*n-1) \\ Bruno Berselli, Mar 25 2016

[3*n^2+2*n-1 for n in (0..50)] # Bruno Berselli, Mar 25 2016

G.f.: (-1 + 7*x)/(1 - x)^3.
E.g.f.: exp(x)*(-1 + 5*x + 3*x^2).
a(n)  = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n)  = A033428(n) + A060747(n).
a(n)  = A045944(n) - 1 = A056109(n) - 2.
a(-n) = A140676(n-1), with A140676(-1) = -1.
Sum_{n>=0} 1/a(n) = 3*(log(3) - 2)/8 - Pi/(8*sqrt(3)) = -0.564745312278736...
a(n) = Sum_{i = n-1..2*n-1} (2*i + 1). - Bruno Berselli, Feb 16 2018
a(n) = A000290(n+1) + 2*A000290(n) - 2. - Leo Tavares, May 28 2023
Sum_{n>=0} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) + 3/4. - Amiram Eldar, Jul 20 2023

A269501 Subsequence immediately following the instances of n in the sequence is n, n-1, ..., 1, n+1, n+2, ....

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 3, 2, 3, 1, 4, 4, 3, 4, 2, 4, 1, 5, 5, 4, 5, 3, 5, 2, 5, 1, 6, 6, 5, 6, 4, 6, 3, 6, 2, 6, 1, 7, 7, 6, 7, 5, 7, 4, 7, 3, 7, 2, 7, 1, 8, 8, 7, 8, 6, 8, 5, 8, 4, 8, 3, 8, 2, 8, 1, 9, 9, 8, 9, 7, 9, 6, 9, 5, 9, 4, 9, 3, 9, 2, 9, 1, 10, 10, 9, 10, 8, 10, 7, 10, 6, 10, 5, 10, 4, 10, 3, 10, 2, 10, 1

Offset: 0

Franklin T. Adams-Watters, Mar 04 2016

The sequence includes every ordered pair of positive integers exactly once as consecutive terms of the sequence. Through n = k^2, it has every pair i,j with 0 < i,j <= k.
Can be regarded as an irregular triangle where row k contains 1, k, k, k-1, k, k-2, ..., 2, k, with 2n-1 terms.
See A305615 for an essentially identical sequence: a(n) = A305615(n)+1. - N. J. A. Sloane, Jul 03 2018

			The first 3 occurs as a(5), so a(6) = 3, the first term of 3, 2, 1, 4, 5, 6, .... The second 3 is thus a(6), so a(7) = 2. The third 3 is a(8), so a(9) = 1. The fourth 3 is a(12), now we start incrementing, and a(13) = 4.
The triangle starts:
  1
  1, 2, 2
  1, 3, 3, 2, 3
  1, 4, 4, 3, 4, 2, 4
  1, 5, 5, 4, 5, 3, 5, 2, 5

Cf. A003059, A060747 (row lengths), A000326 (row sums), A097291, A269780.

cp's OEIS Frontend

A067998 a(n) = n^2 - 2*n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

Extensions

A140811 a(n) = 6*n^2 - 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A141530 a(n) = 4*n^3 - 6*n^2 + 1.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

Extensions

A048495 a(n) = (n-1)*2^n + 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A141310 The odd numbers interlaced with the constant-2 sequence.

Views

Author

A141530 a(n) = 4n^3 - 6n^2 + 1.

A270710 a(n) = 3n^2 + 2n - 1.